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https://mathoverflow.net/questions/42119 | 12 | Hi,
I want to write a proof that relies on the fact that:
There are Borel Sets $A$ and $B$ contained in $\mathbb{R}$ such that
$A \cap B = \emptyset$ and $\lambda(A \cap (x,y)) = \lambda(B \cap (x,y)) > 0$.
Note that $x < y \in \mathbb{R}$ are arbitrary.
I'm fairly sure this is true, but am having trouble coming up with a construction of such sets and it's driving me up the wall. Can anyone help?
| https://mathoverflow.net/users/10053 | Sets with equal positive measure in every interval | The basic step is to construct a nowhere dense set of positive, and controlled, measure. Then iteratively in each interval where the set is empty, you replace by another such set. See for example a paper by Erdös and Oxtoby, *Partitions of the plane into sets having positive measure in every non-null measurable set*.
**EDIT:** This gives $\lambda (A\cap (x,y))>0$ and $\lambda(B\cap (x,y))>0$ for each interval $(x,y)$. We cannot have equality $\lambda (A\cap (x,y))=\lambda(B\cap (x,y))$ in general (see Mike Hall's answer).
| 8 | https://mathoverflow.net/users/4600 | 42121 | 26,829 |
https://mathoverflow.net/questions/42097 | 12 | Let G be a reductive algebraic group over a field k. Let S be a maximal split torus, Z its centraliser and N its normaliser. The Weyl group W is then defined to be the quotient N(k)/Z(k). Now we cannot hope for W to be realisable as a subgroup of G, but I would like to know how close we can get.
There is a classical result of Tits in the case where G is split which says that for each simple reflection s in the Weyl group, we can find a lift ws in G with the property that these lifts satisfy the braid relations. They do not however square to the identity, instead square to an order 2 element of S, and we get an extension of W by an elementary abelian 2-group embedding in G.
So my question ends up becoming, what generalisation of the above theorem of Tits exists when G is no longer assumed to be split? Ideally I'd get an answer for general reductive G, and if there happens to be a simpler formulation in the quasi-split case, I'd be interested in hearing that too.
| https://mathoverflow.net/users/425 | Best approximation to the Weyl group as a subgroup of a reductive group. | Via Theorem 7.2 in Borel-Tits, Groupes reductifs (1965), Tits's lifting result for the Weyl group also applies in the non-split case (for connected groups).
This theorem states that there exists a split subgroup $F$ of $G$ such that $F$ contains the maximal split torus $S$ of $G$ and intersects each relative root group of $G$ non-trivially. In particular, the Weyl groups of $F$ and $G$ are isomorphic.
| 8 | https://mathoverflow.net/users/3380 | 42123 | 26,831 |
https://mathoverflow.net/questions/42106 | 6 | What is known about spaces of embeddings of contractible manifolds into Euclidean space? I am also curious about the case of small codimension (or even codimension 0). The same question about the configuration spaces in such manifolds.
| https://mathoverflow.net/users/9800 | Embedding theory for contractible manifolds | (This is by far not a complete answer, just an example.) In dimension 4, a [paper of Livingstone](http://nyjm.albany.edu:8000/PacJ/p/2003/209-2-8.pdf) (build on previous work of Lickorish) constructs some (compact with boundary) contractible 4-manifold which embeds in $\mathbb R^4$ in infinitely many (countable) distinct ways. These are distinguished by the fundamental group of the complement.
| 6 | https://mathoverflow.net/users/6205 | 42125 | 26,832 |
https://mathoverflow.net/questions/41804 | 2 | Take an algebraic variety $V$, and its set of smooth functions $C^{\infty}(V)$. One can endow $C^{\infty}(V)$ with a canonical locally convex topology (the seminorms are defined using the local coordinate patches of the variety). With respect to this topology the space is a Frechet space (this means that, amongst other properties, that it is metrisable and complete with respect to the metric).
I have two questions:
(1) If $O(V)$ is the set of regular functions of $V$, is $O(V)$ dense in $C^{\infty}(V)$ with respect to the Frechet topology?
(2) Can one caracterize these locally convex topologies on $O(V)$ that are induced by a differential structure?
| https://mathoverflow.net/users/1095 | Varieties, Frechet Completions, and Regular Functions | First, $V$ needs to be the type of variety for which the ring of regular functions $O(V)$ separates points. If it doesn't, then that establishes an equivalence relation on $V$ and you might as well pass to the quotient in which points are separated. Thus $V$ is an affine algebraic variety. Moreover, if $V$ is complex, then you have to take its realification to have any hope of approximating smooth functions; the complex regular functions are all holomorphic and would at best approximate the space of holomorphic functions. So, we can suppose that $V$ is an affine real algebraic variety, or in scheme-speak, the real points of an affine real algebraic variety.
If $V$ is compact, then the Stone-Weierstrass theorem directly tells you that $O(V)$ is dense in $C^\infty(V)$. Okay, you have to extend the Stone-Weierstrass theorem to derivatives, but in finite dimensions you can do that. And you have to consider derivatives at singularities of $V$ if you want to allow singularities. But I don't think that there is any real problem with that either.
If $V$ is not compact, then I have a little trouble interpreting the question, but I think that it comes to the same thing. Every algebraic variety is locally compact in the analytic topology, and when you say "local coordinate patches", the reasonable interpretation is coordinate patches whose closures are compact. If $X$ is any suitably tame locally compact space and you make a Fréchet topology on $C^\infty(X)$ by exhausting $X$ by a sequence of compact subsets, then you can extend the Stone-Weierstrass theorem by diagonalization.
I don't know what is meant by characterizing topologies on $O(V)$ that come from differentiable structure. I can't think of another equally natural Fréchet topology on all of $C^\infty(V)$, other than to take the sup of each derivative on each compact set. But, if you are willing to work with only part of $C^\infty(V)$, and if you are willing to give $V$ a Riemannian metric, then you can use $L^p$ norms instead of sup norms. Or you can make a Banach space instead of a Fréchet space by making some algebraic combination of the seminorms instead of listing them separately. I think that the resulting space is typically smaller than all of $C^\infty(V)$, but it can still be big enough to include all of $O(V)$. It is then a different locally convex topology on $O(V)$ because it has a different completion.
(This answer doesn't feel all that creative, so maybe the intended question is different?)
| 2 | https://mathoverflow.net/users/1450 | 42129 | 26,834 |
https://mathoverflow.net/questions/42127 | 12 | I learned Bezout's Theorem in class, stated for plane curves (if irreducible, sum of intersection multiplicities equals product of degrees). What is the proper general statement, for projective varieties of degree n?
I think it is something like: If finite, the sum of multiplicities equals the product of degrees.. else the (dimension? degree? sums over irreducible components?) of the intersection is less than or equal to the difference in degrees.
Help is appreciated!
| https://mathoverflow.net/users/10054 | generalization of Bezout's Theorem? | Dear unknown, the most straightforward generalization of Bézout's theorem might be the following. Consider $\mathbb P^n $, projective space over the field $k$, and
$n$ hypersurfaces $H\_1,...,H\_n$in general position in the sense that their intersection is a finite set. Then, calling $h\_i$ their local equations, Bézout says
$$\sum dim\_k \mathcal O\_{\mathbb P^n,P\_i}/(h\_1,...,h\_n) =\prod deg (H\_i) $$
The dimension on the left hand side is, of course, to be interpreted as the multiplicity with which to count the point $P\_i$, seen as a fat point i.e. a zero-dimensional non-reduced scheme.
A related, more abstract point of view is the description of the Chow ring of $\mathbb P^n$ as $CH^\ast (\mathbb P^n)=\mathbb Z[x]/(x^{n+1})$ ( where $x$ is the class of a hyperplane in $\mathbb P^n$). From this point of view we have the following version of Bézout. Consider $r$ cycles $\alpha\_1,...,\alpha\_r$ on $\mathbb P^n$ with $\alpha\_i \in CH^{d\_i}(\mathbb P^n) $ and $d\_1+...d\_r \leq n$, . Then
$$deg \prod {\alpha\_i} =\prod deg (\alpha\_i)$$
the product of the $\alpha\_i$'s on the left being calculated in the Chow ring and the degree $deg (\alpha) $ of a cycle $\alpha \in CH^d (\mathbb P^n)$ being the integer $t$ such that $\alpha =t . x \in CH^d(\mathbb P^n)=\mathbb Z .x $.
This is only the tip of the iceberg: a definitive answer would require a book. Fortunately that book exists and has been written, to our eternal gratitude, by Fulton: *Intersection theory, volume 2 of Ergebnisse der Mathematik und ihrer
Grenzgebiete. 3. Folge.
Springer-Verlag, Berlin, second edition, 1998.*
| 26 | https://mathoverflow.net/users/450 | 42131 | 26,836 |
https://mathoverflow.net/questions/42088 | 1 | If you do a linear regression: $||Ax - e ||^2$, where e is iid Gaussian, mean 0 and variance 1, then your answer is $x\_{hat} = (A' A)^{-1} (A' \* e)$ and the covariance of $x\_{hat}$ is $(A' A)^{-1}$
Now, what if I add the linear inequality constraints $Bx > c$? There are algorithms that find the answer for a given $e$, but what is the covariance matrix?
It seems like a non-trivial problem:
<http://www.gurulib.com/_user_manual_file/pic_1247578519497.pdf>
However, the author seems to give up:
"A much more interesting problem is to analyze a properly truncated variance-covariance matrix of $b\*$. However, it is beyond the scope of this paper."
Of course, I can do a Monte-Carlo simulation, but a closed-form solution would be better. Any hint or reference?
Thanks in advance,
Tony
| https://mathoverflow.net/users/10043 | Inequality-constrained linear-regression, what is the covariance of the estimator? |
>
> Well, the vector $x$ is random right?
>
>
>
It's a parameter, so therefore fixed (yet unknown): the estimator $\hat x$ is a random variable. I would agree with Brian that a covariance matrix will not be all that useful the constraints will mean that the estimator will tend to concentrate around the edges, where a lot of the asymptotic machinery breaks down.
Personally, I reckon a Bayesian approach would be better, as the inequality constraints can be easily built into the prior.
| 1 | https://mathoverflow.net/users/8019 | 42135 | 26,839 |
https://mathoverflow.net/questions/36797 | 31 | The following little trick was introduced by E. Trost
(*Eine Bemerkung zur Diophantischen Analysis*,
Elem. Math. 26 (1971), 60-61). For showing that a diophantine equation
such as $x^4 - 2y^2 = 1$ has only the trivial solution, assume that
$a^4 - 2b^2 = 1$; then the quadratic equation $a^4 t^2 - 2b^2 t - 1 = 0$
has the rational solution $t = 1$, hence its discriminant $4b^4 + 4a^4$
must be a square. Thus $a = 0$ or $b = 0$ by Fermat's Last Theorem for
exponent $4$.
Trost gave a few other nice applications of this trick, but I have not seen
it in any textbook on number theory. My questions:
1. Was Trost's trick noticed by anyone (before or after Trost)?
2. Are there any other cute applications?
| https://mathoverflow.net/users/3503 | Trost's Discriminant Trick | T. Nagell in [*Norsk Mat. Forenings Skrifter.* **1**:4 (1921)]
shows that, for an odd prime $q$, the equation
$$
x^2-y^q=1 \qquad (\*)
$$
has a solution in integers $x>1$, $y>1$, then $y$ is even and $q\mid x$.
In his proof of the latter divisibility he uses a similar trick as follows.
Assuming $q\nmid x$ write ($\*$) as
$$
x^2=(y+1)\cdot\frac{y^q+1}{y+1}
$$
where the factors on the right are coprime (they could only have
common multiple $q$). Therefore,
$$
y+1=u^2, \quad \frac{y^q+1}{y+1}=v^2, \quad x=uv,
\qquad (u,v)=1, \quad \text{$u,v$ are odd}.
$$
Using these findings we can state the original equation in the form
$x^2-(u^2-1)^q=1$, or
$$
X^2-dZ^2=1 \quad\text{where $d=u^2-1$}.
$$
The latter equation has integral solution
$$
X=uv, \quad Z=(u^2-1)^{(q-1)/2},
$$
while its general solution (a classical result for this particular Pell's equation)
is taken the form $(u+\sqrt{u^2-1})^n$. It remains to use the binomial theorem in
$$
X+Z\sqrt{u^2-1}
=(u+\sqrt{u^2-1})^n
$$
(for certain $n\ge1$) and simple estimates to conclude that this is not possible.
To stress the use of similar trick: instead of showing insolvability of $x^2-(u^2-1)^q=1$,
we assume that a solution exists and then use $d=u^2-1$ to produce a solution $X,Z$ of $X^2-dZ^2=1$;
finally, the pair $X,Z$ cannot solve the resulting Pell's equation.
(Of course, it is hard to claim that this is exactly Trost's trick,
as here is a dummy variable but no discriminants, except the one for
Pell's equation. Trost's trick is less trickier to my taste. $\ddot\smile$)
Note that Nagell's result was crucial for showing that ($\*$) does not have
integral solutions $x>1$, $y>1$ for a fixed prime $q>3$. This was shown in
a very elegant way, using the Euclidean algorithm and quadratic residues
by Ko Chao [*Sci. Sinica* **14** (1965) 457--460], and later reproduced
in Mordell's *Diophantine equations*. The ideas of this proof are in the
heart of Mihailescu's ultimate solution of Catalan's conjecture. A much
simpler proof of Ko Chao's result, based on a completely different (nice!) trick,
was given later by E.Z. Chein [*Proc. Amer. Math. Soc.* **56** (1976) 83--84].
| 16 | https://mathoverflow.net/users/4953 | 42142 | 26,841 |
https://mathoverflow.net/questions/42141 | 7 | I asked this question on [stats.stackexchange.com](https://stats.stackexchange.com/questions/1441/example-of-a-stochastic-process-that-is-1st-and-2nd-order-stationary-but-not-str) a little while back but didn't get an answer. It was suggested that I post it here at the time. There appears to be some migratory problems going on over there. Hopefully, this question is seen as appropriate.
Let $X(t)$ be a stochastic process. We say that $X(t)$ is *Nth-order stationary* if for every set of ''times'' $t\_1, t\_2, \dots, t\_N$ we have that the joint cumulative density functions
$$F\_{X(t\_1),X(t\_2),\dots,X(t\_N)} = F\_{X(t\_1 + \tau),X(t\_2 + \tau),\dots,X(t\_N + \tau)}$$
for all ''time shifts'' $\tau$. This is quite a strong condition, it says that the joint statistics don't change at all as time shifts.
For example, a 1st order stationary process is such that $F\_{X(t\_1)} = F\_{X(t\_2)}$ for all $t\_1$ and $t\_2$. That is, the $X(t)$ are all identically distributed. It is quite easy to see that a 1st order stationary process need not be 2nd order stationary. Simply assign a correlation structure to say $X(t)$, $X(t+1)$, $X(t+2)$ that *does not* correspond to a (symmetric) Toeplitz matrix. That is, in vector form, the covariance matrix of $[ X(t), X(t+1), X(t+3)]$ could be given as
$$\left[\begin{array}{cc}
\sigma^2 & a & b \newline
a & \sigma^2 & c \newline
b & c& \sigma^2
\end{array}\right]$$
for $a,b,c$ distinct. This is now not 2nd order stationary because $E[X(t)X(t+1)] = a$ and, time shifting by 1 we have $E[X(t+1)X(t+2)] = c \neq a$.
In a similar way (presumably) a process that is 1st and 2nd order stationary need not be 3rd order stationary and this leads to my question:
>
> Does somebody have a nice example of a stochastic process that is both 1st and 2nd order stationary, but not 3rd order stationary?
>
>
>
**Motivation:** Some of the material I have been expected to teach recently has included stochastic processes and I feel this is a gap in my knowledge. Although no student has actually asked this question yet, I think it is natural enough for it to be asked at some point and I would like to have a neat and simple example.
| https://mathoverflow.net/users/5378 | A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary | How about this:
Consider $X\_1,X\_2$ i.i.d. with $\mathbb{P}(X\_i = 1) = \mathbb{P}(X\_i = 0) = 0.5$. Now define $X\_3 = X\_1 + X\_2 (\text{mod }2)$. Notice that $X\_3$ is independent from $X\_1$ and from $X\_2$ individually. However the three variables $X\_1,X\_2,X\_3$ are not jointly independent.
Now consider a sequence $X\_1,X\_2,X\_3,\ldots$ such that each triplet $(X\_{3n+1},X\_{3n+2},X\_{3n+3})$ has the same distribution as $(X\_1,X\_2,X\_3)$ and all these triplets are independent.
With this one has that all $X\_n$ have the same distribution as $X\_1$, so that the process is 1 stationary. Also, all pairs $X\_i,X\_j$ with $i\neq j$ are independent and hence have the same distribution (this gives 2 stationarity). However the triplet $(X\_2,X\_3,X\_4)$ has a different distribution from $(X\_1,X\_2,X\_3)$ (in fact, $X\_2,X\_3$ and $X\_4$ are jointly independent).
| 10 | https://mathoverflow.net/users/7631 | 42150 | 26,844 |
https://mathoverflow.net/questions/42104 | 3 | I am trying to calculate $E(\int\_0^T {W\_s ds})$, where $W\_s$ is a standard Brownian motion.
Now two approaches I can think of:
1) Take a partition of $[0,T]$. Calculate $E(\sum {W\_{t\_i}(t\_{i+1} - t\_i)})$ and take the limit as you shrink the size of the partition.
2) Calculate $\int\_0^T { E(W\_s)ds }$.
However, for approach 1), its not clear what function would dominate the absolute value of the terms inside the E() for all possible partitions, and that it would have a finite expectation. So, interchanging limit and expectation is dicey.
For approach 2), Fubini's theorem would require me to know a-priori that $E(\int\_0^T {|W\_s|ds})$ is finite, and I don't see how I could show that.
