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https://mathoverflow.net/questions/21088 | 4 | I am given a triple of positive integers $a,b,c$ such that $a \geq 1$ and $b,c \geq 2$.
I would like to find an upper bound for $a+b+c$ in terms of $n = ab+bc+ac$. Clearly $a+b+c < ab+bc+ac = n$.
Is there any sharper upper bound that could be obtained (perhaps asimptotically)?
| https://mathoverflow.net/users/1737 | Upper bound for a+b+c in terms of ab+bc+ac | This is (mostly) a pretty routine optimization problem. The methods of (for example) a standard calculus class are enough to tell you that $a+b+c$ will be largest when two of $a,b,c$ are as small as possible and the third is whatever it has to be. So if you don't care whether the variables are integers, take $a=1,b=2,c=(n-2)/3$.
Thus if $n$ has the form $3k+2$, the optimum is achieved in integers. Take $a=1,b=2,c=k$, and then we have $a+b+c=3+\frac{n-2}{3}=\frac{n+7}{3}$.
So $\frac{n+7}{3}$ is an upper bound, and it actually gives the correct answer infinitely often.
| 12 | https://mathoverflow.net/users/2559 | 21098 | 13,982 |
https://mathoverflow.net/questions/21090 | 6 | Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$-algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, are there other ones? Does the smooth structure of $M$ endow $A$ with additional structure such that $M$ can be completely recovered from $A$? In other words, is there some kind of smooth Gelfand-duality?
| https://mathoverflow.net/users/2841 | smooth Gelfand-duality | Perhaps I should post this as an answer (even if I don't really know that theory): in
Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824 [Google Books Preview](http://books.google.it/books?id=79bCwXbVnHAC&printsec=frontcover&dq=C-infinity+differentiable+spaces&source=bl&ots=xjtnnn8bSq&sig=BnSizsqq1t7wT8deQfeE--Ts4Ps&hl=it&ei=8w_DS92DFo7U7APZiu2tCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q&f=false)
a theory of "$C^{\infty}$ -differentiable spaces" is developped, and it would be something like the smooth analog to (possibly singular and nonreduced) complex analytic spaces.
| 4 | https://mathoverflow.net/users/4721 | 21099 | 13,983 |
https://mathoverflow.net/questions/21087 | 8 | Call a subset of $\mathbb{N}$ *primitive-recursively enumerable (p-r.e.)* if it is empty or an image of a primitive recursive function. I feel like a lot must be known about the poset of such sets ordered by inclusion, but I am unable to dig up references. Concretely, I would like to know whether there exists a p-r.e. set whose complement is not p-r.e.
The answer is affirmative if there is a complete set (in the sense of many-to-one reducibilities) that is enumerated by a primitive recursive function. My hunch is that such a set exists, but cannot come up with one.
| https://mathoverflow.net/users/1176 | Is there a "primitive-recursively enumerable" set whose complement is not such? | There is a stronger result: *Every r.e. set is primitive r.e. in your sense.*
Short proof: Kleene's Normal Form Theorem.
Longer proof: Let *S* be an r.e. set, assumed WLOG nonempty; fix *a* ∈ *S*, and fix an algorithm *e* where *S* is precisely the range of the function computed by *e*.
Consider the following algorithm: Given the input pair (*n*, *M*), run *e* on input *n* for *M* steps. If it gives an output by then, output whatever *e* outputs; otherwise output *a*.
The functions which set up the initial state of computation, advance a state by one step, and extract the output from a final state, are all p.r. Thus the above algorithm defines a p.r. function, and it is easy to check that its range is *S*.
Edit: Cutland's *Computability* is a decent resource for these questions.
| 8 | https://mathoverflow.net/users/4133 | 21101 | 13,984 |
https://mathoverflow.net/questions/9785 | 13 | Let $R$ be a local normal domain, and let $P \in Spec (R)$. It is well known that $Cl(R) \to Cl(R\_P)$ is surjective. However, I do not know any example where $Cl(R)$ is torsion-free, but $Cl(R\_P)$ is not (we consider $(0)$ to be torsion-free). So:
If the class group of $R$ is torsion-free, must all the class groups of local rings of $R$ torsion-free? What if $R$ is not local?
I do not have a motivation for this, but it just seems an intriguing question.
| https://mathoverflow.net/users/2083 | Does torsion-freeness of class group localize? | Perhaps something like the following works (I have not checked all the details):
Let $C$ be a smooth plane conic and let $Y$ be the projective cone over $C$. Then $Cl(Y) = \mathbb{Z}$ but the class group of the local ring of the vertex of $Y$ is $\mathbb{Z}/2$. Let $X$ be affine cone over $Y$. Then the class group of the local ring
$R$ of the vertex of $X$ is $\mathbb{Z}$ but it seems that the class group of $R$ localised at the prime ideal corresponding to the cone over the vertex of $Y$ is $\mathbb{Z}/2$.
EDIT
The above is wrong as pointed out by Hailong Dao in his comment. I try to fix it below:
Let Y be as above i.e. the singular quadric in $\mathbb{P}^3$ given by the equation $x^2 + y^2 +z^2 = 0$. It may be viewed as the toric surface given by the complete fan with rays passing through $(1,0)$, $(0,1)$ and $(-1,-2)$. Then $Cl(Y) = \mathbb{Z}$ and $Pic(Y)$ is of index $2$ in $Cl(Y)$. Let $Y'$ be the blowup of $Y$ at a non-singular torus fixed point. We may view
Y as the surface obtained from the fan for $Y$ by adding the ray through the point $(1,1)$. Let $\pi:Y' \to Y$ be the blowup map and let $E$ be the exceptional divisor.
Then $Cl(Y') \cong \mathbb{Z} \oplus \mathbb{Z}$ and $Cl(Y')/\mathbb{Z}E = Cl(Y)$.
Let $H$ be an ample divisor on $Y$. Then for $n >> 0$, $H':= n\pi^\*(H) - E$ is an ample divisor on $Y'$ (this is true for the blowup of a point on any surface). Note that $Cl(Y')/\mathbb{Z}H' \cong \mathbb{Z}$, so it is torsion free. SInce $Y'$ is a projective toric surface and $H'$ is an ample divisor, it follows that $H'$ is very ample and gives a projectively normal embedding of $Y'$ in $\mathbb{P}(H^0(Y',\mathcal{O}(H')))$.
As before, we now let $X$ be the cone over $Y'$ and let $R$ be the local ring of the vertex. We let $P$ be the prime ideal corresponding to the cone over the singular point of $Y'$.
($Cl(Y)$ and $Cl(Y')$ can be computed by hand or using the toric description I gave and the results in Fulton, Toric Varieties, Sections 3.3, 3.4; the fact that an ample divisor on a projective toric surface is very ample is an Exercise at the bottom of p.70.)
Note that by letting $R$ be the coordinate ring of $Y' \backslash D$ , where $D$ is a general divisor linearly equivalent to $H'$ (so not containing the singular point) one gets a normal 2-dimensional (non-local) ring with $Cl(R)= \mathbb{Z}$ and with a prime ideal $P$ such that $Cl(R\_P)=\mathbb{Z}/2$ . In all of the above one can replace $2$ by any integer $n>1$ (by considering the projective cone over the rational normal curve on degree $n$, or, in the toric description, replacing $(-2,-1)$ by $(-n,-1)$.
| 3 | https://mathoverflow.net/users/519 | 21106 | 13,989 |
https://mathoverflow.net/questions/20956 | 26 | I could have made this question very brief but instead I've maximally gone the other way and explained a huge amount of background. I don't know whether I put off readers or attract them this way. The question is waay down there.
Let $f$ be a cuspidal modular eigenform of level $\Gamma\_0(N)\subseteq SL\_2(\mathbf{Z})$ (for example $f$ could be the weight 2 modular form attached to an elliptic curve) and let $p$ be a prime. In the theory of modular forms, one Hecke operator at $p$ is singled out, namely $T\_p$, sometimes called $U\_p$ if $p$ divides $N$, and defined by the double coset attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$. Now $f$ is an eigenform for $T\_p$, and $f$ has an eigenvalue for this operator---a Galois-theoretic interpretation of this eigenvalue is that it is (modulo fixing embeddings of $\overline{\mathbf{Q}}$ in $\overline{\mathbf{Q}}\_\ell$ and $\mathbf{C}$) the trace of the geometric Frobenius on the inertial invariants of the $\ell$-adic representation attached to $f$, for $\ell\not=p$ a prime.
Now here is a very naive question that I don't know the answer to, and I really should, and I'm sure it's very well-known to people who do this sort of stuff. Say $N=p^rM$ with $M$ prime to $p$. One can approach the theory of Hecke operators entirely locally. Let $K:=U\_0(p^r)$ denote the subgroup of $GL\_2(\mathbf{Z}\_p)$ consisting of matrices whose bottom left hand entry is $0$ mod $p^r$. Now there is an "abstract Hecke algebra" of locally left- and right-$K$-invariant complex-valued functions on $G:=GL\_2(\mathbf{Q}\_p)$ with compact support. As a complex vector space this algebra has a basis consisting of the characteristic functions $KgK$ as $KgK$ runs through the double cosets of $K$ in $G$. But this space also has an algebra structure, given by convolution.
If $r=0$ then $K$ is maximal compact, and the structure of this Hecke algebra is well-known and easy. Via the Satake isomorphism, the abstract Hecke algebra is isomorphic to $\mathbf{C}[T,S,S^{-1}]$, with $S$ and $T$ independent commuting polynomial variables. The interpretation is that $T$ is the usual Hecke operator $T\_p$ attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $S$ is the matrix attached to $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$. One doesn't always see this latter Hecke operator explicitly in elementary developments of the theory because it acts in a very dull way---it acts by scalars on forms of a given weight and level $\Gamma\_0(N)$, typically (depending on normalisations) as the scalar $p^{k-2}$ on forms of weight $k$. In particular the "abstract Hecke algebra" doesn't give us any more information than that which classical texts explain, as it's generated by $T\_p$, $S\_p$ and $S\_p^{-1}$.
The next case is $r=1$ and this case I also understand. The abstract Hecke algebra now is non-commutative, "because of oldforms": I don't think the operators attached to $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $\left(\begin{array}{cc}0& p\\ 1&0\end{array}\right)$ (that is, the operators corresponding to these double coset spaces) commute, but if $f$ has level $Mp$ and is old at $p$ then we should be working at level $M$, and if it's new at $p$ then we get two invariants---the $T\_p$ (or $U\_p$) eigenvalue, which is classical, and the $w$-eigenvalue, which is the local sign for the functional equation. Again both of these numbers are classical and a lot is known about them. I am pretty sure that the abstract Hecke algebra in this case is generated by these operators $T\_p$, $w$, and the uninteresting $S\_p$ and $S\_p^{-1}$, the latter two still acting by scalars on forms of a given weight. Am I right in thinking that these operators generate the local Hecke algebra? I think so.
The next case is $r=2$ and this I am not 100 percent sure I understand. The classical theory gives us $T\_p$, $S\_p^{\pm1}$ and $w$. Note that on a newform of level $\Gamma\_0(Mp^2)$, $T\_p$ is zero in this situation, $S\_p$ acts by a scalar, and $w$ is some subtle sign which people have clever ways of working out.
---
Finally then, the question! Let $K$ be the subgroup of matrices in $GL\_2(\mathbf{Z}\_p)$ consisting of matrices for which the bottom left hand corner is $0$ mod $p^2$. Let $H$ denote the abstract double coset Hecke algebra of compactly supported $K$-bivariant functions on $GL\_2(\mathbf{Q}\_p)$.
>
> Is this abstract Hecke algebra generated (as a non-commutative algebra) by the characteristic functions of $KgK$ for $g$ in the set {$\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$, $\left(\begin{array}{cc}0& p^2\\ 1&0\end{array}\right)$, $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$, $\left(\begin{array}{cc}p^{-1}& 0\\ 0&p^{-1}\end{array}\right)$}?
>
>
>
In the language I've been using in the waffle above: modular forms of level $p^2$ have an action of the Hecke operators $T\_p$, $w$, and the invertible $S\_p$. Are there any more, lesser known, Hecke operators that we're missing out on?
| https://mathoverflow.net/users/1384 | Are there any Hecke operators acting on an elliptic curve with additive reduction that I don't know about? | Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma\_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle.
Let's write $K\_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}\_2(\mathbb{Z}\_p)$.
---
CLAIM: For any $n > 0$, the Hecke algebra $H(G // K\_n)$ is generated by the double cosets $K\_n x K\_n$ for $x$ in the set
$S = \left\{
\begin{pmatrix} 1 & 0 \\\ 0 & p \end{pmatrix},
\begin{pmatrix} p & 0 \\\ 0 & p \end{pmatrix},
\begin{pmatrix} p^{-1} & 0 \\\ 0 & p^{-1} \end{pmatrix},
\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 0 & -1 \\\ 1 & 0 \end{pmatrix},
\begin{pmatrix} a & 0 \\\ 0 & 1 \end{pmatrix}
\right\}$
where *a* is your favourite generator of $\mathbb{Z}\_p^\times$.
(These correspond, classically, to $T\_p$, $S\_p$, $S\_p^{-1}$, a "twisting operator at p", something close to the Atkin-Lehner $w$, and the diamond operator $\langle a \rangle$.)
---
Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K\_n g K\_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K\_0$.
We now write
$K\_n\ g\ K\_n = K\_n\ k\ x\ k'\ K\_n$
$ = K\_n\ k\ K\_n\ x\ K\_n\ k'\ K\_n$ (using the normality of $K\_n$ in $K\_0$)
$ = [K\_n\ k\ K\_n]\ [K\_n\ x\ K\_n]\ [K\_n\ k'\ K\_n]$
The first and last terms are obviously in the subalgebra $H(K\_0 // K\_n)$ of $H(G // K\_n)$, which is isomorphic to the group algebra of the finite group $K\_0 / K\_n = {\rm GL}\_2(\mathbb Z / p^n)$. This is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}\_2(\mathbb{Z}\_p)$.
Meanwhile, the middle term is in the subalgebra of $H(G // K\_n)$ generated by $X$, and it's easy to see that for $x, y \in X$ we have $K\_n\ x\ K\_n\ y\ K\_n = K\_n\ xy\ K\_n$. Hence this subalgebra is just the monoid algebra of $X$, which is generated by the first three elements of $S$.
Now, as for your original question, the subgroup $U\_0(p^2) \subseteq {\rm GL}\_2(\mathbb{Z}\_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K\_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K\_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.
| 12 | https://mathoverflow.net/users/2481 | 21108 | 13,990 |
https://mathoverflow.net/questions/21110 | 35 | I only know of one proof of Hilbert 90, which is very smart if not magical. See for example <http://hilbertthm90.wordpress.com/2008/12/11/hilberts-theorem-90the-math/>
Does anyone know of a more intuitive proof or know a good way to view the proof?
I have accepted the answer by Emerton, great thanks as well to David Speyer and Brian Conrad.
| https://mathoverflow.net/users/2701 | Is there a natural way to view the proof of Hilbert 90? | Here is a proof of Hilbert's Theorem 90 in the case of cyclic extensions which
I think is fairly conceptual. The key point (which is also at the heart of Grothendieck's
very general version in terms of flat descent) is that if we want to verify that
a linear transformation has a certain eigenvalue (in our particular case, the eigenvalue
of interest will be 1), we can do so after extending scalars.
The set-up: we have a cyclic extension $L/K$, with Galois group generated by $\sigma$,
and an element $a \in L$ of norm 1. We want to find $b \in L$ such that $a = b/\sigma(b)$.
As in David Speyer's answer, rewrite this as the equation $a\sigma(b) = b$.
The map $b \mapsto a\sigma(b)$ is a $K$-linear transformation of the $K$-vector space
$L$, and we want to show that it has a fixed point, i.e. that it has $1$ as an eigenvector.
Well, we can verify this after extending scalars (the eigenvectors of a matrix don't
change if we enlarge the ground field), and so we tensor up with $L$ over $K$.
Now $L\otimes\_K L \cong L\times\cdots \times L$, an isomorphism of $L$-algebras,
and under this isomorphism the action of $\sigma$ on the left just becomes the cyclic permutation of factors on the right. (To see the isomorphism, write $L = K(\alpha),$
as we may by the primitive element theorem. If $f(X)$ is a minimal polynomial of
$\alpha$ over $K$, then $L \cong K[X]/f(X),$ and so $L\otimes\_K L \cong L[X]/f(X).$
But over $L$, the polynomial $f(X)$ splits as $f(X) = (X-\alpha\_1)\cdots (X-\alpha\_n),$
where the $\alpha\_i$ are all the conjugates of $\alpha$. Choosing the labelling
appropriately, we may assume that $\alpha\_i = \sigma^{i-1}(\alpha)$. Then
$L[X]/f(X) = L[X]/(X-\alpha\_1)\cdots (X-\alpha\_n) \cong L\times\cdots \times L,$
and $\sigma$ does indeed just permute the factors.)
Under the isomorphism $L\otimes\_K L \cong L\times\cdots \times L,$
the base-change of our linear transformation $b \mapsto a \sigma(b)$ is given by
$(b\_1,\ldots,b\_n) \mapsto (a b\_n, \sigma(a) b\_1, \ldots, \sigma^{n-1}(a) b\_{n-1}).$
This transformation has the obvious non-zero fixed vector
$(1,\sigma(a),\sigma(a)\sigma^2(a),\ldots,\sigma(a)\ldots\sigma^{n-1}(a)).$
(Remember that Norm$(a) = 1$, and so the last entry is also just $a^{-1}$.)
Thus our original linear transformation (before extending scalars) has a non-zero fixed vector as well,
as required.
How does this relate to Brian Conrad's comment? Well, the preceding argument
generalizes massively to Grothendieck's theory of faithfully flat descent, which
in particular shows that any quasi-coherent sheaf in the flat topology in fact
arises from a Zariski sheaf. That may sound quite complicated, but what the argument
amounts to is precisely what we used in the preceding argument: If $A \rightarrow B$
is a faithfully flat map of rings, and we want to study the "spectral theory"
of a linear operator on an $A$-module, we can do so after extending scalars to $B$.
(Of course, one has to be precise about what "spectral theory" means when we are
working over rings that aren't fields. This is where faithfully flat comes in:
it is the condition that extending scalars from $A$ to $B$ is exact, and takes
non-zero modules to non-zero modules; this turns out to be exactly the right
generalization of the more naive notion we used above, that extending scalars
preserves the eigenvalues of a matrix.)
Finally, here is an aside about the relation with Galois cohomology:
In cohomological language, Hilbert's Theorem 90 is the statement that $H^1(Gal(L/K), L^{\times}) = 0$
for any finite Galois extension of fields $L/K$. To recover the statement involving
norms, one proceeds as follows: if $Gal(L/K)$ is cyclic, with generator $\sigma$,
and the norm of $a \in L$ equals 1,
then $\sigma \mapsto a$ determines a $1$-cocyle on $Gal(L/K)$ with values in $L^{\times}$.
By the vanishing of $H^1$, this must be a coboundary, which means that there exists $b$
such that $a = \sigma(b)/b.$
The cohomological statement (which, as Brian Conrad pointed out, is still a very special
case of Grothendieck's general theory) can be proved by the same extension of scalars argument as above.
| 38 | https://mathoverflow.net/users/2874 | 21117 | 13,995 |
https://mathoverflow.net/questions/21114 | 9 | This is a reference request question. I would like to know more on the structure of low dimensional nilpotent lie algebras. I heard that up to dimension 6 there are only finitely many isomorphism classes, and every such algebra admits a gradation with only positive degrees (see <http://en.wikipedia.org/wiki/Graded_Lie_algebra>).
Do you know of any source where I can find the corresponding proofs?
| https://mathoverflow.net/users/1049 | Low dimensional nilpotent Lie algebras | Classification of nilpotent Lie algebras in characteristic 0 is an old problem,
with a lot of literature. For the dimensions up to 6 there is a finite list.
Among the many relevant papers on MathSciNet, I'll list just a few:
MR2372566 (2009a:17027) 17B50 (17B20 17B30)
Strade, H. (D-HAMBMI)
Lie algebras of small dimension.
Lie algebras, vertex operator algebras and their applications, 233–265, Contemp. Math., 442,
Amer. Math. Soc., Providence, RI, 2007.
MR0498734 (58 #16802) 17B30
Skjelbred, Tor; Sund, Terje
Sur la classification des alg`ebres de Lie nilpotentes. (French. English summary)
C. R. Acad. Sci. Paris S´er. A-B 286 (1978), no. 5, A241–A242.
MR855573 (87k:17012) 17B30
Magnin, L. (F-DJON-P)
Sur les alg`ebres de Lie nilpotentes de dimension 7. (French. English summary) [Nilpotent
Lie algebras of dimension 7]
J. Geom. Phys. 3 (1986), no. 1, 119–144.
MR1737529 (2001i:17010) 17B30 (17B05)
Tsagas, Gr. (GR-THESS-DMP)
Classification of nilpotent Lie algebras of dimension eight.
J. Inst. Math. Comput. Sci. Math. Ser. 12 (1999), no. 3, 179–183.
EDIT: This is a somewhat random sample (I'm not a specialist), but these papers recall results
for low dimensions and have many references to older literature. The reviews in Math
Reviews (MathSciNet) are helpful to look at, if you have access.
There is also a fairly
modern book, which is very high-priced and probably difficult to access:
MR1383588 (97e:17017)
Goze, Michel(F-HALS); Khakimdjanov, Yusupdjan(UZ-AOS)
Nilpotent Lie algebras.
Mathematics and its Applications, 361. Kluwer Academic Publishers Group, Dordrecht, 1996. xvi+336 pp. ISBN: 0-7923-3932-0
17B30 (17-02 17B40 17B56)
| 17 | https://mathoverflow.net/users/4231 | 21118 | 13,996 |
https://mathoverflow.net/questions/17189 | 29 | The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content.
If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that $a^{2b} + b^{2a}\le 1$.
| https://mathoverflow.net/users/3818 | Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$? | Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.
Assume that $a>b$. Put $t=a-b=1-2b$.
Step 1:
$$
\begin{aligned}
a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right]
\\
&\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right]
\\&
=1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right]
\\
&=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right)
\end{aligned}
$$
(in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)
Step 2.
We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$.
For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.
Step 3:
Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.
We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality
$$
tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1}
$$
to prove.
Now, note that, according to Step 2,
$$
\begin{aligned}
&\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\\
&=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1}
\end{aligned}
$$
Multiplying by $t^{k+1}$ and adding up, we get
$$
t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1}
$$
which is exactly what we need.
The end.
P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$.
$$
\begin{aligned}
&\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2
\\
&\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\}
\end{aligned}
$$
Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story.
| 37 | https://mathoverflow.net/users/1131 | 21126 | 14,000 |
https://mathoverflow.net/questions/21081 | 7 | Let $X$ be a regular scheme which is flat over $\mathbf{Z}$. The arithmetic Grothendieck group $\hat{K}(X)$ is defined to be the quotient of $\hat{G}(X)$ by $\hat{G}^\prime(X)$. This is actually quite a length definition which I added below for the sake of completeness.
In the classical case, for $X$ any noetherian scheme, the Grothendieck group $K\_0(X)$ is defined to be the Grothendieck group of the category of vector bundles on $X$. That is, one applies the notion of a Grothendieck group for an additive subcategory of an abelian category. (In our case the abelian category is the category of coherent sheaves on $X$.) This means just modding out by short exact sequences.
I would like to know if there is a categorical type of definition for this group. Thus, first one needs to decide what kind of categories we're talking about (objects are pairs in some sense) and then the notion of exact sequence should coincide in some sense with the below definition.
Probably there is no such thing. I just ask this question in order to understand the arithmetic Grothendieck group better.
**Note**. Let me sketch the definition of the arithmetic Grothendieck group as given in Faltings. In the above $\hat{G}(X)$ is the direct sum of "the free abelian group generated by all vector bundles which have a hermitian metric on $X\_{\mathbf{C}}$ which is invariant under complex conjugation $F$" and the abelian group $\widetilde{A}^\ast(X)$. Here $\widetilde{A}^\ast(X)$ is generated by all $p$-forms $\alpha^p$ such that $F^\ast \alpha^p = (-1)^p \alpha^p$. Furthermore, $\hat{G}^\prime(X)$ is the subgroup generated by elements of the form $E\_2 - E\_1-E\_3 - \widetilde{ch}(E)$, where $E$ is the short exact sequence $$0\rightarrow E\_1 \rightarrow E\_2 \rightarrow E\_3 \rightarrow 0$$ and $\widetilde{ch}(E)$ is the *secondary Chern form*.
| https://mathoverflow.net/users/4333 | Is there a category-theoretic definition of the arithmetic Grothendieck group | The classical group $K\_0$ can also be thought of as consisting of equivalence classes of chain complexes of vector bundles, such that the *exact* sequences represent the zero of $K\_0$ --- and furthermore every complex is equivalent to a two-term complex, a.k.a. a bundle morphism; the graded tensor product of complexes also gives a ring structure on $K\_0$. If you like, classical $K\_0$ is a categorical interpolation between the Euler Characteristic and the homology of a chain complex. The present construction looks like a refinement of that idea, where instead of representing a *trivial* element, an exact sequence is equivalent to a particular (sum of) differential form(s).
| 2 | https://mathoverflow.net/users/1631 | 21131 | 14,003 |
https://mathoverflow.net/questions/21142 | 2 | To assuage my conscience over an unsourced statement in a paper I'm writing:
I am looking for an example of a commutative algebra over the complex numbers having a maximal ideal of codimension >1, or a statement of inexistence.
| https://mathoverflow.net/users/5316 | Maximal ideal of codimension >1 | If the codimension is finite, then there is no such thing. If the codimension can be infinite, then yes, because there are infinite dimensional complex division algebras which are simple, like $\mathbb C(t)$.
| 6 | https://mathoverflow.net/users/1409 | 21144 | 14,011 |
https://mathoverflow.net/questions/21109 | 9 | Is there a non-trivial class $S$ of smooth Deligne-Mumford stacks over a base $B$ with the property that if $\mathcal{X}, \mathcal{Y} \in S$ have isomorphic coarse moduli spaces (assumed to exist) then $\mathcal{X} \cong \mathcal{Y}$? If $B = Spec(k)$, $k$ a field (of characteristic $0$ if necessary), can one take $S$ to be the class of all irreducible, smooth, separated DM stacks with trivial inertia in codimension $\leq 1$?
| https://mathoverflow.net/users/519 | Stacks determined by their coarse moduli spaces | Yes, the class of all smooth, separated DM stacks over a field of characteristic $0$, with trivial inertia in codimension at most $1$, over a field of characteristic $0$, has the propery you want. The point is that every moduli space of such a stack has quotient singularities; and every variety with quotient singularities is the moduli space of a unique such stack. I believe that this was first proved in Angelo Vistoli: Intersection theory on algebraic stacks and on their moduli spaces. Invent. Math. 97 (1989), no. 3, 613-670, Proposition 2.8 (uniqueness is not stated there, but it follows from the proof).
| 18 | https://mathoverflow.net/users/4790 | 21146 | 14,013 |
https://mathoverflow.net/questions/21134 | 5 | Suppose $K$ is a local field and $L$ a finite cyclic extension of $K$. By Hilbert 90, we know that if an element $a$ in $ L$ such that $N\_{L / K}(a) =1 $ then $ a = b / \sigma(b) $ for some $b$ in $L$ and $ \sigma$ a generator of the Galois group.
My question: Suppose $N\_{L / K}(a) \simeq 1 $, is there some $b$ such that $ a \simeq b /\sigma(b) $?
I guess I could have try to make the question more precise, but there seems to be some merits in leaving it a bit vague.
I have accept the the answer by Paul Broussous, which address the situation when the extension is unramified. It is because that is what I need. I am still curious whether something can be done when the extension is totally ramified?
| https://mathoverflow.net/users/2701 | Is there any approximated version of Hilbert 90? | Yes such approximated versions of Hilbert 90 do exist. But you need some technical conditions.
For instance assume that $L/K$ is unramified of degree $d$ and that
$a\in {\mathfrak o}\_{L}^{\times}$.
Then you condition writes
$N\_{L/K}(a)\equiv 1$ modulo $\mathfrak{p}\_{K}^{n}$,
for some $n>0$ (I assume that this
is what you mean by $\simeq$). This may be rewritten
$N\_{L/K}(a)=1$ in $U\_{L}/U^{n}\_L$, where $U$ denotes a
unit group. So the map
$$\sigma^u\mapsto a\sigma (a) \cdots \sigma^u (a)$$
defines a $1$-cocycle of ${\rm Gal}(L/F)$ in $U\_L /U^n\_L$ (here $\sigma$ denotes the Frobenius substitution).
So what you want is that this cocycle is split. In fact we have $H^{1}({\rm Gal}(L/K), U\_{L}/U^{n}\_{L})=1$. This is proved by a standard filtration argument: this is implied by
$$H^{1}({\rm Gal}(L/K), U\_L /U^{1}\_L ) = H^{1}({\rm Gal}(k\_L /k\_K ), k\_{L}^{\times})=1$$
and
$$H^{1}({\rm Gal}(L/K), U\_{L}^{i}/U\_{L}^{i+1})=H^{1}({\rm Gal}(k\_L /k\_K ), k\_L )=1$$
here $k$ denotes a residue field. You can find the detail of the proof in, I think, Serre's 'Local fields' or Cassels-Fröhlich's 'Algebréaic Number Theory'.
| 8 | https://mathoverflow.net/users/4767 | 21156 | 14,020 |
https://mathoverflow.net/questions/21161 | 15 | Artin's presentation of **braid group on three strands** is:
$$ B\_3 = \langle l,r : lrl = rlr \rangle $$
where you should think of "$l$" as the positive crossing between the left and middle strands and "$r$" as the positive crossing of the right and middle strands:
```
| | | | | |
\ / | | \ /
l = \ | r = | \
/ \ | | / \
| | | | | |
```
Then there is a surjection $B\_3 \to S\_3$ given by $l \mapsto (12)$ and $r \mapsto (23)$. ($S\_3$ is the symmetric group on three letters: it is generated by $(12)^2 = 1 = (23)^2$ and the braid relation above.) The **pure braid group** $PB\_3$ is the kernel of this surjection.
The **six-crossing braid** $b = lr^{-1}lr^{-1}lr^{-1}$ is an element of the pure braid group. Let $N$ be the minimal normal subgroup of $B\_3$ that contains $b$. Certainly $N \subseteq PB\_3$.
>
> **Question:** Do we have $N = PB\_3$?
>
>
>
### Motivation
The motivation for my question comes from a neat trick that Conway showed us years ago. It leads to a more nuanced question than what I asked that I will pose as its own question if the answer above is "no". My memory is that at the time Conway did not know the answer to the more nuanced question, which suggests that the answer above cannot be "yes".
Take a long and reasonably thin rectangle of paper, and score it with two cuts, so that you have three strips of paper that are connected at both ends, so that you end up with a (framed, oriented, ...) "theta graph":
```
|--------|
| |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| |
|--------|
```
Then with some finagling, you can in fact "tie" the six-crossing braid in those three strands, without ripping the paper further, by passing the bottom through itself a few times. (The trick is that it's easier to unbraid than to braid, so make your paper long enough that you can put $bb^{-1}$ into it, and then unbraid the $b^{-1}$.) Put another way: you can first put a hair-tie at the end of your ponytail, and then braid your hair.
The harder question is to characterize all braids that you can put on the above "theta graph". The following facts are essentially obvious:
* Any "braiding" of the theta graph is pure.
* The set of braidings of the theta graph is a subgroup of $B\_3$.
* The set of braidings of the theta graph is closed under conjugating by arbitrary braids.
Therefore, the set $T$ of braidings of the theta graph is a normal subgroup of $B\_3$, with $N \subseteq T \subseteq PB\_3$. In particular, a positive answer to the question above characterizes $T$.
| https://mathoverflow.net/users/78 | Is the pure braid group on three strands generated as a normal subgroup of the braid group by the six-crossing braid? | No.
Let $P\_n$ be the pure braid group on $n$ strands. Forgetting the first strand gives a projection $P\_3\to P\_2$. The kernel is the free group $F\_2=\pi\_1(\mathbb{C}\setminus
\{2,3\},1)$ given by moving the first strand around the other two while they are held in place. Since $P\_2$ is just $\mathbb{Z}$, we have a short exact sequence
$1\to F\_2\to P\_3\to \mathbb{Z}\to 1$
Your element $b$ is contained in the kernel of this sequence (see the picture below) so its normal closure is contained in $F\_2\lt P\_3$. For an appropriate basis $\langle x,y\rangle=F\_2$, the element $b$ represents $xyx^{-1}y^{-1}$.
```
\ | |
\ |
\ |
\ |
| |
| |\
| | |
| /
| /
| /
/ |
/ |
/ |
| | |
\| |
| |
| \
| \
| \
| | |
| |/
| |
|/ |
| |
| |
/ | |
```
In fact, removing *any* of the three strands results in a trivial 2-strand braid (as can be seen from the picture). Closing up the ends of the braid thus results in a 3-component link so that if any one component is removed, the other two are unlinked. This is the well-known [Borromean rings](http://en.wikipedia.org/wiki/Borromean_rings). There are three different surjections $P\_3\to \mathbb{Z}$, given by forgetting each strand separately, and the group of [Brunnian braids](http://en.wikipedia.org/wiki/Brunnian_link#Brunnian_braids) is exactly the intersection of their kernels. Since $xyx^{-1}y^{-1}$ normally generates the commutator subgroup of $F\_2$, the normal closure $N$ of $b$ will be exactly the group of Brunnian braids. In particular, $N$ is an infinite-rank free group.
| 17 | https://mathoverflow.net/users/250 | 21166 | 14,027 |
https://mathoverflow.net/questions/21152 | 33 | A **magma** is a set $M$ equipped with a binary operation $\* : M \times M \to M$. In abstract algebra we typically begin by studying a special type of magma: groups. Groups satisfy certain additional axioms that "symmetries of things" should satisfy. This is made precise in the sense that for any object $A$ in a category $C$, the invertible morphisms $A \to A$ have a group structure again given by composition. An alternate definition of "group," then, is "one-object category with invertible morphisms," and then the additional axioms satisfied by groups follow from the axioms of a category (which, for now, we will trust as meaningful). Groups therefore come equipped with a natural notion of representation: a representation of a group $G$ (in the loose sense) is just a functor out of $G$. Typical choices of target category include $\text{Set}$ and $\text{Hilb}$.
It seems to me, however, that magmas (and their cousins, such as non-associative algebras) don't naturally admit the same interpretation; when you throw away associativity, you lose the connection to composition of functions. One can think about the above examples as follows: there is a category of groups, and to study the group $G$ we like to study the functor $\text{Hom}(G, -)$, and to study this functor we like to plug in either the groups $S\_n$ or the groups $GL\_n(\mathbb{C})$, etc. on the right, as these are "natural" to look at. But in the category of magmas I don't have a clue what the "natural" examples are.
**Question 1:** Do magmas and related objects like non-associative algebras have a "natural" notion of "representation"?
It's not entirely clear to me what "natural" should mean. One property I might like such a notion to have is an analogue of Cayley's theorem.
For certain special classes of non-associative object there is sometimes a notion of "natural": for example, among not-necessarily-associative algebras we may single out Lie algebras, and those have a "natural" notion of representation because we want the map from Lie groups to Lie algebras to be functorial. But this is a very special consideration; I don't know what it is possible to say in general.
(If you can think of better tags, feel free to retag.)
**Edit:** Here is maybe a more focused version of the question.
**Question 2:** Does there exist a "nice" sequence $M\_n$ of finite magmas such that any finite magma $M$ is determined by the sequence $\text{Hom}(M, M\_n)$? (In particular, $M\_n$ shouldn't be an enumeration of all finite magmas!) One definition of "nice" might be that there exist compatible morphisms $M\_n \times M\_m \to M\_{n+m}$, but it's not clear to me that this is necessarily desirable.
**Edit:** Here is maybe another more focused version of the question.
**Question 3:** Can the category of magmas be realized as a category of small categories in a way which generalizes the usual realization of the category of groups as a category of small categories?
**Edit:** Tom Church brings up a good point in the comments that I didn't address directly. The motivations I gave above for the "natural" notion of representation of a group or a Lie algebra are in some sense external to their equational description and really come from what we would like groups and Lie algebras to do for us. So I guess part of what I'm asking for is whether there is a sensible external motivation for studying arbitrary magmas, and whether that motivation leads us to a good definition of representation.
**Edit:** I guess I should also make this explicit. There are two completely opposite types of answers that I'd accept as a good answer to this question:
* One that gives an "external" motivation to the study of arbitrary magmas (similar to how dynamical systems motivate the study of arbitrary unary operations $M \to M$) which suggests a natural notion of representation, as above. This notion might not look anything like the usual notion of either a group action or a linear representation, and it might not answer Question 3.
* One that is "self-contained" in some sense. Ideally this would consist of an answer to Question 3. I am imagining some variant of the following construction: to each magma $M$ we associate a category whose objects are the non-negative integers where $\text{Hom}(m, n)$ consists of binary trees with $n$ roots (distinguished left-right order) and $m$ "empty" leaves (same), with the remaining leaves of the tree labeled by elements of $M$. Composition is given by sticking roots into empty leaves. I think this is actually a 2-category with 2-morphisms given by collapsing pairs of elements of $M$ with the same parent into their product. An ideal answer would explain why this construction, or some variant of it, or some other construction entirely, is natural from some higher-categorical perspective and then someone would write about it on the nLab!
| https://mathoverflow.net/users/290 | Do non-associative objects have a natural notion of representation? | Since magmas in general don't have much structure, we can't reasonably expect a representation to preserve much structure. We can therefore define a left representation of a magma $M$ to be a set $V$ equipped with a map $M \times V \to V$. We do the analogous thing for general nonassociative algebras. Serge Lang liked to describe a notion of left regular representation of an algebra $A$, which is just the linear map $A \to \operatorname{End} (A)$ that takes an element to the linear transformation it induces by left multiplication. As expected, this map is a homomorphism if and only if the algebra is associative.
There are special cases of nonassociative algebras that admit good notions of representation, and in the cases I know, these arise from operads that have "good relationships" with the associative operad. The standard example is the natural map from the Lie operad to the Associative operad that yields the forgetful functor from associative algebras to Lie algebras. This functor admits the universal enveloping algebra functor as left adjoint. There is a formalism of enveloping operads, which generalizes this case. The upshot is that these special cases have a lot more structure than a simple composition law, so we can demand more from a representation (namely, that it respect the operad structure as manifested through the universal enveloping algebra).
| 13 | https://mathoverflow.net/users/121 | 21172 | 14,029 |
https://mathoverflow.net/questions/21171 | 43 |
>
> **Possible Duplicate:**
>
> [Cohomology and fundamental classes](https://mathoverflow.net/questions/1489/cohomology-and-fundamental-classes)
>
>
>
Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi\_\*[N]\in H\_\*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every homology class of $M$ can be represented by a submanifold in this manner, however for some special cases it is.
For example, for $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold. Another example is when $M$ an Euclidean configuration space.
My questions are:
1) Under what circumstances can every homology class of $M$ be represented by a submanifold and
2) What are some examples of manifolds who have homology classes not representable in this manner?
| https://mathoverflow.net/users/5323 | When is a Homology Class Represented by a Submanifold? | Here are a few simple answers to the question you asked:
1. Every class in $H\_{n-1}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to S^1$ representing the Poincare dual in $H^1(M;Z)=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected.
2. Similarly, every class in $H\_{n-2}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to CP^\infty$ representing the Poincare dual in $H^2(M;Z)=[M,CP^\infty]$, homotop $f$ into a finite skeleton, say $CP^N$, and take the preimage of $CP^{N-1}$.
3. Transversality says that if you can represent $x\in H\_k(M)$ by a *map* of a smooth manifold (e.g. elements in the image of the Hurewicz map, or by Thom) , and $2k < n$, then you can represent it by an embedded submanifold (as Andy mentions above). For example, any class in $H\_1(M)$ for $dim(M)\ge 3$. With care you can also make this work for $2k=n$, and there are techniques available in the "metastable" range (no triple points) involving generalizations of Whitney's trick and other ways to replace double points.
| 31 | https://mathoverflow.net/users/3874 | 21198 | 14,044 |
https://mathoverflow.net/questions/21200 | 12 | Brauer's permutation lemma states that any two permutation matrices are conjugate in $\mathrm{GL}(n,\mathbb{C})$ if and only if they are conjugate in the symmetric group, i.e., they have the same cycle type (we can replace $\mathbb{C}$ by any field of characteristic zero).
Another way of stating it is that any two permutation representations of a finite cyclic group that are conjugate as linear representations in characteristic zero are in fact equivalent as permutation representations.
My question: for what classes of finite groups does this result hold? i.e., for what classes of finite groups is it true that any two permutation representations that are conjugateas linear representations in characteristic zero are in fact equivalent as permutation representations?
I think some counterexamples involving Mathieu groups exist, so this is not true for all finite groups. What I was looking for are theorems of the form that it is true for all groups satisfying some well-studied property or in some well-identified class.
One application of this would be as follows: suppose $G$ is a finite group such that the quotient of the automorphism group of $G$ by the group of class-preserving automorphisms of $G$ satisfies the above condition. (Class-preserving automorphisms are automorphisms that preserve each conjugacy class; for a finite group, this is equivalent to preserving all the characters). Then, the orbit sizes under the action of the automorphism group of $G$ are the same for the set of conjugacy classes and the set of (equivalence classes of) irreducible representations.
| https://mathoverflow.net/users/3040 | Brauer's permutation lemma — extending to some other finite groups? | There are many group actions on sets which are linearly equivalent but not equivalent as actions. In fact, every group other than the cyclic group has one. This follows from some easy linear algebra:
* the number of irreducible reps over $\mathbb{Q}$ is the number of conjugacy classes of cyclic subgroups of $G$, (**EDIT**:there are at most this many since any two elements which generate conjugate cyclic groups have the same character in a rational representation; on the other hand, the characters of the inductions of the trivial from any set of cyclic groups, no two of which are conjugate, are linearly independent, so there are at least this many) and
* the number of non-isomorphic transitive G-sets is the number of conjugacy classes of subgroups.
Thus, there must be an integer valued linear combination of transitive actions which has trivial character. Moving all the actions with negative coefficients to the other side of the equality, we get two different actions with the same character, and thus isomorphic representations.
I actually wrote [a paper](https://arxiv.org/abs/math/0610205) about this a few years back, which I think is a reasonable starting place for the subject, which actually has quite a long history, and a reasonably extensive literature.
| 11 | https://mathoverflow.net/users/66 | 21204 | 14,045 |
https://mathoverflow.net/questions/21199 | 5 | A theorem by S. Garrison states that if $G$ is a finite solvable group and $|cd(G)| = 4$ then $dl(G)\leq |cd(G)|$ (the Taketa inequality, which is conjectured to hold for all finite solvable groups). So far I have been unable to find a proof of this theorem anywhere. The only references I have seen are to Isaacs' book on character theory (where he only mentions that it has been proven by S. Garrison), and to the Ph.d thesis of S. Garrison (which has not been published, so not much help there). oes anyone know where one might find the proof?
| https://mathoverflow.net/users/4614 | Reference request: A theorem by S. Garrison | A new proof was published in:
Isaacs, I. M.; Knutson, Greg. "Irreducible character degrees and normal subgroups."
J. Algebra 199 (1998), no. 1, 302–326.
[MR1489366](http://www.ams.org/mathscinet-getitem?mr=1489366)
[DOI:10.1006/jabr.1997.7191](http://dx.doi.org/10.1006/jabr.1997.7191)
This was extended to cd(G)=5 in:
Lewis, Mark L. "Derived lengths of solvable groups having five irreducible character degrees. I."
Algebr. Represent. Theory 4 (2001), no. 5, 469–489.
[MR1870501](http://www.ams.org/mathscinet-getitem?mr=1870501)
[DOI: 10.1023/A:1012706718244](http://dx.doi.org/10.1023/A:1012706718244)
It mentions that "Because of the length and complexity of his argument, Garrison never published this result." and has some other useful comments.
| 7 | https://mathoverflow.net/users/3710 | 21215 | 14,051 |
https://mathoverflow.net/questions/21207 | 9 | Faber's perfect pairing conjecture states that the tautological ring $R^\*$ of the moduli space $\mathcal{M}\_g$ of curves of genus $g$ behaves as if it were the rational cohomology of a closed, oriented manifold of dimension $g-2$. Specifically, $R^{g-2}$ is rank one, and multiplication into this degree gives a perfect pairing between $R^k$ and $R^{g-2-k}$.
My understanding is that it is known (through work of Looijenga, Faber, and Pandharipande) that $R^{g-2} = \mathbb{Q}$, but the perfect pairing part hasn't been proven (though it has been verified in low genus cases). I'd like to know:
1. Why might Faber have conjectured this to be the case? What is it about $R^\*$ that suggests that it might satisfy Poincare duality?
2. If true, what sort of applications does this have (to our understanding of $\mathcal{M}\_g$, for instance)?
| https://mathoverflow.net/users/4649 | Applications of Faber's conjecture | 1. Numerical evidence, from computing the cases $g=2,3,\dots$, eventually $g\le 15$, and seeing the symmetry in the numbers $\dim R\_g^n$. I recall Carel saying he made the conjecture when $g$ was still pretty low, maybe 6. For any $g$, there is an algorithm computing $\dim R^n\_g$ in finite time, that Faber came up with.
2. That's not so clear. But that's a very mysterious property. The search for the meaning is on.
| 6 | https://mathoverflow.net/users/1784 | 21222 | 14,057 |
https://mathoverflow.net/questions/21168 | 17 | This continues my question about [smooth Gelfand-duality](https://mathoverflow.net/questions/21090/smooth-gelfand-duality). In the book
>
> Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824
>
>
>
it is shown that $M \mapsto C^\infty(M)$ is a fully faithfull contravariant functor from the category of manifolds (smooth, separable and without boundary) to the category of $\mathbb{R}$-algebras. Isn't this nice? It would be even more nice if there is an algebraic description of the essential image of this functor, so that we have an antiequivalence of categories between manifolds and certain $\mathbb{R}$-algebras. Thus my question is:
* Which $\mathbb{R}$-algebras $A$ are isomorphic to $C^{\infty}(M)$ for some manifold $M$?
Of course, you could just formulate that $Spec\_r(A)=Hom(A,\mathbb{R})$ with the obvious structure sheaf is a manifold and that the canonical map $A \to C^{\infty}(M)$ is an isomorphism in terms of the ring structure of $A$. But this does not seem to be handy at all. I want some nontrivial purely algebraic formulation. If possible avoiding structure sheaves at all.
Here are some necessary conditions:
* If $f \neq g$ in $A$, then there is some $\mathbb{R}$-homomorphism $\phi : A \to \mathbb{R}$ such that $\phi(f) \neq \phi(g)$. In particular, $A$ is reduced.
* For every $p \in Spec\_r(A)$ with corresponding maximal ideal $m\_p$, then the maximal ideal $\overline{m\_p}$ of $A\_{\mathfrak{m}\_p}$ is finitely generated, say by elements $f\_1,...,f\_n$, and the canonical map
$\mathbb{R}[t\_1,...,t\_n] / (t\_1,...,t\_n)^{r+1} \to A\_{\mathfrak{m}\_p} / \overline{m\_p}^{r+1}, t\_i \mapsto f\_i$
is an isomorphism for all $r \geq 0$.
* With the notation above, the canonical map $A/m\_p^{r+1} \to A\_{\mathfrak{m}\_p} / \overline{m\_p}^{r+1}$ is an isomorphism.
* The function $Spec\_r(A) \to \mathbb{N}, p \to \dim\_\mathbb{R} \mathfrak{m}\_p/{\mathfrak{m}\_p}^2$ is locally constant.
Are they sufficient [**no, see Michael's answer**]? Finally [**solved by Dmitri's answer**]:
* How can we characterize the algebras (at least within all the $C^{\infty}(M)$'s), that come from *compact* manifolds?
You might admit that "$Spec\_r(A)$ is compact with the Gelfand topolgy" is not a satisfactory answer ;-).
**Addendum**: At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods. I don't claim that this applies to my problem. But at least I invite you to think about it. The properties of the algebras above are just an approximation. Even if we add some of the conditions in the answers (such as $\cap\_{r} \overline{m}\_p^{r+1} \neq 0$), it would be a great surprise that the conditions are sufficient. But I'm not convinced of the contrary as soon someone provides a counterexample. It is fun trying to deduce some of the differential geometric theorems such as IFT from the properties above (if $A \to B$ is an isomorphism in one tangent space, then it is a local isomorphism). Perhaps a first step is to characterize the local rings $C^{\infty}\_p(\mathbb{R}^n)$.
| https://mathoverflow.net/users/2841 | How to classify the algebras C^∞(M)? | There is a strongly geometric characterization of those algebras which arise as $C^\infty(M)$ for $M$ compact and orientable, recently proved by Connes, see
[here](http://arxiv.org/abs/0810.2088). This has come up on MO before, e.g. in Joel Fine's answer to this question:
[Algebraic description of compact smooth manifolds?](https://mathoverflow.net/questions/5344/algebraic-description-of-compact-smooth-manifolds)
Like the proofs of most major theorems in differential topology, Connes's approach invokes (a) Riemannian metrics, and (b) hard analysis. Spectral geometry is not my area, so this will be an amateurish explanation...
If $M$ is a compact smooth manifold, $C^\infty(M)$ is represented faithfully on the Hilbert space $H$ of $L^2$ sections of any hermitian vector bundle $S$. If $D$ is a first-order differential operator acting (unboundedly) on $H$ then we can recover the projective $C^\infty(M)$-module of sections $C^\infty(M;S)$ within $H$ as $\bigcap\_{k>0}{dom(D^k)}$ (these domains will actually by Sobolev spaces, I believe). The algebraic counterpart of being first-order is that $[[D,f],g]=0$ for any $f,g \in C^\infty(M)$.
$D$ has particularly nice properties when it's elliptic. There's no canonical elliptic operator over a smooth manifold until one chooses a Riemannian metric; there's then the signature operator $D=d+d^\ast$ acting on the complexified differential forms. This is an example of a Dirac operator (its square is a Laplacian - this is a condition on the symbol of the operator). As such, it's formally self-adjoint, Fredholm, and its (real) spectrum has known growth rate depending on $\dim(M)$.
Connes (see Theorem 11.4) shows that a commutative $\mathbb{R}$-algebra $A$ arises as $C^\infty(M)$ for a smooth manifold structure on the space $M$ Gelfand-dual to $A$ provided that it's part of a "spectral triple" $(A,H,D)$ of the right kind. This means that $A$ should act on a Hilbert space $H$ carrying an unbounded symmetric operator $D$ satisfying various properties. I've hinted at some of these; the most sophisticated property is an "orientation" condition invoking a Hochschild cycle $c\in Z\_{\dim M}(A,A)$. This cycle is something like a volume form, and from its components Connes rebuilds local charts.
| 10 | https://mathoverflow.net/users/2356 | 21230 | 14,063 |
https://mathoverflow.net/questions/20980 | 5 | It is well known that for a closed hyperbolic 3-manifold $M$ the rank of $\pi\_1(M)$ is bounded above by some universal constant $K$ times the volume of $M$. Using similar methods, i.e. the thick-thin decomposition of $M$, one can also show that the Heegaard genus of $M$ is bounded above by a universal constant times the volume of $M$. (I believe that Thurston showed this first, though I am not sure as to how.)
I am looking to construct a sequence of (closed) hyperbolic 3-manifolds, say {$M\_n$} such that the volume grows linearly in the Heegaard genus of $M\_n$. That is, I am trying to show that a linear bounded on Heegaard genus in terms of volume is the best one can do. So far, I am having some trouble constructing such an example.
Does anyone have a good method or reference for constructing such an example? Also, another approach to the problem would be wonderful (short of solving the rank versus Heegaard genus conjecture of hyperbolic 3-manifolds, of course).
| https://mathoverflow.net/users/2788 | Heegaard genus in hyperbolic 3-manifolds | Agol says: Take a 3-manifold group that maps onto a free group, and take induced covers (the rank and volume will both grow linearly). See [this paper of Lackenby](http://ams.org/mathscinet-getitem?mr=2218779).
| 2 | https://mathoverflow.net/users/66 | 21240 | 14,069 |
https://mathoverflow.net/questions/21238 | 1 | Would someone be able to point me to a good resource explaining step by step the process for solving inhomogenous recurrence relations? (ie something of the form $ a\_n = \sum{{b\_i}{a\_{n-i}}} + f(n)$ )
| https://mathoverflow.net/users/2973 | Inhomogenous recurrence relations | One standard method is generating functions. Set $A(t)=\sum\_{n=0}^\infty a\_n t^n$
and $B(t)=\sum\_{i=1}^\infty b\_i t^i$. Then
$$A(t)=a\_0+B(t)A(t)+\sum\_n f(n)t^n$$
so that
$$A(t)=(1-B(t))^{-1}\left(a\_0+\sum\_n f(n)t^n\right).$$
For an excellent text on generating functions, see Herb Wilf's
*generatingfunctionology*:
<http://www.math.upenn.edu/~wilf/DownldGF.html> .
| 6 | https://mathoverflow.net/users/4213 | 21241 | 14,070 |
https://mathoverflow.net/questions/21205 | 6 | Are there nef divisors D on a complex projective manifold X such that $h^0(X,D)$ is less than or equal to $\dim X$?
Edit: In fact I'm interested in nef line bundles D, not just divisors.
| https://mathoverflow.net/users/nan | Nef divisors with few global sections | Here's an interesting example from complex algebraic surfaces: the so called Godeaux surface. This is a surface, $S$ on which the canonical bundle is $ample$ and yet $h^{0}(S,K\_{S})=0$.
To construct such an $S$ we start with a quintic, $S'$ in $\mathbb{P}^{3}$ defined by the Fermat form:
$X\_{0}^{5}+X\_{1}^{5}+X\_{2}^{5}+X\_{3}^{5}=0$
The group $\mathbb{Z}\_{5}$ acts on $S'$ by:
$[X\_{0},X\_{1},X\_{2},X\_{3}]\rightarrow [X\_{0},\zeta X\_{1},\zeta^{2}X\_{2},\zeta^{3}X\_{3}]$
Where in the above $\zeta$ is a primitive fifth root of unity. One readily checks that this action is fixed point free so we get a smooth quotient $S=S'/\mathbb{Z}\_{5}$.
We will compute the invariants of $S$ by first computing those of $S'$ and then taking the quotient. The Lefschetz theorem on hyperplane sections tells us that $h^{1}(S')=0$. Next by adjunction, $K\_{S'}=H$ is a hyperplane section of $S'$. Applying the exact sequence of sheaves:
$ 0\rightarrow O\_{\mathbb{P}^{3}}(H-S')\rightarrow O\_{\mathbb{P}^{3}}(H)\rightarrow O\_{S'}(H)\rightarrow 0 $
we obtain the exact sequence of cohomology groups:
$ 0=H^{0}(\mathbb{P}^{3},-4H)\rightarrow H^{0}(\mathbb{P}^{3},H)\rightarrow H^{0}(S',K\_{S'})\rightarrow H^{1}(\mathbb{P}^{3},-4H)=0$
It follows that $h^{0}(S',K\_{S'})=h^{0}(\mathbb{P}^{3},H)=4$ and thus the holomorphic Euler characteristic of $S'$ is:
$\chi(S')=1-h^{1}(S')+h^{0}(S',K\_{S'})=5$
Now, since $S$ is a quotient of $S'$ by a free action of a finite group, the holomorphic forms on $S$ are just the $\mathbb{Z}\_{5}$ invariant forms on $S'$. In particular $h^{1}(S)=0$. Further, $\chi(S)=\frac{1}{5}\chi(S')=1$. Thus we have:
$h^{0}(S,K\_{S})=\chi(S)+h^{1}(S)-1=0$
So the canonical bundle of $S$ has no sections.
On the other hand, the canonical bundle of $S$ is manifestly ample. Indeed, $\pi\_{1}(S)=\mathbb{Z}\_{5}$ so $S$ is irrational. And:
$K\_{S}^{2}=\frac{1}{5}K\_{S'}^{2}=\frac{1}{5}\mathrm{deg(S')}=1 >0$
So a sufficiently high power of $K\_{S}$ maps $S$ birationally onto its image. It is not hard to see that for sufficiently high powers this birational map must be an isomorphism so $K\_{S}$ is ample.
| 11 | https://mathoverflow.net/users/5124 | 21254 | 14,078 |
https://mathoverflow.net/questions/21243 | -1 | Has anyone worked out a formal, general-enough definition of what is 'useful', so that it could reflectively be used in mathematics? I am aware of the work in [utility theory](http://en.wikipedia.org/wiki/Useful) from economics (but originally from the Bernoullis and improved by von Neumann, so very much 'mathematical'). Such a formalization should be adequate to decide if a particular definition (or theorem) is considered 'useful'.
Note that I fully expect *utility* to be a relative notion, in other words I don't expect anything to be 'universally useful'. I have some tentative definitions, but before I spend too much time working this out, I would like to know if this has already been done mathematically (as the work of economists on this is [expectedly] too biased towards economic utility).
A concrete example: 20 years ago, elliptic curves would have been considered 'not useful' in the context of cryptography, now it is considered 'useful'. **This can be made completely formal**. [In other words, my question is about what has been done before, not a discussion of what this is, which if off-topic for MO].
| https://mathoverflow.net/users/3993 | Formal definition of 'useful' ? | It seems to me that you have two questions here.
First, you inquire about a formal account of "usefulness". I believe that this is already provided by the formal mathematical accounts of utility in utility theory. The concept of utility in that theory is extremely flexible, not limited to economics or any other specific endeavor. Thus, it seems able to provide for any account of "usefulness" you may have in mind. Let's just say that the utility provided by a given thing is equal to the "usefulness" you had in mind for it.
Your second question is more directly aimed at analyzing the usefulness of various specific mathematical ideas. For this question, I'm not sure that what is lacking is a formal definition of usefulness. After all, even if one knows a lot of formal utility theory, it doesn't help you to find out which flavor of ice cream your child likes best. Rather, what one would seem to want is ways of measuring various specific measurable aspects of that utility function. Thus, it is a problem of measurement, rather than formal theory. In the case of measuring the importance of utility or usefulness of various mathematical theorems or definitions, several people have suggested a page-rank type calculation, based on citation statistics, which I find interesting.
Another approach to this second question is the one I described in my answer to the question [here](https://mathoverflow.net/questions/17964/is-there-a-known-way-to-formalise-notion-that-certain-theorems-are-essential-on/18141#18141), which is to analyze the mathematical relationships between the various theorems of mathematics, over a very weak base theory. This subject is known as [Reverse Mathematics](http://en.wikipedia.org/wiki/Reverse_mathematics), and one of the most surprising conclusions (not at all obvious) of this research effort is that the great majority of classical mathematical theorems (and contemporary ones as well) fall into one of five equivalence classes. That is, most theorems turn out to be logically equivalent to one of the big five. This kind of analysis may lead you to abandon what might otherwise have been a tempting principle: that logically equivalent theorems should be equally useful.
| 3 | https://mathoverflow.net/users/1946 | 21256 | 14,080 |
https://mathoverflow.net/questions/21258 | 4 | Notations : Suppose V is a closed contact compact manifold with contact form $\alpha$, of dimension 2n+1. Consider the symplectic sub-bundle $ \xi \subset TV $ given by $ \xi=$ ker($\alpha$). So $ \xi \rightarrow V $ is a vector bundle of rank $2n$ admitting the symplectic structure $d\alpha$. Also consider the Reeb vector field "$R$" determined by $\alpha$ and corresponding flow $\phi\_t$. Assume that there is a finite number of simple periodic orbits of Reeb flow and all of them are non-degenerate.
Here is the question :
In general there are plenty of almost complex structure on the symplectic bundle $\xi$ but is there any one which is invariant under Reeb flow, $\phi\_t$ ?
I believe the answer will be no in general due to complicated behavior of $\phi\_t$, but
I wish that I am not right.
| https://mathoverflow.net/users/5259 | Looking for almost complex structure on a contact manifold invariant under flow of Reeb vector field !? | In general there is no invariant complex structure.
Let $\gamma$ be a closed orbit of the Reeb field. Consider a linearization $A$ of the Poincare return map along $\gamma$. $A$ is not, in general, a realification of a complex operator (with respect to any arbitrary complex structure). For example, as far as I remember, its Lefschetz number det(1−A) could be negative, which is impossible for a realification of a complex operator.
| 8 | https://mathoverflow.net/users/2823 | 21268 | 14,086 |
https://mathoverflow.net/questions/21217 | 0 | The following question naturally originates from [this question](https://mathoverflow.net/questions/21168/how-to-classify-the-algebras-cm)
and [this one](https://mathoverflow.net/questions/21090/smooth-gelfand-duality).
While the usual $C^{0}$ Gelfand duality involves a topology on the function algebras considered (it relates compact Hausdorff topological spaces to unital $C^{\*}$-algebras, which in particular are Banach algebras), why the "smooth Gelfand duality" seems, according to what I understood from the above questions, to see *only* the "pure" algebraic structure of certain algebras over $\mathbb{R}$ ?
**Edit:** I've just read the introduction of [this](http://books.google.it/books?id=79bCwXbVnHAC&printsec=frontcover&dq=C-infinity+differentiable+spaces&source=bl&ots=xjtnnn8bSq&sig=BnSizsqq1t7wT8deQfeE--Ts4Ps&hl=it&ei=8w_DS92DFo7U7APZiu2tCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q&f=false). The topology actually enters the picture, but not in the form of a structure of topological algebra on the function spaces that locally model those $C^{\infty}$-differentiable spaces; it enters the picture when [defining](http://books.google.it/books?id=79bCwXbVnHAC&printsec=frontcover&dq=C-infinity+differentiable+spaces&source=bl&ots=xjtnnn8bSq&sig=BnSizsqq1t7wT8deQfeE--Ts4Ps&hl=it&ei=8w_DS92DFo7U7APZiu2tCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q&f=false) a "differentiable algebra" as the quotient of the algebra of smooth functions on $\mathbb{R}^n$ by a *Fréchet-closed* ideal.
But a question still stands: would it be possible to define compact Hausdorff topological spaces in the analogous way?
Perhaps the answer is "no because of a lack of a universal local model $C^{0}(...)$", but "yes in the case of topological *manifolds*".
Does it make sense?
| https://mathoverflow.net/users/4721 | Why "smooth Gelfand duality" does not involve a topology on the algebras? | We don't need a topology because the proofs show that the space can be recovered without a topology. I think it's just that simple. Perhaps this is not satisfactory for you?