How can any of these approaches be fixed, if at all? Or is there another way to solve the problem?
| https://mathoverflow.net/users/10049 | Expectation of time integral of Wiener process | For approach 2, Fubini's theorem works just as well if you show
$$ \int\_0^T E|W\_s|ds < \infty $$
which is easy. Indeed, perhaps even easier is to note
$$ \int\_0^T E(|W\_s|^2)ds = \int\_0^T s ds = \frac{1}{2}T^2 < \infty$$
and use Jensen/Hölder/Cauchy-Schwarz.
| 5 | https://mathoverflow.net/users/4832 | 42156 | 26,847 |
https://mathoverflow.net/questions/42147 | 2 | I need a name for a regular category where the inverse image maps have right adjoints.
If $\mathcal C$ is a regular category, then the poset of subobjects $\mathsf{Sub}(X)$ of any object $X$ is a semilattice and the inverse image map of any arrow $f:X\to Y$ has a left adjoint $\exists\_f:\mathsf{Sub}(X) \to \mathsf{Sub}(Y)$. If $\mathcal C$ is a Heyting category, then the inverse image map $f$ also has a right adjoint $\forall\_f:\mathsf{Sub}(X) \to \mathsf{Sub}(Y)$. But Heyting categories also have all finite coproducts and I want a name for regular categories that just have those right adjoints.
Do you know if this category of categories already has a name? Can you suggest a name?
Update: Heyting categories or logoses need not have all finite coproducts, but posets of subobjects are lattices, where I only need semilattices.
| https://mathoverflow.net/users/3603 | name my cat: regular categories where inverse images also have right adjoint | From Freyd and Scedrov's book *Categories, Allegories*: a **logos** is a regular category in which $Sub(A)$ is a lattice for each object $A$, and in which the inverse-image operation $f^\*: Sub(B) \to Sub(A)$ has a right adjoint for each morphism $f: A \to B$ (page 117).
| 4 | https://mathoverflow.net/users/2926 | 42164 | 26,851 |
https://mathoverflow.net/questions/42153 | 13 | Yesterday my first work in mathematics was sent to a publisher, and of course I'm interested in its usefulness. But I know, that sometimes it is hard to get a paper, it is not available for free. I hope my paper will not be of that kind. I hope it will be simple to find it and to read it, I hope a few hundred people will read it. Therefore I have this question. When do mathematicians usually lose their right to share their papers on the internet, why some of them doesn't like to do this and which ways are there to overcome these "sharing difficulties"?
| https://mathoverflow.net/users/8134 | Which rights do mathematicians usually have on their published works and how do they use them? | Read any contract you sign carefully, otherwise you may lose rights you wanted to keep.
In 2000, CRC Press sued Eric Weisstein because he posted free updates to the web of a mathematics book he had written and published with them.
Usually non-commercial publishers like the AMS allow you to keep any rights you want, while commercial publishers often try to get the rights for themselves (but will usually allow you to keep them if you make a fuss).
| 14 | https://mathoverflow.net/users/51 | 42175 | 26,858 |
https://mathoverflow.net/questions/42190 | 0 | I am interested in solving the following equations:
$f(x) + \int\_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = 0$
and
$f(x) + \int\_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = u(x)$
when $u(x)$ is the unknown function defined on $[0^+,\infty)$ and all other functions are known and are assumed to have convenient differentiability and other reasonable properties. In particular, both lower and upper bounds $\alpha(x)$ and $\beta(x)$ are not constants.
I know that when both bounds are constant they belong to Fredholm equations, and when only one bound is constant they belong to Volterra equations, where plenty of literature work exists.
However I am not clear about the case when both bounds are variables. Can the equations defined above be conveniently translated into Fredholm or Volterra theory? If they cannot, what is the theory for these type of equations?
Thanks!
| https://mathoverflow.net/users/9950 | how to solve general integral equations with both variable lower and upper bounds | If $\alpha$ and $\beta$ are bounded functions, with $a\le\alpha\le\beta \le b$, then your equations are of standard type since they can be written as
$$f(x)+\int\_a^b H(x,t) u(t) dt =0 $$
where
$$ H(x,t)=K(x,t)\chi\_{[\alpha(x),\beta(x)]}(t). $$
Here $\chi\_A$ denotes the characteristic function of the set $A$.
If $\alpha,\beta$ are unbounded, then we need some thinking...
| 1 | https://mathoverflow.net/users/7294 | 42193 | 26,867 |
https://mathoverflow.net/questions/42181 | -1 | Hi,
Do you think the following limits are correct?
$\displaystyle\lim\_{d\to\infty}\frac{\sum\limits\_{k=1}^{d} {\phi(N) \choose k} {d-1 \choose k-1}}{\phi(N)^d}=0$
$\displaystyle\lim\_{N\to\infty}\frac{\sum\limits\_{k=1}^{d} {\phi(N) \choose k} {d-1 \choose k-1}}{\phi(N)^d}=c$
I plotted the equations and guessed the results according to the graphs but I could not prove them mathematically by myself. Any hints would be appreciated. Graphs are as follows:
<http://deniz.cs.utsa.edu/plots/d_vs_Eq.jpeg>
<http://deniz.cs.utsa.edu/plots/N_vs_Eq.jpeg>
Thanks,
| https://mathoverflow.net/users/10062 | Limit involving the totient function and combination | We have $$\sum\_{k = 1}^d \binom{\phi(N)}{k} \binom{d - 1}{k - 1} = \binom{d + \phi(N) - 1}{d}$$ (this is the Vandermonde identity). Thus, with $N$ fixed the numerator of your fraction is polynomial in $d$ and so the first result follows (except for $N = 1, 2$).
Edited to add:
Okay, and the second result follows by the same analysis, since $\phi(N) \to \infty$ as $N \to \infty$. In particular, the resulting constant is $\frac{1}{d!}$.
| 13 | https://mathoverflow.net/users/4658 | 42194 | 26,868 |
https://mathoverflow.net/questions/42176 | 18 | Shortly after his work on the foundations of geometry David Hilbert turned his attention to finding a suitable statement of the Dirichlet principle, from which to prove the Riemann mapping theorem and vindicate the topological program for complex analysis. Based on comments made in letters to Frege a major motivation for Hilbert's foray into geometry and independence proofs was to investigate the Archimedean axiom. Specifically, Hilbert mentions (to Frege) Dehn's dissertation on the Archimedean axiom and Legendre's theorem. This leads me to think that conformal mappings were on Hilbert's mind and to guess that Weierstrass's counterexample somehow concerned the Archimedean property. But I can't find anything in the secondary history/philosophy of math literature that quite puts all the pieces of the puzzle together--that Weierstrass had a counter-example is mentioned but details are skirted--and qua philosopher I'm bumping up against my mathematical horizons in piecing it together myself.
| https://mathoverflow.net/users/2833 | What was Weierstrass's counterexample to the Dirichlet Principle? | Weierstrass simply observed that not every problem in the calculus of variations would have a solution. He considered the example
$$D[y]=\int\_{-1}^{1}x^2\left(\frac{d y}{dx}\right)^2dx\to \min,$$
where the functional $D[y]$ is minimized over continuous functions having piecewise continuous first derivatives in $[-1,1]$ and satisfying the boundary conditions
$y(-1)=0$, $y(1)=1$. He proved that although there is a minimizing sequence $y\_n=y\_n(.)$ which makes $D[y\_n]$ arbitrarily small, the minimal value of zero is never actually attained.
Weierstrass's example called into question the *a priori* validity of Dirichlet's principle. However, it did not completely refute the specific applications of Dirichlet's principle to boundary value problems for Laplace's equation developed by Green, Dirichlet, Riemann and others. It simply implied that the particular result required by Riemann would need a formal proof, which Riemann had not provided. For that reason some people refer to this example as Weierstrass's critique rather than Weierstrass's counterexample.
The story is briefly discussed in [*"A History of Analysis"*](http://books.google.co.uk/books?id=CVRZEXFVsZkC&printsec=frontcover&dq=A+history+of+analysis&source=bl&ots=ibDC2FGy1i&sig=frWcc8aQWudQ9weQCKHMmUm6mKo&hl=en&ei=Eke3TLTOHcvFswbrw-iSCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q&f=false) edited by Hans Niels Jahnke.
| 25 | https://mathoverflow.net/users/5371 | 42195 | 26,869 |
https://mathoverflow.net/questions/41958 | 30 | I think most students who first learn about (finite) groups, eventually learn about the possibility of classifying certain finite groups, and even showing certain finite groups of a given order can't be simple (I'm pretty sure every beginning algebra text has some exercises like this). Up to order 1000, I think there is one that is considered by far the most difficult: 720.
Does anyone know of a proof that there are no simple groups of order 720, which **avoids** showing, via contradiction, such a group would be $M\_{10}$? [To clarify the **avoids** part, the proof sketched by Derek Holt [here](http://sci.tech-archive.net/Archive/sci.math/2006-12/msg07456.html), while very nice, would not qualify.]
I should also disqualify the inevitable reference to Burnside's article on this very topic, which I am fairly confident is flawed (or, at the very least, incomplete).
| https://mathoverflow.net/users/1446 | No simple groups of order 720? | You can only really ask for a proof that avoids a particular fact or construction, if that fact or construction is difficult enough or distinct enough from the thing that you're proving. By that principle, the comments imply that it would be enough to show that Derek Holt's subgroup $\langle a,b,c,e \rangle$ has index 2 in the candidate group $\langle a,b,c,d,e \rangle$. In fact Derek's proof is good: It is easy to show that $\langle a,b,c,e \rangle$ is isomorphic to $L\_2(9) = \text{PSL}(2,\mathbb{F}\_9)$, which plainly has order 360. ($L\_2(9)$ is also isomorphic to $A\_6$, but this fact is not needed.) Once you know that, you can also quickly construct $M\_{10}$ as well, since you can show that the extra generator $d$ normalizes $L\_2(9)$, and that $d^2 \in L\_2(9)$.
In that Usenet posting, Derek gave these expressions:
$$\begin{matrix} a &=& (2\; 3\; 4)(4\; 6\; 7)(8\; 9\; 10) \\
b &=& (2\;5\;8)(3\;6\;9)(4\;7\;10) \\
c &=& (3\;5\;4\;8)(6\;7\;10\;9) \\
e &=& (1\;2)(5\;8)(6\;7)(9\;10) \end{matrix}.$$
Recall that $\mathbb{F}\_9 = (\mathbb{Z}/3)[i]$. You can define a bijection
$$\alpha:\mathbb{F}\_9 \cup \{\infty\} \to \{1,2,\ldots,10\}$$
by the formula
$$\alpha(x+iy) = 2+x+3y \qquad \alpha(\infty) = 1,$$
using the gauche embedding $\mathbb{Z}/3 = \{0,1,2\} \subseteq \mathbb{Z}$.
Then it is easy to check these expressions (using more normal arithmetic in $\mathbb{F}\_9$):
$$a(z) = z+1 \qquad b(z) = z+i \qquad c(z) = iz \qquad e(z) = 1/z.$$
So, that's $L\_2(9)$.
| 20 | https://mathoverflow.net/users/1450 | 42199 | 26,873 |
https://mathoverflow.net/questions/42186 | 38 | There are many equivalent ways of defining the notion of compact space, but some require some kind of choice principle to prove their equivalence. For example, a classical result is that for $X$ to be compact, it is necessary and sufficient that every ultrafilter on $X$ converge to a point in $X$. The necessity is easy to prove, but the sufficiency requires a choice principle to the effect that every filter can be extended to an ultrafilter.
Some years ago I heard from a very good categorical topologist that many, perhaps most of the useful properties of compact spaces $X$ readily flow from the fact that for every space $Y$, the projection map $\pi: X \times Y \to Y$ is closed. Of course that is a very classical consequence of compactness which can be left as an exercise to beginners in topology, and I was struck by the topologist's assertion that you could in fact use this as a *definition* of compactness, and that this is a very good definition for doing categorical topology. (I am still not sure what he really meant by this, but that's not my question.)
My own proof that this condition implies compactness goes as follows. Let $Y$ be the space of ultrafilters on the set $X$ with its usual compact Hausdorff topology, and suppose the projection $\pi: X \times Y \to Y$ is a closed map. Let $R \subseteq X \times Y$ be the set of pairs $(x, U)$ where the ultrafilter $U$ converges to the point $x$. One may show that $R$ is a closed subset, so the image $\pi(R)$ is closed in $Y$. But every principal ultrafilter (one generated by a point) converges to the point that generates it, so every principal ultrafilter belongs to $\pi(R)$. Now principal ultrafilters are dense in the space of all ultrafilters, so $\pi(R)$ is both closed and dense, and therefore is all of $Y$. This is the same as saying that every ultrafilter on $X$ converges to some point of $X$, and therefore $X$ is compact.
I was at first happy with this proof, but later began to wonder if it's overkill. Certainly it uses heavily the choice principle mentioned above, and my question is whether the implication I just proved above really requires some form of choice like that.
| https://mathoverflow.net/users/2926 | Does "compact iff projections are closed" require some form of choice? | Martin Escardó wrote a very nice note ["Intersections of compactly many open sets are open"](http://www.cs.bham.ac.uk/~mhe/papers/compactness-submitted.pdf) which you might want to read.
| 25 | https://mathoverflow.net/users/1176 | 42208 | 26,878 |
https://mathoverflow.net/questions/42201 | 13 | Let $M$ be the double-torus with a hyperbolic Riemannian metric. The geodesic flow on the unit tangent bundle $T^1M$ has many invariant Borel probability measures. In particular there are closed geodesics projecting to each non-trivial homotopy class of $M$, and one can support an invariant probability measure on each of these. Also one can take convex linear combinations of these invariant measures.
My question is the following: Are these the only invariant measures of zero metric (Kolmogorov-Sinai) entropy?
More generally, what are the zero entropy invariant probability measures of an Anosov geodesic flow?
Also I'm interested in the same question for shifts on finitely many symbols (i.e. What are the zero entropy invariant measures?).
Besides references giving an answer, other related references are of course very welcome.
| https://mathoverflow.net/users/7631 | What are the zero entropy invariant measures for an Anosov geodesic flow? | There are lots of zero entropy invariant probability measures, many more than just the obvious ones supported on periodic orbits. As you suggest in the question, one can understand the general case by just considering what happens for symbolic systems.
**Explicit example:** Let $\alpha$ be irrational and let $a\_n$ denote the fractional part of $n\alpha$. Consider the sequence in $\Sigma\_2 = \{0,1\}^\mathbb{Z}$ given by $x\_n = 0$ if $0 \leq a\_n < 1/2$, and $x\_n=1$ if $1/2\leq a\_n<1$. Let $X\subset \Sigma\_2$ be the orbit closure of $x=(x\_n)$; then there is an entropy-preserving isomorphism between the space of invariant measures for the shift map $\sigma\colon X\to X$ and for the irrational rotation $R\_\alpha\colon S^1 \to S^1$. The latter preserves Lebesgue measure on the circle and is uniquely ergodic with zero entropy, so $X$ supports exactly one invariant probability measure $\mu$, which comes from Lebesgue and has zero entropy. Now $\mu$ is a shift-invariant probability measure on $\Sigma\_2$ that has zero entropy but is not supported on a periodic orbit.
**General result:** In fact, the above construction is representative of a general phenomenon. As RW points out in his answer, you can get lots of zero entropy measures on shift spaces by taking generating partitions for zero entropy transformations. You can even get more, using the Jewett-Krieger embedding theorem (see Petersen's "Ergodic Theory" or Denker, Grillenberger, and Sigmund's "Ergodic Theory on Compact Spaces"), which lets you find a closed shift-invariant subset of the shift space that has the desired measure as its only shift-invariant probability measure. So there's a lot there.
| 7 | https://mathoverflow.net/users/5701 | 42214 | 26,883 |
https://mathoverflow.net/questions/42215 | 25 | The classic example of a non-measurable set is described by [wikipedia](http://en.wikipedia.org/wiki/Vitali_set#Construction_and_proof). However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} /\mathbb{Q}$.
"Since each element intersects [0,1], we can use the axiom of choice to choose a set containing exactly one representative out of each element of R / Q."
>
> Is it possible to construct a non-measurable set (in $\mathbb{R}$ for example) without requiring the A.o.C.?
>
>
>
| https://mathoverflow.net/users/3121 | Does constructing non-measurable sets require the axiom of choice? | In the 1960's, Bob Solovay constructed a model of ZF + the axiom of dependent choice (DC) + "all sets of reals are Lebesgue measurable." DC is a weak form of choice, sufficient for developing the "non-pathological" parts of real analysis, for example the countable additivity of Lebesgue measure (which is not provable in ZF alone). Solovay's construction begins by assuming that there is a model of ZFC in which there is an inaccessible cardinal. Later, Saharon Shelah showed that the inaccessible cardinal is really needed for this result.
| 35 | https://mathoverflow.net/users/6794 | 42220 | 26,886 |
https://mathoverflow.net/questions/42222 | 3 | Hi all,
Sorry if this question is not the right level for mathoverflow, but I already tried math.stackexchange and received no answers.
Suppose that $\mathcal{E}$ is a well-pointed elementary topos, that $X$ and $Y$ are objects of $\mathcal{E}$, and that $F$ is a function which maps global elements $p: 1 \to X$ to global elements $F(p): 1 \to Y$ (here $1$ is the terminal object of $\mathcal{E}$). Does there exist a (necessarily unique) arrow $f: X \to Y$ in $\mathcal{E}$ such that $fp = F(p)$ for all $p$? Equivalently, is any object in a well-pointed topos the coproduct over its global elements of $1$? It's easy to show that the answer is "yes" if the coproduct exists since the induced map $\coprod\_{p \in \Gamma X} 1 \to X$ is iso. But I don't know whether the coproduct exists in general.
| https://mathoverflow.net/users/7842 | Do functions defined on global elements give rise to arrows in a well-pointed topos? | Here's a counterexample. Take $\mathcal E$ to be the topos of sets and functions of some countable model of ZFC (or a suitable weaker set theory, if you're worried that ZFC might be inconsistent). This is a well-pointed topos. The natural-number object $N$ in $\mathcal E$ has a countable infinity of global elements. So the number of functions (in the real world, not in the countable model) from global elements of $N$ to global elements of $N$ is the cardinal of the continuum. Only countably many of these correspond to morphisms in $\mathcal E$ from $N$ to $N$, because these morphisms are elements of your countable model.
| 5 | https://mathoverflow.net/users/6794 | 42223 | 26,887 |
https://mathoverflow.net/questions/42219 | 13 | This is a question my son Bob asked me. For some sets it is relatively easy
to test for membership but a lot more difficult to find members, and for others
the reverse is true. Here is an elementary example to get the idea across. An
$m \times n$ real matrix $M$ defines a linear map $x \mapsto M x = y$, from
${\mathbb R}^n $ to ${\mathbb R}^m $. It is easy to test if $x$ is in the kernel;
just compute $M x$ and see if it is zero, but to find an $x$ in the kernel you
must solve $M x = 0$ which is more computationally intensive. Conversely it is
easy to find an element in the range; just choose any $x$ and compute $M x$;
but to test if $y$ is in the range you must solve $M x = y$. Does anyone know
if there is a standard name for this distinction or for sets of these two types?
| https://mathoverflow.net/users/7311 | Is there a name for sets for which it is easier to test membership than to find members---and vice versa? | This phenomenon occurs both positively and negatively in
many parts of logic, but to my knowledge, there is no
particular adjective that is always used in such
situations.