Continuing George's comment, there is an important result in the theory of $C^\\*$-algebras: the norm is unique. The reason is that every $C^\\*$-homomorphism is of norm $\leq 1$. In particular, the category of $C^\\*$-algebras is a full subcategory of the category of $\mathbb{C}$-algebras with involution and basically I could also ask for a pure algebraic characterization of the essential image... In particular, the category of compact hausdorff spaces is a full subcategory of the opposite category of commutative $\mathbb{C}$-algebras with involution. Thus we have the same situation as in the smooth Gelfand duality.
| 3 | https://mathoverflow.net/users/2841 | 21271 | 14,088 |
https://mathoverflow.net/questions/21267 | 39 | A number field $K$ is said to be [monogenic](http://en.wikipedia.org/wiki/Monogenic_field) when $\mathcal{O}\_K=\mathbb{Z}[\alpha]$ for some $\alpha\in\mathcal{O}\_K$. What is currently known about which $K$ are monogenic? Which are not? From Marcus's *Number Fields*, I'm familiar with the proof that the cyclotomic fields are monogenic, and for example that $\mathbb{Q}(\sqrt{7},\sqrt{10})$ is not monogenic (it is exercise 30 of chapter 2), but because Marcus eschews anything local, I haven't seen any of the perhaps more natural proofs of these results.
If $K$ is monogenic, is there an effective method of determining those $\alpha\in\mathcal{O}\_K$ for which $\mathcal{O}\_K=\mathbb{Z}[\alpha]$?
More generally, what is known about the minimal number of generators of $\mathcal{O}\_K$ as a $\mathbb{Z}$-algebra? That is, can we determine, or at least put non-trivial bounds on, the minimal $m$ such that $\mathcal{O}\_K=\mathbb{Z}[\alpha\_1,\ldots,\alpha\_m]$ for some $\alpha\_i\in\mathcal{O}\_K$? We know that any $\mathcal{O}\_K$ has an integral basis of $n=[K:\mathbb{Q}]$ elements, so certainly $m\leq n$ (I'm considering that trivial).
| https://mathoverflow.net/users/1916 | Which number fields are monogenic? and related questions | Zev, when $[K:{\mathbf Q}] > 2$, finding all $\alpha$ which are ring generators for ${\mathcal O}\_K$ is a hard problem in general: there are only finitely many choices modulo the obvious condition that if
$\alpha$ works then so does $a + \alpha$ for any integer $a$. In other words, up to adding an integer there are only finitely many possible choices -- which could of course mean there are no choices.
Here is a nice example: what are the possible ring generators for the integers of ${\mathbf Q}(\sqrt[3]{2})$? We know a basis for the ring of integers is $1, \sqrt[3]{2}, \sqrt[3]{4}$, so a ring generator over $\mathbf Z$ would, up to addition by an integer, have the form $\alpha\_{x,y} = x\sqrt[3]{2} + y\sqrt[3]{4}$ for some integers $x$ and $y$ which are not both 0. The index of the ring ${\mathbf Z}[\alpha\_{x,y}]$ in the full ring of integers is the absolute value of the determinant of the matrix expressing $1, \alpha\_{x,y}, \alpha\_{x,y}^2$ in terms of $1, \sqrt[3]{2},\sqrt[3]{4}$, and after a computation that turns out to be $|x^3 - 2y^3|$. We want this to be 1 in order to have a ring generator, which means we have to find all the integral solutions to the equation $x^3 - 2y^3 = \pm 1$. Well, that's a pretty famous example of an equation with only finitely many integral solutions. Up to sign the only solutions are $(1,0)$ and $(1,1)$, so $\alpha\_{x,y}$ is $\sqrt[3]{2}$ or $\sqrt[3]{2} + \sqrt[3]{4}$ up to sign (and then addition by an integer).
Here's a more general cubic exercise, just to put the previous example in some perspective (among concrete examples). Let ${\mathbf Q}(\alpha)$ be a cubic field where $\alpha^3 + b\alpha + c = 0$ for integers $b$ and $c$.
a) Show for $x, y \in {\mathbf Z}$ not both 0 that $[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]] = |x^3 + bxy^2 + cy^3|$. Therefore if $1,\alpha,\alpha^2$ is known to be a ${\mathbf Z}$-basis of the ring of integers, finding all other ring generators besides $\alpha$, up to addition by integers, amounts to solving $x^3 + bxy^2 + cy^3 = \pm 1$ in integers.
b) It is natural to guess from part a that if $\alpha^3 + a\alpha^2 + b\alpha + c = 0$ and $x, y \in {\mathbf Z}$ are not both 0 the index
$[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]]$ should be $|x^3 + ax^2y + bxy^2 + cy^3|$. Decide if that natural guess is right!
In general, finding all possible ring generators (modulo addition by an integer) for the ring of integers in a number field amounts to solving some norm-form equation equal to $\pm 1$, and beyond the quadratic case that kind of equation will have just a finite number of integral solutions. A place to look for further discussion is Narkiewicz's massive tome on algebraic number theory: pp. 64--65 and especially p. 80. It turns out the question of finiteness of the number of possible ring generators up to addition by an integer goes back to Nagell. The general case was settled by Gyory in 1973; see MathSciNet MR0437489.
There's actually a whole book on this theme: Diophantine Equations and Power Integral Bases by István Gaál, Birkhauser, 2002.
Update in 2018: to address your question about finding a ring of integers needing many generators as a $\mathbf Z$-algebra (not just as a $\mathbf Z$-module), see my answer at [Explicit family of number rings $\mathcal{O}\_{K\_n}$ requiring $n$ generators?](https://mathoverflow.net/questions/311133/).
| 45 | https://mathoverflow.net/users/3272 | 21284 | 14,097 |
https://mathoverflow.net/questions/21247 | 4 | I am trying to use the Chebotarev Density Theorem to say something about the Galois groups of a class of polynomials. To be more precise, by factoring a polynomial mod some prime p, I want to show that there is an element in the Galois group of the polynomial with a certain cycle structure. Unfortunately my knowledge of algebraic number theory is rather thin!
In the statement of the theorem, it is required that p be unramified. My question is this: if I can prove that there are no repeated factors in the factorization of the polynomial mod p, does it matter that I don't know whether p divides the discriminant in general or not?
In other words, is the requirement that p be unramified purely to avoid repeated factors, or is there more to it than that?
| https://mathoverflow.net/users/4078 | Ramified primes in the Chebotarev Density Theorem | Adam, the requirement that $p$ be unramified in the number field is to explain the existence of an element (really, conjugacy class) in the Galois group with a certain cycle structure on the roots of a generator for the number field. The way this element of the Galois group is constructed requires algebraic number theory, but it can be translated into a more elementary-sounding proposition about factoring a polynomial mod $p$ at the expense of giving up on being able to apply the result to a few primes for which the method really does work at a more technical level.
If $K = {\mathbf Q}(\alpha)$ and $\alpha$ is an algebraic integer with minimal polynomial $f(x)$ in ${\mathbf Z}[x]$, the elementary proposition is that if $p$ is a prime number such that $f(x) \bmod p$ is a product of distinct irreducibles with degrees $d\_1,\dots,d\_r$ then there's an element of the Galois group of the Galois closure of $K/{\mathbf Q}$ whose cycle structure on $\alpha$ and its ${\mathbf Q}$-conjugates consists of disjoint cycles of length $d\_1,\dots,d\_r$.
The more advanced proposition, which makes no reference to polynomials mod $p$, is that if $p$ is a prime number unramified in $K$, so
necessarily $p{\mathcal O}\_K = {\mathfrak p}\_1\cdots {\mathfrak p}\_r$ for some distinct primes ${\mathfrak p}\_i$ with residue field degrees $d\_i$, then there is an element of the Galois group of the Galois closure of $K/{\mathbf Q}$ whose permutation action on $\alpha$ and its ${\mathbf Q}$-conjugates is a product of disjoint cycles with lengths $d\_i$.
The link between the elementary and advanced propositions is: $\text{disc}(f) = [{\mathcal O}\_K:{\mathbf Z}[\alpha]]\text{disc}(K)$. This equation implies that if $f(x) \bmod p$ has distinct irreducible factors then $p$ doesn't divide $\text{disc}(f)$ and therefore also doesn't divide the discriminant of $K$, so $p$ is unramified in $K$. Moreover, $p$ doesn't divide that ring index, which implies that the shape of the factorization of $p{\mathcal O}\_K$ matches the shape of the factorization of $f(x) \bmod p$. So under the condition that $f(x) \bmod p$ has distinct irreducible factors the elementary and advanced propositions are both applicable (their hypotheses are both satisfied) and lead to the same conclusion: both propositions imply the existence of an element of the Galois group with the same cycle structure as a permutation of the ${\mathbf Q}$-conjugates of $\alpha$. Primes at which the elementary proposition hold are always primes at which the advanced proposition holds, but not conversely: there can be primes $p$ which are unramified in $K$ (that is, $p$ doesn't divide $\text{disc}(K)$) but the reduced polynomial $f(x) \bmod p$ has multiple irreducible factors (that is, $p$ divides $\text{disc}(f)$), so the advanced proposition can be applied to this prime $p$ but the elementary one can not.
Incidentally, for a ramified prime $p$ in $K$ with ramification indices $e\_1,\dots,e\_r$ and respective residue field degrees $d\_1,\dots,d\_r$, it's natural to ask if there might be an element of the Galois group of the Galois closure of $K/{\mathbf Q}$ whose permutation action on the ${\mathbf Q}$-conjugates of $\alpha$ is a product of disjoint cycles where there are $e\_i$ cycles of length $d\_i$ for all $i$. I have a copy of a letter Serre sent to Thomas Hawkins in 2000 which outlines a method to give a counterexample where $[K:{\mathbf Q}] = 6$. That means this naive attempt to extend the Galois group existence technique to ramified primes doesn't generally work.
Here is an explicit example: $K = {\mathbf Q}(a)$ where $a^6 - 35a^4 + 3a^2 - 225 = 0$.
This field has degree 6 and Galois group $S\_4$ over ${\mathbf Q}$. This Galois group acts on the roots in the way $S\_4$ naturally permutes the 6 two-elements subsets of {1,2,3,4}: this is an embedding of $S\_4$ into $A\_6$, which will be important.
Using PARI, 3 factors in the integers of $K$ as $P^2Q$ where $P$ and $Q$ both have residue field degree 2. Now *if* there were a "corresponding" element of the Galois group of $K$ over ${\mathbf Q}$ as dreamed above, it would permute the 6 roots as a product of three disjoint 2-cycles. But alas, that is not an even permutation of the roots and I already said the Galois group is $S\_4$ acting on the roots as a subgroup of $A\_6$, so entirely by even permutations. Thus we have a contradiction so there is no such "dream automorphism" associated to 3 in the Galois group.
By the way, this degree 6 polynomial did not come out of nowhere: it is related to a 3-adic approximation of another polynomial, but that connection would take longer to describe than I wish to write about here, as this answer is already pretty long.
| 12 | https://mathoverflow.net/users/3272 | 21289 | 14,099 |
https://mathoverflow.net/questions/21290 | 17 | Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$?
I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery).
| https://mathoverflow.net/users/5337 | What's an example of a transcendental power series? | If $k$ has characteristic zero, then $\displaystyle e^t = \sum\_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any purported algebraic relation satisfied by $e^t$.
**Edit:** Let me fill in a few details. Given a polynomial $P$ in $e^t$ of degree $d$ where each coefficient is a polynomial in $k[t]$ of degree at most $m$, the possible terms that appear in any formal derivative of $P$ lie in a vector space of dimension $(m+1)(d+1)$, so by taking at least $(m+1)(d+1)$ formal derivatives we obtain too many linear relationships between the terms $t^k e^{nt}$. The coefficient of $e^{dt}$ in particular eventually dominates all other coefficients.
| 26 | https://mathoverflow.net/users/290 | 21291 | 14,100 |
https://mathoverflow.net/questions/21286 | 11 | In his book *Commutative Ring Theory*, Matsumura proves that if a local ring is equidimensional, and a quotient of a regular local ring, then its completion is equidimensional.
What is an example of a local ring which does not admit such a presentation?
| https://mathoverflow.net/users/1594 | A local ring not a quotient of a regular local ring | A source available online is this [paper](https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-70/issue-none/Examples-of-bad-Noetherian-local-rings/nmj/1118785530.full) "Examples of bad Noetherian rings" by Marinari (example 2.1).
The reason many of these types of construction work is because of the following vague and counter-intuitive phenomemon:
*It is usually easier than we think for a complete local ring to be a completion of a Noetherian ring with certain properties.*
For example, there is this amazing [theorem](https://www.jstor.org/stable/2154327) by Heitmann that most complete local ring of depth at least $2$ is a completion of a UFD !
So back to Marinari's paper, the example is as follows: start with some local Artinian ring $(Q,m)$ such that $Q$ is not Gorenstein. Then $Q[[X]]$ is complete, and one can find a local domain $R$ such that $\hat R=Q[[X]]$. Now if $R$ is a quotient of a regular local ring, then the comletion of $R$ is generically a complete intersection. But $Q[[X]]$ is not even generically Gorenstein, since $Q$ is not.
| 15 | https://mathoverflow.net/users/2083 | 21304 | 14,105 |
https://mathoverflow.net/questions/21121 | 14 | The homology of an infinite loop space, which represents a spectrum, is an algebra over the Dyer-Lashof algebra (see for example Cohen-Lada-May's Springer volume, or for part of the story the more accessible Luminy notes of Bisson-Joyal). Has anyone used this to construct a spectral sequence converging under some assumptions to $[X,Y]$, the homotopy classes of infinite-loop-maps between $X$ and $Y$, which starts with some kind of derived (Ext/Tor) maps between their homology in the category of algebras over the Dyer-Lashof algebra? Have any calculations been done with such a spectral sequence?
| https://mathoverflow.net/users/4991 | Dyer-Lashof based spectral sequence for homotopy classes of maps between infinite loop spaces (spectra). | This might not quite be what you're looking for, Dev, but you should check out Paul Goerss and Mike Hopkins' "Multiplicative ring spectra project," on Paul's [webpage](http://www.math.northwestern.edu/~pgoerss/). They construct such a spectral sequence using Andre-Quillen cohomology in "Moduli spaces of commutative ring spectra," and "Andre-Quillen (co-)homology for simplicial algebras over simplicial operads." A relevant theorem would be 4.3 in the first reference, which gives the spectral sequence.
Though this doesn't use Dyer-Lashof operations, they appear in section 6 (especially Prop 6.4) where Goerss and Hopkins give a second spectral sequence which computes the $E\_2$ term of the original spectral sequence. The new $E\_2$ term is given in terms of an $Ext$ functor in the category of unstable modules over the Dyer-Lashof algebra.
They use this machinery to show in section 7 that the space of $E\_\infty$ maps between Lubin-Tate spectra is homotopically discrete. If you're looking for computations using these spectral sequences, that's a great place to start.
| 9 | https://mathoverflow.net/users/4649 | 21305 | 14,106 |
https://mathoverflow.net/questions/21310 | 2 | Given a planar graph $G=(V,E)$ with vertices $V$ and edges $E$, call $\bar G = (V,\bar E)$ a non-planar extension of $G$ if $\bar G$ is non-planar and $E \subset \bar E$.
I'm interested in minimal non-planar extensions in the sense that if $\bar G$ is a non-planar extension of $G$, there is no non-planar extension of $G$ that is a subgraph of $\bar G$.
I first wondered whether these minimal extensions could be unique, but this is quickly disproved by the existence of maximally planar (also called triangulated) graphs. I refine this question slightly:
(1) Is the minimal non-planar extension of an arbitrary planar graph unique up to isomorphism?
(2) What if we define minimality as having the smallest number of edges for one of these extensions?
and also:
(3) Do these extensions mean anything interesting for representations of algebraic objects (eg. groups) on them?
| https://mathoverflow.net/users/5312 | Minimal Non-planar Extensions of a Graph | (1) and (2) Take the union of a $K\_5$ missing an edge and a $K\_{3,3}$ missing an edge. This graph has 2 minimal non-planar extensions, which are obviously not isomorphic.
| 6 | https://mathoverflow.net/users/3684 | 21311 | 14,109 |
https://mathoverflow.net/questions/21295 | 37 | Let $(\mathcal{M},g)$ be a $C^{\infty}$-Riemannian manifold. A basic fact is that $g$ endows the manifold $\mathcal{M}$ with a metric space structure, that is, we can define a distance function $d:\mathcal{M}\times\mathcal{M}\longrightarrow\mathbb{R}$ (the distance between two points will be the infimum of the lengths of the curves which join the points)
which is compatible with the topology of $\mathcal{M}$. Of course $d$ is continuous function, but what can we say about the differentiability of $d$?, is it smooth?. If not, Is there some criterion to know when it is?
Thanks in advance.
| https://mathoverflow.net/users/5069 | Smoothness of distance function in Riemannian Manifolds | As others mentioned, you have to remove the diagonal of $M\times M$ or square the distance function. Then, for a complete $M$, the answer is the following.
The distance function is differentiable at $(p,q)\in M\times M$ if and only if there is a *unique* length-minimizing geodesic from $p$ to $q$. Furthermore, the distance function is $C^\infty$ in a neighborhood of $(p,q)$ if and only if $p$ and $q$ are not conjugate points along this minimizing geodesic.
Thus, the function is smooth everywhere if and only if $M$ is simply connected and the geodesics have no conjugate points. This property has numerous equivalent reformulations, including the following
* for every pair of points, there is a unique minimizing geodesic between them;
* for every pair of points, there is a unique geodesic between them;
* every geodesic is minimizing;
* the exponential map at every point $p\in M$ is a diffeomorphism from $T\_pM$ to $M$.
In general, the distance function has *one-sided* directional derivatives everywhere. This derivative has a nice description in the case when you fix $p\in M$ and study the function $f=d(p,\cdot)$. Namely let $q\in M$, $q\ne p$, and denote by $\vec{qp}$ the set of initial velocity vectors (in $T\_qM$) of unit-speed minimizing geodesics from $q$ to $p$. Then, for a vector $v\in T\_qM$, the one-sided derivative $f'\_v$ of $f$ in the direction of $v$ is
$$ f'\_v=\min\{-\langle v,\xi\rangle:\xi\in \vec{qp}\} . $$
This follows from the first variation formula and holds not only in Riemannian manifolds but also in Alexandrov spaces. It is not hard to derive the above differentiablity properties from this.
I don't have a textbook reference for this precise formulation in the Riemannian case, but any book that covers Berger's lemma about geodesics realizing the diameter probably has directional derivatives as a sublemma. For Alexandrov spaces, the standard reference is Burago-Gromov-Perelman's paper. An intro-level proof (not in a full generality) can be found in (a shameless advertisement follows) "A course in metric geometry" by Burago, Burago and myself, section 4.5.
| 64 | https://mathoverflow.net/users/4354 | 21316 | 14,110 |
https://mathoverflow.net/questions/21315 | 9 | On a Kahler manifold, the different Laplacians are compatible: $\Delta\_d=2\Delta\_{\bar{\partial}}=2\Delta\_{\partial}$.
Are there non-Kahler Hermitian manifolds where the above identity holds?
| https://mathoverflow.net/users/nan | Non-Kahler manifolds where the different Laplacians are compatible | Hermitian manifolds $M$ where $$\Delta\_d f=2\Delta\_{\bar{\partial}} f=2\Delta\_{\partial} f$$ holds for every smooth function $f$ on $M$ are called *balanced*.
For more information, you can search for "balanced hermitian manifolds", [Here](http://cdsweb.cern.ch/record/422889/files/), for instance, is a paper that reviews their basic properties and conditions to be Kahler.
| 12 | https://mathoverflow.net/users/2384 | 21317 | 14,111 |
https://mathoverflow.net/questions/17641 | 8 | Suppose that C is a fusion category (over the complex numbers) and that Z(C) is its Drinfel'd center. By definition an object in Z(C) consists of an object V in C together with a collection of half-braidings $V \otimes W \rightarrow W \otimes V$ for every object W in C satisfying some naturality conditions. Hence there is a restriction functor R:Z(C)->C given by forgetting the half-braiding. Adjoint to this is an induction functor I:C->Z(C).
A Theorem of Etingof-Nikshych-Ostrik says that $R(I(V)) = \bigoplus\_X X \otimes V \otimes X^{\\*}$. In particular, we see that I(V) is $\bigoplus\_X X \otimes V \otimes X^{\\*}$ (where X ranges over simple objects up to isomorphism) together with some particular choice of half-braidings. I'm pretty sure I know what those half-braidings are. In particular there's a nice picture (the X,Y summand of the half-braiding with W is a sum over diagrams with a trivalent vertex connecting W to X\* and Y\* and another trivalent vertex connecting X and Y to W, where the two vertices range over dual bases).
What I really would like is a reference that explains this so that I don't have to write it up myself. The only description I know is in ENO's "On Fusion Categories" where it's written in terms of weak Hopf algebras.
The motivation is removing any mention of weak Hopf algebras from the construction in Section 5 (about cyclotomicity of certain Drinfel'd centers) in Scott and my [Noncyclotomic Fusion Categories](http://arxiv.org/abs/1002.0168). It turns out that the diagram description above can be slightly modified (in a way suggested to me by Ben Webster) in order to give a description of I(V) where V is an object in a non-split fusion category over an arbitrary field.
| https://mathoverflow.net/users/22 | Is there a source for a diagrammatic description of the induction functor C->Z(C)? | In the meanwhile a reference has appeared: Theorem 2.3 in <http://arxiv.org/abs/1004.1533>
| 2 | https://mathoverflow.net/users/1035 | 21320 | 14,113 |
https://mathoverflow.net/questions/21314 | 15 | Motivated by [this question](https://mathoverflow.net/questions/21290/whats-an-example-of-a-transendental-power-series), I remembered a question I was curious about sometime which I am sure has some easy and nice example for it as well, which I just can't think of for some reason. I want an example of a power series that is not differentially algebraic. A differential algebraic power series is a series $f(t)$ satisfying an equation $P(t,f(t),f'(t),\ldots,f^{(k)}(t))=0$ for some $k$ and some polynomial $P$ in $k+2$ variables.
**Update:** examples in the comments below ($\sum t^{n^n}$, $\sum t^{2^n}$) make me ask a refinement (of a sort) for the original question: these examples are reminiscent of all those Liouville-flavoured examples of transcendental numbers, - I wonder if there is a Liouville-flavoured proof, stating that if the polynomial P is of given (multi)degree, some inequality holds that is obviously impossible for the series above?
**Update 2:** there are quite a few examples now, and I am tempted to accept the $\sum t^{n^n}$ answer since the example itself is easy and it came together with an easy explanation. I wonder what are other general approaches besides the ones that are exhibited in answers here (looking at p-adic norms of coefficients and looking at powers of $t$ with nonzero coefficients).
| https://mathoverflow.net/users/1306 | An example of a series that is not differentially algebraic? | You'd be better off in characteristic zero, for $f^{(p)}(t)=0$ in characteristic $p$. Then the sea of zeroes example $\sum\_n t^{n^n}$ will do the trick. For large enough $n$, there will be a cluster of non-zeroes in degrees $kn^n-m$ for small (and bounded) $k$ and $m$, "reachable" only by products of $(t^{n^n})^{(s)}$, the same $n$, bounded $s$. Their vanishing will give infinitely many linear relations on the coefficients of $P$, which you can explicitly write down and see that there are no nonzero solutions on coefficients.
| 7 | https://mathoverflow.net/users/5301 | 21321 | 14,114 |
https://mathoverflow.net/questions/11183 | 6 | I have a N-point metric space defined by the pairwise distance matrix. I want to encode these N points with binary strings, i.e. each point will be mapped to a vertex in a hypercube. The lengths of the edges on the hypercube could be different for different dimensions. The hypercube basically is a hyper-rectangle.
Now the questions are the following:
1. Given the dimension of the hyper-rectangle, what is the lower bound of the distortion to the original metric space?
2. How to achieve that, i.e., the lengths of the edges, the vertices for each point?
3. Is the optimal embedding P or NP?
$A = (P,C), |P| = N, C\in [0,1]^{N\times N}$, find a mapping $f:P \rightarrow \times\_{j=1}^D $ { $0,l\_j $ }, $l\_j > 0$.
such that for any $\frac{1}{\mu} C\_{ij} \le |f(P\_i)-f(P\_j)| \le \mu C\_{ij} $,
where $\mu \sim \Omega(g(D,N))$, is a polynomial function.
Thanks a lot!
| https://mathoverflow.net/users/2013 | How can I embed an N-points metric space to a hypercube with low distortion? | Y. Bartal has studied a related problem of embedding metric spaces to *hierarchically separated trees*. With $1 < \mu$ being a fixed real number, a $\mu$-HST is equivalent to the set of corners of a rectangle whose edges are of length $c, c\mu^{-1}, c\mu^{-2}, \dots, c\mu^{1-D}$ with the $l\_\infty$-metric. That is, if you think of the space as the set of bit sequences of length $D$, the distance of two sequences is $c\mu^{-j}$ if they first differ in bit $j$.
Now in your question you didn't ask for $\infty$-metric, but for this set of points, it doesn't really matter which metric you take because the distortion between this and either the $l\_1$ or $l\_2$ metric is bounded by a constant (if you fix $\mu$ but $D$ can vary).
(This metric can be considered a graph metric on a special tree, that is, one where the points are some (but not necessarily all) vertexes of a tree graph with weighted edges, and the distance is the shortest path. This is where "tree" in the name comes from.)
Now Bartal's result in [1] basically says that you can embed any metric space *randomly* to a $\mu$-HST with distortion at most $\mu(2\ln n+2)(1+\log\_\mu n)$ where $n$ is the number of points. (Also, this embedding can be computed with a randomized polynomial algorithm.)
For this, you need to know what a distortion $\alpha$ random embedding $f$ means. It means that for any two points $d(x,y) < d(f(x),f(y))$ is always true and that the expected value of $d(f(x),f(y))$ is at most $\alpha d(x,y)$. For many applications, this is just as good as a deterministic embedding with low distortion. In fact, you can make a deterministic embedding with low distortion from it by imagining the metric $d^\* $ on the original space where $d^\*(x,y) = E(d(f(x), f(y))$, but this notion isn't too useful because the resulting metric does not have nice properties anymore (it's not HST). Indeed, I believe the randomness is essential here as I seem to remember reading somewhere that you can't embed a cycle graph (with equal edge weights) to a tree graph with low distortion.
Anyway, this might not really answer your question. Firstly, $D$ (the number of dimensions of the rectangle) is not determined in advance, but that's not a real problem because if you have $D$ significantly different distances in the input metric then you need at least that large a $D$ for any embedding; and with this embedding you don't need a $D$ larger than $\log\_\mu (\Delta/\delta)$ where $\Delta$ and $\delta$ are the largest and smallest distances in the input. The real problem is that you seem to want to know a deterministic embedding, and the highest possible distortion necessary in that case, which this really doesn't tell. For example, a cycle graph with an even number $n$ of vertexes can of course be embedded isometrically to a cube of dimension $n/2$.
Survey [2] has some more references.
[1]: Yair Bartal, On Approximating Arbitrary Metrics by Tree Metrics. *Annual ACM Symposium on Foundations of Computer Science*, 37 (1996), 184–193.
[2]: Piotr Indyk, Jiří Matoušek, Low-distortion embeddings of finite metric spaces. Chapter 8 in *Handbook of Discrete and Computational Geometry*, ed. Jacob E. Goodman and Joseph O'Rourke, CRC Press, 2004.
| 3 | https://mathoverflow.net/users/5340 | 21322 | 14,115 |
https://mathoverflow.net/questions/21257 | 5 | I am interested in examples where the [Shooting Method](http://en.wikipedia.org/wiki/Shooting_method) has been used to find solutions to systems of ordinary differential equations that are either
* reasonably large systems, or
* the search algorithm in the shooting parameters is somewhat prohibitive because of the nature of the solutions, or
* both of the above.
Any references, descriptions, recent progress, folklore, in the ballpark would be of interest. Feel free to interpret "reasonably large" subjectively if necessary.
| https://mathoverflow.net/users/3623 | What is state of the art for the Shooting Method? | I don't know about state-of-the-art and I'm not sure if this is the kind of thing you were looking for...however in my two first papers I've used the shooting method in a parameter space that was originally too big (4 and 6 dimensions if I recall correctly) and the problem was that with randomly chosen parameters the numerical solver would not reach the other end of the domain and so I could not use a root-finding algorithm to search for the correct initial conditions.
The problem was there were unstable directions in the ODE and thus even with the correct initial conditions, the numerical noise would grow so large that you would not reach the other side.
My solution was to find more natural variables to use (using an algebraic similarity solution that satisfies the boundary conditions) and to rewrite the system in terms of the new variables. In the new variables the similarity solution is a fixed point and one can reach this fixed point only via its **stable manifold**, which had a lower dimension than the original space (in my case...). This allowed the root-finding algorithm to kick in and find a solution.
OK, This was a little vague. Here are the two papers (shameless plug):
<http://arxiv.org/abs/0711.0730>
<http://arxiv.org/abs/0711.0734>
(Added later:)
Recently I've been working on another problem that has highly unstable directions and there I use the *collocation* method, which (AFAIK) basically amounts so splitting the domain into many smaller part, doing shooting on each part, and trying to get the pieces to match up. If the problem is linear, this is a simple linear problem, if the problem isn't linear you need a non-linear root finder. I didn't write the code for the collocation, Matlab does it for me...look up BVP4C or BVP5C.
In writing this answer, I looked for "collocation method" online and found very little that seemed relevant. So I can only refer you to the Matlab function. perhaps someone else can find a reference that is relevant here.
| 7 | https://mathoverflow.net/users/3578 | 21324 | 14,116 |
https://mathoverflow.net/questions/21298 | 12 | An N-subset $\{x\_1,\dots,x\_N\}$ of a compact set $X\subset \mathbb R^d$ is called a set of *Fekete points* (named after [Michael Fekete](http://en.wikipedia.org/wiki/Michael_Fekete)) if it maximizes the product $$\prod\_{1\le k<j\le N}|x\_k-x\_j|\qquad (1)$$ among all such $N$-tuples. When $X\subset \mathbb C$, one can express this product in terms of the Vandermonde determinant. In this case Fekete points are of particular interest in approximation theory (as interpolation nodes). Generally, they have no explicit form and must be found numerically. But there are exceptions.
When $X$ is a circle, Fekete points are equally spaced. This is well-known and can be proved like this: we may assume $x\_k=\exp(i\theta\_k)$ with $0=\theta\_1<\theta\_2<\dots <\theta\_N<2\pi$. For a fixed integer $m$, $1\le m\le \lfloor N/2 \rfloor$, consider the product $\Pi(m)=\prod\_{k=1}^N|x\_{k+m}-x\_{k}|$ with indices taken mod $N$. Since each point $x\_k$ appears in just two terms of this product, it is not hard to see that $\Pi(m)$ is maximal when $x\_k$ are equally spaced. (Indeed, $\log \sin$ is a concave function.)
Now let $X$ be the union of two concentric circles of radii $r$ and $R$. Assume that $N=2n$ is even and that exactly $n$ of the points lie on each circle. With this additional assumption we are not quite in the Fekete problem anymore, but we have an obvious candidate for the maximum -
>
> Under the assumptions of the preceding paragraph, is the product (1) maximized by equally spaced, interlaced points? E.g., by
> $$x\_k=\begin{cases} r\exp(2\pi i k/n), \quad k=1,\dots,n \\
> R\exp(\pi i (2k-1)/n), \quad k=n+1,\dots,2n \end{cases} \qquad (2)$$
>
>
>
Presumably, a solution would consist of two steps.
1. Show that the maximum is attained by interlaced points, i.e., when ordered by the
argument, the points alternate between the two circles.
2. Show that for interlaced points, maximum of (1) is attained when they are equally spaced.
One can hope that step 2 is doable with standard calculus tools, since there should be no other local maxima for interlaced points. But Step 1 seems to require a more imaginative approach.
Why do I find this question interesting? There is a natural way to distribute N points on a circle (i.e., at equal distances), and this configuration is known to give the solution for many extremal problems, such as the Fekete problem, isoperimetric N-gon problem, etc. There is also a natural way to distribute **two** sets of N points -- namely, by formula (2), but I don't know of any nontrivial problem for which the optimality of such configuration has been established.
| https://mathoverflow.net/users/2912 | How to best distribute points on two concentric circles? | (This was too long to fit in a comment.)
For subsets of the complex plane, a slightly different perspective is that the square of the quantity you want to maximize is the absolute value of the discriminant of the monic polynomial whose roots are precisely the points. Up to a factor depending only on the degree, the discriminant of a monic polynomial is the product of the values of the polynomial at the roots of the derivative. Also note that in the configuration you suggested, the derivative of the polynomial is almost the optimal polynomial for points on a circle: maybe you can prove that the derivative of an optimal polynomial is close to an optimal polynomial?
| 2 | https://mathoverflow.net/users/4344 | 21325 | 14,117 |
https://mathoverflow.net/questions/21327 | 4 | It is an easy exercise to show that the Euclidean plane cannot be partitioned into round circles (note however that it is possible to do so for $\mathbb{R}^3$). It seems almost obvious that it is not possible to partition the plane into Jordan curves either.
However, I am not able to design a proof that does not use the choice axiom. With choice, assume you have such a partition $J$ and define on $J$ a partial ordering: $j<k$ if the curve $j$ is contained in the interior of $k$. Any decreasing chain $j\_n$ has a lower bound: if $K\_n$ is the closure of the interior of $j\_n$, then $K\_n$ is a decreasing sequence of compact sets, thus there is some point $x$ in the intersection. Then the curve of $J$ that contains $x$ is a lower bound of $(j\_n)$. Now Zorn Lemma ensures that there is a minimal element $j$ in $J$. But this is obviously impossible since $j$ would have a non-empty interior, therefore containing another curve of $J$.
The question is therefore the following: can we prove that there exist no partition of the plane into Jordan curves without assuming the Choice axiom?
| https://mathoverflow.net/users/4961 | Is it still impossible to partition the plane into Jordan curves without choice? | It is possible to change your argument so that the choice is over countable set; hope this is good enough. Namely, topology on the plane has countable base (say, circles at rational points with rational radii); let's index this base as $U\_1,\dots,U\_n,\dots$; your argument can be used to construct a sequence of Jordan curves $C\_1,\dots,C\_n,\dots$ such that $C\_{i+1}$ is contained in the interior of $C\_i$, and $U\_i$ is not contained in the
interior of $C\_i$.
| 4 | https://mathoverflow.net/users/2653 | 21330 | 14,119 |
https://mathoverflow.net/questions/21328 | 4 | Consider a tree with k nodes and for each node v the vector **l**v = (lv0, lv1, ..., lvk-1) with lvd the number of leaves (!) with distance d to v. I wonder whether two nodes v, w with **l**v = **l**w are [conjugate](https://mathoverflow.net/questions/10807/conjugate-vertices-and-distinguishing-properties) (I guess they are). Can anyone help me to prove this - or give a counter example?
| https://mathoverflow.net/users/2672 | A distinguishing node property in trees? | I have a counterexample. It is not enough just to count leaves, since this doesn't take into account the number of possible ways to arrive at those leaves.
Consider the graph below.
```
A - B - C - D - E - F
| |
G H
|
I
```
I think the vector for C and D both is 002200000, since they each have two leaves at distance 2 and two leaves at distance 3. But they are not conjugate, since C has degree 3 and D has degree 2.
I think this might be a minimal counterexample.
| 8 | https://mathoverflow.net/users/1946 | 21335 | 14,123 |
https://mathoverflow.net/questions/21340 | 3 | Let $A$ be a Dedekind domain, $K:=\text{Frac}(A)$ and $L/K$ finite so that the integral closure $B$ of $A$ in $L$ is Dedekind. If $A$ is a PID, for example, then there exists an integral basis : $B$ is free over $A$, and this basis gives a basis of $L/K$.
In particular, if $A=\mathbb{Z}$ or $\mathbb{Z}\_p$ (or even the ring of integers of a finite extension of $\mathbb{Q}\_p$), and $L/K$ finite then an integral basis exists.
I'm looking for a non-example ; $A$ Dedekind, $L/K$ finite for which there does not exist an integral basis. I suspect that taking $A$ to be the ring of algebraic integers of a finite extension of $\mathbb{Q}$ which is a UFD but not PID might do the trick.
Highly pathological examples (e.g. outside the realm of number fields) are very much welcome !
| https://mathoverflow.net/users/4399 | Non-existence ofintegral basis of integral closure in a finite extension of Frac(A), A Dedekind. | Watch out: just because $A$ is a PID does not make $B$ a free $A$-module. You need to know that $B$ is finitely generated over $A$ to conclude $B$ has an $A$-basis when $A$ is a PID. If $L/K$ were *separable* then using discriminants you can stuff $B$ inside a finitely generated $A$-module so $B$ is finite free if $A$ is a PID.
If we drop the separability condition on $L/K$ in this corollary, then the whole discriminant
argument breaks down and in fact $B$ need not be a finitely generated $A$-module, even if $A$ is a PID! For an example, see Exercise 11 on p. 205 of Borevich and Shafarevich's "Number Theory".
I wrote up a course handout with examples in number fields where the top ring of integers is not a free module over the bottom ring of integers: look at
<http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf>. Note that the bottom ring having class number greater than 1 is a necessary condition for such a phenomenon to occur but it's not sufficient. An example illustrating that is at the end of the handout.
| 8 | https://mathoverflow.net/users/3272 | 21348 | 14,132 |
https://mathoverflow.net/questions/21287 | 13 | I am looking for references talking about different ways to prove flag variety $G/B$ is projective variety. Now I have some in mind:
1. There is a proof in Humphreys Linear algebraic groups, he first prove $G/S$ is a complete algebraic variety hence a projective variety, where $S$ is a Borel subgroup of $G$ of largest possible dimension. Then he used the Borel Fixed Point Theorem to obtain the assertion.
2. There is a proof in A.L.Onishchik and E.B.Vinberg Lie groups and algebraic groups, they used Chevalley's theorem to give the unique projective variety structure to the coset variety $G/B$.
3. There is a sketch of the proof in Tanisaki's D-modules, Perverse sheaves and Representation theory, he identified $G/B$(in particular case, suppose $G=GL\_{n}(k)$) with set of flags in $k^{n}$. Then it is clear that one can give the projective variety structure on the set of flags.
4. One can also prove the flag variety can be embedded to Grassmannian variety, Then using Plucker embedding to give projective variety structure to this Grassmannian.
5. I think one can prove the line bundle on $G/B$ is positive and then use Kodaira embedding to prove it is a projective variety. But I did not find a proper expository reference.
Are there any other interesting proof? Any related comments are welcome. Thanks!
Edit: $G$ is an algebraic group, $B$ is Borel subgroup and $k$ is algebraic closed and $char k=0$
| https://mathoverflow.net/users/1851 | How many ways are there to prove flag variety is a projective variety? | I'm not sure this is an answer, but it got too long to be a comment! The projectivity of $G/B$ seems to me to follow from two facts: a) a homogeneous space $G/H$ is always a quasi-projective variety (which is due to Chevalley), and b) the variety $G/B$ must be complete.
The first fact clearly has nothing to do with Borel subgroups but it would be maybe interesting to ask about how many proofs we have of it. The idea of the construction is clear: find a representation of $G$ which contains a line $L$ which has $H$ as its stabilizer and then take the orbit of $L$ in $\mathbb P(V)$, but to see that you get a categorical quotient this way takes some more care: you need to use some infinitesimal properties, (which you can tidily say using the Lie algebra of course).
The second fact perhaps depends on what you're willing to assume, but it must come down to the Borel fixed point theorem in some form or other, since this tells you that if $G/H$ is complete, then any solvable subgroup will be contained in a conjugate of $H$, thus Borel subgroups are the only solvable subgroups with a chance of having an associated homogeneous space which is complete.
The argument from Humphreys' book uses the strategy of picking first a maximal solvable subgroup of largest dimension so as to show the orbit has to be closed (essentially because the stabilizer is largest so the orbit has smallest, but even then you use the Borel fixed point and the theorem for $GL\_n$ if I remember correctly). I don't know how Onishchik and Vinberg get around this (if they do). Then, as the question says, you get the general result from the Borel fixed point theorem.
The proof for $GL\_n$ shows that $G/B$ is projective if you take $B$ to be the subgroup of upper triangular matrices, but one still needs to show that this subgroup is a Borel, which again I only know how to do using the Borel fixed point theorem in some form. (And of course you can use the same strategy for other classical groups if you can eyeball a candidate Borel subgroup).
Given that, it's maybe worth pointing out that in all of this you can get away with a weak version of the Borel fixed point theorem: namely if $V$ is a representation of a solvable group $H$, and $X$ is an $H$-stable closed subvariety of $\mathbb P(V)$ then $X$ has an $H$-fixed point. This can be shown just using the Lie-Kolchin theorem (that is, that a representation of a solvable group contains a one-dimensional subrepresentation), which can be proved directly. Since the standard proof of the general Borel fixed point requires you to use something like Zariski's main theorem, this is maybe a noticeable saving.
All of this leads me to wonde how many proofs do we know for i) the fact that for any closed subgroup $H$ the homogeneous space $G/H$ has to be quasi-projective and ii) the Borel fixed point theorem (or some variant)?
| 7 | https://mathoverflow.net/users/1878 | 21352 | 14,135 |
https://mathoverflow.net/questions/21358 | 1 | Let *C, E* be small categories, let *Ĉ* = Set*C*op, and let *F:Ĉ → Ê* be cocontinuous. I think *F* will always have a right adjoint when *C, E* are small, but not necessarily if they're large. Is that right?
| https://mathoverflow.net/users/756 | Does every cocontinuous functor between categories of presheaves on small categories have a right adjoint? | The part "*F* will always have a right adjoint when *C*, *E* are small" is definitely right. Using some mildly overkill machinery: in this case *Ĉ* and *Ê* are locally presentable categories, and the result is then the **adjoint functor theorem for locally presentable categories**:
Theorem: Let C and D be locally presentable categories and F : C → D a functor. Then
1. F has a right adjoint iff F preserves small colimits.
2. F has a left adjoint iff F is accessible (preserves κ-filtered colimits for some κ) and preserves small limits.
(Reference: *Higher Topos Theory* Corollary 5.5.2.9 for the (∞,1)-categorical version)
| 4 | https://mathoverflow.net/users/126667 | 21363 | 14,143 |
https://mathoverflow.net/questions/21361 | 27 | I was reading recently online Peter May's complaints (I'm a fan, you can tell, I'm sure) about teaching the third quarter of the graduate algebra sequence at the University of Chicago. This course focuses on homological algebra and attempts to be as up-to-date as possible. May's conundrum stems from the fact that homological algebra is inexorably tied to algebraic topology and as a result, it's difficult to separate the 2 out in the course completely. May questions whether or not this is in fact a good idea; however, since this an algebra course and not a topology course, he feels compelled to work hard to do this.
That being said, he raises a very good pedagogical problem in the teaching of mathematics, particularly at the graduate level where the better schools are trying to prepare students to enter research as quickly as possible. Mathematics is now a very holistic, intertwined discipline: Algebra increasingly permeates virtually all of mathematics, the study of manifolds now requires very sophisticated analytic tools from differential equations and functional analysis, probability theory now partakes of a considerable amount of harmonic analysis, mathematical physics is now a major player in the construction of new mathematical structures-I could go on and on, but you get the idea.
So here's the question: Is the old model of keeping the subdisciplines of mathematics separate in coursework for the sake of focus obsolete? I know a lot of mathematicians in recent decades have begun to draw from various disciplines in constructing the first year graduate sequences of most universities; Columbia is one local example. The question is really are they going far enough? The problem of course is that when you begin weakening those artificial barriers, you run the risk of them collapsing altogether and you ending up with a hodgepodge of theory and methods that seems to have no focus.
So anyone want to comment on what the solution here might be from their own experiences as both teachers and students? How far should courses go in being interrelated? And does this lead to better prepared graduate students for the research level?
| https://mathoverflow.net/users/3546 | The Interrelationship Problem Of Modern Mathematics- How To Deal With it In First Year Graduate Courses? | Of course you should show students, taking into account their backgrounds, that the material they are learning in one course is relevant elsewhere. It makes it clearer to the students that topics they are studying have wide usefulness. At the same time, if you know the students don't have a background to appreciate the technicalities coming from other disciplines (not everyone in algebra has had algebraic topology), then you may have to restrict yourself only to making some broad general remarks, although maybe one or two *special worked examples* from the other disciplines would be accessible without a lot of machinery.
When I discussed characters in an algebra course, I explained a little about Fourier series both for context (otherwise the concept can seem rather far-out) and so they'd see that the otherwise idiosyncratic theorems on characters are related to properties of Fourier series.
I don't think such discussions in a first-year course are going to make the students better researchers, but it will make them better appreciate what they are supposed to be learning.
| 15 | https://mathoverflow.net/users/3272 | 21366 | 14,145 |
https://mathoverflow.net/questions/21347 | 2 | Say I have a black box generating data samples, and I want to estimate the parameters of the black box from the samples.
The black box works like this: it has a parameter **m** (a real number), and to generate a value **v**, it first generates v0 according to a normal distribution (with mean *m* and variance 1), and if v0 is positive it returns v0, if not it returns 0.
So my data samples will be a bunch of zeroes and positive real numbers.
So my question is, from a sample, how do I estimate **m**?
And what kind of mathematical tools do I use to reason about this?
To me, this looks like a straightforward case of bayesian probability, where I would use p(samples|m) to get p(m|samples) and have some prior on the distribution of m.
So uncle Bayes would say: $p(m|series) = p(series|m) \* p(m) / p(series)$
Since the samples are independant, $p(samples|m) = \prod p(sample|m)$
...but some of those $p(sample|m)$ are "discrete probabilities" (when the value is 0), and some are continuous probabilities! Can I muliply them just like that?
(Same goes for calculating $p(samples)$)
Can someone help me clear the confusion?
| https://mathoverflow.net/users/5363 | Estimating the mean of a truncated gaussian curve | OK, let me fully address the question since there is no easy way out. The
normal approach is to maximize the "likelihood" of the data under the
parameter. The key question here is how to define likelihood for a
mixed distribution. Let's use the standard approach as our guide.
Parameter estimation is usually based on the idea that we want to
choose parameters that make our data "the most likely." For a discrete
probability distribution, we interpret this to mean that our data is
the most probable. But this breaks down in the case of continuous
probability distributions, where, no matter our choice of parameters,
our data has probability zero.
Statisticians thus replace the probability with the probability
density for continuous distributions. Here is the justification for
this. Instead of actually having a set of numbers drawn from the
probability distribution, you have a highly accurate
measurement---say, your sequence $\{x\_i\}$ for $i = 1,\dots,n$ tells
you that the true value of the (still unknown) sequence $\{g\_i\}$ satisfies $|x\_i -
g\_i| < \varepsilon$ for all $i$. When $\varepsilon$ is sufficiently
small, the replacement
$$
\mathbb{P}(|x\_i - g\_i|) < \varepsilon )\approx \varepsilon p\_{g}(x\_i)
$$
is very accurate, where $p\_g$ is the pdf of $g\_i$. Assuming that
your sequence is iid, we are led to the approximation
$$
\mathbb{P}(|x\_i - g\_i| < \varepsilon \text{ for all } i)
\approx \varepsilon^n \prod\_{i=1}^n p\_g(x\_i).
$$
We thus choose the pdf from our family which maximizes the right
hand side of the above equation, reproducing the standard maximum
likelihood method.
Now the question is, what do we do with mixed distributions? When
there is a mass at a point $x\_i$, that is $\mathbb{P}(x\_i=g\_i) > 0$,
our first approximation is incorrect; for very small $\varepsilon$, we
have the approximation
$$ \mathbb{P}(|x\_i - g\_i| < \varepsilon) \approx \mathbb{P}(x\_i = g\_i)
$$
If we let $\mathcal{N}$ be the index set of the "massless" samples, we can approximate
the probability of our data as
$$ \mathbb{P}(|x\_i - g\_i| < \varepsilon) \approx \varepsilon^n
\prod\_{i \in \mathcal{N}} p\_g(x\_i) \prod\_{i \notin \mathcal{N}} \mathbb{P}(x\_i = g\_i).
$$
where $n$ is the number of elements in $\mathcal{N}$. That is, we can reasonably define our maximum likelihood estimate
for a parameter $m$ as the value of the parameter that maximizes
$$
\prod\_{i \in \mathcal{N}} p\_g(x\_i) \prod\_{i \notin \mathcal{N}} \mathbb{P}(x\_i = g\_i).
$$
In your case, it is fairly simple to write down the value of the likelihood function above. First, note that
$$\mathbb{P}(x=0) = \frac{1}{\sqrt{2\pi}} \int\_{-\infty}^{-m}
e^{-x^2/2}dx.$$
For $x>0$, you have the standard Gaussian pdf
$p\_g(x) = \tfrac{1}{\sqrt{2\pi}} e^{-(x-m)^2/2}$.
I won't do any more here; suffice it to say that the standard approach
to maximizing the likelihood involves taking the logarithm of the likelihood function
and setting its derivative to zero. You will probably get a
transcendental equation that you will need to solve numerically.
| 2 | https://mathoverflow.net/users/4261 | 21371 | 14,148 |
https://mathoverflow.net/questions/21370 | 10 | Basically, I'm aware of "splitting principles" for the following three objects (which are all isomorphic modulo torsion).
**1**. The Chow group a la Fulton.
**2**. The classical Grothendieck group of vector bundles or coherent sheaves.
**3**. The $\gamma$-graded Grothendieck group.
I was just wondering where the idea of "the splitting principle" comes from. I'm guessing somewhere in topology when one wanted to define Chern classes and show some properties. But I don't know.
And above that, is there some more general way of looking at this? I know there is a theorem that connects higher K-groups with Chow groups in a sense. So I ask, is there a way of deducing the splitting principle for one of the above objects from the other? (It's easy if we want to do this modulo torsion, of course.)
| https://mathoverflow.net/users/4333 | Where does the splitting principle come from and does it generalize | We can think of the splitting principle as a *condition* on a "cohomology theory" (of some sort) $E^\*$, coming about when working with Chern classes for instance, and then ask: When does $E^\*$ satisfy this condition? First, let's make the condition more precise and reformulate it:
**Condition 1:** Given $X$ and a vector bundle $V$ on $X$, there exists $f: X' \to X$ such that $f^\* V'$ has a filtration with subquotients line bundles, and $f^\*: E^\*(X) \to E^\*(X')$ is injective.
But there is a *universal* choice for $X'$, namely the *flag variety of $V$*: $p: Fl(V) \to X$. Any $f: X' \to X$ with $f^\* V'$ filtered with line bundle subquotients will factor through $p$, and so we're really just asking if $p^\*: E^\*(X) \to E^\*(Fl(V))$ is injective.
**Condition 1':** For all $X$ and $V$, $p^\*: E^\*(X) \to E^\*(Fl(V))$ is injective.
At this point there are two ways this answer can go, depending on ones tastes:
1. $Fl(V)$ is a very geometric object over $X$, so we might as well ask that we actually have a formula for $E^\*(Fl(V))$ in terms of $E^\*(X)$. If $E^\*$ is "reasonable" (i.e., has Chern classes giving rise to a "projective bundle formula") then iteratively applying the projective bundle formula will give such a thing, and in fact show that $E^\*(X)$ is a direct summand of $E^\*(Fl(V))$.
2. (My favorite:) There's a nice way of strengthening Condition 1' that also holds in all reasonable cases, and that looks rather natural. You can ask that $Fl(V) \to X$ behave like a "covering", i.e. that
(**Condition 2:**)
$$ E^\*(X) \to E^\*(Fl(V)) \to E^\*\left(Fl(V) \times\_X Fl(V)\right) $$
is an equalizer diagram. (So not only is pullback injective, but you can identify its image...) (In fact, in reasonable cases it'll be a split equalizer diagram, related to the direct summand thing above.)
If your question is one of proof + generalization (which I think it is), rather than vague motivation, then I haven't addressed it yet:
In topology. one can show that any complex-oriented cohomology theory (i.e., one with Chern classes for line bundles) $E^\*$ has a projective bundle formula, satisfies all the conditions, etc.
In more-algebro-geometric contexts, you could deduce the Chow + K-theory (I don't know anything about the $\gamma$-filtration) statements by either
1. Constructing $c\_1$ + proving a projective bundle formula, and then feeding this into a general argument using these to prove the rest.
2. Going to the universal example of *algebraic cobordism* and then deducing the results for Chow + K-theory from the known relationships between them and algebraic cobordism. (Though this second approach is not so great, since those relationships hold under much more stringent hypotheses than are necessary to run the argument.)
One could also ask to generalize this in another direction, replacing vector bundles and $Fl(V)$ by more general $G$-bundles and their associated $G/B$-bundles. In general, that's a more complicated story...
| 21 | https://mathoverflow.net/users/1921 | 21381 | 14,154 |
https://mathoverflow.net/questions/21382 | 16 | Let $\mathcal C,\otimes$ be a monoidal category, i.e. $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is a functor, and there's a bit more structure and properties. Suppose that for each $X \in \mathcal C$, the functor $X \otimes - : \mathcal C \to \mathcal C$ has a right adjoint. I will call this adjoint (unique up to canonical isomorphism of functors) $\underline{\rm Hom}(X,-) : \mathcal C \to \mathcal C$. By general abstract nonsense, $\underline{\rm Hom}(X,-)$ is contravariant in $X$, and so defines a functor $\underline{\rm Hom}: \mathcal C^{\rm op} \times \mathcal C \to \mathcal C$. If $1 \in \mathcal C$ is the monoidal unit, then $\underline{\rm Hom}(1,-)$ is (naturally isomorphic to) the identity functor.
Then there are canonically defined "evaluation" and "internal composition" maps, both of which I will denote by $\bullet$. Indeed, we define "evaluation" $\bullet\_{X,Y}: X\otimes \underline{\rm Hom}(X,Y) \to Y$ to be the map that corresponds to ${\rm id}: \underline{\rm Hom}(X,Y) \to \underline{\rm Hom}(X,Y)$ under the adjuntion. Then we define "composition" $\bullet\_{X,Y,Z}: \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$ to be the map that corresponds under the adjunction to $\bullet\_{Y,Z} \circ (\bullet\_{X,Y} \otimes {\rm id}) : X \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to Z$. (I have supressed all associators.)
>
> **Question:** Is $\bullet$ an associative multiplication? I.e. do we have necessarily equality of morphisms $\bullet\_{W,Y,Z} \circ (\bullet\_{W,X,Y} \otimes {\rm id}) \overset ? = \bullet\_{W,X,Z} \circ ({\rm id}\otimes \bullet\_{X,Y,Z})$ of maps $\underline{\rm Hom}(W,X) \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$? If not, what extra conditions on $\otimes$ are necessary/sufficient?
>
>
>
| https://mathoverflow.net/users/78 | If "tensor" has an adjoint, is it automatically an "internal Hom"? | It is associative. Consider the evaluation cube drawn [here](http://ejenk.com/misc/internal-hom.pdf). Four of the faces commute by definition of the composition map, and one by functoriality of the tensor product. The commutativity of these five faces implies that any of the maps $W \otimes \operatorname{Hom}(W, X) \otimes \operatorname{Hom}(X, Y) \otimes \operatorname{Hom}(Y, Z) \to Z$ are equal, so by adjunction, the two composites of compositions are equal.
| 13 | https://mathoverflow.net/users/396 | 21385 | 14,156 |
https://mathoverflow.net/questions/21367 | 63 | I just taught the classical impossible constructions for the first time, and in finding my class a reference for the transcendence of pi, I found a dearth of distinct proofs. In particular, those that I read all require the existence of infinitely many primes, which strikes me as extraneous. Is there a known proof that requires only knowledge that I would "expect", namely, integral calculus to get your hands on the actual constant and algebraic properties of polynomials in connection with the assumption that the constant is algebraic?
| https://mathoverflow.net/users/5373 | Proof that pi is transcendental that doesn't use the infinitude of primes | The infinitude of primes (more precisely, the existence of arbitrarily large primes) might actually be necessary to prove the transcendence of $\pi$. As I explained in an [earlier answer](https://mathoverflow.net/questions/19857/has-decidability-got-something-to-do-with-primes/20944#20944), there are structures which satisfy many axioms of arithmetic but fail to prove the unboundedness of primes or the existence of irrational numbers. Shepherdson presented a simple method for constructing such models, I will present such a model where $\pi$ is rational!
The Shepherdson integers $S$ consist of all Puiseux polynomials of the form
$$a = a\_0 + a\_1T^{q\_1} + \cdots + a\_kT^{q\_k}$$
where $0 < q\_1 < \cdots < q\_k$ are rationals, $a\_0 \in \mathbb{Z}$, and $a\_1,\dots,a\_k \in \mathbb{R}$. This is a discrete ordered domain, where $a < b$ iff the most significant term of $b-a$ is positive; this corresponds to making $T$ infinitely large. This ring $S$ satisfies open induction axioms
$$\phi(0) \land \forall x(\phi(x) \to \phi(x+1)) \to \forall x(x \geq 0 \to \phi(x))$$
where $\phi(x)$ is a quantifier free formula (possibly with parameters). So the ring $S$ satisfies the same basic axioms as $\mathbb{Z}$, but only a very limited amount of induction. In the field of fractions of $S$, $\pi$ is equal to the ratio $\pi T/T$. In other words, $\pi$ is a rational number!
Is $\pi T/T$ really $\pi$? The integers form a subring of $S$, and if $p,q \in \mathbb{Z}$ then $p/q < \pi T/T$ in $S$ if and only if $p/q < \pi$ in $\mathbb{R}$. So $\pi T/T$ defines the same Dedekind cut as $\pi$ does, which is a very accurate description of $\pi$. Indeed, any proof of the transcendence of $\pi$ must ultimately be based on the comparison of $\pi$ and its powers with certain rational numbers, which $\pi T/T$ will accomplish just as well as the real number $\pi$. However, the usual definitions of $\pi$ are not easily formalizable in this basic theory, so there is much room for debate here and I wouldn't claim that $\pi T/T$ satisfies all reasonable definitions of $\pi$. Shepherdson only presented this argument for real algebraic numbers like $\sqrt{2}$, which have a finitary description in this theory and leave little room for debate. In any case, the conclusion to draw from this is that basic arithmetic with open induction does not suffice to prove that $\pi$, or *any* other real number, is irrational (never mind transcendental).
What about primes? In the ring $S$, the only primes are the ones from $\mathbb{Z}$. Although there are infinitely many primes in $S$, it is not true that there are arbitrarily large primes. For example, there are no primes larger than $T$. Thus $S$ is a model where the unboundedness of primes fails and so does the irrationality of $\pi$. This only shows that basic arithmetic with open induction does not suffice to prove either result. A possible line of attack to show that the unboundedness of primes is necessary to prove the transcendence of $\pi$ would be to show that the minimum amount of induction necessary to prove that $\pi$ is transcendental also suffices to prove the unboundedness of primes. Unfortunately, I do not know how much induction is necessary to prove the transcendence of $\pi$. (And the minimum amount of induction necessary to prove the unboundedness of primes is still an open problem.)
---
Well, here is a partial answer, which is a bit of a bummer. There is another Shepherdson domain $S\_0$ similar to the above where $\pi$ is transcendental over $S\_0$ and $S\_0$ does not have arbitrarily large primes. This shows that the transcendence of $\pi$ does not imply the unboundedness of primes over basic arithmetic with open induction. The ring $S\_0$ is the subring of $S$ where the coefficients of the Puiseux polynomial are restricted to be algebraic numbers. The unboundedness of primes fails in $S\_0$ because the real algebraic numbers form a real closed field just like $\mathbb{R}$. The number $\pi$ is transcendental over $S\_0$ because it is transcendental over the field of real algebraic numbers.
This is not entirely surprising since open induction is a very weak base theory and the Shepherdson type rings are very pathological. To constrain such pathologies Van Den Dries suggested requiring that the domain is integrally closed in its field of fractions; he called such domains *normal* but I don't know if this is standard terminology. Neither $S$ nor $S\_0$ are normal. More convincing examples would be normal discrete ordered domains. The methods of Macintyre and Marker (*Primes and their residue rings in models of open induction*, [MR1001418](http://www.ams.org/mathscinet-getitem?mr=1001418)) suggest that normal analogues of $S$ and $S\_0$ might exist.
The conclusion that I draw from this is that open induction is probably too weak a base theory to study this question. Stronger base theories run into the difficulty that it is still not known just how little induction is necessary to prove the unboundedness of primes. The next reasonable candidate is bounded-quantifier induction (IΔ0), which is not known to imply the unboundedness of primes. Using the Euler product $\pi^2/6 = \prod\_p (1-p^{-2})^{-1}$ looks promising, but so far I can only make sense of this product in IΔ0 + Exp which is known to prove the unboundedness of primes.
| 124 | https://mathoverflow.net/users/2000 | 21389 | 14,158 |
https://mathoverflow.net/questions/21373 | 9 | This is a refinement of my (naive, poorly asked) question [here](https://mathoverflow.net/questions/17951/what-tensor-product-of-chain-complexes-satisfies-the-usual-universal-property). The reference for my question is [Baez and Crans, HDA6](http://math.ucr.edu/home/baez/hda6.pdf).
Background: category objects, etc.
----------------------------------
Let $\mathcal V$ be a category. A **category object** internal to $\mathcal V$ consists of the following data and properties:
1. Objects $C\_0,C\_1 \in \mathcal V$ and morphisms $s,t: C\_1 \to C\_0$ and $i: C\_0 \to C\_1$.
2. Such that $s\circ i = t\circ i = \text{id}\_{C\_0}$ and the pull-back $C\_1 \underset{C\_0}{\times} C\_1 = C\_1 \underset{\displaystyle ^{\searrow\!\!^{\scriptstyle s}} {C\_0} ^{^{\scriptstyle t} \!\!\swarrow} }{\times} C\_1 $ exists.
3. A morphism $m: C\_1 \underset{C\_0}\times C\_1 \to C\_1$ such that $s\circ m = s\_R$ and $t\circ m = t\_L$, where $s\_R: C\_1 \underset{C\_0}\times C\_1 \to C\_0$ is the "$s$" projection from the right factor, and similarly for $t\_L$.
4. And such that the obvious "associativity" square (two ways to get from $C\_1 \underset{C\_0}\times C\_1 \underset{C\_0}\times C\_1$ to $C\_1$) and "identity" triangles (three ways to get from $C\_1 = C\_1 \underset{C\_0}\times C\_0 = C\_0 \underset{C\_0}\times C\_1$ to $C\_1$) commute.
For example, a category object in $\mathcal V = {\rm SET}$ is a small category.
For this post, I will be interested in $\mathcal V = {\rm VECT}$, the category of vector spaces over your favorite field. I will call your favorite field "$\mathbb R$".
A category object in ${\rm VECT}$ is a **2-vector space**.
2-vector spaces are relatively mild things. Indeed, it turns out that in $\rm VECT$ the data and properties of 1-2 above uniquely determine a map $m$ satisfying 3-4.
By the general yoga known as "commutativity of internalization", a 2-vector space is the same as a "vector space object in $\rm CAT$". More precisely, let $\rm CAT$ be the category of small categories. Then it makes sense to talk about "field objects" — like a category object, a field object consists of some objects, some maps, some pull-backs, and some more maps, and some commuting diagrams. In particular, by thinking of $\mathbb R$ as a discrete category ($\mathbb R\_0 = \mathbb R = \mathbb R\_1$ and $s = t = i = {\rm id}$), it is in fact a field object. Then with some more diagrams, we can talk about "vector space objects over $\mathbb R$" internal to $\rm CAT$, and it is straightforward to check that these are the same as 2-vector spaces.