* In classical *computability theory*, the first phenomenon does not
occur. If one can computably test membership in a set, in
the usual Turing sense, then one can computably generate
an instance, simply because one can computably enumerate
all objects in the domain of discourse, and systematically
test them. This is related to the classical fact that if the graph of a
function is decidable, then the function is computable.
Thus, to my mind, the phenomenon is intimately
wrapped up with the ability to effectively enumerate, in the relevant sense, the
objects in the domain of discourse.
* The converse situation, however, does occur in computability theory, and is a central phenomenon. Namely, there are sets of natural numbers whose members can be systematically generated---so the set is computably enumerable---but whose membership test is not computable. These are exactly the sets that are c.e. but not computable. Examples would include the halting problem (the set of programs $e$ halting on trivial input) and many other examples. It is easy to generate many halting programs---one can systematically enumerate them---but impossible to test in general if a given program halts. There is an intensively-studied hierarchy of Turing degrees instantiated by c.e. sets that are not decidable.
* In *complexity theory*, there is a sense in which there are
negative examples. One can imagine a polynomial-time
decidable set $A$, all of whose members are very large,
and hence difficult to produce. To make the problem
precise, however, one should really have a sequence $A\_n$
of sets such that membership $x\in A\_n$ is polynomial
time decidable in $(x,n)$---that is, uniformly in
$n$---but such that there is no polynomial time
computable function $f$ such that $f(n)\in A\_n$. Such an
example is provided simply by the sets $A\_n$ consisting
of all numbers at least $2^n$. Given a pair $(x,n)$, it
is polynomial-time decidable in $(x,n)$ whether $x\geq
2^n$, but there is no polynomial function exceeding
$n\mapsto 2^n$.
Similar examples would be provided by any sequence of sets
$A\_n$, all of whose members were very large in comparison
with $n$, but such that the membership problem $x\in A\_n$
is easily decided.
* In various sorts of higher computability theory, there
are additional negative instances. For example, with the
theory of infinite time Turing machines, there are
infinite time decidable sets of reals with no computable members.
Indeed, the Lost Melody Theorem asserts precisely that there are infinite
time decidable singletons $\{c\}$, such that the real
$c$ is not writable by any infinite time Turing machine.
That is, there are reals $c$, such that it is decidable
by infinite time Turing machines whether a given real $x$
is $c$ or not, by no such machine can produce $c$ on its
own. This seems to be the essence of your phenomenon. (The
``lost melody'' terminology arises from the situation,
where a person is able to recognize a given melody when someone else
sings it, but is unable to sing it on their own.)
* In descriptive set theory, one would look at whether a
set of reals at a given level in the descriptive
set-theoretic hierarchy has members at that same level.
This is false in general, although there are special circumstances
(some involving large cardinal hypotheses) in which instances
of it are true. One way to look at it
is as a Choice principle: given a subset $A$ of the plane
$\mathbb{R}\times\mathbb{R}$, can one find a function $f$
of the same complexity with $\text{dom}(f)=\text{dom}(A)$ such that
$(x,f(x))\in A$ for all $x\in\text{dom}(A)$? This problem is also
known as the *uniformization* problem.
* In a more general set theoretic setting, it is natural to consider
the situation of ordinal-definable sets. Does every
non-empty ordinal definable set contain an
ordinal-definable member? This turns out to be equivalent
to the assertion known as $V=HOD$, which is independent of
ZFC, as explained in the edited version of [this MO
answer](https://mathoverflow.net/questions/10413/definable-collections-without-definable-members-in-zf/10415#10415).
The reason is that the set of non-ordinal-definable sets of
minimal rank is ordinal definable.
| 14 | https://mathoverflow.net/users/1946 | 42232 | 26,892 |
https://mathoverflow.net/questions/42192 | 8 | In search for a Machian formulation of mechanics I find the following problem. In Machian mechanics absolute space does not exists, and the only real entities are the relative distances between the particles. As a consequence, the configuration space of a N-particle system is the set of the distances on a set of N elements. Actually these distances are usually required to be isometrically embeddable in $\mathbb{R}^3$. But if absolute space does not exists, this requirement appears to be not appropriate. The natural generalization it therefore to admit any possible distance as physically acceptable, and to find a preferred way to derive a 3-geometry, possibly non-flat, form a generic distance.
To be more specific, consider the following simple example. Let A be a metric space with 3 elements. There are infinitely many bi-dimensional riemaniann manifolds (surfaces) is which A can be isometrically embedded. There is however a preferred embedding, namely the embedding into a plane. The existence of a preferred embedding defines a preferred value for the angles between the geodetics joining the points, which in this case are simply the angles of the triangle defined by the distance between the points.
Suppose now that A has four point. In general this metric space cannot be isometrically embedded in a 2-plane. The problem therefore is the following: is there a preferred isometric embedding of this metric space in a 2-surface, or equivalently, there is a preferred way for defining the values of the angles between the geodetics?
In more forma way, the problem is the following: is there a preferred isometric embedding of a finite metric space in a riemaniann manifold of given dimension?
| https://mathoverflow.net/users/9584 | Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension | Your question is not well stated.
In particular I did not understand why embedding into a plane is "preferred embedding".
Here are some associations...
**4-point case.**
A generic 4-point metric space can be isometrically embedded into two different model planes (i.e. simply connected surfaces of constant curvature $K$, which is eiter sphere, Euclidean plane or Lobachevsky plane depending on sign of $K$).
Thus you have two values of curvature $K\_1\le K\_2$ associated to (almost) any 4-point metric space.
In this case the metric space can be isometrically embeded into a model 3-space of curvature $K\_1\leq K\leq K\_2$.
In fact for any 4-point metric space $M$ there is an subinterval $\mathbb I\_M$ of $[-\infty,\infty)$ such that $M$ can be isometrically embedded into a model 3-space of any curvature $K\in \mathbb I\_M$.
(We assume that model space of curvature $-\infty$ is an $\mathbb R$-tree.)
Nearly all this was discovered by A. Wald in 1936 or so.
| 6 | https://mathoverflow.net/users/1441 | 42233 | 26,893 |
https://mathoverflow.net/questions/42235 | 4 | Hey everyone!
Lately I remembered an exercise from an algebra class from Jacobson's book: Prove that if an element has more than one right inverse then it has infinitely many, Jacobson attributes this excercise to Kaplansky. Regardless of the solution I began to wonder:
Does anybody know any explicit examples of rings that have this property of having elements with infinitely many (or, thanks to Kaplansky, multiple) right inverses? Is the same true for left inverses?
I came across an article from the AMS Bulletin that studied this topic but skimming through it I could not find an explicit example, sorry I cant remember the author.
Anyways, thanks and good luck!
| https://mathoverflow.net/users/9187 | Rings with right inverses | Let $M$ be a module (over some ring) such that $M$ is isomorphic to $M\oplus M$, for example an infinite-dimensional vector space over a field. Let $R$ be the ring of endomorphisms of $M$. Let $f\in R$ be projection of $M\oplus M$ on the first factor composed with an isomorphism $M\to M\oplus M$. Then $f$ has as many right inverses as there are homomorphisms $M\to M$.
| 6 | https://mathoverflow.net/users/6666 | 42237 | 26,894 |
https://mathoverflow.net/questions/42139 | 8 | Let there be $n$ points on a unit circle. It is known they come from "normal" distribution around particular unknown direction (i.e. sum of 2 "normal" distributions on circle - one centered at point $p$ and the other at its opposite $-p$).
What is the best way to estimate this direction? By best I mean an algorithm that is a. analytical, b. efficient and c. simple.
| https://mathoverflow.net/users/10032 | Estimating direction from a distribution on a circle | This seems too simple to be true, but, combining some of the ideas posted earlier,
I think you could just interpret the vectors as complex numbers and take the RMS.
Squaring will turn the bimodal distribution into a unimodal one.
Then the square roots of the mean should give a good estimate of the modes of the original distribution.
| 10 | https://mathoverflow.net/users/10075 | 42240 | 26,897 |
https://mathoverflow.net/questions/42236 | 3 | Alexander's Theorem guarantees that every oriented link is the closure of some braid. In other words, the map
$$ \displaystyle \coprod\\_n \mathcal B\_n\longrightarrow \{\text{ oriented links }\} $$
is surjective. One algorithm (I'm actually not sure that it's Alexander's original demonstration) involves choosing a basepoint in the complement of a diagram for the link and applying Reidemeister moves until the resulting diagram winds around the point in a consistent direction, at which point the diagram is manifestly the closure of a link.
>
> If we begin with a positive diagram for a link, then do we necessarily obtain a positive braid?
>
>
>
| https://mathoverflow.net/users/813 | Is a positive link the closure of a positive braid? | Rudolph proved that [positive links are strongly quasipositive](http://www.ams.org/mathscinet-getitem?mr=1734423). [This paper](http://www.ams.org/mathscinet-getitem?mr=894383) might also be relevant, which allows one to create a braid with the same number of Seifert circles and writhe. However, Yamada's algorithm doesn't seem to preserve positivity either.
| 5 | https://mathoverflow.net/users/1345 | 42253 | 26,905 |
https://mathoverflow.net/questions/42247 | 0 | I have Edwards and Titmarsch books on Riemann zeta function with me. I could not find (maybe I did not read through that carefully), but are there results similar to the form like the one given below:
> Is there a non-trivial zero $\rho$ starting from which $(\Im(\rho)\log(2)/2\pi )$ runs only through integer values
>
I would love any elaboration or links on this topic.
Thanks,
Roupam Ghosh
| https://mathoverflow.net/users/2865 | Curious about a question on zeta zeros? | Such a conjecture is false.
EDIT: A simpler argument - a more precise asymptotic for the number of
zeroes $N(t)$ of imaginary part $\le t$ (counted with multiplicity) is
$$N(t) = \frac{t}{ \pi} \log \frac{t}{2 \pi e} + o(\log t),$$
This is enough to show that, for any fixed $\epsilon > 0$,
$$N(t + \epsilon) - N(t) \sim \frac{\epsilon}{\pi} \log t,$$
and thus, for sufficiently large $t$, and for any interval of length
$\epsilon$, there are zeroes (whose imaginary part lies) in this interval,
which also implies the conjecture is false. THIS DOES NOT USE GRH.
PS: Scott Carnahan helpfully remarks that the wikipedia article points out that Littlewood noticed that the difference in the imaginary parts of the zeros tends to zero as $t
\rightarrow \infty$ (presumably by exactly using this asymptotic result of von Mangoldt above). Personally I prefer mathematics rather than an appeal to authority, but apparently that is not enough for some.
REMARK: Dear Wadim, please read this again, and realize that it DOESN'T USE GRH. The estimate of zeros (which was basically known by Riemann) is about zeroes in the CRITICAL STRIP (real part in $[0,1]$) not the CRITICAL LINE. Having done this, you can delete all your comments, I'll edit this answer, and we can all pretend it never happened. (In fact, I'll make this community wiki so you can delete this remark yourself.)
| 6 | https://mathoverflow.net/users/nan | 42254 | 26,906 |
https://mathoverflow.net/questions/42204 | 0 | I have been reading the paper - "[Introduction to Quantum Fisher Information](http://arxiv.org/abs/1008.2417)".
In section 1.2 the author talks about the linear map $\mathbb{J}\_D$, which he defines as follows:
Let $D \in M\_n$ be a positive invertible matrix. The linear mapping $\mathbb{J}^f\_D:M\_n \to M\_n$ is defined by the formula
$\mathbb{J}\_D^f=f(\mathbb{L}\_D\mathbb{R}^{-1}\_D)\mathbb{R}\_D$
where $f:\mathbb{R}^+\to\mathbb{R}^+$,
$\mathbb{L}\_D(X)=DX$ and $\mathbb{R}\_D(X)=XD$.
Also, the author points out that the inverse mapping of $\mathbb{J}\_D^f$ is the given by
$\frac{1}{f}(\mathbb{L}\_D\mathbb{R}^{-1}\_D)\mathbb{R}\_D^{-1}$
Then he gives the following example, of which I am able to get the first part but not the inverse map.
*Example 1*: If $f(x)=(x+1)/2$, then
$\mathbb{J}\_DB=\frac{1}{2}(DB+BD)$.
$\mathbb{J}\_DB=\int\_0^\infty \text{exp}(-tD/2)B\text{exp}(-tD/2)dt$
Can you tell me how does he get the required formula for the inverse map? Also, I would like to know what is the meaning of the map $\mathbb{J}\_D$? What does the integral of the linear map represent? Can you also provide me some reference book where I could look for these topics?
| https://mathoverflow.net/users/10071 | Linear Mapping and integration | One way to think of the operation ${J}\_D^f$ is as follows: $M\_n$ has the structure of a Hilbert space by the inner product $\langle X, Y \rangle = {\rm Tr}(X^\* Y) = {\rm Tr}(Y X^\*)$. Then the transformations $L\_D R\_{D}^{-1}$ and ${R}\_D$ are positive operators on this Hilbert space. Since they commute, you may take the functional calculus $f(s) \otimes g(t) \mapsto f(L\_D R\_{D}^{-1}) g({R}\_D)$ from the space of functions on ${\rm Sp}(L\_D R\_D^{-1}) \times {\rm Sp}(R\_D)$ into the space of bounded operators on $M\_n$ (since the spectrum ${\rm Sp}(L\_D{R}\_{D^{-1}})$ of $L\_D$ is a finite subset of ${R}$, there is no need to worry about the regularity of $f$. Ditto for $g$). Then ${J}\_D^f$ is simply the image of $f(s) \otimes t$. Since the functional calculus is an algebra homomorphism, the inverse of ${J}\_D^f$ is represented by $(1/f(s)) \otimes (1/t)$, which equals $\frac{1}{f}(L\_D{R}\_D^{-1}){R}\_D^{-1}$. When $E \subset M\_n$ is the joint eigenspace for $L\_D {R}\_{D^{-1}}$ and ${R}\_D$ with the corresponding eigenvalues $a b^{-1}$ and $b$ (i.e. $L\_D$ has eigenvalue $a$ on this subspace), ${J}\_D^f$ acts by $f(a/b)b$ on $E$.
Any book containing the spectral theory of self adjoint operators on Hilbert spaces will do, like Pedersen's Analysis Now (GTM 118).
About the computation of $\mathbb{J}\_D$ for the case of $f(t) = \frac{t+1}{2}$, the integral formula follows from the identity $\int\_0^\infty exp(-t a / 2) exp(-t b / 2) = \frac{2}{a+b} = \frac{2}{ab^{-1}+1} \frac{1}{b}$.
Edit: I made a stupid mistake in the first version, and this is the corrected version. Sorry for the change of notation from $\mathbb{L}\_D$ to $L\_D$, etc. I somehow couldn't make it to work.
| 0 | https://mathoverflow.net/users/9942 | 42259 | 26,910 |
https://mathoverflow.net/questions/42217 | 22 | I'm absolutely new to this stuff I'm asking about, so I hope this is not nonsense.
If X is a smooth scheme over a perfect field k, I can study its motivic cohomology in the sense of Voevodsky and Morel.
More precisely, to $\mathbb{Z}$ there is associated an Eilenberg-MacLane T-Spectrum. I'll write $H^{p,q}(X,\mathbb{Z})$ for the motivic cohomology of $X$ defined by Hom into this Spectrum.
One fact for smooth $X$ now is that
$$ H^{2p,p}(X,\mathbb{Z}) = CH^p(X). $$
Now I've got two questions:
1) Can I feed something singular into this Voevodsky-Morel machine?
2)If the answer to 1 is yes, are there some maps from $ H^{2p,p}(X,\mathbb{Z})$ to $CH^p(X)$? Maybe defined by some spectral sequence? I'm really interested just in some maps, no isomorphisms.
| https://mathoverflow.net/users/473 | Motivic Cohomology vs. Chow for singular varieties? | The answer to question 1) is yes. However, Chow groups do not form what we should call a cohomology theory, but are part of a Borel-Moore homology theory. This ambiguity comes from the fact that, by Poincaré duality, motivic cohomology agrees with motivic Borel-Moore homology for smooth schemes (up to some reindexing), while they are quite different for non-regular schemes. You have the same phenomena in K-theory: K-theory of vector bundles (which defines a kind of cohomology) agrees with K-theory of coherent sheaves (which defines the corresponding Borel-Moore homology) for regular schemes. That said, they are plenty of ways to extend motivic cohomology into a cohomology theory for possibly non-singular schemes; they all agree in char. 0, or if you work with rational coefficients. As you seem to be interested by Chow groups, it seems that what you want is motivic Borel-Moore homology for possibly non-singular schemes (which agrees with Bloch's higher Chow groups, whence gives classical Chow groups for the appropriate degrees).
To define motivic Borel-Moore homology of $X$ (separated of finite type over a field $k$ of char. 0, unless you tensor everything by $\mathbf{Q}$, or admit resolution of singularities in char. $p$), you may proceed as follows: there is a motive with compact support $M\_c(X)$ in $DM(k)$, and we define
$H^{BM}\_i(X,\mathbf{Z}(j))=Hom\_{DM(k)}(\mathbf{Z}(j)[i],M\_c(X)).$
If you prefer the six operations version, for $f:X\to Spec(k)$ a (separated) morphism of finite type, we have
$H^{BM}\_i(X,\mathbf{Z}(j))\simeq Hom\_{SH(k)}(f\_!f^\*\Sigma^\infty(Spec(k)\_+)(j)[i],H\mathbf{Z})$
where $H\mathbf{Z}$ denotes the motivic Eilenberg-MacLane spectrum in $SH(k)$. To answer your question 2), the link with Bloch's higher Chow groups is that
$H^{BM}\_i(X,\mathbf{Z}(j))\simeq CH^{d-j}(X,i-2j)$
for $X$ equidimensional of dimension $d$, and that, for any separated $k$-scheme $X$,
$H^{BM}\_{2j}(X,\mathbf{Z}(j))\simeq A\_j(X)$
where $A\_j(X)$ stands for the group of $j$-dimensional cycles in $X$, modulo rational equivalence.
| 29 | https://mathoverflow.net/users/1017 | 42268 | 26,913 |
https://mathoverflow.net/questions/42263 | 8 | There are different strategies for numbering displayed math. The most common are
1. Number only the formulas you reference to. It makes your paper more clean and gives more freedom to the editor (i.e. making this math inline).
2. Number all displayed math. Even if you don't reference your formulas, take care of those who will read your paper and will want to reference them.
3. Number only very important formulas and the formulas you reference to.
I am thinking, which strategy to choose and I want to make an informed choice.
>
> Which important strategies have I missed? What are pitfalls and benefits of the strategies, described above? How important are these benefits/pitfalls?
>
>
>
Ideas from the answers below:
strategy 4. Check with the journal you are planning to publish your contribution.
The similar question was asked [here](https://mathoverflow.net/questions/35223/should-one-use-above-and-below-in-mathematical-writing). The most popular answer was "[Use strategy 2: number all your displayed math to help the reader to reference your formulas](https://mathoverflow.net/questions/35223/should-one-use-above-and-below-in-mathematical-writing/35229#35229)". However it is not clear for me if it is ok to have a 40-page paper with 150 numbered formulas. It seems a bit crazy to see "it follows from (146)". Also it was noted, that this strategy violates "[Checkov's gun principle](http://en.wikipedia.org/wiki/Chekhov's_gun)", i.e. fills the paper with irrelevant details (numbers). However the question, stated there was a bit different. I think that that question was completely answered, but not this. For me it is still unclear, whether it is ok to have formula (146) (or formula (1353)) or not. The question made community wiki, so you are free to improve it.
| https://mathoverflow.net/users/8134 | The main ideas in choosing the strategy for numbering displayed math | Fiktor, many journals explicitly indicate option (1). Some also ask to use single/double numeration, the list of references in alphabetical/appearance order, and many other things. Therefore,
>
> Strategy (4): check with the journal you are planning to publish your contribution.
>
>
>
As for (2) and (3), are you happy if somebody else cites your ms in the form [1, Eq. (7.15)]? Wouldn't it be better if there is no number and he needs to reproduce your formula with link to your original source?
Strategy (2) is standard for physical journals, so if you are more on the math physics side, choose (2) or (3) without doubts.
| 3 | https://mathoverflow.net/users/4953 | 42270 | 26,914 |
https://mathoverflow.net/questions/42272 | 27 | I realise that I lack some intuition into how a curve (or surface, or whatever) looks geometrically, from just looking at the equation. Thus, I sometimes resort to some computer program (such as *Mathematica*) to draw me a picture. The problem is, all these programs require input of the form $y=f(x)$, whereas my curve might be something like $y^3+x^3-6x^2 y=0$, and transforming this into the former form is not always easy, and always misses some information. So, my question:
Are there any programs that can take an equation ($p(x,y)=0$, say) as input and return a graph of its zero-set?
**Update:** So, lots of good answers, I wish I could accept them all. I'll accept Jack Huizenga's answer, for the reason of personal bias that I already have Mathematica available.
| https://mathoverflow.net/users/1481 | Are there any good computer programs for drawing (algebraic) curves? | Mathematica does this just fine. You're looking for the command ContourPlot, as in
ContourPlot[y^3+x^3-6x^2y==0,{x,-5,5},{y,-5,5}].
A more serious issue is that if you're trying to do algebraic geometry over $\mathbb C$, the real picture isn't always terribly enlightening.
| 18 | https://mathoverflow.net/users/7399 | 42273 | 26,916 |
https://mathoverflow.net/questions/42276 | 21 | Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x\_1, \ldots, x\_n$ algebraically independent such that $R$ is integral over $k[x\_1, \ldots, x\_n]$.
Does one have a similar statement, under good assumptions, if $k$ is not a field but a ring ? In this discussion, I am also interested by geometric explanations.
| https://mathoverflow.net/users/2330 | Noether's normalization lemma over a ring A | <http://www.math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf>
Supplementary notes from Mel Hochster's commutative algebra class. They discuss, in particular, a generalization of Noether normalization to integral domains.
| 20 | https://mathoverflow.net/users/1353 | 42281 | 26,921 |
https://mathoverflow.net/questions/42283 | 6 | **Definition 1:** Suppose $B$ is a $C^\* $-algebra. $A$ is massive $C^\* $-subalgebra of $B$ iff
1. $A$ is a subalgebra of $B$;
2. for each irreducible representation $\pi$ of $B$ representation $\pi|\_A$ is irreducible;
3. if representations $\pi$ and $\pi'$ aren't (unitary) equivalent then $\pi|\_A$ and $\pi'|\_A$ aren't equivalent too.
**Definition 2:** Suppose $B$ is a $C^\* $-algebra. $A$ is massive $C^\* $-subalgebra of $B$ iff inclusion $i\colon A\to B$ is epic in the category of $C^\* $-algebras, i.e. for any $C^\* $-algebra $D$ and any two homomorphisms $g\_1,g\_2\colon B\to D$ we have $g\_1\circ i=g\_2\circ i \Rightarrow g\_1=g\_2$.
It's not hard to prove, that these definitions are equivalent. But I don't know the answer to the following question:
>
> Does there exist an example of massive proper $C^\*$-subalgebra?
>
>
>
There is a theorem in the book of Dixmier "C\*-algebras" (French original was published in 1969), that if $A$ is a massive $C^\* $-subalgebra of postliminal $C^\* $-algebra $B$ then $A=B$. What's for the general case? Is it an open problem?
| https://mathoverflow.net/users/8134 | Does there exist any massive proper $C^*$-subalgebra? | It was proved by Karl H. Hofmann and Karl-H.Neeb [here](https://doi.org/10.1007/BF01270691 "Hofmann, K.H., Neeb, K.H. Epimorphisms of C^*-algebras are surjective. Arch. Math 65, 134–137 (1995). zbMATH review at https://zbmath.org/?q=an:0829.46044") (see [here](https://arxiv.org/abs/funct-an/9405003) for the preprint), that epimorphisms in the category of $C^{\star}$-algebras are surjective.
| 8 | https://mathoverflow.net/users/8176 | 42289 | 26,923 |
https://mathoverflow.net/questions/42249 | 8 | I have often heard that Lie groups classify geometry. For example that $O(n)$ is about real manifolds, $U$ is about almost complex manifolds, $SO(n)$ about orientable real manifolds and so on.
I have also heard that manifolds can be defined very generally by patching local pieces via a pseduogroup of morphisms.
My questions are
1) Does this pseduogroup relate to the Lie group?
2) How do Lie groups classify geometry?
3) Is there a geometry for every Lie group or only some of them?
and maybe even
4) Does this classification seem work for algebraic (i.e. non-differential) geometry?
I am wondering whether this Erlangen program classification is actually formalized, or only serves as a slogan or principle.
| https://mathoverflow.net/users/nan | How do Lie groups classify geometry? | [R.W.Sharpe, Differential Geometry - Cartan's Generalization of Klein's Erlangen Program](http://books.google.com/books?id=Ytqs4xU5QKAC&pg=PA173&dq=cartan%27s+generalization+of+klein%27s+erlangen&hl=it&ei=aGK4TPfXLIPKswbmsfyqDQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCYQ6AEwAA#v=onepage&q&f=false)
| 3 | https://mathoverflow.net/users/4721 | 42290 | 26,924 |
https://mathoverflow.net/questions/40860 | 10 | Note: in this post, every topological group under consideration is assumed to be Hausdorff.
Given a locally compact abelian group, one can construct its dual group, i.e. its group of (unitary) characters. This turns out to be a locally compact abelian group with respect to the compact-open topology. Pontryagin duality then tells us that there is a natural isomorphism of topological groups between the original group and its double dual. Thus, if we know only the unitary characters of the group (along with their algebraic and topological structure), we can recover the original group (up to isomorphism) by taking its dual.
Similarly, given the irreducible unitary representations of a compact group, one can construct a compact group out of them which turns out to be isomorphic to the original group (the Tannaka duality theorem). Thus the irreducible unitary representations of a compact group contain enough information to recover the original group.
How much of this remains true for general locally compact groups? By the Gelfand-Raikov theorem, such groups have many irreducible unitary representations (enough to separate points). The question is: can one associate a canonical group structure and/or a topological structure to the irreducible representations of the group (or their equivalence classes) so that one recovers the original group (along with its topology) up to isomorphism?
If the answer is "no", can this be remedied by considering more general representations of the group, e.g. representations which aren't necessarily unitary?
I'm not sure if these questions have an easy answer, so I should mention that I'm mostly interested in the cases where the group is a connected nilpotent Lie group or a connected semisimple Lie group.
| https://mathoverflow.net/users/7392 | Which groups can be recovered from their unitary dual? | The nicest way of phrasing it is the following. Let $\mathcal H$ be the category of Hilbert spaces with unitary maps between them. For each locally compact group $G$, one can define a functor
$$Rep\_G : {\mathcal H} \to Top$$
with $Rep\_G(H) = \hom(G,U(H))$, where the space of homomorphisms is endowed with the compact-open topology (with respect to the strong operator topology on $U(H)$). Obviously, the functor $Rep\_G$ is compatible with sums and tensor products of Hilbert spaces. Note that
$Rep\_{\mathbb Z}(H)=U(H).$
Consider now any such functor $F: {\mathcal H} \to Top$
and set
$$D(F) = Nat\_{\otimes,\oplus}(F,Rep\_{\mathbb Z}),$$
i.e. all natural transformations of functors which are compatible with the tensor-product and the sum. $D(F)$ is a group since $Rep\_{\mathbb Z}(H)=U(H)$ is a group for each Hilbert space $H$. It is also a topological group in a natural way.
Now, there is a natural map $\iota\_G : G \to D(Rep\_G)$ which is given by
$\iota\_G(g)(\pi) = \pi(g)$, where $\pi \in hom(G,U(H))$. So just as in the case of Pontrjagin duality, there is a natural bi-dual. A bit of work (relying on results of Takesaki and Gel'fand-Raikov (which you have mentioned)) shows that $\iota\_G$ is a topological isomorphism for all locally compact topological groups.
I studied the analogous question which arises if one restricts everything to the category of finite-dimensional Hilbert spaces (see [here](http://arxiv.org/abs/1003.4093)). This sometimes goes under the name Chu duality, but is not so extensively studied. Everything works for locally compact abelian groups and compact groups by Pontrjagins result and the Tannaka-Krein theorem. However, for finitely generated discrete groups, interesting things happen. First of all, it is trivial to observe that the analogous map
$$\iota\_G : G \to D\_{fin}(Rep\_G)$$
is injective if and only $G$ is maximally almost periodic (by a result of Mal'cev iff $G$ is residually finite). Moreover, and this is more difficult, $\iota\_G$ is an isomorphism if and only if $G$ is virtually abelian. In particular, for $G={\mathbb F\_2}$, the map
$\iota\_{\mathbb F\_2}$ from $\mathbb F\_2$ to $D\_{fin}(Rep\_{\mathbb F\_2})$
is not surjective. This is a bit surprising as there are no natural candidates of elements in
$D\_{fin}(Rep\_{\mathbb F\_2})$, which do not lie in the image.
| 8 | https://mathoverflow.net/users/8176 | 42294 | 26,927 |
https://mathoverflow.net/questions/33931 | 5 | Suppose X is a normal projective complex variety, (X, $\Delta$) is a klt pair and f : X $\to$ Z is a Mori fiber space given by a contraction of an extremal ray for this pair. Here I mean that the relative dimension is at least 1. Do we know anything about the singularities of Z? Z is normal more or less by construction and we do know that if X was Q-factorial then so is Z. Can we say anything more? Eg: is there a $\Delta'$ on Z with (Z, $\Delta'$) klt? Does Z have rational singularities? What if $\Delta$ = 0, is Z terminal? Canonical?
| https://mathoverflow.net/users/7756 | Possible singularities of the base of a Mori fiber space | Under the assumption that $X$ is $\mathbb{Q}$-factorial, section 5 of the paper <https://arxiv.org/pdf/math/0606666> addressed this issue, which was also proved earlier in Ambro's paper. Basically, if you assume the pair $(X,\Delta)$ is klt, so is the base $(Z,\Delta\_Z)$ for some $\Delta\_Z$. As the example of Prokhorov shows, this is optimal, namely, the base may not be terminal (canonical) even you assume the $(X,\Delta)$ is. For lc case, I think dlt modification+perturbation reduce the question to the klt case.
| 6 | https://mathoverflow.net/users/10083 | 42301 | 26,931 |
https://mathoverflow.net/questions/42174 | 20 | (**Edit:** I've realized that there was an error in my reasoning when I was convincing myself that these two formulations are equivalent. Hailong has given a beautiful affirmative answer to my first question in the case of finite type modules over a noetherian commutative ring. Mariano has given a slick negative answer to the question for non-finite-type modules. Greg has given a beautiful negative answer to my "alternative formulation" even in the finite type case over a noetherian commutative ring. I'm accepting Hailong's answer since that's the one I imagine people will be most immediately interested in if they find this question in the future.)
Suppose we're working the category of modules over some ring $R$. Suppose a module $E$ is an extension of $M$ by $N$ in two different ways. In other words, I have two short exact sequences
\begin{array}{ccccccccc}
0&\to &N&\xrightarrow{i\_1}&E&\xrightarrow{p\_1}&M&\to &0\\
& & \wr\downarrow ?& & \wr\downarrow ?& & \wr\downarrow ?\\
0&\to &N&\xrightarrow{i\_2}&E&\xrightarrow{p\_2}&M&\to &0
\end{array}
>
> Must there be an isomorphism between these two short exact sequences?
>
>
>
---
Alternative formulation
-----------------------
$Ext^1(M,N)$ parameterizes extensions of $M$ by $N$ modulo isomorphims *of extensions*. Suppose I'm interested in parameterizing extensions of $M$ by $N$ modulo *abstract isomorphisms* (which don't have to respect the submodule $N$ or the quotient $M$). One obvious thing to note is that there is a left action of $Aut(M)$ on $Ext^1(M,N)$, and that any two extensions related by this action are abstractly isomorphic. Similarly, there is a right action of $Aut(N)$ so that any two extensions related by the action are abstractly isomorphic.
>
> Does the quotient set $Aut(M)\backslash Ext^1(M,N)/Aut(N)$ parameterize extensions of $M$ by $N$ modulo abstract isomorphism?
>
>
>
Note: I'm **not** asking whether all abstract isomorphisms are generated by $Aut(M)$ and $Aut(N)$. They certainly aren't. I'm asking whether for every pair of abstractly isomorphic extensions *there exists* some isomorphism between them which is generated by $Aut(M)$ and $Aut(N)$.
| https://mathoverflow.net/users/1 | Can a module be an extension in two really different ways? | It is worth noting some very interesting cases when the answer is yes. An amazing [result by Miyata](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-7/issue-1/Note-on-direct-summands-of-modules/10.1215/kjm/1250524308.full) states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus N$, any exact sequence
$ 0 \to M \to E \to N \to 0$ must split!
This holds true slightly more generally, when $R$ is (not necessarity commutative) module-finite over a Noetherian commutative ring. Also, the statement holds for finitely generated pro-finite groups, see Goldstein-Guralnick, J. Group Theory 9 (2006), 317–322.
Added: in fact, this [paper by Janet Striuli](http://www.math.unl.edu/%7Ejstriuli2/research/module2.pdf) may be useful for you. She addressed the question: if two elements $\alpha, \beta \in \text{Ext}^1(M,N)$ give isomorphic extension modules, how close must $\alpha, \beta$ be? Her Theorem 1.2 extend Miyata's result (let $I=0$).
| 20 | https://mathoverflow.net/users/2083 | 42309 | 26,936 |
https://mathoverflow.net/questions/38931 | 2 | As an exercise, I'm trying to show that for an $(n-1)$-connected space $L$ with $\pi=\pi\_n(L)$, the map $\iota\_L:L\rightarrow K(\pi,n)$ associated to the fundamental class $\iota\_L\in H^n(L;\pi)$ induces an isomorphism $\pi\_n(L)\rightarrow \pi\_n(K(\pi,n))$. After playing with this for a while, I've found that this feels so tautological that it *has* to be true, although that doesn't quite constitute a proof. I'm supposed to use only basic definitions / first principles. Presumably, these are:
1. The universal coefficient theorem yields $H^n(L;\pi)\cong Hom(H\_n(L),\pi)$. The Hurewicz homomorphism $h:\pi \rightarrow H\_n(L)$ is an isomorphism, and its inverse $h^{-1}\in Hom(H\_n(L),\pi)$ corresponds to $\iota\_L\in H^n(L;\pi)$.
2. We have a canonical bijection $H^n(L;\pi) \cong [L,K(\pi,n)]$. Any $[f]\in [L,K(\pi,n)]$ corresponds to $f^\*\iota\in H^n(L;\pi)$, where $\iota\in H^n(K(\pi,n);\pi)$ is the fundamental class of $K(\pi,n)$ (which is associated to its identity map). This is actually a group isomorphism if we add maps on the right side by using the fact that $K(\pi,n)=\Omega K(\pi,n+1)$ (or at least $\simeq$, although what does $K(\pi,n)$ even mean really). I'd imagine that this is canonical too, but I don't know for sure.