Background: tensor products
---------------------------
I know of two natural approaches to define "tensor products":
1. Define a notion of "bilinear map", such that the assignment $X,Y,Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$ is contravariant in the first two spots and covariant in the last. Then set $X\otimes Y$ to be the object (if it exists) that represents the function $Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$.
2. Define a notion of "internal hom", i.e. a (nice) functor $\underline{\rm Hom}: \mathcal V^{\rm op} \times \mathcal V \to \mathcal V$. For each $X\in \mathcal V$, define the functor $-\otimes X$ by declaring that it is left adjoint to $\underline{\rm Hom}(X,-)$.
Approach 1 is the way that tensor products are introduced in grade school. Approach 2 is I think more standard in the real world. We can implement each in the case of 2-vector spaces:
1. The trick for approach 1 is that the notion of "bilinear" depends on more than just the category. So realize the category of 2-vector spaces as the category of vector spaces objects in $\rm CAT$. Recall that a morphism of 2-vector spaces is simply a morphism of underlying categories so that some diagrams commute. Then we can say the following. Let $X,Y,Z$ be 2-vector spaces. Then a morphism of underlying categories $X \times Y \to Z$ is **bilinear** if a bunch of diagrams commute (these diagrams refer to the vector-space-object structures of $X,Y,Z$, and are precisely the diagrams that you learned in grade school). By reproducing the proofs from vector spaces internal to $\rm SET$, this in fact defines a functor $\otimes$.
2. Given 2-vector spaces $X,Y$, there is a category whose objects are linear functors $X \to Y$ and whose morphisms are linear natural transformations of functors, and this category has a natural structure as a 2-vector space. Moreover, the corresponding notion of $\underline{\rm Hom}$ is correctly functorial, and has an adjoint. So this approach defines a functor $\otimes$.
However:
>
> The two "tensor products" defined in 1-2 above do not agree.
>
>
>
Writing 2-vector spaces $X = (X\_1 \rightrightarrows X\_0)$ and $Y = (Y\_1 \rightrightarrows Y\_0)$ as category objects in $\rm VECT$, approach 1 gives $(X\otimes Y)\_a = X\_a \otimes Y\_a$ for $a=0,1$, with the tensor products of the structure maps. Approach 2 also has $(X\otimes Y)\_0 = X\_0 \otimes Y\_0$, but $(X\otimes Y)\_1 \cong $
$$ X\_0 \otimes Y\_0 \oplus \text{coker}\Bigl( \bigl(\ker(X\_1 \overset s \to X\_0) \otimes \ker(Y\_1 \overset s \to Y\_0)\bigr) \overset{t\otimes{\rm id} - {\rm id}\otimes t}\longrightarrow \bigl( X\_0 \otimes \ker(Y\_1 \overset s \to Y\_0) \oplus \ker(X\_1 \overset s \to X\_0) \otimes Y\_0\bigr) \Bigr)$$
The structure maps are: $s$ is the projection onto the first factor, and $t$ is the sum of the same projection and the map ${\rm id}\otimes t + t\otimes {\rm id}$ from the second factor (it is well-defined out of the cokernel).
(There is probably a way to simplify the above description. The trick is that, as explained in HDA6, the category of 2-vector spaces is equivalent as a category to the category of 2-term chain complexes, and this is the natural "internal Hom" over there.)
Anyway, a dimension count shows that these two "tensor" constructions are inequivalent in general.
Hence:
Question
--------
Does the tensor product of 2-vector spaces given in "approach 1" above — the tensor product defined as representing "bilinear functors" — under this tensor product, does the functor $\otimes X$ have a right adjoint?
| https://mathoverflow.net/users/78 | What functor is adjoint to the tensor product of 2-vector spaces? | I'll denote your category of 2-vector spaces by 2Vect. By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on Δ≤1 where Δ≤1 denotes the full subcategory of Δ on the objects [0] and [1]. Therefore, colimits in 2Vect are computed objectwise under this identification. So the functor – ⊗ X certainly has a right adjoint **Hom**(X, –) (by the adjoint functor theorem for locally presentable categories). This adjunction also respects the Vect enrichment.
To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing Δi for the 2-vector space (Δi)j = HomΔ≤1([j], [i]) • ℝ, we have hom2Vect(Δi, X) = Xi as vector spaces, where hom denotes the Vect-enriched Hom. So
**Hom**(X, Y)0 = hom(Δ0, **Hom**(X, Y)) = hom(Δ0 ⊗ X, Y) = hom(X, Y)
since Δ0 happens to be the unit for this ⊗, but
**Hom**(X, Y)1 = hom(Δ1 ⊗ X, Y)
will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).
| 5 | https://mathoverflow.net/users/126667 | 21390 | 14,159 |
https://mathoverflow.net/questions/21386 | 1 | Given $f(0) = 1-c$, what is the non-recursive $f(n)$ that satify the following equation
$f(n)-\frac{1}{2}f(n-1)f(n)+f(n-1) = 1$
for n = 1,2,3,...?
| https://mathoverflow.net/users/5066 | What is the nonrecursive formula for the following implicit function? | $$ f(n) = 2{\frac {1-c+ \left( -1 \right) ^{n}-c \left( -1 \right) ^{n}-\sqrt
{2}c+ \left( -1 \right) ^{n}\sqrt {2}c}{2-\sqrt {2}-\sqrt {2}c+2
\left( -1 \right) ^{n}+ \left( -1 \right) ^{n}\sqrt {2}+ \left( -1
\right) ^{n}\sqrt {2}c}} $$
which is a non-trivial pattern to spot! [I used a CAS] The thing to notice is that one can transform this first-order recurrence to the constant coefficient 2nd order linear recurrence
$$a(n+2)-2a(n) = 0, a(0) = 1, a(1) = -1-c$$
and then transform back. This can be done as the equation is of Riccati type.
I guess I 'cheated' in that I traced through the CAS's process of solving to extract the above information from it. Though if you don't know what you're looking for, it can be rather difficult information to extract...
| 6 | https://mathoverflow.net/users/3993 | 21391 | 14,160 |
https://mathoverflow.net/questions/21397 | 20 | I am teaching a topics course on Riemann Surfaces/Algebraic Curves next term. The course is aimed at 1st and 2nd year US graduate students who have have taken basic coursework in algebra and manifold theory, but may not have had much expose to algebraic geometry. I will loosely follow the book [Introduction to Algebraic Curves](http://rads.stackoverflow.com/amzn/click/0821845373) by Griffiths. In particular, I hope to spend a minimum amount of time developing basic machinery (e.g. sheaf theory) and to start doing concrete geometry (e.g. canonical models of curves of genus up to 4) as soon as possible.
My question is: what are some good concrete, accessible geometric topics in Riemann Surface/Curve theory that aren't in the standard textbooks?
Let's say that the standard textbooks are the book I mentioned and those discussed:
[here.](https://mathoverflow.net/questions/1916/riemann-surfaces)
| https://mathoverflow.net/users/5337 | What should be taught in a 1st course on Riemann Surfaces? | Good question. I bet you'll get many interesting answers.
About two years ago I taught an "arithmetically inclined" version of the standard course on algebraic curves. I had intended to talk about degenerating families of curves, arithmetic surfaces, semistable reduction and such things, but I ended up spending more time on (and enjoying) some very classical things about the geometry of curves. My lecture notes for that part of the course are available [here](http://alpha.math.uga.edu/%7Epete/8320notes6.pdf):
Some things that I found fun:
1. Construction of curves with large gonality. For instance, after having given several examples of various curves, it occurred to me that I hadn't shown them a non-hyperelliptic curve in every genus g >= 3, so then I talked about trigonal curves, and then...Anyway, there is a very nice theorem here due to Accola and Namba: suppose a curve $C$ admits maps $x,y$ to $\mathbb{P}^1$ of degrees $d\_1$ and $d\_2$. If these maps are independent in the sense that $x$ and $y$ generate the function field of the curve (note that this must occur for easy algebraic reasons when $d\_1$ and $d\_2$ are coprime), then the genus of $C$ is at most $(d\_1-1)(d\_2-1)$.
I sketched the proof in an exercise, which was indeed solved in a problem session by one of the students.
2. Material on automorphism groups of curves: the Hurwitz bound, automorphisms of hyperelliptic curves, construction of curves with interesting automorphism group.
3. Weierstrass points, with applications to 2) above.
| 14 | https://mathoverflow.net/users/1149 | 21399 | 14,167 |
https://mathoverflow.net/questions/21383 | 7 | Cauchy's Arm Lemma is used in the proof of Cauchy's Rigidity Theorem for convex polyhedra. The Lemma states that in the plane or on the sphere that if all but one of the side lengths of two convex polygons $P$ and $P'$ are the same, and the angles formed by the remaining sides of P are less than or equal to those of the remaining sides of $P'$, then the ommitted side length from $P$ is less than or equal to the omitted side length for $P'$, with equality occurring iff the angles are all the same. I know of a couple of extensions of this theorem (a nice presentation of this sort of thing is available in O'Rourke's paper [O'R01]). The Lemma can also be used to show that convex linkages may be (unsurprisingly) straightened.
I'm seeking other applications of this Lemma, particularly those that might be suitable for use as exercises in an advanced undergraduate course, or other applications in the theory of polyhedra.
[O’R01] Joseph O’Rourke, An extension of Cauchy’s arm lemma with application to curve development, Discrete and computational geometry (Tokyo, 2000), Lecture Notes in Comput. Sci., vol. 2098, Springer, Berlin, 2001, pp. 280–291. MR MR2043660 (2004m:52052)
| https://mathoverflow.net/users/4490 | Applications of Cauchy's Arm Lemma | The question is a little too general to have a single answer. Cauchy's arm lemma is a basic technical result in rigidity theory. One reason it has a name is because it is intuitively obvious, but the most natural proof Cauchy originally came up with is false.
Now, other than Cauchy theorem, it has few direct consequences. A.D. Alexandrov developed a whole family of results similar in nature (say, with angles preserved but polygon lengths extended), which proved new results on rigidity. This was, e.g. his way of extending the Minkowski uniqueness theorem in $\Bbb R^3$ to polytopes with (say) equal normals and perimeters of faces. You can read it in Alexandrov's famous [monograph.](https://doi.org/10.1007/b137434 "A.D. Alexandrov. Convex Polyhedra. Springer Monographs in Mathematics, Springer Berlin, Heidelberg, 2005") Sabitov made a [historical study](https://doi.org/10.1023/B:SIMJ.0000035837.80962.0a "Sabitov, I.K. Around the Proof of the Legendre–Cauchy Lemma on Convex Polygons. Siberian Mathematical Journal 45, 740–762 (2004)") behind the original Cauchy lemma and proved a number of related results as well.
Let me emphasize that from a modern point of view, the Arm Lemma is a "relative" version of the discrete [four vertex theorem](https://en.wikipedia.org/wiki/Four-vertex_theorem), which is a fundamental result in its own right. I explain the background, connections and variations in [my book](https://www.math.ucla.edu/%7Epak/book.htm "Igor Pak, Lectures on Discrete and Polyhedral Geometry, 2010"), sections 21-23. Note that this is not the only "relative result" of this type: in addition to Alexandrov's lemmas, see Tabachnikov's theorem in 21.6. See also numerous historical and other references mentioned there. In conclusion, much of this is relatively easy but not straightforward. I think at least some of these results fit well to an undergrad course.
| 7 | https://mathoverflow.net/users/4040 | 21400 | 14,168 |
https://mathoverflow.net/questions/21401 | 23 | I have the opportunity to prepare a research poster for a non-mathematical, yet scientifically savvy audience, and I want to do it well. I have asked a few mathematicians, and I have heard the following sound advice:
* Use interesting graphics.
* Elaborate on possible applications to other scientific fields.
Although easier for applied mathematicians, this will be ok as I study subfactors, which have connections to quantum physics and statistical mechanics, and planar algebras, which provide great graphics. But there are practical questions as well:
* How can I use LaTeX to make a poster?
* How can I avoid using mathematical symbols and technical language?
What makes a good math research poster? What are some good ways to target a non-mathematical audience? Does anyone have examples and/or templates using LaTeX? What other advice would you give a mathematician who has never made a research poster before?
A few asides:
* I'm not sure if this is community wiki. I'm more than happy to click the box if requested.
* Feel free to retag this question as you see fit.
* I will answer my own question after the poster presentation, and I will have all of my materials available online.
| https://mathoverflow.net/users/351 | How do you make a good math research poster for a non-mathematical audience? | For technical question 1 I would recommend [beamerposter](http://www-i6.informatik.rwth-aachen.de/~dreuw/latexbeamerposter.php).
| 10 | https://mathoverflow.net/users/290 | 21402 | 14,169 |
https://mathoverflow.net/questions/20593 | 3 | I am trying to understand Horrocks's [construction of vector bundles](http://dl.dropbox.com/u/3849644/Construction%20of%20Bundles.pdf). However I have been stuck on the proof the first theorem in the paper.
In the paper, a trivial bundle is a direct sum of Hopf bundles $\mathcal{O}(p)$.
Theorem: Let $E$ be a vector bundle without a trivial direct summand. Then there exist a trivial bundle $T$ such that $E\oplus T$ has a filtration
$$E\oplus T=F^0 \supseteq F^1\supseteq F^2\supseteq\cdots\supseteq F^N=0$$
with $F^i/F^{i+1}$ a twisted exterior power of the tangent bundle $T\_{\mathbb{P}^n}$.
Here is his proof:
Take a resolution $L$ of the dual of $E$ by trivial sheaves which is exact as a resolution of graded modules. The dual $L^\*$ can be dismantled into Koszul complexes.
Here are my questions.
How to break up $L^\*$ into Koszul complexes?
What are the Koszul complexes?
Where does the $T$ come from?
Thanks in advance!
| https://mathoverflow.net/users/2348 | Proof of a Theorem in the paper "Construction of bundles on P^n" by Horrocks | That is a pretty terse proof! Let me give an outline of a proof that I know. First, one could deduce the statement from a more general:
**Theorem 1**: Let $R$ be a regular local ring, $E$ be a reflexive $R$-module locally free on $U\_R$, the punctured spectrum such that $E$ has no free direct summand. Then one can find a free module $T$ and a filtration:
$$E\oplus T = F\_0 \supseteq F\_1 \supseteq \cdots F\_N =0$$
with $F\_i/F\_{i+1}$ a syzygy of $k=R/m$.
Why is this local statement implies what you want?
Let $A=k[x\_0,\cdots, x\_n],m=(x\_0,\cdots,x\_n), X=Proj(A)=\mathbb P^n, R=A\_m$. There is natural functor from the category of vector bundles on $X$ to that of vector bundles on $U\_R$, which is the same as the category of reflexive $R$-modules which are locally free on $U\_R$. This is used by Horrocks all the time and is explained in Section 9 of his [paper](http://plms.oxfordjournals.org/cgi/pdf_extract/s3-14/4/689): "Vector bundles on punctured spectrum of a regular local ring".
A proof of Theorem 1 can be found in Chapter 5 (theorem 5.2) of the book "Syzygy" by Evans-Griffith. A brief outline in case you can't find the book:
As suggested in the paper you quoted, one starts with a minimal resolution of $E^\*$. Then dualizing gives a complex (remember that $E^{\*\*} \cong E$ as $E$ is reflexive):
$0 \to E \to L\_0 \to L\_1 \cdots $
whose cohomologies are $Ext^i(E^\*,R)$. Let $i>0$ be the smallest number such that $X=Ext^i(E^\*,R) \neq 0$ Break the l.e.s in to the exact sequences:
$0\to E \to L\_0 \to L\_1 \cdots \to L\_i \to N \to 0 (\*)$
and $0 \to X \to N \to N/X \to 0$. Now build free resolutions for $X$ and $N/X$ and map them onto $(\*)$ as in [Horseshoe Lemma](http://en.wikipedia.org/wiki/Horseshoe_lemma), stopping at the spot $E$, one gets a s.e.s:
$0 \to B \to E\oplus T \to C \to 0$
here $T$ is free and $C$ is a syzygy of $X = Ext^i(E^\*,R)$. Repeat if necessary and you have a filtration whose quotient are syzygies of various $ Ext^i(E^\*,R)$. But each of this $Ext$ modules has finite length (as $E$ is locally free on $U\_R$), so they can be filtered by copies of $k$. Now use the same trick to build a finer filtration whose quotients are syzygies of $k$. Since $R$ is regular, the resolution of $k$ is the Koszul complex, answering your second question.
| 4 | https://mathoverflow.net/users/2083 | 21420 | 14,181 |
https://mathoverflow.net/questions/21415 | 44 | Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules. Then $\mathfrak g\text{-rep}$ comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over $\mathbb C$. Moreover, $\mathfrak g\text{-rep}$ is symmetric monoidal with duals, and the forgetful functor preserves all this structure. By Tannaka-Krein duality (see in particular the excellent paper [André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991](http://www.maths.mq.edu.au/~street/CT90Como.pdf)), from this data we can reconstruct an affine algebraic group $\mathcal G$ such that $\mathfrak g \text{-rep}$ is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of $\mathcal G$.
However, [it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group](https://mathoverflow.net/questions/124/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-an-algebraic-group). So it is not true that $\mathcal G$ is, say, necessarily the simply-connected connected Lie group with Lie algebra $\mathfrak g$, or some quotient thereof. So my question is:
>
> Given $\mathfrak g$, what is an elementary description of $\mathcal G$ (that avoids the machinery of Tannaka-Krein)?
>
>
>
For example, perhaps $\mathcal G$ is some Zariski closure of something...?
| https://mathoverflow.net/users/78 | What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra? | Rather than an answer, this is more of an anti-answer: I'll try to persuade you that you probably don't want to know the answer to your question.
Instead of some exotic nonalgebraic Lie algebra, let's start with the one-dimensional Lie algebra $\mathfrak{g}$ over a field $k$. A representation of $\mathfrak{g}$ is just a finite-dimensional vector space + an endomorphism. I don't know what the affine group scheme attached to this Tannakian category is, but thanks to a 1954 paper of Iwahori, I can tell you that its Lie algebra can be identified with the set of pairs $(\mathfrak{g},c)$ where $\mathfrak{g}$ is a homomorphism *of abelian groups* $k\to k$ and $c$ is an element of $k$. So if $k$ is big, this is huge; in particular, you don't get an algebraic group. (Added: $k$ is algebraically closed.)
By contrast, the affine group scheme attached to the category of representations of a semisimple Lie algebra $\mathfrak{g}$ in characteristic zero is the simply connected algebraic group with Lie algebra $\mathfrak{g}$.
In summary: this game works beautifully for semisimple Lie algebras (in characteristic zero), but otherwise appears to be a big mess. See arXiv:0705.1348 for a few more details.
| 31 | https://mathoverflow.net/users/930 | 21429 | 14,188 |
https://mathoverflow.net/questions/21245 | 24 | The Erdős-Ko-Rado theorem talks about how large an intersecting set system (a set of pairwise intersecting sets) can be if the size of the base set is fixed. I'm interested about intersecting set systems where the base set is not fixed, but the size of the sets is bounded. I can prove the following lemma (see proof below).
**Lemma 1.** For every natural number $k$ there is a natural number $N(k)$ such that for every set $C$ each of whose elements are sets of size at most $k$, if every two element of $C$ has a common member, then there is a kernel $A$ which is a set of size at most $N$ so that every two element of $C$ also has a common member that's in $A$.
I'd like to know if this lemma is known in some literature, and whether you can give me a simpler proof for it than mine.
I'd also like to know what bound you can give on $N(k)$. An exact bound is probably hard and not too interesting, but I'd like to get the order of magnitude, say whether you can make $N(k)$ a polynomial of $k$. My proof only gives $N(k) = 2^{O(k^2)}$, so anything with a smaller order of magnitude would be nice. (I know that $N(k)$ has to be $\Omega(k^2)$. You can show this by chosing a prime $q$ between $k/2-1$ and $k-1$ and then letting $C$ be the set of lines of a finite projective plane of order $q$.)
There's also a strengthening of the lemma, which follows easily from my proof and can be useful.
**Lemma 2.** For every natural number $k$ there is a natural number $N^{\ast}(k)$ such that for every set $C$ each of whose elements are sets of size at most $k$, if every two element of $C$ has a common member, then there is a kernel $A$ which is a set of size at most $N^{\ast}$ so that if $Y \in C$ and $X$ is a set that intersects every element of $C$ *and $X$ has at most $k$ elements* then $X \cap Y \cap A$ is nonempty.
**Update:** the original phrasing of lemma 2 was wrong, I added the condition that $|X|\le k$.
I'm asking the same questions as above for this stronger version, and also whether it follows easily from the first lemma.
**Proof** of lemma 1.
Fix $k$. We will use induction on $p$ to show the existence of a set $A\_p$ such that the size of $A\_p$ is bounded by a constant natural number depending only on $k$ and $p$ (but not $C$), and that for every $X \in C$ either $p \le |X \cap A\_p|$ or the intersection $X \cap Y \cap A\_p$ is non-empty for every $Y\in C$. This is enough because $A = A\_{p+1}$ satisfies the conditions of the lemma (in fact even $A = A\_p$ would work). The case of $p = 0$ is trivial, because $A\_0$ can be the empty set.
Now suppose we have found $A\_p$ and we want to construct $A\_{p+1}$. Now sort the elements of $C$ in equivalence classes such that two element is equivalent if their intersection with $A\_p$ is equal. There are at most as many such classes as subsets of $A\_p$ (or even subsets with at most $k$ elements), which is a constant bound because the size of $A\_p$ is bounded by a constant. Now chose a single element from each equivalence class and let $B$ be the set of these elements. Let $A\_{p+1} := A\_p \cup \bigcup\_{Y\in B} Y$.
Thus all we have to prove is that for every $X \in C$ either $X \cap A\_{p+1}$ has at least $p+1$ elements or it intersects every element of $C$. From the induction hypothesis we know that $X \cap A\_p$ either has at least $p$ elements or intersects with every element of $C$. If it's the latter, we're done, because $X \cap A\_{p+1}$ is a superset of $X \cap A\_p$, so let's now assume the former: $X \cap A\_p$ has at least $p$ elements. Now if $X \cap A\_{p+1}$ intersects all elements of $C$ then we're done, so we can also assume that there is a $Z \in C$ such that $X \cap A\_{p+1} \cap Z$ is empty. Now consider the class of $Z$ in the equivalence we defined above, that is, all sets $Y$ for which $Y \cap A\_p = Z \cap A\_p$, and let $Y$ be the representant element we chose from this class for the construction. This means that $Y \in B$ thus $Y \subset A\_{p+1}$. Now $X$ and $Y$ has a common element, say $x$. Now it's not possible that $x \in A\_p$, because by our second assumption $X \cap Y \cap A\_p = X \cap Z \cap A\_p \subset X \cap Z \cap A\_{p+1}$ is empty. But then $X \cap A\_{p+1}$ has the at least $p$ elements of $X \cap A\_p$ from our first assumption (because $A\_p \subset A\_{p+1}$), and the extra element $x$ which is not in $X \cap A\_p$, so it has at least $p + 1$ elements, which completes our proof.
| https://mathoverflow.net/users/5340 | Pairwise intersecting sets of fixed size | This isn't an answer but simply a way of thinking about the question that I quite like. (Later: see below for an attempted proof.)
Let's regard each set in our collection as a vertex of a graph, and let's join two vertices if and only if the corresponding sets intersect. That doesn't sound very interesting, since we are hypothesizing that every pair of sets intersects. But that's fine -- we get the complete graph. The interest comes in what we can say about how the graph is built up. For each point x in the ground set, we can define a clique in the graph: its vertices are all sets from our collection that contain x. This gives us a system of cliques whose union is the whole graph. What else do we know about these cliques? We know that each vertex is contained in exactly k of them. And what do we want to prove? That we can choose N(k) of our cliques that cover the complete graph.
It feels more natural to relax the problem and just insist that each vertex is contained in *at most* k of the cliques.
In the reverse direction, if we have a bunch of cliques that cover the complete graph, we can associate with each vertex the set of cliques that contain that vertex, and the condition that the sets intersect is precisely the condition that the cliques cover all the edges of the graph. So it's a trivial reformulation, but I find it a helpful alternative picture of what is going on.
Added later: here is an attempt to improve the bound to $k^{Ck}$. I think it works but have not 100% checked.
Take a minimal subcollection S of the cliques that covers the complete graph, and suppose that S contains $N$ cliques.
Because S is minimal, for each clique in S there is some edge contained in just that clique and no other clique from S. Since each vertex is in at most k cliques, no vertex is contained in more than k of these edges. So we can find a collection of at least N/k disjoint edges, each of which is contained in just one clique from S. Let M be the number of edges in this set.
At this point I'm going to be very sketchy. I want to prove that M is at most $k^{Ck}$ by showing that if it's bigger than that, then I'm going to have to have a vertex that's contained in more than k cliques.
Let the M disjoint edges be called $x\_1y\_1,...,x\_My\_M$, and let $K\_i$ be the clique that contains $x\_i$ and $y\_i$. Let's from two cliques $L\_i$ and $M\_i$ from $K\_i$, one obtained by removing $x\_i$ and one by removing $y\_i$. Then between them $L\_i$ and $M\_i$ cover all the edges that $K\_i$ covers, apart from the edge $x\_iy\_i$. So we'll be done if we can show that some vertex has to be contained in more than 2k of the cliques $L\_i$ and $M\_i$, together with other cliques that don't contain any of the edges $x\_iy\_i$.
So our assumption now is that we have a bunch of cliques, and for each i no clique contains both $x\_i$ and $y\_i$, and yet all other pairs $x\_iy\_j$ or $y\_ix\_j$ are joined in at least one of the cliques. We want to show that some vertex is contained in at least 2k cliques.
In particular for each i and j (not equal) we must have a clique that joins $x\_i$ to $y\_j$ or $y\_i$ to $x\_j$ (since in fact we must do both). But no clique ever contains both $x\_i$ and $y\_i$ or both $x\_j$ and $y\_j$.
Given any clique $K$ from the new collection, the set of pairs ij such that $K$ joins $x\_i$ to $y\_j$ or $y\_i$ to $x\_j$ is a bipartite graph. (Its vertex sets are the set of i such that $K$ contains $x\_i$ and the set of i such that $K$ contains $y\_i$.) So we would like to cover a complete graph with M vertices with bipartite graphs, in such a way that no vertex is contained in more than 2k of those bipartite graphs.
An averaging argument (this is the sketchy bit) should show that we can discount bipartite graphs with a vertex set that is smaller than cM/k, since they do not contribute enough to the average degree. But if we just use bipartite graphs with vertex sets of size at least cM/k, then we can use the following standard argument. Let $X\_1$ be one of the vertex sets of the first bipartite graph. Then none of the edges inside $X\_1$ are covered. So by induction we know that the number of vertices in $X\_1$ is at most M(k-1) (where the M that we are trying to bound is M(k)). But it's also at least cM(k)/k, so we get a bound of $(k/c)^k$ for M.
Added yet later: I think I may be able to use similar ideas to prove a lower bound. Take $2^k$ pairs $x\_sy\_s$, where each s is a vertex of the discrete k-dimensional cube. For each i between 1 and k form a clique by joining every $x\_s$ such that $s\_i=0$ to every $y\_s$ such that $s\_i=1$, and also the other way round. Then for every s not equal to t we cover both the edges $x\_sy\_t$ and $x\_ty\_s$. Also, we never cover the edge $x\_sy\_s$, and the number of cliques is 2k. One more thing ... each vertex is contained in precisely 2k cliques so far. We now add in all the cliques of size 2 with vertex sets $x\_s,y\_s$. So now each vertex is in precisely 2k+1 cliques. So we have a minimal system of at least $2^k$ cliques, with each vertex in 2k+1 of them. Thus, the bound for the original problem (if my reasoning is correct) is at least exponential.
| 10 | https://mathoverflow.net/users/1459 | 21435 | 14,194 |
https://mathoverflow.net/questions/20890 | 3 | Recall that a morphism of rings $R\to S$ is called (essentially) *smooth* if it is formally smooth and (essentially) finitely presented.
(Note: $R\to S$ is *essentially finitely presented* provided that $S$ is the localization of some finitely
presented $R$-algebra $T$ at some multiplicative system $A \subset T$, that is, $S=A^{-1}T$.)
In class, our professor said that working with smooth or essentially smooth morphisms yields an effectively equivalent theory. This motivates my question: Is there a general technique to lift results from the smooth case to the essentially smooth case?
Edit: According to Mel, every essentially smooth morphism *is* a localization of a smooth morphism. However, this direction is much more involved than the other direction, which is immediate from the definitions. Anyway, this would be the answer to the question.
| https://mathoverflow.net/users/1353 | Lifting results from smooth maps to essentially smooth maps. | It seems that what you are looking for is theorem 5.11 [here](http://arxiv.org/abs/0809.1201). See also example (e) on section 5.12. Also if you don't feel like reviewing from EGA you can look at section 1.5 of "Introduction to algebraic stacks" by A. Canonaco which I think covers the relevant facts (including 17.5.1)
| 4 | https://mathoverflow.net/users/2384 | 21436 | 14,195 |
https://mathoverflow.net/questions/20915 | 1 | Let $X$ be a scheme. Consider the preorder of locally closed immersions into $X$ (considered, say, as a full subcategory of $Sch/X$). Is it complete or cocomplete? That is, are there infima or suprema?
Ok it's easy to see that every nontrivial locally closed subscheme of $\mathbb{A}^1\_\mathbb{C}$ contains only finitely or cofinitely many rational points of $\mathbb{A}^1\_{C}$. Therefore we should restrict to **finite** infima or suprema.
I guess everything works fine for infima if we restrict to seperated schemes. Take the ideal sheaves in the open subschemes which correspond to the closed immersions, restrict them to their intersection, take the smallest quasi coherent ideal which contains them and consider the corresponding closed subscheme of the intersection.
| https://mathoverflow.net/users/2841 | is the preorder of locally closed immersions complete? | The maximum may not exist in general. Take X=Spec k[T,U] the affine plane, A the complement of the vertical line L passing through the origin (A=Spec k[T,U,1/T]) and B the origin (Spec k[T,U]/(T,U)). Then, the maximum C of A and B in the ordered set of subschemes of X does not exist. If it was the case, then C should be some subscheme of X. Such are usually described as closed subschemes in some open subset of X, but can also be described as open subsets in some closed subschemes Z of X (I cannot find the reference in general, but it is certainly true when the schemes are noetherian). As U is (schematically)-dense in X, this Z can be nothing else as X itself, then C should be an open subset of X containing A and B, that is the open complement of a finite set of closed points of the line L except the origin, in which case it is clear that we could find an open subset of X containing A and B and strictly contained in C, which would give a contradiction.
| 3 | https://mathoverflow.net/users/5383 | 21450 | 14,206 |
https://mathoverflow.net/questions/21424 | 40 | I am writing an exam for my students, and the topic is intro knots theory. I have no idea how to put knots into the file, but I know many MO users who can draw amazing diagrams in their papers.
Can someone please provide some hints on what can be used, preferably with some example codes? I do not need complicated diagrams, just some simple knots and links with few crossings. Thanks in advance.
| https://mathoverflow.net/users/2083 | How to draw knots with Latex? | Knotinfo has .png files of all knots of 12 crossings or less.
<http://www.indiana.edu/~knotinfo>
| 46 | https://mathoverflow.net/users/3874 | 21456 | 14,209 |
https://mathoverflow.net/questions/21458 | 6 | Prompted by this [question](https://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex) I would like to ask the community how they convert their mathematics into pdf files. In any given procedure for converting mathematics into pdf I am interested in two issues: first typographical quality of text and of mathematical formulas and second production and placement of figures and labels within figures.