3. We have a map $[L,K(\pi,n)]\rightarrow Hom(\pi,\pi)$ given by $[f]\mapsto f\_\#$. Presumably the idea is to show that the image of $\iota\_L$ is an isomorphism, but I can't tell if I'm just complicating the question by phrasing it in these terms.
I think my problem is that I don't really understand the defining way Eilenberg-MacLane spaces work. I have an intuitive picture of the Hurewicz homomorphism, and so I guess I have an intuitive picture of its inverse: it takes a homology class in degree $n$ and realizes it as the image of a bunch of based $n$-spheres (which is made possible by the Hurewicz theorem). We can look at this more or less as a cochain in $C^n\_{cell}(L;\pi)$, and this is the element $\iota\_L\in H^n(L;\pi)$. But then I have no idea how to actually turn this into a map $\iota\_L : L\rightarrow K(\pi,n)$, or whether I'm even supposed to.
Because I'd still like to work this out myself to whatever extent I can, a good hint (if one exists) is worth more to me than a straight-up answer. Of course, I'm happy with either. Thanks!
| https://mathoverflow.net/users/303 | How can I show that the map L-->K(\pi_n(L),n) representing the fundamental class of an (n-1)-connected space is an isomorphism on \pi_n? | Here is a solution, based on Jeffrey's comment:
Let $f:L\rightarrow K(\pi,n)$ be the map classifying $\iota$. We will show that the induced map $f\_\*:H\_i(L)\rightarrow H\_i(K(\pi,n))$ is isomorphic for $i \lt n+1$ and epimorphic for $i=n+1$. This will imply by Whitehead's theorem that the same is true for $f\_\# :\pi\_i(L)\rightarrow \pi\_i(K(\pi,n))$ (assuming $n \gt 1$).
First, $f\_\*$ is automatically an isomorphism for $i \lt n$, since in that case by the Hurewicz theorem $H\_i(L)$ and $H\_i(K(\pi,n))$ are both trivial.
Next, in dimension $n$, we use the fact that $f^\*:H^n(K(\pi,n);\pi)\rightarrow H^n(L;\pi)$ satisfies $f^\*\iota\_n = \iota$, where $\iota\_n$ is the fundamental class of $K(\pi,n)$. By the universal coefficient and Hurewicz theorems, we may regard $H^n(K(\pi,n))=Hom(H\_n(K(\pi,n)),\pi)$ and $H^n(L)=Hom(H\_n(L),\pi)$, and the fundamental classes both correspond to isomorphisms (namely, the inverses of the respective Hurewicz homomorphisms). In general, if a group homomorphism $\varphi:B\rightarrow A$ induces $\varphi^\#:Hom(A,C)\rightarrow Hom(B,C)$ and $\varphi^\#$ takes an isomorphism $\psi\_1:A\rightarrow C$ to an isomorphism $\psi\_2:B \rightarrow C$, then $\varphi$ itself must be an isomorphism, because $\psi\_2 = \varphi^\# (\psi\_1)=\psi\_1\varphi$ and so $\varphi = \psi\_1^{-1}\psi\_2$. Here, since $f^\*:Hom(H\_n(K(\pi,n)),\pi)\rightarrow Hom(H\_n(L),\pi)$ is induced from $f\_\*:H\_n(L)\rightarrow H\_n(K(\pi,n))$ and takes an isomorphism to an isomorphism, then $f\_\*:H\_n(L)\rightarrow H\_n(K(\pi,n))$ itself is an isomorphism.
Lastly, by the Hurewicz theorem, since $K(\pi,n)$ is $n$-connected, the Hurewicz homomorphism $h:\pi\_{n+1}(K(\pi,n))\rightarrow H\_{n+1}(K(\pi,n))$ is an epimorphism, but $\pi\_{n+1}(K(\pi,n))=0$ by definition so $H\_{n+1}(K(\pi,n))=0$ as well. Hence $f\_\*: H\_{n+1}(L)\rightarrow H\_{n+1}(K(\pi,n))$ is automatically epimorphic.
| 1 | https://mathoverflow.net/users/303 | 42328 | 26,945 |
https://mathoverflow.net/questions/42331 | 12 | Every symplectic form on a manifold $M^n$ determines a De Rham cohomology class in $H^2(M)$ (often a nontrivial class), and this in turn determines a class in $H\_{n-2}(M)$. What in general can be said about this class? For example, over the rationals this class is represented by a submanifold of $M$; is it possible to explicitly describe such a submanifold in terms of the symplectic structure?
If there is a nice answer to this question, does it also shed light on the Poincare duals of $\omega^2$, $\omega^3$, etc?
| https://mathoverflow.net/users/4362 | What is the Poincare dual of a symplectic form? | One of the big advances in symplectic topology in the 90s was Donaldson's theorem that when the symplectic class is integral, high multiples of its dual are represented by symplectic submanifolds.
These submanifolds behave like hyperplane sections in algebraic geometry; for instance, they satisfy the Lefschetz hyperplane theorem. They form the fibres of "symplectic Lefschetz pencils". Their intersections can be made to give symplectic submanifolds dual to multiples of wedge-powers of $\omega$.
Imagine first that $M$ is a compact complex manifold, $L\to M$ a hermitian, holomorphic line bundle, whose Chern connection has curvature $-2\pi i\omega$, a closed $(1,1)$-form. Then the zero-set of a $C^\infty$ section $s$ of $L^{\times k}$, if cut out transversely, is dual to $k[\omega]$. If $\omega$ is positive, the Kodaira embedding theorem then tells us that $L$ is ample: its high powers have enough holomorphic sections to embed $M$ into projective space. If $M$ is merely symplectic, with $-2\pi i\omega$ the curvature of some unitary connection in a hermitian line bundle, we can choose an almost complex structure $J$ on $M$ and consider transverse sections $s\_k$ of $L^{\otimes k}$ for which, asymptotically, the $(0,1)$-part of $\nabla s\_k$ along $s\_k^{-1}(0)$ is much smaller than the $(1,0)$-part. Then $s\_k^{-1}(0)$ will not quite be a $J$-holomorphic submanifold, but for $k \gg 0$ its tangent spaces will be close enough to being $J$-linear that it will still be a symplectic submanifold.
References: S. K. Donaldson, "Symplectic submanifolds and almost-complex geometry", J. Differential Geom. Volume 44, Number 4 (1996), 666-705; "Lefschetz pencils on symplectic manifolds", J. Differential Geom. Volume 53, Number 2 (1999), 205-236.
These papers are brilliant both geometrically and analytically: the analysis is mostly low-tech but extremely subtle.
| 24 | https://mathoverflow.net/users/2356 | 42334 | 26,947 |
https://mathoverflow.net/questions/42344 | 6 | I have no idea how to achieve this, any help would be greatly appreciated and very useful to me.
I have a loop in some computer code, that loops through every single combination of 7 on bits in a 64 bit integer.
For example,
```
Permutation 1: 00...001111111
Permutation 2: 00...010111111
Permutation 3: 00...011011111
```
etc.
These evaluate to the decimal numbers:
```
Permutation 1: 00...001111111 = 127
Permutation 2: 00...010111111 = 191
Permutation 3: 00...011011111 = 223
```
In total there are:
```
621,216,192
```
Combinations (64 choose 7).
Given any decimal/binary number (it doesn't matter which type), that is guaranteed to be a valid permutation value (we don't need to worry about 128 being passed in for example), how can I calculate which permutation number this is?
IE:
```
whatPermutation(127) = 1;
whatPermutation(191) = 2;
whatPermutation(223) = 3;
```
etc.
Any help would be brilliant, again, I have no idea where to start.
| https://mathoverflow.net/users/9667 | Convert integer to permutation number | Suppose that your 64 bit number is $n=a\_{63} a\_{62} ... a\_0$. Let $i\_7>i\_6>\cdots>i\_1$ be those indices $j$ for which $a\_j=1$. Then
$$ whatPermutation(n)=1+{i\_7\choose 7} +{i\_6\choose 6}+\cdots+{i\_1\choose 1}. $$
For instance, if $n= 00\cdots 110001011011$, then
$$ whatPermutation(n)= 1+{11\choose 7}+{10\choose 6}+{6\choose 5}+{4\choose 4}+{3\choose 3}+ {1\choose 2}+{0\choose 1} $$ $$ = 549. $$
| 17 | https://mathoverflow.net/users/2807 | 42347 | 26,954 |
https://mathoverflow.net/questions/42353 | 2 | Let $V$ be a vector space with dimension greater than 1 over the field $F$ and $Sim = \{(f\in \operatorname{Lin}(V))\mapsto gf(g^{-1}) : g\in \operatorname{GL}(V)\}$, ie $Sim$ is the set of all similarity transforms on $\operatorname{Lin}(V)$. Let $Aut$ be the set of all automorphisms of the $F$-algebra $\operatorname{Lin}(V)$.
What is $Aut\setminus Sim$? In particular, is it empty?
| https://mathoverflow.net/users/nan | Are all automorphisms of Lin(V) given by similarity transforms? | Let $M$ denote the $F$-algebra $\hom(V, V)$. An $F$-algebra homomorphism $M \to \hom(V, V)$ is tantamount to an $M$-module structure on $V$ in the category of $F$-vector spaces. But the category of $M$-modules is equivalent to $Vect\_F$ (there is a Morita equivalence between the categories given by the functor $- \otimes\_F V: Vect\_F \to Mod\_M$). That is to say, every $M$-module is isomorphic to a direct sum of copies of $V$, and in particular, any module structure on $V$ is isomorphic to the standard $M$-module structure on $V$.
So, if $\mu': M \otimes\_F V \to V$ is any module structure, and $\mu: M \otimes\_F V \to V$ denotes the standard structure, there is an isomorphism $g: V \to V$ such that $g \circ \mu' = \mu \circ (M \otimes\_F g)$. Putting all this together, any $F$-algebra *homomorphism* $M \to \hom(V, V)$ is obtained by conjugating the standard homomorphism (the identity) by an isomorphism $g: V \to V$.
Edit: If this proof seems too categorical, there is a proof in slightly alternative language given on page 401 of Bilinear algebra: An Introduction to the algebraic theory of quadratic forms by Szymiczek (Google books [here](http://books.google.com/books?id=CcM8_iiGPxAC&pg=PA401&lpg=PA401&dq=automorphisms+of+a+matrix+algebra&source=bl&ots=Ba5Y2F3AbT&sig=qX8ef4eQVkLoFN-_5CtHGX4s8h8&hl=en&ei=7hm5TJyVOs6s8AauyfH-Dg&sa=X&oi=book_result&ct=result&resnum=7&ved=0CDQQ6AEwBjgK#v=onepage&q=automorphisms%20of%20a%20matrix%20algebra&f=false).
| 2 | https://mathoverflow.net/users/2926 | 42355 | 26,959 |
https://mathoverflow.net/questions/42332 | 3 | I don't know anything about graph theory and I was wondering about something :
If you draw two parallel rows of n points in $\mathbb R^4$ and link each point with all the points in the opposite row except the one right in front of it, and then allow the points to move where ever, what do you get ?
n=2 gives two lines
n=3 gives a hexagon
n=4 gives a cube
n=5 ...
Is there a general object class apart the one described just the way I did ?
| https://mathoverflow.net/users/5375 | On a special kind of graph connectig n point to n points. | These are the [crown graphs](http://en.wikipedia.org/wiki/Crown_graph).
| 4 | https://mathoverflow.net/users/440 | 42358 | 26,960 |
https://mathoverflow.net/questions/42356 | 6 | Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.
Recall that a ring $R$ is said to have regular fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes\_S \frac{S\_\mathfrak{q}}{\mathfrak{q}S\_\mathfrak{q}}=R\otimes\_S \kappa(\mathfrak{q})$ is regular.
Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity of the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?
| https://mathoverflow.net/users/1353 | Does regularity of a prime ideal in the fibre imply regularity of the prime? | First note that it is not true that regularity above $\mathfrak q$ implies regularity of the fibre.
For example, consider the map $\mathbb k[s] \to \mathbb k[t]$ given by $s \mapsto t^2$.
Each prime in $k[t]$ is regular, and so in particular each prime of $k[t]$ above
the prime $(s)$ is reqular. (In fact there is just one of them, namely $(t)$.) On the other hand, the fibre over $(s)$ is the ring $k[t]/(t^2)$, which is a non-regular
local ring.
If you want an arithmetic example instead, consider the inclusion
$\mathbb Z \subset \mathbb Z[i]$, and take the prime $\mathfrak q = (2)$ downstairs,
which has a unique prime $(1+i)$ lying over it. Again, every prime in the PID $\mathbb Z[i]$
is regular, but the fibre $\mathbb Z[i]/2 = \mathbb Z[i]/(1+i)^2$ is a non-regular local
ring.
(The general phenomenon is that a map $X \to Y$ between smooth spaces can have
non-smooth fibres: these occur at the critical points of the map.)
As for your question, your asking if the fact that a map $X\to Y$ has
a smooth fibres over some point implies that the target is smooth at that point.
This is also false in general.
Consider for example the identity map
$k[t^2,t^3] \to k[t^2,t^3]$. The fibre over $(t^2,t^3)$ is just $k$, which is a regular local ring. But $K[t^2,t^3]$ is not regular at $(t^2,t^3).$ (Of course this example is cheap, but its existence foreshadows the existence of many other counterexamples, for example for any etale map $k[t^2,t^3] \to R$, of which there are many non-trivial examples,
as well as my trivial example.)
On the other hand, when the base is Spec $\mathbb Z$, which is very nicely behaved
(regular, Noetherian, excellent, perfect residue fields, ... ), if the map
$X \to $ Spec $\mathbb Z$ is flat and of finite type (e.g. arising from an inclusion
$\mathbb Z \subset R$ of the form you envisage) then having regular (and hence smooth) fibre
at one point implies being smooth in a neighbourhood of that point, and a smooth map over
a regular base has a regular total space --- thus $X$ will be regular in a neighbourhood of the regular fibre. In particular, the points of the regular fibre will themselves be regular in $X$.
In particular, in your special case $\mathbb Z \subset R$, the answer is "yes".
Edit: I should note that in your situation, where $R$ is a subring of a number field,
this "yes" is easily proved directly: one combines the fact that $R/\mathfrak q$ is
regular with the fact that it is *a priori* finite (in cardinality) to see that
it is a product of finite field extensions of $\mathbb Z/\mathfrak q$, and hence
that the completion of $R$ at $\mathfrak q$ is a product of DVRs, and hence that
$R$ is a DVR --- and thus regular --- after localization at each prime above $\mathfrak q$.
The point of the more highbrow explanation above is to indicate how one thinks about
such questions geometrically --- which is normally the easiest way to see what should
be true and what should be false for these kinds of questions.
| 6 | https://mathoverflow.net/users/2874 | 42360 | 26,962 |
https://mathoverflow.net/questions/42329 | 4 | It's clear that the axiom of replacement can be used to construct very large sets, such as
$$
\bigcup\_{i=0}^\infty P^i N,
$$
where $N$ is the natural numbers. I assume that it can be used to construct sets much lower in the Zermelo hierarchy, such as sets of natural numbers, but I don't know of an example. Is there an easy example? (Just to be clear, I mean an example that requires the use of replacement, not just one where you could use replacement if you wanted to.)
I would guess you can cook up an example using Borel determinacy, since that involves games of length $\omega$, but it would be great if there was an even more direct example.
Also, I'd be curious to know for any such examples at what stage they first come along in the constructible universe. $\omega + 1$? The first Church-Kleene ordinal? Some other ordinal I've never heard of?
| https://mathoverflow.net/users/3711 | Replacement and Sets of Natural Numbers | This probably isn't what you are looking for, but one can write down an explicit Diophantine equation for which ZFC proves that it has a solution, but ZFC minus replacement does not (assuming it is consistent). Namely, use Godel encoding and the solution of Hilbert's 10th problem to write down a Diophantine equation whose only solutions encode proofs of the consistency of "ZFC minus replacement." One wants a "naturally occurring" example instead, but it's hard to say what that means.
*(Edit: The following is corrected thanks to Andres' comments)*
For instance, I think the answer to Ricky's formulation in the comments is α = ω+1, but again probably not for the reason you expect. Namely, we can prove in ZFC that ZFC-Repl has a *countable* transitive model. To do this we start from an arbitrary transitive model (such as $V\_{\omega+\omega}$) and apply the downward Lowenheim-Skolem theorem to find a countable submodel. This countable submodel may no longer be transitive, but it is still well-founded, so by Mostowski's collapsing lemma it is isomorphic to an (also countable) transitive model.
Since ZFC-Repl has a countable transitive model, $V\_{\omega+1}$ (being uncountable) cannot be a subset of all such transitive models. But $V\_\omega$ is the set of hereditarily finite sets, which I think have to be in any transitive model since each of them can be uniquely characterized by a formula.
| 1 | https://mathoverflow.net/users/49 | 42364 | 26,965 |
https://mathoverflow.net/questions/42339 | 3 | Let $T$ be a rooted tree with root $r$. Say an ordering $v\_1,\ldots,v\_n$ of the vertices of $T$ is a *search order* if $v\_1=r$ and for all $2 \leq i \leq n$, there is $j < i$ such that $v\_j$ is the parent of $v\_i$. In other words, parents are explored before their children in the order.
For a given search order $v\_1,\ldots,v\_n$, let $w(v\_i)=\max(j:v\_iv\_j \in E(T))-\min(j:v\_jv\_i \in E(T))$. The max is the time the last child of $v\_i$ is explored, and the min is the time the parent of $v\_i$ is explored. Say the width of the order is $\max(w(v\_i):1 \leq i \leq n)$, and say the width of $T$ is the minimum width of an ordering of $T$.
Is anything known about the width? Is it a known concept under another name? Have any theorems been proved about it? Any equivalent characterizations/definitions? Any useful bounds~~, perhaps in terms of the maximum degree of $T$~~?
**Edit**: This is the directed bandwidth, as David Eppstein points out below. I'm still interested in any bounds -- perhaps some upper bound with a simple form, perhaps even with an approximation guarantee?
| https://mathoverflow.net/users/3401 | Theorems about the directed bandwidth of a rooted tree? | What you call width is known more specifically as "directed bandwidth". It's known to be NP-complete even for trees; see [Complexity Results for Bandwidth Minimization, M. R. Garey, R. L. Graham, D. S. Johnson and D. E. Knuth, SIAM Journal on Applied Mathematics Vol. 34, No. 3 (May, 1978), pp. 477-495](http://www.jstor.org/stable/2100947). The hardness of the undirected case for trees (without the parent-child ordering constraint) is detailed in section 8 of the paper, and in section 9 they claim without giving a detailed proof that it's also hard in the directed case.
| 4 | https://mathoverflow.net/users/440 | 42365 | 26,966 |
https://mathoverflow.net/questions/42322 | 0 | Is every action from an amenable group amenable on a unital $C^\*$-algebra?
| https://mathoverflow.net/users/9401 | Is every action from an amenable group amenable on a unital $C^*$-algebra? | Yes it is. It follows from Theorem 3.3 of [1] and the fact that the trivial action of an amenable group on $\mathbb{C}$ is amenable. More modern reference is [2] (in particular Section 4.3).
[1] C. Anantharaman-Delaroche. Systèmes dynamiques non commutatifs et moyennabilité. Math. Ann., 279(2):297–315, 1987.
[2] N. P. Brown and N. Ozawa. C\*-algebras and finite-dimensional approximations, volume 88 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2008.
| 3 | https://mathoverflow.net/users/9942 | 42367 | 26,967 |
https://mathoverflow.net/questions/42366 | 4 | Let $V$ be a vector space (over $\mathbb C$, but I don't think it matters), and $m: V\otimes V \to V$ a "multiplication" that is associative and commutative (but I do not demand that it is unital). Is it possible that $m$ is an *isomorphism* $V\otimes V \overset \sim \to V$? Yes: $V$ can be zero-dimensional, or $V$ can be one-dimensional and $m$ non-zero.
Since $0$ and $1$ are the only finite solutions to $v^2 = v$, any other example must have $\dim V = \infty$. But there are many $\infty$s and many possible maps, and although I am sure that there are some examples, I am having trouble writing one down. Hence:
>
> What is an example of an infinite-dimensional vector space $V$ and an isomorphism $m: V\otimes V \overset\sim\to V$ that is associative and commutative? Or is this impossible?
>
>
>
| https://mathoverflow.net/users/78 | Can a commutative, associative "multiplication" on an infinite-dimensional vector space be an isomorphism? | It is impossible. Let $x$ and $y$ be linearly independent vectors in $V$. Then $x \otimes y \neq y \otimes x$ but $m(x \otimes y) = m(y \otimes x)$.
| 10 | https://mathoverflow.net/users/121 | 42368 | 26,968 |
https://mathoverflow.net/questions/42274 | 1 | This is just an extension of my previous question [Tightness of probabilty distributions](https://mathoverflow.net/questions/41259/tightness-of-probabilty-distributions)
Let $\mathcal{P}(\mathbb{N})$ be the set of all PMF's on $\mathbb{N}=\{1,2,\dots \}$. Let $E$ be a convex subset of $\mathcal{P}(\mathbb{N})$ and $Q\notin E$. Let $\alpha>1$ and $\beta=\frac{1}{\alpha}$.
Let us suppose that $s:=\displaystyle \sup\_{P' \in E'}\sum P'(x)^{\beta}Q'(x)^{1-\beta} >0$, where $P'(x)=\frac{P(x)^{\alpha}}{\sum P(x)^{\alpha}}$,and $E'=\{P':P\in E\}$.
My problem is to find a convex $E$ such that $E'$ is closed (with respect to the total variation metric) but $s$ is not attained in $E'$.
By making use of the example given by Bill Bradley in the above mentioned thread, I have the following very close counterexample.
Let $Q=(1,0,0,0,...)$ and let $R\_n$, for $n=2,3,...,$ as $R\_n(1)=\frac{1}{2}-\frac{1}{n}, R\_n(n)=\frac{1}{2}+\frac{1}{n}$ and $R\_n(x)=0$ for all other values and let $E=$ Convex hull of $\{R\_n\}$.
As you can see $s=1$ and is not attained in $E'$ (actually attained at $(1, 0, 0,\dots)$ which is not in $E'$.) This is what I would be happy with.
But the problem here is that $E'$ is not closed also, because if we take $P\_n=\frac{1}{n} \sum\_{i=2}^{n+1}R\_i$, then $P\_n'\to (1, 0, 0, \dots)\notin E'$. But I want $E'$ to be closed.
Can one somehow change the things here a bit and get a counterexample ( i.e., $E$ convex, $E'$ closed but $s$ is not attained)?
Or may the result be true?
| https://mathoverflow.net/users/7699 | Showing non-attainment of supremum | I think that in your assumption, the supremum is actually attained.
Consider the set
$$\hat E:=\{tp \ :\ t\geq0 \ , \quad p\in E\ \} \cap\bar B(0,1;\ell^\alpha).$$
Since $E$ is convex, $\hat E$ is convex too (here $\bar B(0,1;\ell^\alpha)$ denotes the closed unit ball of the sequence space $\ell^\alpha$).
Moreover, we are going to show that the assumption that $E'$ is closed in $\ell^1,$ implies that $\hat E$ is a closed bounded subset of the reflexive space $\ell^\alpha$, thus weakly compact. Indeed, let $u$ belong to the $\ell^\alpha$ norm closure of $\hat E.$ So, there exists a sequence $t\_j\geq0,$ and a sequence $p\_j\in E,$ such that $u \_j:=t \_j\\ p\_j$ converges to $u$ in $\ell^\alpha.$ If $u=0$ then $u\in \hat E$ and there's nothing to prove; otherwise we have (for large $j$) that $p \_j / \| p \_j \| \_\alpha = u \_j / \| u \_j \| \_\alpha $ which converges in $\ell^\alpha$ to $u/\|u\| \_\alpha.$ Hence $p'\_j:=\big(p \_j/\|p \_j \| \_\alpha\big)^\alpha$ converges in $\ell^1$ to $\big(u/\|u \| \_\alpha\big)^\alpha, $ showing that the latter belongs to $E'$, which is $\|\cdot\| \_1$-closed. This implies that for some $p\in E,$ $u$ has the form $\frac{\|u\| \_\alpha}{\|p\| \_\alpha} p,$ so is in $\hat E$.
Now consider $v:=\big(q/\|q\| \_\alpha\big)^{\alpha-1}.$ It is a norm-one element of the dual space $\ell^{\alpha'},$ and your optimization problem can be rewritten as
$$s:=\sup \_{p\in E}\big(\frac{p}{\|p\| \_\alpha}\cdot v \big) =\sup \_{u\in\hat E} (u\cdot v),$$
that is attained by the weak compactness of $\hat E.$
| 2 | https://mathoverflow.net/users/6101 | 42376 | 26,973 |
https://mathoverflow.net/questions/42161 | 19 | **Edit:** I think LMO is correct. Massuyeau has a nice explanation [here.](http://ldtopology.wordpress.com/2010/10/14/a-problem-with-lmo/)
**Edit:** Renaud Gauthier has retracted the claim of an error in the foundations of the LMO construction, and has withdrawn both preprints from arXiv.
Original post follows:
---
In two papers posted to the arXiv in the past few days, Renaud Gauthier claims to have discovered an error in the definition of the framed Kontsevich integral used in the construction of the LMO invariant. I have no reason to doubt him. I looked at these papers some years back and recall that something funny was going on with the normalization under handle-slides. I got the wrong multiple of the normalization factor $\nu$, just as Gauthier does. Gauthier fixes the normalization so that it works, but then remarks that subsequent results depending on this construction need to be carefully checked.
My question is whether anyone knows of results that use the fine details of the definition of the framed Kontsevich integral (or LMO invariant or Aarhus integral) which are now thrown into doubt because of this error.
**Edit:** Here are links to the papers.
[On the foundations of the LMO invariant](http://arxiv.org/abs/1010.2422)
[On the LMO Invariant, the Wheeling Theorem, and the Aarhus Integral](http://arxiv.org/abs/1010.2559)
**Edit 2:**
Moskovich has started a blog post on this. Thanks to Ryan Budney for pointing this out.
[A problem with LMO?](http://ldtopology.wordpress.com/2010/10/14/a-problem-with-lmo/)
| https://mathoverflow.net/users/9417 | Propagation of an error in the LMO invariant? (Revision: I don't think LMO is wrong!) | Having been a part of the LMO story from its beginning, and having read and checked all relevant papers carefully at the time, and having taken part in many cross-checks that the LMO invariant passed (normalization-compatibility with Reshetikhin-Turaev, various explicit computations), and having consulted on email with my collaborators at the time, and having superficially read through Gauthier, my informed guess is that in this particular case of inconsistency the first place to look for a problem is in Gauthier, not in LMO.
| 17 | https://mathoverflow.net/users/8899 | 42380 | 26,977 |
https://mathoverflow.net/questions/42310 | 45 | I have never studied any measure theory, so apologise in advance, if my question is easy:
Let $X$ be a measure space. How can I decide whether $L^2(X)$ is separable?
In reality, I am interested in Borel sets on a locally compact space $X$. I can also assume that the support of the measure is $X$, if it helps...
I cannot even decide at the moment for which locally compact groups $G$ with Haar measure, $L^2(G)$ is separable...
| https://mathoverflow.net/users/5301 | When is $L^2(X)$ separable? | Without loss of generality we can assume that the support of the measure equals $X$
(i.e., the measure is faithful),
because we can always pass to the subspace defined by the support of the measure.
The space $^2(X)$ is independent of the choice of a faithful measure and depends only
on the underlying [enhanced measurable space](/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820) of $X$, i.e., measurable and negligible subsets of $X$.
There is a complete classification of measurable spaces up to isomorphism.
Every measurable space canonically splits as a disjoint union of its
ergodic subspaces, i.e., measurable spaces that do not admit measures
invariant under all automorphisms.
Ergodic measurable spaces in their turn can be characterized using
two cardinal invariants $(m,n)$, where either $m=0$ or both $m≥ℵ\_0$ and $n≥ℵ\_0$.
The measurable space represented by $(m,n)$ is the disjoint
union of $n$ copies of $2^m$, where $2=\{0,1\}$ is a measurable
space consisting of two atoms and $2^m$ denotes the product of $m$ copies of 2.
The case $m=0$ gives atomic measurable spaces (disjoint unions of points),
whereas $m=ℵ\_0$ gives disjoint unions of real lines (alias standard Borel spaces).
Thus isomorphism classes of measurable spaces are in bijection
with functions M: Card'→Card, where Card denotes the class of cardinals
and Card' denotes the subclass of Card consisting of infinite cardinals and 0.
Additionally, if $m>0$, then $M(m)$ must belong to Card'.
The Banach space $^p(X)$ ($1≤p<∞$) is separable if and only if
$M(0)$ and $M(ℵ\_0)$ are at most countable and $M(m)=0$ for other $m$.
Thus there are two families of measurable spaces whose $^p$-spaces are separable:
1. Finite or countable disjoint unions of points;
2. The disjoint union of the above and the standard Borel space.
Equivalent reformulations of the above condition assuming $M(m)=0$ for $m>ℵ\_0$:
1. $^p(X)$ is separable if and only if $X$ admits a faithful finite measure.
2. $^p(X)$ is separable if and only if $X$ admits a faithful $σ$-finite measure.
3. $^p(X)$ is separable if and only if every (semifinite) measure on $X$ is $σ$-finite.
The underlying measurable space of a locally compact group $G$ satisfies the above conditions if and only if $G$ is second countable as a topological space.
The underlying measurable space of a paracompact Hausdorff smooth manifold $M$
satisfies the above conditions if and only if $M$ is second countable, i.e.,
the number of its connected components is finite or countable.
More information on this subject can be found in this answer:
[Is there an introduction to probability theory from a structuralist/categorical perspective?](https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820)
Bruckner, Bruckner, and Thomson discuss separability of $^p$-spaces in Section 13.4 of their textbook Real Analysis:
<http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf>
| 48 | https://mathoverflow.net/users/402 | 42383 | 26,979 |
https://mathoverflow.net/questions/42371 | 6 | The adjoint of the exterior derivarive is defined by
$\delta:=(-1)^k\ast^{-1}d\ast$,
but I need a way which avoids the Hodge $\ast$ operator.
>
> Is there another definition?
>
>
>
For example, for positive definite metric there is the alternative of defining it by
$\int\_M \langle d\alpha,\beta\rangle\_{k+1} d\_{vol} = \int\_M\langle\alpha,\delta\beta\rangle\_k d\_{vol}$
but I don't know if we can apply the Riesz representation theorem to differential forms on semi-Riemannian manifolds, since they do not form a Hilbert space, because the metric is not positive definite.
Can you provide a link or a reference?
| https://mathoverflow.net/users/10095 | Can the adjoint of the exterior derivative in semi-Riemannian geometry be defined without the Hodge * operator? | On any Riemannian manifold, any differential operator has a formal adjoint.
This has nothing to do with functional analysis, it is a pure calculus fact that uses little more than Stokes theorem.
This is because taking adjoints is linear, preserves (or rather reverses)
composition and commutes with multiplication by real-valued functions.
It is also clear how the adjoint of an order 0 operator (i.e. vector bundle homomorphisms) has to be formed.
The last and most important step is to construct the adjoint of a vector field on functions. Once this is done,
the existence of an adjoint is proven, because the above operators generate the algebra of differential operator. This last argument also generalizes
to
differential operators on vector bundles. The formula below shows you also - for free - that the adjoint is a differential operator, which is
something you have to work on if you define adjoints on the (pre) Hilbert-space level.
Let $X$ be a vector field, viewed as a differential operator on compactly supported function. To compute the adjoint of $X$, start with the formula
$\int\_M Lie\_X \alpha = \int\_M d \iota\_X \alpha + \int\_M \iota\_x d \alpha =0$ for any $n$-form $\alpha$ and any vector field $X$. The first equality is the Cartan infinitesimal homotopy formula. The second summand is
zero for degree reasons, the first by Stokes. Now let $\omega$ the volume form and $f,g$ two functions. Therefore
$0= \int\_M Lie (fg\omega)= \int\_M (Xf) g \omega + \int\_M f f (Xg) \omega+ \int\_M fg \cdot div (X) \omega$ by the definition of the divergence.
Hence the adjoint of $X$ is $-X - div (X)$.
So: Once you know how define the volume form on a semi-Riemannian manifold, you know that formal adjoints exists. So you can say that
taking adjoints is a canonical operation on the algebra of differential operators, depending only on the volume form.
The argument above can be easily made into a formula for the adjoint once your operator is given in local coordinates. Whether this formula is useful or enlightening is a matter of taste.
For the exterior derivative, that formula can be written in terms of the Hodge star. This is a (rather simple) theorem. I do not know whether there is any other formula for $d^\*$.
Pseudodifferntial operators also have adjoint, which is proven in a different fashion.
It is more difficult to show that the formal adjoint agrees with the Hilbert space adjoint (and I am not really qualified to say much on it and refer you to a book or a real expert)
| 10 | https://mathoverflow.net/users/9928 | 42384 | 26,980 |
https://mathoverflow.net/questions/42391 | 10 | Dear MO Community,
Let $N$ be a prime, and let $X\_0(N)$ be the classical modular curve over $\mathbb{Q}$. We know ([1]) that, if there are noncuspidal points in $X\_0(N)(\mathbb{Q})$, then $N \in$ {${ \mbox{primes } \leq 19} $} $ \cup $ {37,43,67,163}.
The basic question of this post is:
>
> Are there similar lists of primes when $\mathbb{Q}$ is replaced by a number field $K$? That is, if we fix a general number field $K$, can we determine the primes $N$ for which $X\_0(N)(K)$ has noncuspidal points?
>
>
>
Perhaps in this generality the question is hard, so suppose we restrict from now on to imaginary quadratic $K$. Then [1] gives an approach to the question, but with the following snags:
1. We need to construct an "optimal" quotient of $J\_0(N)$, call it $A$, such that $A(K)$ has Mordell-Weil rank 0;
2. We must restrict ourselves to primes $N$ which are inert in $K$.
Actually, I don't think (1) is a big problem; provided $N \> 48h(K)^3 + 1$, we can take $A = \widetilde{J}$, the Eisenstein quotient. [Speculation : if we took the "winding quotient" instead, maybe we can lower that bound...]
When $N$ splits, then we can construct "CM points" on $X\_0(N)(\mathbb{C})$, but usually they will not be defined over $K$, and even if they are, there will only be a handful of them.
>
> Question: For which $N$ that splits in $K$ do we have points on $X\_0(N)(K)$ that are neither cuspidal nor CM? Is there a way to systematically find these points?
>
>
>
By "systematically", I guess I mean something like the 'isogeny character' approach of [1], where the hunt for the $N$s comes down to when certain congruences are satisfied mod $N$.
Many thanks.
[1]: Mazur, B. "Rational isogenies of prime degree", Inventiones Mathematicae, 1978
| https://mathoverflow.net/users/5744 | Rational Isogenies of Prime Degree | I'm glad to have stumbled on your question -- I'm actually working on something along the lines of what you're asking about now with Eric Larson. If what we think we've proven is true (and don't quote this yet, since we're not yet even done with the write-up,) then in fact you can say something stronger: you cannot have isogenies of order p for any prime p sufficiently large (where "sufficiently large" depends on K), as long as K does not contain the class field of an imaginary quadratic extension in which p splits (this exactly gets rid of primes which are isogenies of a CM curve defined and with CM over K). So this would mean that if K is quadratic imaginary, there can be no p-isogenies for *any* p sufficiently large unless K is one of the finitely many quadratic imaginary fields of class number one (in which case, if our arguments work, this would mean that all but finitely many possible prime degrees of isogeny p split over K.)
If you're interested in what happens to primes that split, you can ask whether, more generally, for *any* number field K, there are only finitely many primes that can be isogenies for a curve over K without CM. This seems harder to prove, and as far as we know, is an open problem (to prove it, you'd need to have a good way of distinguishing CM from non-CM curves, or carefully counting points on $X\_0(N)$). But hypothetically this is also true: it fits into the general framework of Serre's hypothesis that "exceptional primes" of curves without CM are bounded by a constant depending only on $K$.
| 11 | https://mathoverflow.net/users/7108 | 42394 | 26,984 |
https://mathoverflow.net/questions/42298 | 10 | I have always been wondering
>
> why the term "model" is used by mathematicians (especially in mathematical logic) in a conceptually different (even opposite) way than it is used by other scientists, e.g. physicists, biologists, chemists, economists etc.