As a concrete example my current procedure is: latex and bibtex until the references settle down, dvips -o to produce postscript, and then ps2pdf to produce pdf files. I go through postscript in order to make psfrag labels work. I've never been fully happy with the output - in particular label placement inside of figures is difficult.
As a final issue, in the [question](https://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex) referenced, Tilman suggests that pdftex has typographical improvements over latex. I've looked around [on-line](http://www.pragma-ade.com/pdftex/thesis.pdf) and these seem to be margin kerning (hanging punctuation) and glyph scaling (font expansion). How does one use these features? Do they make a difference in practice?
EDIT: After a bit of pain, I've managed to switch from my previous procedure (described above) to the much simpler procedure of using pdflatex. Instead of psfrag I now use Colin Rourke's pinlabel package. I am very happy with pinlabel -- the fonts are exactly what I expect them to be, and the job of labelling is much easier than it used to be. It is still possible to align labels inside of a figure, and they virtually always show up where I intended.
I started using the microtype package, which turns on margin kerning and glyph scaling. I can see that these change the output, but I honestly can't say that the output is better - I guess my eyes aren't that sensitive. One thing to watch out for - pdftex 1.20 threw show-stopping errors when typesetting figure captions. I updated to 1.40 and the problem went away.
Thanks for your suggestions - if other people have other latexing procedures I'd be interested to hear about them.
| https://mathoverflow.net/users/1650 | Typesetting mathematics: how do {\em you} convert text into pdf? | You can do margin kerning (aka “protrusion”) and font expansion in pdfLaTeX simply by loading the package `microtype` (i.e., by adding
>
> `\usepackage{microtype}`
>
>
>
to the preamble). I also suggest using the `tracking` option for small-caps, which increases the space between letters (which is typographically correct, but *only* for small-caps and all-caps text):
>
> `\usepackage[tracking=smallcaps]{microtype}`
>
>
>
For further information (and for several fine-tuning options) you can consult the `microtype` [manual](http://tug.ctan.org/tex-archive/macros/latex/contrib/microtype/).
Edit: Yes, in my opinion and (I think) in a typographer’s opinion, these features do make a lot of difference. Margin kerning and font expansion help pdfLaTeX typeset the text, producing a lot less over/underfull hboxes. Letterspaced small-caps are also more legible and much more aesthetically pleasing.
| 8 | https://mathoverflow.net/users/5304 | 21466 | 14,214 |
https://mathoverflow.net/questions/21444 | 5 | Let $G$ be the reductive group $\operatorname{GSp}\_{4}$. Let $\pi$ be a smooth admissible cuspidal representation of $\operatorname{GSp}\_{4}(\mathbb{A}^{(\infty)})$ of dominant weight. Assume, for caution, that $\pi$ satisfies a multiplicity one hypothesis.
Fix $p$ an odd prime. To $\pi$ is attached a $p$-adic representation $\rho$ of the absolute Galois group of $\mathbb{Q}$ unramified outside a finite set of finite places and such that the characteristic polynomial of the Frobenius morphisms $Fr\ell$ for $\ell$ outside this set coincides with the Euler factor at $\ell$ of the degree 4 $L$-function of $\pi$. This Galois representation occurs in the degree 3 cohomology of the étale cohomology of a Siegel-Shimura variety.
The image of complex conjugation under $\rho$ is semi-simple so can be chosen to be diagonal with eigenvalues 1 and -1. How many $-1$ are there?
| https://mathoverflow.net/users/2284 | What is the image of complex conjugation under Siegel Galois representations? | The preprint "Conjecture de type de Serre et formes compagnons pour $GSp\_4$" by Florian Herzig and Tilouine (available [here](http://math.northwestern.edu/~herzig/ht08-7c.pdf)) indicates that there should be two 1's and two -1's.
EDIT: Having come in to work today, I could look up the more precise statement given in chapter 9 of Tilouine's book "Deformations of Galois representations and Hecke algebras" (incorrectly referenced in the aformentioned article of Herzig–Tilouine). The statement is the following: Let $G=GSp(4)$, let $\pi$ be a regular algebraic cuspidal automorphic representation of $G(\mathbf{A}\_F)$, where $\mathbf{A}\_F$ is the adele ring of a number field $F$. Let $\bar{\rho}$ be the associated mod $p$ Galois representation. Let $v$ be a real place of $F$. Then the $G$-conjugacy class of ${\bar\rho}(c\_v)$ (where $c\_v$ is a complex conjugation at $v$) contains the matrix $\text{diag}(1,1,-1,-1)$.
| 5 | https://mathoverflow.net/users/1021 | 21469 | 14,217 |
https://mathoverflow.net/questions/21471 | 7 | When I was at school I wondered if a surface could locally appear to be a unit sphere, yet `carry on forever'. More formally, my question is:
Can you place a metric of constant curvature +1 on ${\mathbb R}^2$, such that the identity map to ${\mathbb R}^2$ (with standard Euclidean metric) is uniformly continuous?
It is possible to induce such a metric on ${\mathbb R}^2 - {\mathbb Z}^2$, by identifying each unit square with integer vertices, with a hemisphere on the unit sphere.
| https://mathoverflow.net/users/5394 | Can you have a spherical plane? | By Myers's theorem (see <http://en.wikipedia.org/wiki/Myers%27s_theorem>) you have that if the Ricci curvature of a complete $n$-manifold $M$ is bounded below by $(n − 1)k > 0$, then its diameter is bounded by some constant depending on $k$. In particular, it is compact. Therefore if there is such a metric, it cannot be complete.
As pointed out by Sergei Ivanov in the comments below, the bound on the Ricci curvature holds (and therefore compactness follows) if the sectional curvature is bounded below by $k$.
| 10 | https://mathoverflow.net/users/3995 | 21474 | 14,219 |
https://mathoverflow.net/questions/21470 | 12 | In a comment to [this question](https://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice), Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to be shown that choice is required).
Unfortunately, I haven't been able to track down a reference, so if someone could link me to the original result (or provide a short proof) that would be great. I have roughly worked out a proof for myself, and I'd like to check it against the literature.
| https://mathoverflow.net/users/2233 | Subset of the plane that intersects every line exactly twice | By AC, choose a *cardinal* well-ordering of the lines in in the plane and any well-ordering of all the points.
We proceed by transfinite induction.
Suppose $A\_l$ is a set of points, no three colinear, and let $B\_l$ be the set of lines spanned by points of $A\_l$, and let $C\_l=\cup B\_l$. Suppose further that $l'\prec l$ implies $l'\in B\_l$. Note that $|A\_l| \leq |B\_l| < |l|$, so that
$$|C\_l\cap l| = |\cup\_{l'\in B\_l} l\cap l'| < |l| .$$
* If $l\in B\_l$, let $A\_{Sl} = A\_l$.
* If $a\in l \cap A\_l$, take the minimal point $b\in l\backslash C\_l$ and let $A\_{Sl} = A\_l\cup \{b\} $.
* If $A\_l\cap l = \emptyset$, take the minimal two points $a,b\in l\backslash C\_l$, and let $A\_{Sl}=A\_l\cup\{a,b\}$.
* Otherwise if $l'\ $ is a limit ordinal, let $A\_{l'} = \bigcup\_{l\prec l'} A\_l$.
It is easy to check that $A\_{l'}$ has no three points colinear --- they'd have to all be in some $A\_l$ for $l\prec l'$. The final union $\bigcup\_l A\_l$ has exactly two points on each line $l$.
| 8 | https://mathoverflow.net/users/1631 | 21476 | 14,221 |
https://mathoverflow.net/questions/21473 | 6 | If $A$ is an $n\times n$ matrix over a field, and $A^{k} = I$, with $k$ the least positive integer such that this occurs, then must there be some vector $v$ such that $\{v,Av,A^{2}v,\dots,A^{k-1}v\}$ has $k$ distinct elements in it? In other words:
>
> Must every matrix of finite multiplicative order have a regular orbit?
>
>
>
If A has prime power order, $k = p^{m}$, then $A^{p^{m-1}}-I$ is nonzero, so its kernel is proper, and everything outside of that kernel is a vector in a regular orbit. Over a finite field of size $q$, the index of a proper subspace is at least $q$, so we can even just choose (on average) $q$ random vectors to find one in a regular orbit. Over an infinite field, the same idea roughly says any random vector should work, as long as one can make some sort of "uniformly" distributed choice.
If $A$ has order a product of two prime powers, then I am assured this is true by a (special case) of an exercise in Isaacs's Finite Group Theory. I cannot imagine an argument that does not work for arbitrary orders $k$, but I also cannot find a convincing proof even for the product of two prime powers. The sum of vectors in regular orbits of the $p$ - parts of $A$ need not themselves be in regular orbits of $A$. Every matrix (over a finite field) I've tried has a regular orbit.
---
Assuming this is easy, how does one handle the case where $A$ is an automorphism of a finite group $G$, and the order of $A$ is a product of two prime powers? In other words:
>
> Prove every automorphism of order $p^{a}q^{b}$ of a finite group has a regular orbit.
>
>
>
Assuming the first question's answer is "yes", then what goes wrong for arbitrary orders? Isaacs's book gives an example where the general automorphism can fail to have a regular orbit, but it is impossible to compare this until I have at least some idea of why the two-prime case does work.
---
A related version of this question is: regular orbits are quite important in permutation and (finite) matrix groups and are a standard technique in several important (solved and unsolved) problems in modular representation theory.
>
> Is there sort of a gentle introduction that puts these techniques in context?
>
>
>
For any individual paper is clear that what they say works, but my picture of this area is incredibly disjointed and I suspect that is not true for everyone. For instance Khukhro has an excellent book on automorphisms of p-groups with few fixed points, and many finite group theory texts have chapters on fixed-point-free automorphisms and the consequences for the group structure of the group being acted upon. However, I haven't found any "textbook" exposition of regular orbits yet.
| https://mathoverflow.net/users/3710 | Linear algebra and regular orbits | For your first question, I presume you also wish to insist that $k$
be the *least* integer such that $A^k=I$. The matrix $A$ is then similar
over your field to a direct sum $B\_1,\ldots,B\_m$ of companion matrices
of (over your field $F$) factors of $X^k-1$, say $f\_1,\ldots,f\_m$.
Then $F^n$ decomposes as a direct sum of subspaces where $A$ acts
cyclically with generator $v\_i$ annihilated by $f\_i(A)$.
Let $v=v\_1+\cdots +v\_m$. Then
for a polynomial $g$, $g(A)v=0$ if and only if $g(A)v\_i=0$ for all $i$
if and only if $f\_i\mid g$ for all $i$.
Hence $F\mid g$ where $F=f\_1\cdots f\_m$. But then $F(A)u=0$ for all $u$
(so that $F$ is the minimum polynomial of $A$).
If $A^l=I$ where $l < k$ then $F\mid(X^l-1)$ and then $A^l-I=0$,
contrary to hypothesis. So yes, $A$ has a regular orbit.
| 4 | https://mathoverflow.net/users/4213 | 21477 | 14,222 |
https://mathoverflow.net/questions/21484 | 14 | The following questions seem basic, but I can't find them in the literature. I'm also unable to think of a counterexample.
Let $A$ be a local Cohen-Macaulay ring of dimension $d$.
1. Let $I$ be an ideal generated by $r$ elements. Is is true that the depth of $A/I$ is at least $d-r$?
2. Let $Q$ be a minimal prime of $A$. Is it true that $A/Q$ is also Cohen-Macaulay?
| https://mathoverflow.net/users/1594 | Two questions about Cohen-Macaulay rings | Nice questions! The answers are no in both cases, although the examples are more interesting than one would expect.
1) Even when $A$ is regular, one can always find an ideal $I$ with $3$ generators such that $A/I$ has depth $0$. This is due to a very nice [result](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=42575&vfpref=html&r=81&mx-pid=399074) by Bruns, which says you can construct $3$-generated ideal with all kinds of homological patern. The details are explained in this [answer](https://mathoverflow.net/questions/9352/depth-zero-ideals-in-the-homogenized-weyl-algebra/9357#9357).
2) Let $R=k[X^4,X^3Y,XY^3,Y^4]$. Then $R$ is a domain of dimension $2$ which is not Cohen-Macaulay. So one can write $R=S/Q$, where $R=k[a,b,c,d]$ and $Q$ is a prime ideal of height 2. Take $(f,g)$ to be a regular sequence in $Q$ and let $I=(f,g)$. Then $A=S/I$ is Cohen-Macaulay (being a complete intersection), but $Q$ is a minimal prime of $A$ and $R=S/Q$ is not CM.
| 18 | https://mathoverflow.net/users/2083 | 21488 | 14,228 |
https://mathoverflow.net/questions/21492 | 11 | I have two naive questions about stacks.
>
> 1) Is it possible to define stacks in the Zariski topology?
>
>
>
Presuming you can:
>
> 2) If a stack has a coarse moduli, and the coarse moduli space is a scheme, then does that mean that your stack is a stack in the Zariski topology?
>
>
>
In general, I am trying to understand why a new notion of open cover is necessary if all I am interested in is remembering stabilizers. Certainly this is too simple a mind-set, so feel free to enlighten me.
| https://mathoverflow.net/users/1231 | Stacks in the Zariski topology? | 1.) It's possible to define stacks on ANY category equipped with a Grothendieck topology (such a category with a topology is called a site). In particular, this holds true for the Zariski site. Moreover, there is always a way to define an "Artin stack"- these are those stacks which arise as torsors for a groupoid object in your site. Outside of algebraic geometry, these give rise to notions of topological and differentiable stacks, for instance.
EDIT: As long as groupoid objects exist in your category.
2.) As in Harry's post, any stack which is a stack in a site which is finer than the Zariski topology is also a stack in the Zariski topology.
To address your general question as to "why a new notion of open cover is necessary if all I am interested in is remembering stabilizers", you should learn a bit about Grothendieck topologies. I'll make a couple remarks:
i) If all you cared about were stabilizers, then you wouldn't need to use any covers at all; ordinary fibered categories would do the trick!
Indeed, take a group object in your site acting an object, and take the action groupoid- it is a groupoid object. Look at the pseudo-functor which assigns each object of your site the groupoid of maps into this groupoid object (considering the object as a groupoid with all identity arrows). This remembers the stabilizers for this action.
ii) (subcanonical) Grothendieck topologies are a choice of a type of cover for your objects, in such a way that this object is the colimit of these covers, AND "this is important to remember". This is a little imprecise, so, allow me to elaborate via an example from topology:
Let $U\_i$, $i\in I$ be an open cover of a space X. Then, continuous maps from X to another space Y are in bijection with with continuous maps $f\_i:U\_i \to Y$ which agree on their intersection. This is just saying that X is the colimit of this open cover. Instead, we can view this a property of the presheaf $Hom(blank,Y)$ represented by $Y$ on the category of topological spaces (for you set theorists, choose a Grothendieck universe).
For any $X$ and any open cover $U\_i$, $i\in I$ of $X$, (let $Hom(blank,Y)=F$)
the natural map $F(X) \to \varprojlim \left[{\prod{F(U\_i)}} \rightrightarrows {\prod{F(U\_{ij})}}\right]$
is a bijection.
If $F$ is any presheaf, this is just saying $F$ is a sheaf. Since this is NOT true for an ARBITRARY presheaf $F$, X is no longer the colimit of its open covers in the full category of all presheaves. The same argument holds for all fibred categories- it's only true if we restrict to STACKS (and $X$ then becomes the weak colimit of this cover, but, never mind).
The reason you add the condition for descent for covers, is so that "all maps into your stack from a space are continuous". More precisely, and more generally, it's so that maps from a space, scheme, whatever you site is, into a stack can be determined by mapping out of elements of some covering of your object in a way that glues (for stacks, rather than sheaves, they don't need to AGREE on the intersection, but, agree up to an invertible 2-cell, plus some coherency conditions).
Combining these ideas, if you have a group acting on an object, the pseudo-functor produced by the action groupoid is rarely a stack with respect to your topology, but you can stackify it, and then it will become one and still remember all the stabilizers. I hope this helps!
| 11 | https://mathoverflow.net/users/4528 | 21500 | 14,234 |
https://mathoverflow.net/questions/21427 | 5 | I'm wondering whether the following ideas/questions give rise to an existing body of research. (Accordingly: please suggest appropriate changes to the tags!)
***Preamble***
We consider polynomials f ∈ ℤ[x] with roots in ℝ, and for each polynomial f, the **principal root** is the real root with the largest magnitude. In the case of two roots of equal magnitude, we take the positive one.† So, for instance, √5 is the principal root of x2 − 5, the golden ratio φ is the principal root of x2 − x − 1, and −φ is the principal root of x2 + x − 1.
[ † **Edit**: *Previously, I had defined the principal root to be the maximal one; I revised this definition based on remarks by Kevin Buzzard below. This also motivates some revisions to the questions I ask.* ]
It's tempting to think that we could represent algebraic numbers by (minimal) polynomials over ℤ for which they are the principal root. We implicitly do this with rational numbers all the time: *a*/*b* denotes the real number which is the principal root of *b*x − *a*, and can formally be defined in such terms. This is also precisely what we do with algebraic integers: n√*b* is defined to be the principal root of xn − *b*, at least for *b* non-negative.
This approach seems problematic for negative algebraic numbers such as φ−1, whose minimal polynomial is x2 + x − 1, which is the same as for −φ; thus every polynomial for which φ−1 is a root will also have the (larger in magnitude) root −φ. A similar problem arises for −√5, of course. But let us focus on algebraic numbers which are the principal roots of their minimal polynomials.
***Questions***
For two irredicible polynomials f,g ∈ ℤ[x] with roots in ℝ, let *u,v* ∈ ℝ be their principal roots.
1. Are there broad classes of polynomials f and g, including ones of degree 2 or more, of course, for which there is a "simple" formula (e.g. involving no recursive functions more complicated than sums, products, exponents, and "well-known" number sequences) for the minimal polynomials of *u* + *v*, *uv*, or *uv*−1?
2. Does there exist such a "simple" formula for *some* polynomial (not necessarily irredicuble) for which *u* + *v*, *uv*, or *uv*−1 is the principal root?
3. If such questions are a proper subject of some body of research or well-studied theory: what is the name of the associated field of mathematics? (E.g. is this a special topic of Galois theory?)
Note that it is unlikely that we can obtain any sort of satisfactory answer for obtaining a minimal polynomial for *u* − *v* ; for instance, if *u* = √2 and *v* = √3, then the minimal polynomial of all four numbers ± √3 ± √2 have the same minimal polynomial, x4 − 26x2 + 145. For similar reasons, it is unlikely that there is a complete solution for arbitrary sums of principal roots: if *u* is a principal root of f(x), then −*u* is a principal root of f(−x), which differs from f in the case that f is not an even function of x. Thus, some differences of principal roots may also be expressed as sums of principal roots.
The difficulties described above and in the preamble suggest that a clean and elegant theory is unlikely; but I'm hoping that there are interesting classes of algebraic numbers which may be treated in this way.
| https://mathoverflow.net/users/3723 | Transformations of integer polynomials under combinations of their roots | If $u$ is a root of $f$ and $v$ is a root of $g$ then $u+v$ is a root of the resultant of $f(x-y)$ and $g(y)$ (for the purpose of calculating the resultant, we take these as polynomials in $y$). The resultant is just a determinant of a matrix whose entries are all coefficients of the polynomials, so it would seem to satisfy your request for a simple formula.
There are similar formulas for polynomials with $uv$ as a root, or $uv^{-1}$.
| 3 | https://mathoverflow.net/users/3684 | 21519 | 14,245 |
https://mathoverflow.net/questions/21513 | 19 | Let $K$ be a number field and let $\mathcal O\_K$ be the ring of integers. Following [this paper](https://www.emis.de/journals/JTNB/2005-3/article02.pdf "Division-ample sets and the Diophantine problem for rings of integers") of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O\_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.
Recently [Mazur and Rubin](https://arxiv.org/pdf/0904.3709v2 "Ranks of twists of elliptic curves and Hilbert's Tenth Problem") have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $ШT\_2$).
If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:
>
> Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?
>
>
>
It would appear that Byeon, Jeon, and Kim have done so in [this paper (probably need an institutional login)](https://doi.org/10.1515/CRELLE.2009.060 "Rank-one quadratic twists of an infinite family of elliptic curves"). Vatsal obtains a weaker result [here](https://personal.math.ubc.ca/%7Evatsal/research/der.PDF "Rank-one Twists of a Certain Elliptic Curve") that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.
>
> Can anyone do the above **WITHOUT** invoking the proven part of the BSD rank conjecture or assuming any conjectures?
>
>
>
| https://mathoverflow.net/users/4872 | Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD? | I don't think that this should be too hard: take a simple family of curves, such as
$y^2 = x^3 + px$ or something similar, and choose $p$ from a certain set of residue classes to guarantee that the 2-Selmer group has rank 1. You can complete the proof either by invoking rather deep constructions using Heegner points, or finding a family for which conditions such as $p = a^4 + b^2$ (there are infinitely many primes of this form) give you a global point (choose your family in such a way that $b^2 = px^4 - y^4$ occurs as a
principal homogeneous space in your standard 2-descent; see e.g. Silverman's book).
Sorry for being a little bit vague - a hard disk crash currently prevents me from looking at my own notes.
| 16 | https://mathoverflow.net/users/3503 | 21534 | 14,254 |
https://mathoverflow.net/questions/21483 | 5 | I have a very particular situation involving a (non-exact) complex $K$ of coherent sheaves on a nonsingular projective variety $X$, and I need to compute the hypercohomology of the complex. The associated spectral sequence is highly degenerate. Of course, in degenerate cases, one hopes that there are techniques for getting at the hypercohomology rapidly. In the most degenerate case, wherein every sheaf in the complex is acyclic, then the hypercohomology of the complex is the cohomology of the spaces of global sections (with respect to the maps induced by the complex's differential, say, $d$):
$\mathbb{H}^q(X,K^\bullet)\cong H^q\_d(\Gamma(X,K^\bullet))$.
The problem with my situation is that the sheaves are not quite acyclic. I will be more formal now about what I am dealing with. I have a non-exact complex $(K^\bullet,d)$ of three sheaves:
$K^1\stackrel{d}{\rightarrow}K^2\stackrel{d}{\rightarrow}K^3$.
Here, $K^1$ has no nonzero sheaf cohomology except for $H^1(K^1)$, whilst $K^2$ and $K^3$ have *nonzero zero-th and first cohomology* and all higher cohomology vanishes.
I would think that there must be a way of dealing with such a specialized, degenerate situation, but I haven't found it yet.
Please feel free to specialize this further, if it helps (e.g. torsion-free instead of only coherent ones, or even vector bundles). Also, feel free to make this more general: I am only restricting myself to three sheaves because that is precisely the problem I am faced with. In similar spirit, if giving $K^1$ a nonzero $H^0$ doesn't harm the chances of finding a reasonable solution, then by all means do so. (This would give the problem at hand a uniform description: *hypercohomology of a complex of sheaves in which each sheaf has nonvanishing zero-th and first cohomology, and vanishing cohomology elsewhere*.)
If there is a specific reference for where this situation is worked out (as I'm sure it must be and I probably just haven't looked hard enough), then please do pass it along.
Thanks!!
| https://mathoverflow.net/users/5395 | Question about hypercohomology / spectral sequence of a complex of "almost-acyclic" sheaves | Ok, so what will the spectral sequence give you? This is a very easy exercise, but since Altgr is not experienced, here is the solution. The term $E\_2^{p,q}$ is the $p^{\rm th}$ cohomology group of the complex $\mathrm{H}^q K^{\bullet}$ (the complex of groups obtained by applying the $q^{\rm th}$ cohomology functor to the complex $K^\bullet$). Then one has equalities
$$\mathbb H^i (K^\bullet) = 0 \quad{\rm for}\ i \neq 2,3,4$$
$$\mathbb H^4 (K^\bullet) = E\_2^{3,1}$$
and an exact sequence
$$
0 \to E\_2^{2,0} \to \mathbb H^2 (K^\bullet) \to E\_2^{1,1} \to E\_2^{3,0}
\to \mathbb H^3 (K^\bullet) \to E\_2^{2,1} \to 0
$$
The homomorphism $E\_2^{1,1} \to E\_2^{3,0}$ is the only differential at the $E\_2$ level that can possibly be non-zero.
Without further information, there is nothing else one can say, other than this: working with hypercohomology without knowing spectral sequences is like driving nails into a wall without a hammer.
| 7 | https://mathoverflow.net/users/4790 | 21536 | 14,256 |
https://mathoverflow.net/questions/21512 | 17 | A colleague asked me this question recently. Every injective continuous map between manifolds of the same (finite) dimension is open - this is Brouwer's Domain Invariance Theorem. Is the same true for complete boundaryless Alexandrov spaces (of curvature bounded below)?
Alexandrov spaces are manifolds almost everywhere, and their singularities have special structure. In dimensions 1 and 2 there are no topological singularities (all Alexandrov spaces are manifolds). Higher dimensional singularities have a sort of inductive description: every point $x$ in an $n$-dimensional Alexandrov space has a neighborhood homeomorphic to the cone over an $(n-1)$-dimensional connected(!) Alexandrov space $\Sigma\_x$ which carries a metric of curvature $\ge 1$. The last property implies that $\Sigma\_x$ is compact and its fundamental group is finite.
For example, in dimension 3 the only type of singularity is the cone over $RP^2$. In dimension 4 there are cones over lens spaces, $\mathbb R\times Cone(RP^2)$ and maybe other beasts.
There is a similar purely topological question: is Domain Invariance Theorem true for "almost manifolds"? An "almost manifold" is a pseudo-manifold obtained from $n$-simlices by gluing their $(n-1)$-dimensional faces in pairs - that is, there are exactly two $n$-simplices adjacent to each $(n-1)$-simplex, and there are no extra identifications between lower-dimensional faces. I'm not sure that all Alexandrov spaces admit triangulations, but if they do, they are "almost manifolds" (of a special kind).
| https://mathoverflow.net/users/4354 | Is there Domain Invariance for Alexandrov spaces? | The following lemma from *[Grove--Petersen, A radius sphere theorem](http://www.math.psu.edu/petrunin/papers/alexandrov/radius_sphere_theorem+.pdf)* does the trick.
>
> **Lemma 1.** Let $X$ be a compact Alexandrov space without boundary. Then $X$ has a fundamental class in Alexander-Spanier cohomology with $\mathbb Z\_2$ coefficients;
> i.e. $\bar H^n(X,\mathbb Z\_2 ) = \mathbb Z\_2$.
>
>
>
---
**Why:** First note that it is true for compact spaces --- in this case the map has $\mathbb Z\_2$-degree one. Moreover, in this case the same it true for any continuous map which is injective around one point in the target.
Now let $X$ and $Y$ be $m$-dimensional Alexandrov spaces,
$\Omega\subset X$ be an open subset and $f:\Omega\to Y$ be an injective continuous map and $y=f(x)$.
One can use $f$ to construct a continuous map between sperical suspensions over spaces of directions $\mathbb S\,\Sigma\_x\to \mathbb S\,\Sigma\_y$ which is injective around one point in the target --- take a small sherical neghborhood $W\ni y$ and collapse everything outside of $W$ to the south pole of $\mathbb S\,\Sigma\_y$.
(We can do it sinse small spherical neighborhood of a point $x$ in an Alexandrov space is homeomorphic to cone over space of directions at $x$.)
---
The same is true for the second question --- link of any pseudomanifold is a pseidomanifold, thus it has $\mathbb Z\_2$-fundamental class.
| 7 | https://mathoverflow.net/users/1441 | 21537 | 14,257 |
https://mathoverflow.net/questions/6764 | 10 | Consider the following: Let $A$ be a commutative ring, let $M$ be an $A$-module. When is the functor from $A$-algebras to Sets given by $R \mapsto R \otimes M$ representable by an $A$-scheme?
Unless I've made a mistake, this is always be an fpqc sheaf. When $M$ is a finitely generated free A-module, then $\mathrm{Spec}( \mathrm{Sym}^\bullet M^\*)$ does the trick.
| https://mathoverflow.net/users/788 | When is tensoring with a module representable by a scheme? | When $A$ is noetherian and $M$ is finitely generated, Nitin Nitsure showed that the functor is representable if and only if $M$ is projective (see <http://arxiv.org/abs/math/0308036>).
| 4 | https://mathoverflow.net/users/4790 | 21539 | 14,258 |
https://mathoverflow.net/questions/21521 | 8 | The Haag-Kastler approach to quantum field theory (QFT) is one of the oldest approaches to rigorously define what a QFT is, it deals with nets of operator algebras: You start with a spacetime and assign von Neumann algebras (or $C^\*$-algebras, but my question is about the von Neumann algebra situation only) to certain subsets of the spacetime subject to certain axioms.
I am interested in results about the Murray-von Neumann classification of these algebras, i.e. which kind of factors can occur in the central decomposition (the decomposition of the algebra as a direct integral of factors).
To make this more precise, here is an example: One can define vacuum representations on Minkowski spacetime, for details please see
[Haag-Kastler vacuum representation](http://ncatlab.org/nlab/edit/Haag-Kastler+vacuum+representation) on the nLab.
A net of a vacuum representation is said to satisfy duality, or be a dual net, if one has $\mathcal{M}(\mathcal{O}') = \mathcal{M}'(\mathcal{O})$, put in words: the algebra of the causal complement of a bounded open set $\mathcal{O}$ is the commutant of the algebra associated with $\mathcal{O}$. Then it is a theorem that algebras associated to diamonds can only have factors of type $III\_1$ in their central decomposition.
1. Is the assumption of duality necessary or is causality enough? Is Haag duality enough? (Haag duality means that the duality condition does not have to hold for all algebras associated to bounded open regions, but for diamonds only).
2. What are the necessary assumptions to deduce that algebras associated to diamonds are a factor of type $III\_1$, i.e. have trivial center? What are the necessary assumptions to get that these algebras are hyperfinite?
3. Are there similar results about the factor decomposition of algebras associated to more general subsets than diamonds of the Minkowski space, like open bounded subsets?
4. Are there similar results about more general spacetimes, like globally hyberbolic ones?
| https://mathoverflow.net/users/1478 | Murray-von Neumann classification of local algebras in Haag-Kastler QFT | There is a nice overview about algebraic quantum field theory by Halverson and Müger, which covers some of the stuff I mention below and can be found at
<http://arxiv.org/abs/math-ph/0602036>
Concerning your question(s): Having a factor of type III means that every (non-zero) projection in it is Murray-von Neumann equivalent to the identity. In the Haag-Kastler approach Borchers came up with a slightly weaker notion of "type III-ness", which is sometimes called 'property B'.
**Definition:** Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras on some common Hilbert space $H$. $\mathcal{A}$ has *property B* if for any two double cones $\mathcal{O}\_1$ and $\mathcal{O}\_2$ such that $\overline{\mathcal{O}\_1}\subset \mathcal{O}\_2$ and for any non-zero projection $E \in \mathcal{A}(\mathcal{O}\_1)$ there is an isometry $V \in \mathcal{A}(\mathcal{O}\_2)$, such that $VV^\* = E$ and $V^\*V= 1$ (in other words: $E$ is equivalent to $1$ in the algebra $\mathcal{A}(\mathcal{O}\_2)$).
The point is the following theorem proven by Borchers in
A remark on a theorem of B. Misra, H.J. Borchers, Communications in Mathematical Physics, Volume 4 (5), page 315-323.
**Theorem:** Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras satisfying *microcausality*, the *spectrum condition*, and *weak additivity*. Then the net $\mathcal{A}$ satisfies property B.
So, what are these notions?
* *microcausality* means that the algebras associated to spacelike separated double cones commute
* *additivity* means that for any double cone $\mathcal{O}$ the set of all $\mathcal{A}(\mathcal{O} + x)$ for $x \in T$, where $T$ denotes the translation group generates the associated (universal) $C^\*$-algebra (I am not quite sure, what weak means in the statement).
* *the spectrum condition* means there is a subset $T\_+ \subset T$ with $T\_+ \cap (-T\_+) = \{0\}$ and the spectrum of the unitary representation of $T$ is contained in $T\_+$.
Since these are all motivated by physics, they are all kind of natural to demand for nets of von Neumann algebras in AQFT.
Anyway there are also some results concerning the type of local algebras. For example, if you take wedge shaped regions $W$ in Minkowski spacetime, then it is shown in
On the Net of von Neumann algebras associated with a Wedge and Wedge-causal Manifolds, H.J. Borchers, <http://www.lqp.uni-goettingen.de/papers/09/12/09120802.html>
that the local algebra associated to $W$ is actually of type $III\_1$ for nets satisfying similar assumptions motivated by physics.
Furthermore there is the paper
The Universal Structure of Local Algebras, Buchholz, D., D'Antoni and Fredenhagen, K., Communications in Mathematical Physics 111 (1), page 123-135
in which it is shown that if you assume that your net is derived from a Wightman QFT and satisfies asymptotic scale invariance and nuclearity, then the local algebras associated to double cones are of the form $\mathfrak{R} \otimes Z(\mathcal{O})$, where $\mathfrak{R}$ is the unique hyperfinite type $III\_1$-factor and $Z(\mathcal{O})$ is the center of the local algebra $\mathcal{A}(\mathcal{O})$.
So, this and the fact that in many examples of nets you can check directly that the local algebras are of type $III\_1$ is the justification for physicists to assume this to be the case in everything physically relevant.
| 6 | https://mathoverflow.net/users/3995 | 21551 | 14,264 |
https://mathoverflow.net/questions/21555 | 16 | Modular forms could be defined for arbitrary subgroups of the modular group, and I have read that this is done in some papers, but every definition of a modular form I have seen has been with respect to congruence subgroups.
| https://mathoverflow.net/users/4692 | Why are modular forms (usually) defined only for congruence subgroups? | At a certain level, it's mostly a matter of (i) terminology and (ii) reading the right books. Technically the word "modular" in modular forms refers to the "modular group $SL\_2(\mathbb{Z})$".
In Miyake's book *Modular Forms*, he defines an **automorphic form** with respect to an arbitrary Fuchsian group $\Gamma$ (i.e., a discrete subgroup of $SL\_2(\mathbb{R})$). Then he goes on to say (p. 114) that "Automorphic functions and forms for modular groups are called *modular functions* and *modular forms* respectively." Despite the title, plenty of the book deals with the general case, or with the special case of Fuchsian groups associated to quaternion algebras, which do not yield modular forms according to his definition.
In Shimura's book *Introduction to the Arithmetic Theory of Automorphic Functions* he defines (pp. 28-29) automorphic functions and forms with respect to an arbitrary Fuchsian group of the first kind (i.e., finite hyperbolic covolume). The phrase "modular forms" is sometimes used in his book, but doesn't appear to get a formal definition.
These are, to my mind, the two most standard and authoritative references on "modular forms", and they both entertain the concept of a modular form with respect to a rather general Fuchsian group, whatever they want to call it.
On the other hand, there are reasons for restricting to Fuchsian groups which are **arithmetic** (which is a technical term here) and **of congruence type**. A theorem of Margulis shows that arithmeticity is equivalent to having a sufficiently rich theory of Hecke operators, which is highly important in number-theoretic applications. Similarly, being arithmetic of congruence type puts you in the realm of Shimura varieties, and gives you a rich theory of models of the Riemann surfaces defined over various abelian number fields. On the other hand, it is a famous consequence of Belyi's theorem that every algebraic curve over $\mathbb{Q}$ can be uniformized by a finite-index subgroup of $SL\_2(\mathbb{Z})$ (generally of non-congruence type). So if one is interested in the "special" arithmetic properties of modular curves, it makes sense to restrict to congruence type.
Indeed, continuing work of John Voight and myself indicates that the congruence type condition is even more arithmetically significant than the arithmeticity [sic!]. We define congruence subgroups of non-arithmetic Fuchsian triangle groups and derive some of the arithmetic applications (using techniques from group theory and the arithmetic theory of branched coverings) that are parallel to those satisfied by the usual modular curves. See
[http://alpha.math.uga.edu/~pete/triangle-091309.pdf](http://alpha.math.uga.edu/%7Epete/triangle-091309.pdf)
Note that this work is not yet finished, to my consternation. (Mea culpa. **Mea** culpa.)
| 22 | https://mathoverflow.net/users/1149 | 21556 | 14,267 |
https://mathoverflow.net/questions/21565 | 2 | In "Infinite dimensional analysis, A hitchhikers guide" by Aliprantis and Border, they write that these 2 classes of topologies "by and large include everything of interest".
@Pete Clarke: I was asking if it holds for mathematics in general, and not just for functional analysis. From the answers I get the impression that their statement is a fairly good first approximation.
@Gerald Edgar: I thought I read in Mathoverflow that the Zariski topology can be regarded as a weak topology (at least in some cases).
| https://mathoverflow.net/users/4692 | Is it true that the only interesting topologies are metric topologies and weak topologies? | Picking up on Gerald's interpretation of the question (namely, that it really focusses on infinite dimensional vector spaces) then I say: absolutely not!
For example, piecewise-smooth paths in some Euclidean space has a topology that is neither of these (it's an uncountable inductive limit of Frechet spaces). (Not that I particularly recommend this space!)
Close to Gerald's comment, "dual-Frechet" spaces (that is, the dual of a Frechet space with the **strong** topology) have very nice properties, almost as nice as Frechet spaces themselves. This class includes distributions with the "right" topology.
And that's the point, really. If you're only interested in, say, distributions for what they can say about compactly supported functions, then the weak topology is probably fine. However, if you are interested in distributions *in their own right* then the weak topology is very unlikely to be okay.
Here's an example from my research: I like infinite dimensional manifolds, and I quite like loop spaces. To construct the Dirac operator on a loop space, I needed to put an inner product on the **cotangent** bundle. So I needed, in effect, a continuous injective map $(L\mathbb{R}^n)^\* \to H$ ($H$ being some standard Hilbert space). I can't do this with the weak topology any continuous map from $(L\mathbb{R}^n)^\*$ with the weak topology to a normed vector space has to factor through a finite dimensional space. With the strong topology, though, it was no problem.
So, deal with metric and weak topologies if you like; but **real** analysts use the strong topology[1].
[1] Not sure what complex analysts use.
| 13 | https://mathoverflow.net/users/45 | 21569 | 14,277 |
https://mathoverflow.net/questions/21527 | 7 | **Background:** It is possible (see e.g., [this](http://www.cs.cmu.edu/%7Elblum/PAPERS/TuringMeetsNewton.pdf)) to define a Turing machine over an arbitrary ring. It reduces to the classical notion when the ring is $\mathbb{Z}\_2$; the key difference is that elementary algebraic computations are allowed to be performed in one step (and one has infinite precision). It is possible to define analogs of classical complexity classes (e.g. $\mathrm{P}\_R, \mathrm{NP}\_R, \mathrm{BPP}\_R$, etc. with respect to a ring $R$). Sometimes the ring may be required to be ordered, and the machine is allowed to test for ordering (it is always allowed to test for equality).
It should, therefore, be possible to define a class $\mathrm{IP}\_R$ of problems that can be solved by a polynomial-time interactive proof system over $R$ with error probability $\leq \frac{1}{3}$, a class $\mathrm{PSPACE}\_R$ of problems that can be solved in polynomial space by a (deterministic) Turing machine. (Probably this has already been done, but I just haven't been able to find it.) When $R = \mathbb{Z}\_2$ (i.e. the classical case), it is a known result that $\mathrm{IP}=\mathrm{PSPACE}$.
**Question:** Does the same work over other rings $R$?
Perhaps there might be a problem, since the only proof I've seen (e.g. in Sipser's *Introduction to the Theory of Computation*) uses the $\mathrm{PSPACE}$-completeness of $TQBF$, and I don't know whether that would work over an arbitrary ring.
| https://mathoverflow.net/users/344 | Does IP = PSPACE work over other rings? | I think PSPACE does not make much sense over arbitrary rings (it is better to consider classes such as PAR).
Consider $R=\mathbb Z$ or $R=\mathbb Q$ with order. You can encode any finite collection of numbers in a single number and extract elements back in constant space. So you can emulate a potentially infinite memory in constant space (or linear, if you count the size of input). So anything decidable at all is in $PSPACE\_R$.
What $IP\_R$ is depends on what kind of probablity distributions the verifier is allowed to use. In any case, I cannot imagine that problems like "is the given number a power of two?" (with input of size 1!) can be solved by IP in constant time.
In the classic Blum-Shub-Smale case (that is, $\mathbb R$ with order), I suggest the problem "is the given number an integer?". It is trivially solvable in constant space and I'm sure it is not in (constant time) IP due to its infinutely many connected components.
| 5 | https://mathoverflow.net/users/4354 | 21573 | 14,281 |
https://mathoverflow.net/questions/21313 | 8 | [[UPDATE: This work has now been published at SIAM J Discrete Math.: [Formulae for the Alon–Tarsi Conjecture](http://epubs.siam.org/sidma/resource/1/sjdmec/v26/i1/p65_s1).]]
By equating two formulae (one congruence by Glynn (1) (which has just appeared) and one unpublished formula) for the number of even Latin squares minus the number of odd Latin squares, we find the following result.
For odd primes $p$ we have \[\sum\_{A \in B} (-1)^{\sigma\_0(A)} \equiv 1 \pmod p\] where $B$ is the set of $(p-1) \times (p-1)$ $\\,(0,1)$-matrices whose determinant is indivisible by $p$ and $\sigma\_0(A)$ is the number of zeroes in $A$. It happens to be true for $p=2$ also (but it does not follow from Glynn's result).
>
> Is this result already known? If so, it would provide an alternate proof of Glynn's result.
>
>
>
To illustrate, consider when $p=3$. The (0,1)-matrices whose determinants are indivisible by $p$ are
```
01 10 01 10 11 11
10 01 and 11 11 01 10
```
So the sum becomes $+2-4=-2 \equiv 1 \pmod 3$.
It is equivalent to the congruence \[\sum\_{A \in C} (-1)^{\sigma\_0(A)} \det(A)^{p-1} \equiv 1 \pmod p\] where $C$ is the set of all $(p-1) \times (p-1)$ $\\,(0,1)$-matrices (via Fermat's Little Theorem).
(1) Glynn, D., 2010. [The conjectures of Alon-Tarsi and Rota in dimension prime minus one](http://epubs.siam.org/sidma/resource/1/sjdmec/v24/i2/p394_s1). SIAM J. Discrete Math., 24 (2010), 394-399.
| https://mathoverflow.net/users/2264 | (0,1)-matrix congruence: is it known? | Darij's proof was wrong but both his claims (that it is next to trivial and that it requires some long formulae) were actually correct. Here is the demonstration.
To start with, we'll count the sum over degenerate (in $\mathbb Z\_p$) matrices instead because the sum over all matrices is clearly $0$. Also, I'll prefer to use $\sigma\_1$ instead of $\sigma\_0$ as the exponent of $(-1)$ (they have the same parity if $p>2$). Now, let us observe that the degenerate matrices are exactly those that kill some non-zero vector $x\in\mathbb Z\_p^{p-1}$. Moreover, the total number of vectors they kill is $-1$ (I'm counting modulo $p$, of course). Thus, we can just cound the sum over all $x\ne 0$ of the "signatures" of matrices that kill $x$. Now, a matrix kills $x$ if and only if each row kills $x$, so this sum is just the sum of signatures of all rows killing $x$ to the power $p-1$. The sum of row signatures over all possible rows is $0$, so we can again replace it with the sul over all rows not killing $x$ to the power $p-1$, which is the same as the sum over all matrices not killing $x$ in any row. Now, if we count modulo $p$, this sum can be written as
$$
\begin{aligned}
&\sum\_x\sum\_{\varepsilon}\prod\_{i,j}(-1)^{\varepsilon\_{ij}}\prod\_i\left(\sum\_j\varepsilon\_{ij}x\_j\right)^{p-1}
\\
&=\sum\_{x,\varepsilon}\sum\_\lambda\frac{(p-1)!^{p-1}}{\prod\_{i,j}\lambda\_{jj}!}\prod\_{i,j}(-1)^{\varepsilon\_{ij}}\varepsilon\_{ij}^{\lambda\_{ij}}x\_j^{\lambda\_{ij}}
\\
&=\sum\_{x,\lambda}\frac{(p-1)!^{p-1}}{\prod\_{i,j}\lambda\_{jj}!}\prod\_{i,j}x\_j^{\lambda\_{ij}}(0^{\lambda\_{ij}}-1)
\end{aligned}
$$
($\lambda$ here runs over all matrices with non-negative entries such that $\sum\_j\lambda\_{ij}=p-1$, $x$ runs over non-zero vectors in $\mathbb Z\_p$ and $\varepsilon$ runs over all $0,1$ matrices)
Now we use the deep formula $0^0=1$, which tells us that the only $\lambda$ for which the product is not $0$ is the one consisting of all ones. This reduces the sum to
$$
(p-1)!^{p-1}\sum\_x\prod\_j x\_j^{p-1}=(p-1)!^{p-1}\left(\sum\_{a\in\mathbb Z\_p}a^{p-1}\right)^{p-1}\equiv (p-1)!^{p-1}(p-1)^{p-1}\equiv 1\mod p
$$
(we can now include the $0$ vector in the sum, it doesn't contribute anything).
The end.
| 4 | https://mathoverflow.net/users/1131 | 21574 | 14,282 |
https://mathoverflow.net/questions/21552 | 48 | I have read Hartshorne's *Algebraic Geometry* from chapter 1 to chapter 4, so I'd like to find some suggestions about the next step to study arithmetic geometry.
* I want to know how to use scheme theory and its cohomology to solve arithmetic problems.
* I also want to learn something about moduli theory.
Would you please recommend some books or papers? Thank you very much!
| https://mathoverflow.net/users/5274 | Roadmap for studying arithmetic geometry | My suggestion, if you have really worked through most of Hartshorne, is to begin reading papers, referring to other books as you need them.
One place to start is Mazur's "Eisenstein Ideal" paper. The suggestion of Cornell--Silverman is also good. (This gives essentially the complete proof, due to Faltings, of the Tate conjecture for abelian varieties over number fields, and of the Mordell conjecture.) You might also want to look at Tate's original paper on the Tate conjecture for abelian varieties over finite fields,
which is a masterpiece.
Another possibility is to learn etale cohomology (which you will have to learn in some form or other if you want to do research in arithemtic geometry). For this, my suggestion is to try to work through Deligne's first Weil conjectures paper (in which he proves the Riemann hypothesis), referring to textbooks on etale cohomology as you need them.
| 29 | https://mathoverflow.net/users/2874 | 21583 | 14,289 |
https://mathoverflow.net/questions/21578 | 36 | I proposed to a master's student to work, from the exercise in Ghys-de la Harpe's book, on the proof that a finitely generated group $G$ that is quasi-isometric to $\mathbb{Z}$ is virtually $\mathbb{Z}$. However I initially had in mind the result that gives the same conclusion from the hypothesis that $G$ has linear growth.
Do you know of any simple (and elementary, in particular without assuming Gromov's theorem on polynomial growth groups) proof that a group of linear growth is quasi-isometric to $\mathbb{Z}$?
| https://mathoverflow.net/users/4961 | Is there a simple proof that a group of linear growth is quasi-isometric to Z? | Here are two papers dealing with this question:
Wilkie, A. J.(4-MANC); van den Dries, L.(1-IA-S)
An effective bound for groups of linear growth.
Arch. Math. (Basel) 42 (1984), no. 5, 391--396.
20F05
and
Imrich, Wilfried(A-MNT); Seifter, Norbert(A-MNT)
A bound for groups of linear growth.
Arch. Math. (Basel) 48 (1987), no. 2, 100--104.
the second one using ends of groups, as indicated by Sam.
| 10 | https://mathoverflow.net/users/3380 | 21586 | 14,291 |
https://mathoverflow.net/questions/20730 | 5 | Preliminars and notation:
Let $M$ be an $n$-dimensional compact manifold, $T\colon M\rightarrow M$ a diffeomorphism and $( x\_n)\_{n\in\mathbb{Z}}$ a dense orbit under $T$, ($x\_n = T^n(x\_0)$). Let $p\in M$ be another point and define, for $\delta>0$, $B\_n(\delta) = B(p, e^{-n\delta})$.
Question: Is it true that for every $\delta>0$ the set
$A = $( $n\in\mathbb{N} | x\_n \in B\_n(\delta)$ )
has finite cardinality? How can it be proven?
Thank you for the answers!
| https://mathoverflow.net/users/5231 | About dense orbits on dynamical systems | This is called the *shrinking target problem*, and there is a reasonably large literature on it. For hyperbolic dynamical systems we can usually find quite a few pairs $x$, $p$ such that $A$ is infinite for all $\delta$. Indeed, I believe that there are results showing that in certain cases, for any point $z$ and positive real number $\delta>0$, the set of all $x$ such that $d(T^nx,z)<\exp(-n\delta)$ for infinitely many $n \geq 1$ has positive Hausdorff dimension. A good place to start would be the articles "Ergodic theory of shrinking targets" and "The shrinking target problem for matrix transformations of tori", both by Hill and Velani, but there are many results beyond this.
For illustration, here is a nice example in the case where $T$ is a smooth map of the circle which is not a diffeomorphism. I realise that this falls slightly outside the purview of your question, but it is possible to extend this argument to the case of toral diffemorphisms using the technical device of a Markov partition. (I will not attempt this here because it is very fiddly.) Let $X=\mathbb{R}/\mathbb{Z}$ be the circle, let $T \colon X \to X$ be given by $Tx = 2x \mod 1$, and let $d$ be a metric on $X$ which locally agrees with the standard metric on $\mathbb{R}$. Take $p=0 \in X$ and fix any $\delta>0$. Now, the orbit of $x$ is dense if and only if it enters every interval of the form $(k/2^n,(k+1)/2^n)$, if and only if every possible finite string of 0's and 1's occurs somewhere in the tail of its binary expansion.
On the other hand, we have $d(T^nx,0)<2^{-\delta n}$ as long as the binary expansion of $x$ contains a string of zeroes starting at position $n$ and having length $\lceil \delta n \rceil$. I think that it is not difficult to see that we can construct an infinite binary expansion, and hence a point $x$, such that this condition is met for infinitely many $n$, whilst simultaneously meeting the condition that the orbit of $x$ is dense. In particular we can construct a point $x$ for which $A$ is infinite, even for all $\delta$ simultaneously if you like.
| 6 | https://mathoverflow.net/users/1840 | 21588 | 14,292 |
https://mathoverflow.net/questions/21232 | 22 | Suppose that $X$ is a non-singular variety and $Z \subset X$ is a closed subscheme. When is the
blow-up $\operatorname{Bl}\_{Z}(X)$ non-singular?
The blow-up of a non-singular variety along a non-singular subvariety is well-known to be non-singular, so the real question is ``what happens when $Z$ is singular?" The blow-up can be singular as the case when $X = \mathbb{A}^{2}$and $Z$ is defined by the ideal $(x^2, y)$ shows. On the other hand, the example where $Z$ is defined by the ideal $(x,y)^2$ shows that the blow-up can be non-singular.
**Edit:** Based on the comments of Karl Schwede and VA, I think that it would also be interesting to find non-trivial examples of appropriate $Z$'s. I am splitting this off as a [separate question.](https://mathoverflow.net/questions/21627/what-are-non-trivial-examples-of-non-singular-blow-ups) In the comments there, the users quim and Karl Schwede say a bit about what can be said about this question using Zariski factorization.
| https://mathoverflow.net/users/5337 | When is a blow-up non-singular? | Craig Huneke told me about this [paper](https://www.tandfonline.com/doi/abs/10.1080/00927879708825958): "On the smoothness of blow-ups" ([MR1446135](https://mathscinet.ams.org/mathscinet-getitem?mr=1446135), [Zbl 0878.13004](https://zbmath.org/?q=an%3A0878.13004), by O'Carroll and Valla). The title alone seems to suggest it might be useful for you.
| 12 | https://mathoverflow.net/users/2083 | 21592 | 14,294 |
https://mathoverflow.net/questions/21591 | 5 | Here is the essence of a problem I have run in to: I have a finite poset D with a terminal object. If I formally invert all of the morphisms, and add these into my diagram, does the new diagram D' still commute?
I think that the resulting diagram will still commute basically because I have done a lot of examples. Working out a few examples you can see that it basically follows by doing it for the "commutative triangle", and applying this finitely many times. It feels like I should be able to do some kind of messy induction, but I do not really want such a proof cluttering up my work.
Is there a reference I could quote for a result like this? It seems like if it is true it should be a "folk lemma".
Of course, if you have more relaxed criteria for when the result will hold, that would be helpful too.
Also if you know of a conceptual proof which does not fall back on some messy induction, that would be wonderful!
EDIT: An example might help to clarify my question. (How do you draw diagrams on MO?)
```
a-->b
^ ^
| |
c-->d
```
is my poset. b is the terminal object. Now say someone told you that this was actually a subcategory of a larger category, and in that larger category all of the arrows were invertible. Now consider the larger diagram consisting of the 4 original arrows and their inverses. Is this diagram also commutative? Yes! It is just one or two lines of formal manipulation.
| https://mathoverflow.net/users/1106 | When does adding inverses of morphisms preserve commutativity of a diagram? | There is an easy conceptual proof using the fact that the category obtained by formally inverting all the arrows in a category C is equivalent to the fundamental groupoid of the nerve NC of C, and that the nerve of a category with a final object is contractible. Without the assumption of a final object your assertion is false in general, e.g., reverse the arrows from c in your example.
But it should also be easy to prove by induction: for any zigzag of arrows between a and b, the corresponding map in the category with all arrows inverted, when composed with the map from b to the original terminal object, is equal to the map from a to the original terminal object (this is by induction); and so any two maps from a to b in the category with all arrows inverted are equal. In symbols: let me write $t\_x$ for the unique morphism in C from $x$ to the terminal object and $[f]$ for the image of $f$ in the category with all arrows inverted. Suppose $[f\_1]^{\pm 1} \cdots [f\_n]^{\pm 1}$ is a typical map in the category with all arrows inverted with domain $a$ and target $b$. Then the inductive claim is that $[t\_b] [f\_1]^{\pm 1} \cdots [f\_n]^{\pm 1} = [t\_a]$, and so $[f\_1]^{\pm 1} \cdots [f\_n]^{\pm 1} = [t\_b]^{-1} [t\_a]$.
| 5 | https://mathoverflow.net/users/126667 | 21596 | 14,296 |
https://mathoverflow.net/questions/21599 | 35 | Let $\beta \mathbb{N}$ be the Stone-Cech compactification of the natural numbers $\mathbb{N}$, and let $x, y \in \beta \mathbb{N} \setminus \mathbb{N}$ be two non-principal elements of this compactification (or equivalently, $x$ and $y$ are two non-principal ultrafilters). I am interested in ways to "model" the ultrafilter $y$ using the ultrafilter $x$. More precisely,
Q1. (Existence) Does there necessarily exist a continuous map $f: \beta \mathbb{N} \to \beta \mathbb{N}$ which maps $x$ to $y$, while mapping $\mathbb{N}$ to $\mathbb{N}$? To put it another way: does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that $\lim\_{n \to x} f(n) = y$?
Q2. (Uniqueness) Suppose there are two continuous maps $f, g: \beta \mathbb{N} \to \beta\mathbb{N}$ with $f(x)=g(x)=y$, which map $\mathbb{N}$ to $\mathbb{N}$. Is it then true that $f$ and $g$ must then be equal on a neighbourhood of $x$?
I suspect the answer to both questions is either "no" or "undecidable in ZFC", though perhaps there exist "universal" ultrafilters $x$ for which the answers become yes. But I do not have enough intuition regarding the topology of $\beta \mathbb{N}$ (other than that it is somewhat pathological) to make this more precise. (The fact that $\beta\mathbb{N}$ is not first countable does seem to indicate that the answers should be negative, though.)
| https://mathoverflow.net/users/766 | "Transitivity" of the Stone-Cech compactification | The answer to Q1 is no. This has been well studied in set theory; you're basically asking whether any two non-principal ultrafilters on $\mathbb{N}$ are comparable under the Rudin-Keisler ordering. Variations on your question have led to many, many interesting developments in set theory, but your question Q1 is easy to answer by a cardinality argument.
First note that every $f:\mathbb{N}\to\mathbb{N}$ has a unique extension to a continuous function $\bar{f}:\beta\mathbb{N}\to\beta\mathbb{N}$. Any $x \in \beta\mathbb{N}$ has at most $2^{\aleph\_0}$ images through such $\bar{f}$, but there are $2^{2^{\aleph\_0}}$ ultrafilters on $\mathbb{N}$, so there are very many $y \in \beta\mathbb{N}$ which are not images of $x$ through such $\bar{f}$.
The answer to Q2 is also no. Let $y$ be a nonprincipal ultrafilter on $\mathbb{N}$. The sets $A \times A\setminus\Delta$ where $A \in y$ and $\Delta = \{(n,n) : n \in \mathbb{N}\}$ form a filter base on $\mathbb{N}\times\mathbb{N}$. Let $x$ be an ultrafilter on $\mathbb{N}\times\mathbb{N}$ that contains all these sets. The left and right projections $\pi\_1,\pi\_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ both send $x$ to $y$, but they are not equal on any neighborhood of $x$.
However, the answer to Q2 is yes when $x$ is a selective ultrafilter. Recall that $x$ is selective if for every $h:\mathbb{N}\to\mathbb{N}$ there is a set $A \in x$ on which $h$ is either constant or one-to-one. Given $f,g:\mathbb{N}\to\mathbb{N}$ such that $\bar{f}(x) = \bar{g}(x)$ is nonprincipal, then we can find $A \in x$ on which $f$ and $g$ are both one-to-one. In that case, $f\circ g^{-1}$ must be well-defined on some $A' \in x$. Any extension of $f\circ g^{-1}$ to the complement of $A'$ must map $x$ to $x$, which means that $f \circ g^{-1}$ is the identity on some $A'' \in x$. Thus $f$ and $g$ are equal on the neighborhood of $x$ defined by $A''$.
| 43 | https://mathoverflow.net/users/2000 | 21605 | 14,299 |
https://mathoverflow.net/questions/21604 | 11 | I am writing something for the journal of the university on the Lorentz space, and I want to prove that the following definition of the hyperbolic distance on the upper sheet of the hyperboloid $v \cdot v=-1$ is a metric:
*The hyperbolic distance between $u$ and $v$ is the only positive number $\eta (u,v)$ such that*
$$\cosh(\eta(u,v))=-u \cdot v.
$$
The first properties are easy, but all the proofs of the triangle inequality that I have seen seem complicated. In Ratcliffe's Foundations of Hyperbolic Manifolds the lorentzian cross product is used along with many of its properties, and I would like not having to introduce the cross product just for this proof.
I have thought about showing that the hyperbolic angle is the same as the lorentzian arc-length distance, and then showing that the metric given by the arc length satisfies the triangle inequality, but this seems even more tedious. Is there a shorter and nicer proof of this?
| https://mathoverflow.net/users/1619 | Nice proof of the triangle inequality for the metric of the hyperbolic plane | We need to check $\eta(u,v)+\eta(v,w)\ge\eta(u,w)$. Introduce coordinates $x,y,z$ so that the form is $x^2+y^2-z^2$.
First, verify that there is a Lorentz map sending $v$ to $(0,0,1)$. Since it is an isometry, we may now assume that $v=(0,0,1)$. This is the main idea. For added convenience, you may also rotate the $xy$-plane so that the $y$-coordinate of $u$ equals 0.
Next, observe that the formula yields equality in the case when the projections of $u$ and $w$ to the $xy$-plane are endpoints of a segment containing (0,0). This is straigtforward if you write $u=(\sinh a,0,\cosh a)$ and $w=(-\sinh b,0,\cosh b)$ where $a,b\ge 0$.
Finally, rotate $w$ around the $z$-axis until it comes to a position as above. The product $u\cdot w$ grows down (it equals contant plus the scalar product of the $xy$-parts, since $z$-coordinates are fixed). Hence $\eta(u,w)$ grows up while the two other distances stay, q.e.d.
Of course, for writing purposes the last step is just an application of Cauchy-Schwarz for the scalar product in $\mathbb R^2$.
This was about two-dimensional hyperbolic plane, in higher dimensions just insert more coordinates (they will not actually show up in formulae).
| 12 | https://mathoverflow.net/users/4354 | 21608 | 14,300 |
https://mathoverflow.net/questions/21601 | 3 | This is version 2 of a question about the ultimate limits of Tennenbaum's Theorem. The attempt to find these limits by moving up the induction heirarchy, as in Wilmer's Theorem, seems somehow indecisive. I suggested that maybe there is a Theory $T$ extending open induction such that
1) $T$ has a recursively presentable nonstandard model.
2) If the sentence $\phi$ is not provable from $T$, then
$T+\phi$ has no recursively presentable nonstandard model.
François G. Dorais immediately replied that this just amounts to $T$ being complete.
So... What about asking for the maximum $n$ such that the theory of all true (in the integers) all-2 sentences with n existential quantifiers has a recursive nonstandard model?
What is known about this? Is it known that $n<2$?
| https://mathoverflow.net/users/5229 | Ultimate limits of Tennenbaum's Theorem | This is equivalent to saying that $T$ is complete. Let $M$ be a recursively presentable model of $T$. If $T$ is not complete, then we can find $\phi$ which is not provable from $T$ but is true in $M$. Of course, $M$ is then a recursively presentable model of $T+\phi$. If $T$ is complete then $T+\phi$ has no model at all if $\phi$ is not provable from $T$. So the complete theory of a recursively presentable model of open induction will do.
| 4 | https://mathoverflow.net/users/2000 | 21609 | 14,301 |
https://mathoverflow.net/questions/21614 | 3 | My question is coming from the method Reid and Chris suggested in solving the problem [here](https://mathoverflow.net/questions/19116/colimits-in-the-category-of-smooth-manifolds). Help on any point is greatly appreciated!
Question 1. For a real manifold $M$, consider $C^{\infty}(M,\mathbb{R})$. For a point $p\in M$, consider the ideal $I\_p=\{ f : f(p)=0 \}$. Is $I\_p^n$ equal to the set of smooth maps $f$ which have $n-1$st order contact (ref. Golubitsky, Guillemin pg. 37) with the $0$ function?
Question 2. Consider $C(\mathbb{R},\mathbb{R})$ and the ideal $I\_0= \{ f : f(0)=0 \}$. Is it the case that $I\_0^2 = \{ f : f(0)=f'(0)=0\}$? Is it the case that $I\_0^n=\{ f : f(0)=f'(0)=...=f^{(n-1)}(0)=0\}$?
To question 2, the one inclusion is immediate. However the inclusion $ \{ f : f(0)=f'(0)=0\} \subseteq I\_0^2 $ doesn't seem obvious to me right now. It seems like one could perhaps find a function which doesn't agree with it's Taylor series that satisfies the derivative condition, but not be in $I^2\_0$.
I hope this isn't too elementary for MO. Thanks for your help!
| https://mathoverflow.net/users/4517 | Ideals in the ring of smooth endomorphisms of the real line | This is all fairly well-known stuff. In Q1, the ideal $I\_p^n$ is the
set of functions for which the derivatives up to order $n-1$
at $p$ vanish in a local coordinate system centred at $p$.