> And when has this terminology first arisen?
>
>
>
In essentially all of natural sciences we have a "world" (our physical world, a living being, a market...) which hosts specific instances of certain "phenomena" (that involve "objects" of some kind, and "relations" between them), and the goal of the theory is to provide a *model* that describes that phenomena. In this case, a *model* is understood to be some kind of conceptual construction that *abstracts* the common relevent properties of several instances of the phenomenon in question and puts them in a rational (mathematical or not) framework that enables us to draw consequences and predictions about the phenomenon itself. Think e.g. of the "Standard Model" of particle physics.
Also mathematics has a "world", populated by objects (e.g. groups or ordered fields), between which some relations hold. The goal of a mathematical "theory" (at least from the perspective of formal theories in the sense of mathematical logic) is to provide a simple "model" (mind the use of quotation marks) that describes all the instances of certain "phenomena", and it accomplishes this task by a list of *axioms* (e.g. the axioms of group theory) that incapsulate the relevant properties of the objects in question (e.g. being a group).
It is possible that new "species" of the same kind of objects are found, that is, they verify the axioms, i.e. they fall under the description by the same "model".
A natural scientist would call such an object an incarnation (manifestation, realization, example, explicitation, instance...) of the "model" provided by the axioms. Mathematicians, on the contrary, call it a *model* (here in the tecnical sense, hence without quotation marks) for the axioms.
Isn't it strange?
(Edit: there is also another way the word "model" is used in maths, as in "Weierstrass model" or "Néron model", in which cases it is essentially sinonimous with "normal form". This latter use of the word seems to me more consistent with the general natural sciences use)
| https://mathoverflow.net/users/4721 | The use of the word "model" in Mathematical Logic vs the same word in Natural Sciences | * yes, it is strange, because one expects words to have similar meaning even in different contexts (but as others have noted, whether this is strange or not, there are all sorts of words that have multiple meanings, and even contradictory ones).
* yes, I agree in your assessment of how model has two, not contradictory, just oppositely directed meanings. In common language, natural sciences, and statistics, a model is an abstraction, where incidental details are removed, leaving just the bare bones abstract thing of what you care about (a model airplane, a model of population dispersal, a regression model predicting a value from a small set of variables). In mathematical logic, it is the other way around; a model is an -example- of the syntactically presented axioms (the permutations of 3 items is a model of the group axioms, a valuation (an assignment of boolean values) is a model for a propositional statement.
In short, in mathematical logic, there is the abstract theory that has a model as an example, the model fits the theory. In the rest, the model -is- the theory and the phenomena of the world are the examples (which the theory/model are trying to fit).
When this divergence of use started and by who, I can't say. But the logical use is certainly later. Who first started using the concept (but not necessarily the usage) in logic? Was it Goedel? How about Hilbert in 'Foundations of Geometry' where he shows independence of individual axioms by giving different models satisfying the rest (i don't know what terms he uses in the original or if they were common usage at that time for the concept).
| 0 | https://mathoverflow.net/users/3237 | 42397 | 26,985 |
https://mathoverflow.net/questions/42393 | 5 | Pressing the envelope, presumably the best scenario would be a simple proof of the Prime Number Theorem. After all, Wilson’s Theorem gives a necessary and sufficient condition, in terms of the Gamma Function, for a number to be a prime, and Stirling’s Formula specifies the asymptotic behaviour of the Gamma Function.
| https://mathoverflow.net/users/8649 | Has Stirling’s Formula ever been applied, with interesting consequence, to Wilson’s Theorem? | Using Robbins' [1] form of Stirling's formula,
$$\sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n+1))< n!< \sqrt{2\pi}n^{n+1/2}\exp(-n+1/(12n))$$
we get
$$\left\lceil\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-11))\right\rceil$$
$$\le (n-1)!\le$$
$$\left\lfloor\sqrt{2\pi}(n-1)^{n-1/2}\exp(-n-1+1/(12n-12))\right\rfloor$$
which is accurate enough to distinguish prime from composite for $n\le8$. For larger numbers, the error bound is too large.
---
This can be extended further using a modification of Wilson's theorem: for n > 9,
$$\lfloor n/2\rfloor!\equiv0\pmod n$$
if and only if n is composite. This allows testing 10 through 15, plus (with some cleverness) 17.
With tighter explicit bounds and high-precision evaluation, it might be possible to test as high as 100 with related methods: direct evaluation up to 25 and the 'divide by 4' variant of the above for n > 25.
This is not so much 'using a cannon to swat a fly' (using methods more powerful than needed) as it is 'using the space station to swat a fly': the methods must be extremely powerful and accurate to do very little.
---
[1] H. Robbins, "A Remark on Stirling's Formula." *The American Mathematical Monthly* **62** (1955), pp. 26-29.
| 36 | https://mathoverflow.net/users/6043 | 42398 | 26,986 |
https://mathoverflow.net/questions/42401 | 0 | My question is most precisely stated in the title. As an example, if we consider base 10, and k=4, then I am asking, is it possible to have a sequence of length 10^4 + 3, such that each 4 digit number appear exactly once in a consecutive block? This is related to finding a Hamiltonian cycle in a b-regular graph. But one would also like to construct the actual cycle.
| https://mathoverflow.net/users/4923 | how to find a sequence of digits in base b such that each consecutive block of size k appears exactly once? | Google for "generating De Bruijn sequences".
| 5 | https://mathoverflow.net/users/1409 | 42403 | 26,990 |
https://mathoverflow.net/questions/41816 | 13 | Given three points $a,b,c$ in a (geodesic) metric space $X$, one defines a *comparison angle* $\angle(a,b,c)$ by the cosine law:
$$
\angle(a,b,c) = \arccos \frac{|ab|^2 + |ac|^2 - |bc|^2}{2\cdot|ab|\cdot|ac|}
$$
where $|ab|$, $|ac|$ and $|bc|$ are distances in $X$. In other words, $\angle(a,b,c)$ is the angle of a planar triangle with the same side lengths as the triangle $abc$ in $X$.
Given two paths $\xi,\eta:[0,\varepsilon)\to X$ starting at the same point $p=\xi(0)=\eta(0)$, define a *metric angle* $\angle(\xi,\eta)$ by
$$
\angle(\xi,\eta) = \lim\_{t,t'\to 0} \angle(p,\xi(t),\eta(t'))
$$
if the limit exists.
If $X$ is a smooth manifold with a $C^2$ Riemannian metric (which defines a geodesic distance in the usual way) and $\xi,\eta$ are $C^1$ regular curves, then a curvature comparison argument shows that the metric angle is well-defined and equals the Riemannian angle between the velocity vectors $\dot\xi(0)$ and $\dot\eta(0)$. But what if the Riemannian metric is only $C^1$, or even $C^0$? More precisely:
* Let $X$ be a smooth manifold with a $C^1$ Riemannian metric. Does the metric angle exist for every pair of $C^1$ regular curves $\xi,\eta$?
* If not, does every $C^1$ regular curve have a well-defined angle with itself? (Of course, this angle is zero if it exists).
* If the answer is affirmative, what about $C^0$ Riemannian metrics?
**Remark.**
The limit in the definition trivially exists (even in the $C^0$ case) if the ratio $t/t'$ is bounded away from zero and infinity as $t$ and $t'$ go to zero. The trouble is with very "thin" triangles $p,\xi(t),\eta(t')$ where $t\ll t'$ or vise versa.
| https://mathoverflow.net/users/4354 | Metric angles in Riemannian manifolds of low regularity | To Deane Yang, concerning the smoothness of geodesics. These issues have been studied by Hartman
and his coauthors more than 40 years ago. Together with Calabi he claimed to prove
that in a $C^{\alpha}$ continuous metric all geodesics are uniformly $C^{1,\alpha}$.
(I do not have mathscinet access now, but I think the paper was called
"On the smoothness of isometries", or something like that). The paper is not quite correct,
the computations prove that geodesics are only $C^{1,\alpha /2}$
(this was observed at some point by Reshetnyak). The optimal regularity appears in my paper with Asli Yaman
"On Hoelder continuous Riemannian and Finsler metrics" (I apologize for citing my own work).
The starting point of all these computations is a metric characterization of $C^{1,\alpha}$
curves in metric terms (in terms of the distance of the midpoint of a pair of points on the curve and the image of the corresponding midpoint of the interval of the definition). It is contained in the work of Hartman, but is probably much older.
To Sergei Ivanov: The questions of extendibility of geodesics have also been studied by Hartman. I again do not remember the correct reference, but can check it on Monday. I think he constructs example of even $C^{1,\alpha}$ Riemannian metrics, for any $\alpha <1$, that do not have extendible geodesics. (For $\alpha =1$, the metric has curvature bounded from both sides).
Now the main point. I am very sorry and have to apologize for my previous short posting.
I slightly misunderstood your question. But I hope that my answer was still
correct. Is the following more detailed explanation correct?
1. We assume that $t'>>t$ and want to prove that
$d(\xi (t), \eta (t') )=
d(\eta (t'), p) - cos (\alpha ) t + o(t)$. Here $o(t) /t \to 0$ and $\alpha$ is
the "usual Euclidean " angle between $\xi$ and $\eta$.
1. The statement about the uniform "smoothness" of geodesics, show that it suffices to prove the claim in the case, when $\eta$ and $\xi$ are geodesics (since the "usual" angles
between $\eta$ and any geodesic between $\eta (t')$ and $p$ goes to $0$).
2. Going on the geodesic $\eta$ first to the point $\eta (rt)$, with large, but fixed $r$, and then to $\xi (t)$, one obtains the right upper bound for $d (\eta (t'),p)$.
3. In a "general" metric space it would be now difficult to obtain the right lower bound
of $d(\eta (t'),p)$. One could do it, if one knew that the geodesic $\eta$ was extendible beyond $p$. Then one could use a traingle inequality and the distance from $\xi (t)$ and $\eta (-rt)$. I hope I understood your question correctly and this was what you had in mind.
4. Here one can use another trick to obtain a lower bound. Observe namely, that from the
"smoothness" of geodesics the "usual" angle between $\xi$ and any geodesic from
$\eta (t') $ to $\xi (t)$ is very close to $\alpha$ (goes to $\alpha$ , with $t' \to 0$).
Now we just look at the triangle $\xi (t),p, \eta (t')$ from $\xi (t)$ and not from $p$.
The inequality obtained in point 3. above gives us the right upper bound of $d(\eta (t'),p)$.Just a side remark. Your question is very closely related to the question about the
correctness of the first formula of variation. I have thought about that question
and proved that the formula holds true in Riemannian manifolds with Hoelder continuous metrics in "Differentiation in metric spaces". I just hope that the above proof is correct. Other wise the corresponding statement in my paper is wrong as well.
| 8 | https://mathoverflow.net/users/10109 | 42422 | 27,003 |
https://mathoverflow.net/questions/42404 | 2 | Assuming that one needs $k$ quantifiers to express that a graph contains an $k$-cycle, $\lfloor n/2 \rfloor$ counting quantifiers suffice to express that a graph is an $n$-cycle:
*G has exactly $n$ nodes, each node has exactly 2 neighbors, and G doesn't contain a 3-cycle, a 4-cycle, ... and an $\lfloor n/2 \rfloor$-cycle.*
>
> Question: Can be shown, that one needs
> **at least** $\lfloor n/2 \rfloor$ counting
> quantifiers to express that a graph is
> an $n$-cylce? And how?
>
>
>
| https://mathoverflow.net/users/2672 | Expressing a graph property with counting quantifiers | The previous answer showed that quantifier rank $\log\_2(k) \pm 1$ is sufficient and necessary for expressing "A is a $k$-cycle". This implies a lower bound of at least $\log\_2(k)$ quantifiers, but not an upper bound since quantifier rank counts nesting-depth of quantifiers and not the number of quantifiers. However, it is possible to express "A is a $k$-cycle" with $O(\log k)$ quantifiers. Here is an explicit expression:
You start saying that every vertex has degree exactly two. These are $4$ quantifiers. Next we need to say that every two vertices $x$ and $y$ of the graph are connected by a path of length at most $k$. Let $P\_k(x,y)$ denote this predicate. The naive way to express $P\_k(x,y)$ requires $k$ quantifiers. A less naive way could use divide-and-conquer but, if we do it directly without any additional trick, the result would still be $k$ quantifiers. The final solution will be to add a trick to the divide and conquer solution.
The divide and conquer solution goes as follows: you quantify the middle vertex $z$ and then recurse on the halfs $x$-$z$ and $z$-$y$ by requiring that both $P\_{k/2}(x,z)$ and $P\_{k/2}(z,y)$ hold. An easy computation shows that this method produces $k$ quantifiers. To improve and achieve $O(\log k)$ quantifiers we make use of a "reusing trick" that those familiar with computational complexity will recognize from the proof that QBF is PSPACE-complete: you quantify the middle vertex $z$ and then say: for every $u$ and $v$, if $u = x$ and $v = z$, or $u = z$ and $v = y$, then $P\_{k/2}(u,v)$. This recursive formula expresses the same as before but now $P\_{k/2}$ is used recursively only once. This gives the $O(\log k)$ quantifiers that we want.
Finally, in order to complete the formula expressing "A is a $k$-cycle" we need to say that every vertex $x$ is connected by a path of length exactly $k$ to itself. This we can do with $O(\log k)$ quantifiers using the same trick that we used for $P\_k$.
(Note: all this is much easier to argue if $k$ is an exact power of $2$ but can be made to work for any $k$).
| 2 | https://mathoverflow.net/users/9355 | 42423 | 27,004 |
https://mathoverflow.net/questions/42429 | 2 | In the definition of a metric space, replace the triangle inequality by the weaker inequality
*d* (*x*, *z*) ≤ *C* max {*d* (*x*, *y*), *d* (*y*, *z*)},
where C is a positive constant (depending on the "metric", but not on the points *x*, *y*, *z*). Had structures like this ever been studied?
One can associate a more or less natural topology to this "metric", calling a set *open* if every point belonging to the set has a ball of positive radius, centered at this point and contained in the set. But I cannot say much on this topology. For instance, it is not obvious (and may be not true) that a ball is an open set. Neither could I prove that this topology is Hausdorff.
Any information, reference, etc. would be welcome.
| https://mathoverflow.net/users/9937 | weak metric space | Yes, they were introduced in valuation theory by Emil Artin and remain present in many contemporary treatments, including mine: see
[http://alpha.math.uga.edu/~pete/8410Chapter1.pdf](http://alpha.math.uga.edu/%7Epete/8410Chapter1.pdf)
especially Section 1.2 and
[http://alpha.math.uga.edu/~pete/8410Chapter2.pdf](http://alpha.math.uga.edu/%7Epete/8410Chapter2.pdf)
[**Added**: as RW has justly pointed out, my answer here makes sense in the context of valuation theory only. The procedure that I give from passing from a "weak metric" to a metric is not going to work in general, I think, but only in the presence of some additional algebraic structure. If I were the OP, I might not choose this as the accepted answer, or at least not yet.]
The basic idea here is to consider two such guys equivalent if one can be obtained from the other by the operation of raising $d$ to some positive real number power $\alpha$. In such a way, one can make the constant $C \leq 2$ in which case one gets an actual metric. The topology one gets in this way is easily seen to be independent of the choice of $\alpha$.
As it happens, when writing up these notes for a course I taught last spring I also thought a little bit about trying to define "balls" with respect to such a weak metric (i.e., without first renormalizing). I didn't get anywhere with this either.
Finally, I should say that in valuation theory at least, these "weak metrics" (in the valuation theoretic context I called them "Artin absolute values" and then after the course was over decided to change to the terminology to just "absolute values", while retaining the word "norm" for such a guy which was actually a metric) come up as a useful tool but are not really studied on their own or in any deep way: I have yet to see a text on valuation theory where weak norms appear after page 20 or so. Whether they may have wider applicability in some other context, I couldn't say...
| 3 | https://mathoverflow.net/users/1149 | 42435 | 27,008 |
https://mathoverflow.net/questions/41305 | 15 | (Grothendieck) topoi are left-exact reflective subcategories of a category of presheaves. An important class of quasi-topoi (see: <http://ncatlab.org/nlab/show/quasitopos>) arise as the category of concrete sheaves on a concrete site. Concrete sheaves are those sheaves $X$ such that the induced map $Hom(C,X) \to Hom(\underline{C},\underline{X})$ is injective for all objects $C$, where $\underline{C}$ is the underlying set of $C$ and $\underline{X}$ is the value of $X$ on the terminal object. Concrete sheaves are a reflective subcategory of all sheaves. Concrete sheaves are a particularly nice example of a quasi-topos as the resulting quasi-topos is both complete and cocomplete. My question is, is there a way to represent quasi-topoi (or nice ones) as reflective subcategories of a Grothendieck topos (with some condition on the reflector)? (Of course, for this, you'd need the quasi-topos to be complete, since reflective subcategories of complete categories are again complete). More generally, is there some theorem saying that a category is a (possibly non-complete) quasi-topos if and only if it can be embedded into a topos such that the embedding has such-and-such property?
| https://mathoverflow.net/users/4528 | Classification of Quasi-topoi | This is only a partial answer, and you may know it already since I mentioned it recently at the nForum, but for completeness, here it is. Theorem C2.2.13 in *Sketches of an Elephant* shows that the following are equivalent for a category C:
1. C is the separated objects for a Lawvere-Tierney topology on a Grothendieck topos (hence reflective in that topos)
2. C is the category of sheaves for one Grothendieck topology on a small category which are also separated for a second Grothendieck topology
3. C is a locally small, cocomplete quasitopos with a strong-generating set
4. C is locally presentable, locally cartesian closed, and every strong equivalence relation is effective
A category of this sort is called a *Grothendieck quasitopos*. The third characterization seems most similar to what you're looking for. I doubt you can get away without some generating-set condition, since it seems very unlikely that the (complete, cocomplete, locally small) quasitopos of pseudotopological spaces (for example) can be reflectively embedded in a topos.