I'll just give a rough outline of how to reduce to the case where $M=\mathbb{R}^k$.
Take a coordinate neighbourhood $U$ of $p$ where $p$ corresponds to $0$.
It's clear that any function in $I\_p^n$ has Taylor expansion in $U$
with no terms of total degree $< n$. Conversely suppose that $f$ is such
a function. Using the result for $M=\mathbb{R}^k$ we can express
$f$ restricted to $U$ as a finite sum of terms $g\_1\cdots g\_n$
where each $g\_i$ vanishes at $0$. Multiply each $g\_i$ by a smooth function $h$
vanishing on a neighbourhood of $0$ and equalling $1$ on a smaller
neighbourhood. We can regard $h g\_i$ as an element of $I\_p\subseteq C^\infty(M)$.
Then $f$ minus a linear combination of terms of the form
$(hg\_1)\cdots (hg\_n)$ vanishes in a neighbourhood of $0$ and to
complete the proof one has to show that all such functions are also in $I\_p^n$
(which is straightforward).
| 4 | https://mathoverflow.net/users/4213 | 21615 | 14,305 |
https://mathoverflow.net/questions/21616 | 2 | Let A be a pair (i, x)
$H(A)$ <=> program i halts on input x
$P(A)$ <=> (there exists a proof for $H(A)$) $\vee$ (there exists a proof for $\neg H(A)$)
Assume $\forall A: P(A)$ then we can solve the halting problem (write an algorithm that enumerates all proofs and checks whether they proof $H(A)$ or $\neg H(A)$)
Therefore $\exists A: \neg P(A)$.
For such an A:
Assume $H(A)$: Then we have a proof for $H(A)$ (the sequence of steps that lead to the halting state). Therefore $\neg H(A)$.
But now we have a proof for $\neg H(A)$ and therefore $P(A)$ which contradicts the definition of A???
I assume I have a basic misconception about something. Please tell me what went wrong?
Thanks
| https://mathoverflow.net/users/5415 | Provability of termination. Whats wrong with my reasoning? | I believe that the error sneaks in with the phrase "for such an A." It is indeed true that for any $A$ such that $\neg P(A)$ we have that $\neg H(A)$; it follows that to be able to prove $\neg P(A)$ actually requires us to be able to prove $\neg H(A)$. You've substituted the *a priori* knowledge "this is such an $A$" for the proof that this $A$ has the property $\neg P(A)$, but it's the latter of these that you actually need. This avoids any paradox.
| 5 | https://mathoverflow.net/users/4658 | 21618 | 14,307 |
https://mathoverflow.net/questions/21562 | 29 | Almost any mathematical concept has antecedents; it builds on, or is related to, previously known concepts. But are there concepts that owe little or nothing to previous work?
The only example I know is Cantor's theory of sets. Nothing like his concrete manipulations of actual infinite objects had been done before.
| https://mathoverflow.net/users/4692 | What are some mathematical concepts that were (pretty much) created from scratch and do not owe a debt to previous work? | Shannon's work on Information theory. Maybe the math wasn't new but the ideas (such as positing a qualitative metric of information and identifying its relevance to design of communication systems) definitely were.
| 36 | https://mathoverflow.net/users/2878 | 21620 | 14,308 |
https://mathoverflow.net/questions/21543 | 8 | Hi,
I have what I hope is a very simple question related to unfamiliar notation.
I am looking through a maths paper on a topic related to set theory which contains a symbol,
$\uplus$,
and I would like to know how, if at all, it differs from the typical
$\cup$
symbol in terms of its meaning.
The context leads me to believe that it does not in fact differ at all but since I don't even know the name of the symbol other than the latex id that I looked up, I can't seem to confirm that suspicion.
Cheers
edit: it seems that the latex renderer also does not know about this obscure symbol '\cupplus' but '\uplus' does work.
edit2: thanks for all the replies! however, since everyone commented rather than providing an answer it seems I cannot grant the coveted 'answer' status to anyone. the disjoint union makes the most sense.
| https://mathoverflow.net/users/179 | What is the definition of the $\uplus$ symbol? | For the sake of giving a response to tag as an answer:
Some authors use this to denote a disjoint union of sets, i.e., "$A \cup B$ for sets $A, B$ where $A \cap B = \emptyset$."
| 11 | https://mathoverflow.net/users/572 | 21644 | 14,319 |
https://mathoverflow.net/questions/21641 | 11 | What is the maximal number of perfect matchings a graph $G(V,E)$ can have if $|V|$ and $|E|$ are fixed? I am particularly interested in a case when $|E| = c|V|^2$.
| https://mathoverflow.net/users/2641 | Maximum number of perfect matchings in a graph | I think this is exactly the main result of [this recent paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V00-4X9D5FW-2&_user=4423&_coverDate=02%2F28%2F2010&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1298182872&_rerunOrigin=scholar.google&_acct=C000059605&_version=1&_urlVersion=0&_userid=4423&md5=e0f086e2b1298619e290e92b111dc6b8) we just published in *Discrete Mathematics*. Just in case the link doesn't work: this is "Graphs with the maximum or minimum number of 1-factors" by D. Grossa, N. Kahl and J.T. Saccoman. I have read only the abstract. Let me know if this is what you were looking for.
| 13 | https://mathoverflow.net/users/4040 | 21648 | 14,323 |
https://mathoverflow.net/questions/21651 | 33 | For $K$ a compact Lie-group with maximal torus $T$, I'd like to know the cohomology $\text{H}^{\ast}(K/T)$ of the flag variety $K/T$.
If I'm not mistaken, this should be isomorphic to the algebra of coinvariants of the associated root system, according to the "classical Borel picture", as it is often called in the literature (sadly, often without a reference). Unfortunately, Borels original paper is quite long and so I have the following
**Question:** Does anybody know a short proof for the theorem?
For example, can it be proved using a direct computation of the Lie-algebra cohomology $\text{H}^{\ast}({\mathfrak k},{\mathfrak t})$?
As a side-question: Is it correct that for a complex semisimple Lie-group $G$ with Borel $B$, compact real form $K$ and maximal torus $T$ the map $K/T\to\ G/B$ is a homotopy equivalence? What is a good reference for these things?
I'm sorry if this is too elementary for MO, but apart from Borel's original paper I couldn't find good sources.
| https://mathoverflow.net/users/3108 | Cohomology of Flag Varieties | Borel's lengthy 1953 Annals paper is essentially his 1952 Paris thesis. It was
followed by work of Bott, Samelson, Kostant, and others, which eventually answers your
side question affirmatively. For a readable modern account in the setting
of complex algebraic groups rather than compact groups, try to locate a copy of the lecture notes: MR649068 (83h:14045) 14M15 (14D25 20F38 57N99 57T15)
Hiller, Howard,
Geometry of Coxeter groups.
Research Notes in Mathematics, 54.
Pitman (Advanced Publishing Program), Boston, Mass.-London, 1982. iv+213 pp.
ISBN 0-273-08517-4. (This was based on his 1980 course at Yale. Eventually
he left mathematics to work for Citibank.) The identification of the cohomology ring with the coinvariant algebra of the Weyl group has continued to be important for algebraic and geometric
questions, for instance in the work of Beilinson-Ginzburg-Soergel. While Hiller's notes are not entirely self-contained, they are helpfully written. (But note that his short treatment of Coxeter groups has a major logical gap.)
ADDED: In Hiller's notes, Chapter III (Geometry of Grassmannians) is most
relevant. For connections with Lie algebra cohomology, the classical paper
is: MR0142697 (26 #266) 22.60 (17.30)
Kostant, Bertram,
Lie algebra cohomology and generalized Schubert cells.
Ann. of Math. (2) 77 1963 72–144. Nothing in this rich circle of ideas can be made
quick and easy; a lot depends on what you already know.
P.S. Keep in mind that Hiller tends to give explicit details just for the
general linear group and grassmannians, but he also points out how the main
results work in general, with references. I don't know a more modern textbook
reference for this relatively old work. But the intuitive connection between
the Borel picture and the Bott/Kostant cohomology picture is roughly this: The
Lie subalgebra spanned by negative root vectors plays the role of tangent space
to the flag manifold/variety. In the Lie algebra cohomology approach you get an explicit graded picture for each degree in terms of number of elements in
the Weyl group of a fixed length, whereas the Borel description in terms of Weyl group coinvariants makes the
algebra structure of cohomology more transparent. (What I
don't know is whether a simpler proof of Borel's theorem can be derived using Lie algebra cohomology.)
Concerning the relationship between $K/T$ and $G/B$, this goes back to the
work around 1950 on topology of Lie groups (Iwasawa, Bott, Samelson): all the
topology of a connected, simply connected Lie group comes from a maximal compact subgroup. So the two versions of the flag manifold are homeomorphic.
In later times, emphasis has often shifted to treating $G$ as a complex algebraic group, so that $G/B$ is a projective variety. For me the literature is hard to compactify.
One more reference, which treats the Borel theorem in a semi-expository style: MR1365844 (96j:57051) 57T10
Reeder, Mark (1-OK),
On the cohomology of compact Lie groups.
Enseign. Math. (2) 41 (1995), no. 3-4, 181–200. There is some online access [here](http://retro.seals.ch/digbib/voltoc?pid=ensmat-001:1995:41).
| 21 | https://mathoverflow.net/users/4231 | 21659 | 14,328 |
https://mathoverflow.net/questions/21656 | 2 | I know this might be a very elementary question. But I could not find the original definition of Extreme(or Extremal)vectors of some weights $\lambda$(fixed the $w\in W$,where $W$ is Weyl group)
I am looking for definition for Extreme vector for finite dimensional Lie algebra and Affine Lie algebra. I found a paper saying :"Extreme vector verifies Weyl Character formula" What does it mean?
I am looking for reference talking about this concept. Thanks in advance
EDIT: I guess it is just highest weight vector, but I am not sure
| https://mathoverflow.net/users/1851 | What is Extreme/Extremal vector according to some weights | Usually "extremal weight" means a weight in the Weyl group orbit of the highest, and I would interpret "extremal vector" as an element of said weight space.
| 7 | https://mathoverflow.net/users/66 | 21662 | 14,331 |
https://mathoverflow.net/questions/21667 | 41 | This is a somewhat vague question; I don't know how "soft" it is, and even if it makes sense.
[Edit: in the light of the comments, we can state my question in a formally precise way, that is: **"Is the homotopy category of topological spaces a concrete category (in the sense, say, of Kurosh and Freyd)?"**. You may still want to read what follows, for a bit of motivation]
Historically, I'd bet people started to look at concrete metric spaces $(X,d)$ before exploring the utility of the abstraction given by general topological spaces $(X,\tau)$ ($\tau$ is a topology). The heuristic idea captured by the concept of a topological space is endowing a set $X$ with a "geometry" that forgets the rigidity of a hypothetical metric structure $(X,d)$, though retaining the "qualitative" aspects of the geometry given by the metric.
Of course there are non-metrizable topological spaces, but let stick to metrizable ones for the moment. I think it should be possible to see a topological space $(X,\tau)$ as an equivalence class of metric spaces: $(X,[d])$, where $[d]$ is the class of all metrics on the set $X$ that give rise to the same topology. So, an $(X,\tau)$ just has several "rigid" models $(X,d)$, and a morphism of topological spaces $f:(X,\tau) \rightarrow (Y,\tau')$ is given by taking any map $(X,d) \rightarrow (Y,d^{'})$ of "representatives" which is continuous according to the "metric ball" definition.
[Please correct the above picture if it is imprecise or even just wrong!!]
The (perhaps naive) way I have always thought about homotopy is that it is an even "further step" in making the geometry more "qualitative" and less rigid: you can "collapse positive dimensional appendices" of a space as far as they are contractible, and so forth. When trying to make this formal, you consider "homotopy types", which are equivalence classes $[(X,\tau)]$ of topological spaces, where $(X,\tau) \sim (X',\tau')$ if there is a homotopy equivalence $\varphi:(X,\tau) \rightarrow (X',\tau')$. What are morphisms in the homotopy category? Just morphisms $f$ between "representatives", but now you have also to consider them up to homotopy, i.e. you take $[f]$ where $f \sim f'$ if there's a homotopy $\alpha: f \rightarrow f'$.
It's ugly to think of topological spaces as $[(X,d)]$ (or, rather, $(X,[d])$): it's better to use the simpler and more expressive abstraction $(X,\tau)$.
So, **the question is:**
* Is there some kind of "homotopology" $h$ (whatever it is) that one can put on sets $S$ so that each homotopy type $[(X,\tau)]$ is fully described by a "homotopical space" (whatever it is) $(S,h)$ and homotopy classes of morphisms $[f]$ correspond to "$h$-compatible" (whatever it means) set-theoretic maps $F:S \rightarrow S'$?
(I don't even dare asking about the existence of "non-topologizable homotopical spaces" because the above question is already by far too vague!)
| https://mathoverflow.net/users/4721 | Are there any "homotopical spaces"? | [No](http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html).
| 59 | https://mathoverflow.net/users/290 | 21668 | 14,332 |
https://mathoverflow.net/questions/21639 | 3 | Good example teaches sometimes more than couple of theorems. I wonder what are your favourite examples in complex algebraic geometry, the ones that were astonishing for you, the simpler (at least available for people without PhD in AG) the better.
| https://mathoverflow.net/users/5419 | Amazing examples in complex Algebraic Geometry | I will give an example which I was looking for a long time by myself. The question concerns CONTRACTIBILITY OF CURVES ON SMOOTH SURFACES:
Having a smooth (hence projective) surface $X$ and a divisor $E$ when can $E$ be contracted into an algebraic singularity (i.e. such that the quotient $X/E$ is an algebraic surface)?
Background: It is known that a divisor on a smooth projective surface can be contracted in the analytic category if and only if its intersection matrix is negative definite (Grauert). However, the question in the algebraic category is much more subtle. There are criteria (numerical) by Artin for contractibility of bunch of rational curves to a rational singularity but no general ones. The following two examples show that it is not possible to give a general numerical criterion for contractibility (people refer to Zariski, *The theorem of Riemann-Roch for high multiples of an effective divisor on an algebraic surface*. Ann. of Math. (2) 76, 1962, 560--615).
Background for the example: Let $C$ be an elliptic curve and let $P\_0$ be a point in $C$. The linear system $|3P\_0|$ gives an embedding of $C$ into $\mathbb{P}^2$, in particular for some hyperplane section $H$ we have $H\cap C=3P\_0$. The bijection $P\to \mathcal{O}\_C(P-P\_0)$ gives $C$ the group structure from $Pic^0(C)$.
Example 1: Take 12 points $P\_1,\ldots,P\_{12}\in C$. Let $p:X\to \mathbb{P}^2$ be the blowup in these points, denote the proper transform of $C$ by $C'$. Since $C'^2=9-12=-3$, it can be contracted analytically (to a singular point), let $\pi:X\to Y$ be the contraction. We show that for a general choice of $P\_i$'s each divisor on $Y$ meets the singular point, so $Y$ cannot be algebraic.
Suppose $D$ is a divisor on $Y$ which does not meet the singular point. The divisor $B=p\_\*\pi^\*D$ meets the cubic only in the chosen points, so scheme-theoretically for some integers $k\_i$ we have $\sum k\_i P\_i=B\cap C\sim deg B\cdot H\cap C=3deg B\cdot P\_0$. we see that $\sum k\_iP\_i$ is the zero of the group, which does not happen for a general choice of $P\_i$'s. (Note that since $C$ is uncountable, one can easily find $12$ $\mathbb{Z}$-independent points on it).
Example 2: In the example above take for $P\_i$'s the points of intersection of some quartic $Q$ with our cubic. The linear system $\mathfrak{s}$ spanned by $Q$ and quartics of type $C+line$ is transformed birationally by $p$ to a free linear system $p^{-1}\mathfrak{s}$. We see easily that for an irreducible $U\subseteq X$ we have $U\cdot (C'+p^{-1}line)=0$ if and only if $U=C'$, so the new system realizes the contraction $X\to Y$ algebraically.
| 4 | https://mathoverflow.net/users/5419 | 21678 | 14,337 |
https://mathoverflow.net/questions/21627 | 3 | This question arose from the responses to [this question.](https://mathoverflow.net/questions/21232/when-is-a-blow-up-non-singular) The references to the comments of Karl Schwede and VA are to comments made there.
The blow-up of the variety $X=\mathbb{A}^2$ along the closed subscheme $Z$ defined by $(x,y)^2$ is non-singular. As Karl Schwede points out, this example is trivial in the sense that the blow-up along the power of a maximal ideal is naturally isomorphic to the blow-up of the maximal ideal. VA's comment, on the other hand, suggests that perhaps singular closed schemes $Z$ with $\operatorname{Bl}\_{Z}(X)$ non-singular are ubiquitous.
This suggests a question: what are non-trivial examples of a singular closed subscheme $Z$ of a non-singular variety $X$ with $\operatorname{Bl}\_{Z}(X)$ non-singular. Here "non-trivial" means the ideal of $Z$ is not a power of the ideal of a non-singular subvariety.
Particularly interesting would be such a $Z$ such that
$\operatorname{Bl}\_{Z}(X)$
is not isomorphic (as a scheme over $X$) to $\operatorname{Bl}\_{Z'}(X)$ for any non-singular subvariety $Z'$ of $X$.
**Edit:** I have not been able to access the paper "On the smoothness of blow-ups" (MR1446135, by O'Carroll and Valla) yet, but the mathsci review states that they prove that the blow-up of a regular local ring $A$ along an ideal generated by a subset of a regular system of parameters is smooth. Let's also consider those examples to be trivial.
**Edit:** I added "of a non-singular variety" to the title to emphasize that I am interested in examples where the ambient space is non-singular.
| https://mathoverflow.net/users/5337 | What are non-trivial examples of non-singular blow-ups of a non-singular variety? | Suppose that $X = \mathbb{A}^2$. Let $Y$ be the blow up of $X$ at the maximal ideal $(x,y)$ and let $W$ be the blow up of $Y$ at a point on the exceptional divisor of $Y$ over $X$. Of course, the composition $f: W \rightarrow X$ is birational and an isomorphism away from the origin. The fiber of $f$ over the origin is the union of two $\mathbb{P}^1$'s meeting at a single point, but the total space $W$ is non-singular. The map $f$ is identified with the blowup of $X$ along *some* closed subscheme $Z$ of $X$ supported only at the origin. I believe an example of an ideal defining such a $Z$ is $(x^3, xy, y^2)$.
By taking the composition of blowups along smooth centers, there is some ideal sheaf on the base giving the composition in "one step". In theory, you can compute this ideal by tracing through the proof that every birational morphisms is a blow up - but in practice I think this is usually difficult.
| 5 | https://mathoverflow.net/users/397 | 21679 | 14,338 |
https://mathoverflow.net/questions/21682 | 14 | I was told that a polynomial group law on (all of) $\mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group.
Is it true? Where can I find a proof?
A counterexample for open subsets of $\mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law.
| https://mathoverflow.net/users/1049 | Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent? | This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"
| 15 | https://mathoverflow.net/users/4008 | 21689 | 14,342 |
https://mathoverflow.net/questions/21683 | 1 | This question follows the article discussed [here](https://mathoverflow.net/questions/16764/equality-of-the-sum-of-powers)
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Problem
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Suppose we're trying to bound the number of integral solutions to a system of multi-variable polynomials,
say
$$ \sum\_{i=1}^n x\_i^t = \sum\_{i=1}^n y\_i^t, $$
where each $x\_i,y\_i \in \mathbb N$ and for each $t < c$ for some constant $c$.
If we do not put any constrains on the solution,
there are infinitely many possible solutions even when $n=C=1$.
So if we put some constrains on {$x\_i,y\_i$} like $x\_i,y\_i \in$ {$0,1,\ldots,n$},
then how many possible solutions can we get?
Naively there are $O(n^n)$ choices, but it seems highly unlikely that there are many solutions to the system of equations. Is there any exist bound on the number of solutions,
say $O(n^k)$ for fixed $k$ or even better bounds? Are there some well-known approaches to bound the number of solutions of an equation?
Motivation
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This question arose when I'm trying to come up with some reasonable constrains with the equation in [Prouhet-Tarry-Escott Problem](http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_problem).
It seems like if we restrict the maximum value of variables, there aren't many solutions to the equation. I tried to add more constrains to get rid of the already few solutions,
but it seems that there is no direct way making the solution set empty, that is, no possible solutions under such constrains.
So I turn to find some existing bounds for the equation, but sadly nothing occurred.
Can it be still hard to find such results, or there are some theorems like the Fundamental Theorem of Algebra, concerning the number of solutions to a multi-variable equation? Any information is useful. Thank you all!
Edited
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According to Felipe Voloch (Thanks!), the general approach to the question is the Hardy-Littlewood method, which considers the number of solutions to an equal-power Diophantine equation. But it seems that the method gives a lower bound on the number of solutions (is this correct?), rather than an upper bound. Or there are some ways to give upper bounds by the same method?
One more question: How about further restricting the solutions to be prime numbers?
Does this make any difference?
| https://mathoverflow.net/users/4248 | Number of integral solutions to multi-variable polynomials | People studying Waring's problem via the Hardy-Littlewood method often consider this kind of problem. You could start by looking at Vaughan's book, "The Hardy-Littlewood method".
| 3 | https://mathoverflow.net/users/2290 | 21694 | 14,346 |
https://mathoverflow.net/questions/20418 | 8 | In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo & Isola,1996], etc.) define the index of a conditional expectation of a von Neumann algebra M onto a vN-subalgebra N (here, a conditional expectation is a normal, faithful N-N bimodule map fixing the subalgebra pointwise). An inclusion is said to have finite index if there exists a conditional expectation that has finite index. However, in the case where M is finite we might be interested in restricting ourselves to the conditional expectations that preserve some trace on M.
This leads us to the question: For a given (normal, faithful, finite) trace on M, Umegaki gives us a unique trace preserving conditional expectation E:M->N. Are there any nice necessary and sufficient conditions for a conditional expectation to arise in this manner? What if we allow the trace to be semifinite?
Since subfactors give rise to more than one conditional expectation, it is certainly not the case that all conditional expectations come from traces. A necessary condition is that E(xy)=E(yx) whenever x or y is an element of the relative commutant $N^\prime \cap M$. This is not sufficient, however.
| https://mathoverflow.net/users/2085 | When does a conditional expectation preserve some trace? | I would expect a general answer to be difficult, because the set of traces on your von Neumann algebra will depend a lot on the centre of the algebra.
In the case of a factor, the question becomes how to tell if a given expectation is the one that commutes with the trace. Not checking all my facts very carefully, I think that in this case the necessary condition is also sufficient: that is, if $E(xy)=E(yx)$ whenever x is in $N^\prime \cap M$, then $E$ commutes with the trace. This follows from the fact that this condition is equivalent to the modular group of the expectation (see Combes-Delaroche, 1975) being trivial; and in the case of a factor, the modular group characterizes the expectation (Remark 4.12.b in Combes-Delaroche).
| 3 | https://mathoverflow.net/users/3698 | 21712 | 14,358 |
https://mathoverflow.net/questions/21707 | 4 | Consider an Ab-category $\mathcal{D}$ and its pseudo-abelian envelope $\mathcal{C}$. If necessary, one can assume $\mathcal{D}$ to be $k$-linear over a field $k$ and have finite-dimensional hom-spaces. I am interested in conditions on $\mathcal{D}$ that will ensure that $\mathcal{C}$ is abelian. Are there any such nontrivial conditions? (For motivation, note for instance that the categories in, say, [this paper](http://arxiv.org/abs/math/0610552) are defined via such a pseudo-abelianization procedure.)
**Edit:** The answer below discusses the semisimple abelian case. What about the non-semisimple case?
| https://mathoverflow.net/users/344 | When is the pseudo-abelian envelope abelian? | A sufficient (but not necessary) condition is that the endomorphism ring of every object be semi-simple. See [Jannsen](http://www.ams.org/mathscinet-getitem?mr=1150598).
| 6 | https://mathoverflow.net/users/5434 | 21714 | 14,360 |
https://mathoverflow.net/questions/21709 | 21 | For example, I find the first group isomorphism theorem to be vastly more opaque when presented in terms of commutative diagrams and I've had similar experiences with other elementary results being expressed in terms of exact sequences. What are the benefits that I am not seeing?
| https://mathoverflow.net/users/4692 | What are the advantages of phrasing results in terms of exact sequences and commutative diagrams? | Holy cow, go *beyond* the first homomorphism theorem! For example, if you have a long exact sequence of vector spaces and linear maps
$$
0 \rightarrow V\_1 \rightarrow V\_2 \rightarrow \cdots \rightarrow V\_n \rightarrow 0
$$
then exactness implies that the alternating sum of the dimensions is 0.
This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$.
Replace vector spaces and linear maps by finite abelian groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1.
The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are the only way to think about or formulate large chunks of modern mathematics. For instance, you need commutative diagrams to make sense of universal mapping properties (which is the way many concepts are defined or at least most clearly understood) and to understand the opening scene in the movie "It's My Turn".
Here is a nice exercise. When $a$ and $b$ are relatively prime,
$\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes:
$$
\varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}.
$$
You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.
| 62 | https://mathoverflow.net/users/3272 | 21715 | 14,361 |
https://mathoverflow.net/questions/21664 | 5 | I am trying to find out what results are already out there in this direction.
For example, from the Ito-Michler theorem, if $\rho(G)$ is the set of prime divisors of irreducible characters of $G$ and either $\rho(G)\subseteq \pi$ or $|\rho(G)\cap\pi|\leq 1$ then we are guaranteed the existence of a $\pi$-Hall subgroup.
Also, there is a result by J. Thompson that if some prime $p$ divides the degrees of all non-linear irreducible characters of a finite groups, then that group has a normal $p$-complement.
Are there other results like these, which give a relations between some set connected to the degrees of irreducible characters of a group and a set $\pi$ of prime numbers, such that we are guaranteed the existence of a $\pi$-Hall subgroup?
| https://mathoverflow.net/users/4614 | Existence of certain Hall-subgroups based on knowledge of the degrees of irreducible characters | Background:
>
> (Ito–Michler) p does not divide the order of the degree of any (absolutely) irreducible (ordinary) character of G iff G has a normal, abelian, Sylow p-subgroup.
>
>
>
For any subset π of ρ(G)′, it follows that G has a normal, abelian, Hall π-subgroup, and hence also a Hall π′-subgroup. Flipping the roles of π and π′ gives your first existence claim.
Chapter 14 of Isaacs's Character Theory textbook contains this and further results, which I assume are already known. For instance, the classical:
>
> (Blichfeldt) if G has a faithful (possibly reducible, ordinary) character of degree n, then for any π consisting of primes p > n, G has an abelian, Hall π-subgroup.
>
>
>
Blichfeldt's paper below contains the result that G has an abelian, normal, Hall π-subgroup for π a collection of primes p > (n-1)(2n+1). This was refined to:
>
> (Feit–Thompson) if G has a faithful (possibly reducible, ordinary) character of degree n, then for any π consisting of primes p > 2n+1, G has an abelian, normal, Hall π-subgroup.
>
>
>
The more detailed information alluded to in Isaacs's textbook is likely Winter's 1964 paper below. It gives a complete list of groups with non-normal Sylow p-subgroups for p > (2n+1)/3. A non-abelian but normal Sylow π-subgroup is implied by the following:
>
> (Feit) if G has a a faithful, irreducible (ordinary) character of degree p-2 ≥3, then either the Sylow p-subgroup of G is normal, G is a direct product of SL(2,p) and an abelian group, or G is a direct product of 3.Alt(6) and abelian group.
>
>
>
Now each of these results required *faithful* characters. If you are OK with that, then Berkovich ([MR1189113](http://www.ams.org/mathscinet-getitem?mr=1189113)) has a Thompson like hypothesis and an Ito like conclusion (normal Hall π-subgroup containing the nilpotent residual), but it looks at π(G/Z(χ)) too.
The Ito–Michler result is also valid in some sense over finite fields:
>
> (Manz) Every irreducible, p-modular character degree is coprime to p iff G has a normal Sylow p-subgroup.
>
>
> (Manz–Wolf) If G is p-solvable and every irreducible, p-modular character degree is coprime to q, then Oq′(G), the subgroup generated by the Sylow q-subgroups, is solvable of q-length at most 2 (π-length roughly measures how far away from having a normal Hall π-subgroup you are).
>
>
>
(Navarro, et al.) has generalized both the Ito and Thompson results to blocks in [MR1810119](http://www.ams.org/mathscinet-getitem?mr=1810119), [MR1956546](http://www.ams.org/mathscinet-getitem?mr=1956546), and [MR2159761](http://www.ams.org/mathscinet-getitem?mr=2159761).
A more recent article by (Matterei) gives a reasonable survey of results. The way he rephrases results may be quite interesting to you. He also gives the Carter–Hawkes ([MR1199667](http://www.ams.org/mathscinet-getitem?mr=1199667)) results that continue the ideas of Thompson. A recent paper of Dolfi et. al ([MR2469367](http://www.ams.org/mathscinet-getitem?mr=2469367)) derives Ito-like results using the orders of a non-vanishing elements g in G, such that χ(g) ≠ 0 for any any ordinary, irreducible character χ and derives Ito's and Thompson's results as corollaries.
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Blichfeldt, H. F. "On the order of linear homogeneous groups."
Trans. AMS. 4 (1903), 387–397. [JFM 34.0176.02](http://www.zentralblatt-math.org/zmath/en/search/?q=an:34.0176.02) [JSTOR 1986408](http://www.jstor.org/stable/1986408)
Feit, Walter; Thompson, John G. "Groups which have a faithful representation of degree less than (p-1)/2". Pacific J. Math. 11 (1961), 1257–1262. [MR133373](http://www.ams.org/mathscinet-getitem?mr=133373) [euclid.pjm/1103036911](http://projecteuclid.org/getRecord?id=euclid.pjm/1103036911)
Winter, David L. "Finite groups having a faithful representation of degree less than (2p+1)/3." Amer. J. Math. 86 (1964), 608–618. [MR183788](http://www.ams.org/mathscinet-getitem?mr=183788) [DOI: 10.2307/2373026](http://dx.doi.org/10.2307/2373026)
Feit, Walter. "On finite linear groups. II." J. Algebra 30 (1974), 496–506. [MR357634](http://www.ams.org/mathscinet-getitem?mr=357634) [DOI: 10.1016/0021-8693(74)90220-8](http://dx.doi.org/10.1016/0021-8693(74)90220-8)
Manz, Olaf. "On the modular version of Ito's theorem on character degrees for groups of odd order." Nagoya Math. J. 105 (1987), 121–128. [MR881011](http://www.ams.org/mathscinet-getitem?mr=881011) [euclid.nmj/1118780642](http://projecteuclid.org/getRecord?id=euclid.nmj/1118780642)
Manz, Olaf; Wolf, Thomas R. "Brauer characters of q′-degree in p-solvable groups."
J. Algebra 115 (1988), no. 1, 75–91. [MR937602](http://www.ams.org/mathscinet-getitem?mr=937602) [DOI: 10.1016/0021-8693(88)90283-9](http://dx.doi.org/10.1016/0021-8693(88)90283-9)
Mattarei, Sandro. "Retrieving information about a group from its character degrees or from its class sizes." Proc. Amer. Math. Soc. 134 (2006), no. 8, 2189–2195. [MR2213690](http://www.ams.org/mathscinet-getitem?mr=2213690) [DOI: 10.1090/S0002-9939-06-08274-8](http://dx.doi.org/10.1090/S0002-9939-06-08274-8)
| 7 | https://mathoverflow.net/users/3710 | 21716 | 14,362 |
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