What I don't know is whether one can put conditions directly on a reflective subcategory of a topos, analogous to left-exactness of the reflector, to guarantee that it is of this form. The reflector for separated objects preserves finite products and monics, but I have no idea whether that would be sufficient as a characterization.
| 8 | https://mathoverflow.net/users/49 | 42441 | 27,012 |
https://mathoverflow.net/questions/42424 | 6 | Let $M$ be an even dimensional smooth manifold.
I want to find an example $M$ satisfying the following conditions,
1. $M$ admits a Kahler structure.
2. $\omega$ is a symplectic form on $M$.
3. There is no Kahler structure $(M,\omega',J)$ such that $[\omega']=[\omega] \in H^2(M;\mathbb{R})$
(I mean, want to find an example $M$ such that "Kahler cone $\neq$ symplectic cone" with non-empty Kahler cone.)
Thank you in advance.
| https://mathoverflow.net/users/11705 | Examples of symplectic non-Kahler classes. | One sort of example arises from the fact that if one starts with a Kahler form $\omega$ (which represents a class of type (1,1) in the Hodge decomposition by definition of a Kahler form), then if $\phi$ is the real part of any closed form of Hodge type (2,0), $\omega+\phi$ will still be a symplectic form (it tames the complex structure $J$), but won't any longer be Kahler, at least if one regards the complex structure as being fixed--in principle there could be another complex structure with respect to which the form is Kahler. Thus you get examples this way on any Kahler manifold with $H^{2,0}\neq 0$. In the case of Kahler surfaces (symplectic $4$-manifolds) this is equivalent to the geometric genus being nonzero (or, in language more familiar to topologists, $b^+>1$).
In fact, a paper of Draghici (see the last paper listed on [this page](http://www2.fiu.edu/%7Edraghici/research/resint.html)) shows essentially that, on a minimal Kahler surface of general type, if one starts at $\omega$ and goes out sufficiently far on the ray in the direction of $\phi$, then one eventually gets to classes that aren't represented by Kahler forms with respect to *any* complex structure, not just the original one.
There's a different sort of example in a [paper](https://arxiv.org/abs/math/0601540) of T.-J. Li and myself: we observe
that if the Kahler surface $(M,\omega,J)$ contains any smooth J-complex curve (real 2D surface) $C$ of negative self-intersection other than a sphere of square $-1$, then one can obtain symplectic forms in the class $[\omega\_t]=[\omega]+tPD[C]$ (where PD means Poincare dual) for a range of values of $t$ including some large enough that $[\omega\_t]$ evaluates negatively on $C$. So the resulting symplectic form $\omega\_t$ can't even be tamed by $J$. Again, in general $\omega\_t$ might in principle be Kahler after deforming $J$ to some different complex strucutre, but Section 4.1 of that paper gives an example where this is carried out on a rigid surface (i.e. one admitting no deformations of the complex structure).
| 3 | https://mathoverflow.net/users/424 | 42443 | 27,014 |
https://mathoverflow.net/questions/42447 | 1 | Let $\mathbb N\_{n} = \{1,2,\cdots,n\}$.
Let $S$ be of cardinality $n$ where elements of $S$ are integers from $\mathbb N\_{n}$ and at least one element of $S$ is repeated (That is at least one integer from $\mathbb N\_{n}$ is skipped. One can easily find a set $S$ with the property that:
$\displaystyle \sum\_{j \in S}j^{i} = \displaystyle \sum\_{j \in \mathbb N\_{n}}j^{i}$
when $i = 1$. (Example: $n=4$, $S=\{1,1,4,4\}$ has sum $10$, the same as sum of first n consecutive integers)
How about for $i \ge 2$? It is not obvious that higher power sum sets exist due to constraint in the cardinality of $S$ and $\mathbb N\_{n}$. One cannot deny it either? Is there a easy way to tackle some sumset questions?
For $i=2$ it is related to quadratic forms and integer norms. In an integer coordinate system, how many ways can a given integer norm occur when the coordinates are bounded?
| https://mathoverflow.net/users/10035 | Subset higher power sum question (related to quadratic forms) | It's a little easier to state an answer if you let $N\_n=\lbrace0,1,\dots,n-1\rbrace$.
Let $n=2^k$, let $S$ be the multiset of integers with an odd number of ones in binary, each such integer appearing with multiplicity 2. Then it works for all $i\lt k$.
E.g., $k=3$, $S=\lbrace1,2,4,7\rbrace$, each taken twice, you get $1^i+2^i+4^i+7^i=0^i+3^i+5^i+6^i$ for $i=0,1,2$ (where, by convention, $0^0=1$).
If you really need the range to start at 1, just add 1 to everything, take $S=\lbrace2,3,5,8\rbrace$.
This has to do with the Tarry-Escott problem, q.v.
| 2 | https://mathoverflow.net/users/3684 | 42451 | 27,019 |
https://mathoverflow.net/questions/42449 | 17 | How would one approach proving that every real number is a zero of some power series with rational coefficients? I suspect that it is true, but there may exist some zero of a non-analytic function that is not a zero of any analytic function. I was thinking about approaching the problem using arguments of cardinality, but I am unsure about how to begin.
Thank you in advance.
| https://mathoverflow.net/users/nan | How to prove that every real number is a zero of some power series with rational coefficients (if true) | Call your real number $\alpha$. Suppose you have found a polynomial $p$ of degree $n-1$ with rational coefficients such that $|p(\alpha)|\lt\epsilon$. Show you can find a rational $r$ such that $|p(\alpha)-r\alpha^n|\lt\epsilon/2$.
| 29 | https://mathoverflow.net/users/3684 | 42452 | 27,020 |
https://mathoverflow.net/questions/42330 | 4 | I recently learned the following formulation of the Darboux theorem in a class.
Theorem: Suppose $\omega\_t$ is a smoothly varying family of symplectic forms on a closed manifold $M$ such that the cohomology class of $\omega\_t$ is independent of $t$. Then there is a smoothly varying family of diffeomorphisms $F\_t$ of $M$ such that $F\_0$ is the identity and $F\_t^\* \omega\_t = \omega\_0$.
The classical formulation of the Darboux theorem is obtained as a corollary in the following way. The condition that a closed 2-form is nondegerate is an open condition, so given a symplectic form $\omega\_0$ on $M$ there is a neighborhood in the space of representatives of the cohomology class of $\omega\_0$ which consists only of nondegenerate forms. This neighborhood can be taken to be path connected, so the theorem guarantees that for any symplectic form $\omega\_1$ which is sufficiently close to $\omega\_0$ and which belongs the same cohomology class there is automatically a diffeomorphism which pulls back $\omega\_1$ to $\omega\_0$.
My question: is the assumption that $\omega\_1$ is sufficiently close to $\omega\_0$ really necessary? In the proof that I learned, we really do need a path $\omega\_t$ of nondegenerate forms because the idea is to use the Poincare lemma to write $\omega\_t = \omega\_0 + d \beta\_t$ and then obtain a time dependent vector field $X\_t$ satisfying $\iota\_{X\_t}\omega\_t = -\frac{d}{dt}\beta\_t$. To say that $X\_t$ exists we need $\omega\_t$ to be nondegenerate for all time, and the diffeomorphisms $F\_t$ are obtained from the flow of $X\_t$.
I guess the real problem here is that I don't know all that many interesting examples of symplectic manifolds to begin with, and even on those examples that I know I can never produce more than one symplectic structure. Can anyone help?
| https://mathoverflow.net/users/4362 | Can the Darboux theorem be strengthened? | (Sorry I don't have enough reputation to make this a comment.)
I don't fully understand your question, in particular since I don't know which formulations of Darboux's theorem you are concerned with. The version I would describe as the "classical" one is that each point of a symplectic manifold admits a neighbourhood diffeomorphic to a standard symplectic ball in $R^{2n}$. In the proof I know, you need to shrink the size of your neighbourhood twice: the first time, in order to have that $\omega\_1$ and $\omega\_0$ are connected through symplectic forms, and the second time, to guarantee that the flow $\mathbf{F}\_t$ exists to time $t=1$.
Whatever ways you may have of sidestepping one or both of these, you definitely need to shrink the size the neighbourhood somehow. Indeed, the answer to the more global question (how big a neighbourhood can you embed?) is a very subtle one related to symplectic capacities and to symplectic packing problems. For instance, Gromov's non-squeezing theorem shows that in the case of the symplectic cylinder $D^2 \times \mathbb{R}^2$, you can't get a symplectic ball any bigger than the one of radius 1. You can, however, easily smoothly embed a ball of much bigger radius (even in a volume preserving way). When you pull back the symplectic form to this ball, you will not be able to deform it to the standard one, except in a neighbourhood of a given point.
| 5 | https://mathoverflow.net/users/477 | 42456 | 27,023 |
https://mathoverflow.net/questions/42460 | 22 | Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function?
| https://mathoverflow.net/users/5229 | Is a real power series that maps rationals to rationals defined by a rational function? | No. In fact, $p(x)$ can be a complex analytic function with rational coefficients that takes any algebraic number $\alpha$ in an element of $\mathbb{Q}(\alpha)$.
(And everywhere analytic functions are not rational unless they are polynomials).
The algebraic numbers are countable, so one can find a countable sequence of polynomials $q\_1(x), q\_2(x), \ldots \in \mathbb{Q}[x]$
such that every algebraic number is a root of $q\_n(x)$ for some $n$. Suppose that
the degree of $q\_i(x)$ is $a\_i$, and choose integers $b\_i$ such that
$$b\_{n+1} > b\_{n} + a\_1 + a\_2 + \ldots + a\_n.$$
Then consider the formal power series:
$$p(x) = \sum\_{n=0}^{\infty} c\_n x^{b\_n} \left( \prod\_{i=0}^{n} q\_i(x) \right),$$
By the construction of $b\_n$, the coefficient of $x^k$ for $k = b\_n$ to
$b\_{n+1} -1$ in $p(x)$ is the coefficient of $x^k$ in
$c\_n x^{b\_n} \prod\_{i=1}^{n} q\_i(x)$. Hence, choosing the $c\_n$ to be appropriately
small rational numbers, one can ensure that the coefficients of $p(x)$ decrease sufficiently rapidly and thus guarantee that $p(x)$ is analytic.
On the other hand, clearly $p(\alpha) \in \mathbb{Q}[\alpha]$ for every (algebraic)
$\alpha$, because then the sum above will be a finite sum.
With a slight modification one can even guarantee that the same property holds for all derivatives of $p(x)$.
I learnt this fun argument from the always entertaining Alf van der Poorten (who sadly died recently).
| 39 | https://mathoverflow.net/users/nan | 42465 | 27,027 |
https://mathoverflow.net/questions/42466 | 6 |
>
> **Possible Duplicate:**
>
> [A learning roadmap for algebraic geometry](https://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry)
>
>
>
I am a masters student and I want to study algebraic geometry.
Does there exist good good book for selfstudy of algebraic geometry?
| https://mathoverflow.net/users/10118 | Algebraic geometry | Reid's book recommended above (below,depending on your perspective) is certainly your best bet for a ground floor introduction. An older resource that's certainly worth checking out is William Fulton's *Algebraic Curves*.
In connection with Fulton's book,a new resource has recently popped up online that we ALL need to check out and I'm hoping to turn all math majors and graduate students on to it: Last year at MIT, Micheal Artin taught a basic course in algebraic geometry from Fulton's book using only his own course in algebra as prequisites. Artin composed a detailed set of lecture notes and posted most of them in PDF. I'm really hoping one day he turns them into a book-and if it becomes a widely-used resource on the web,he just might. The course and these materials,including a PDF downloadable version of Fulton's book,can be found at <http://math.mit.edu/classes/18.721/>. The fact that Artin is actively seeking email feedback and corrections on the notes strongly suggests he's at least considering turning them into a book-the more feedback he gets,the greater likelihood this will occur. So please-everyone find time to do this,particularly the algebracists and algebraic geometers in here!
More difficult but still very accessible, is the 2 volume second edition of Shafaravich's Algebraic Geometry text.The text is very rigorous,yet very concrete-it has many pictures and examples and builds to the language of schemes rather then throwing the student immediately into these very deep waters. Most of the initial focus is on the "classical" geometry of curves and varieties.
At this point,most people would recommend the classic by Hartshorne,which has given 2 generations of graduate students nightmares.It IS as difficult as people say,but it's very well written and if you're serious about AG,sooner or later,you have to read it. You should be ready to try it after Shafaravich and a good course in commutative algebra (a la Atiyah/MacDonald or Eisenbud).
But before you break your head on that book,there's 2 other options at roughly the same level I'd recommend first. First is Mumford's *The Red Books Of Varieties And Schemes*.This is a very visual yet abstract treatment that I think you'll find much easier going,even though it doesn't cover as much. You'll definitely find Hartshorne a lot easier after Mumford.
The other resource I'd recommend is still very much a work in progress,but it's so beautiful and wonderfully written,i'd be remiss in not recommending it. It's Ravi Vakil's lecture notes, now in thier 2nd or 3rd version of the increasingly popular course he's teaching at Stanford. They can be found in thier most recent iteration at his blog;earlier versions can be found posted at his webpage. I think eventually these notes-after a few more years of polishing-will supercede everything else for graduate students on modern algebraic geometry. They are almost "Hartshorne Explained".They also contain so many insights,it's incredible. Take at look at them,please. You'll thank me later.
That should get you started-good luck!
| 14 | https://mathoverflow.net/users/3546 | 42473 | 27,032 |
https://mathoverflow.net/questions/42454 | 11 | For an example I'm trying to understand, I need to calculate some cohomology group of some $\mathbb Z$-module with coefficients in some other $\mathbb Z$-module (with no interesting actions). (In particular, letting $\mathbb C^\times$ be the multiplicative group of the complex numbers *with the discrete topology*, I would like to know ${\rm H}^2(\mathbb C^\times,\mathbb C^\times)$, which is probably trivial but I have no idea how to show it is.) Having never really learned much cohomology theory, I know some basic definitions and I know how to turn an extension of groups into an exact sequence in cohomology. But this doesn't help me do particular computations unless I at least know the necessary cohomologies for pieces of the group. (For example, I may know that $\mathbb C^\times = S^1 \times \mathbb R$ as discrete groups, but I don't know any $\mathbb H^\bullet(S^1,\mathbb C^\times)$ or $\mathbb H^\bullet(\mathbb R,\mathbb C^\times)$.)
What would be best is some reference book or, as I won't be near a good library for at least two days, a good website that lists sufficiently many cohomology groups and pedagogically explains some ways to do explicit computations to extend their list. Something like (but presumably easier than) the [Knot Atlas](http://katlas.org/wiki/Main_Page) or the [OEIS](http://www.research.att.com/~njas/sequences/). Does such a thing exist?
| https://mathoverflow.net/users/78 | Where can I easily look up / calculate (abelian) group cohomology? | This group is best understood in terms of the universal coefficient formula,
i.e., in terms of the homology of the involved group. Hence, if $A$ is any
abelian group we have $H\_1(A)=A$ and the addition map $A\times A\rightarrow A$
induces a Pontryagin product on homology making $H\_\ast(A)$ a (graded)
commutative algebra. We also have that the square of an element of $H\_1(A)$ is
zero (in $H\_2(A)$). This does not directly follow in the presence of $2$-torsion
but it follows by functoriality; given $a\in H\_1(A)=A$ we have a group
homomorphism $\mathbb Z\rightarrow A$ taking $1$ to $a$ and we are reduced to
showing that $1\cdot1=0\in H\_2(\mathbb Z)$ but $H\_2(\mathbb Z)=0$.
Hence we get an algebra map $\Lambda^\ast A\rightarrow H\_\ast(A)$. This is an
isomorphism in degrees $1$ and $2$ and an isomorphism in all degrees if $A$ is
torsion free. This is proved easily by noting that both sides commutes with
filtered direct limits so that one is reduced to the case when $A$ is finitely
generated and then using the Künneth formula to reduce to the case when $A$ is
cyclic in which case $H\_2(A)=0$ and $H\_i(A)=0$ for $i\geq2$ if $A=\mathbb Z$.
Now, using the universal coefficient formula and the fact that the coefficient
group $\mathbb C^\ast$ is injective we get $H^2(\mathbb C^\ast,\mathbb
C^\ast)=\mathrm{Hom}(\Lambda^2\mathbb C^\ast,\mathbb C^\ast)$. Of course as
abelian group $\Lambda^2\mathbb C^\ast$ is humongous as is the coefficient group
$\mathbb C^\ast$ but they appear naturally in some cases. For instance
$H^2(\mathbb C^\ast,\mathbb C^\ast)$ is the group of characteristic classes of
degree $2$ for complex line bundles with integrable connection (in the smooth
case, otherwise think local systems). Also the related group $\mathbb
R\bigotimes S^1$ appears in the solution of Hilbert's third problem. We also
have that $K\_2(\mathbb C)$ is a quotient of $\Lambda^2\mathbb C^\ast$ and
similar but worse groups (such as $S^2\Lambda^2\mathbb C^\ast$) appear in the
relation with generalisations of Hilbert's third problem (see for instance
J. Dupont: Scissors congruences, group homology and characteristic classes, Nankai
Tracts in Mathematics).
I may misrepresent the feelings of a lot of people if I say that algebraic
topologists are resigned to the appearance of such large abstract groups
whereas algebraic geometers want to believe that they have more structure (such
as $K\_1(-)$ being the multiplicative group scheme) but haven't really been able
to realise this (there are some tantalising results such as Bloch et al's results
on the deformation theory of $K\_2(-)$).
| 16 | https://mathoverflow.net/users/4008 | 42476 | 27,035 |
https://mathoverflow.net/questions/42439 | 7 | The Dedekind eta function $\eta(\tau)$ can be regarded as a formula which assigns a number to a lattice $L \subset \mathbb{C}$. The algorithm is: rotate the lattice so that one of its basis vectors lies along the real axis, then pick another basis vector $\tau$ in the upper half plane, then compute the usual formula
$$
\eta(\tau) = q^\frac{1}{24} \prod\_{n=1}^\infty (1-q^n)
$$
where $q=e^{2 \pi i \tau}$. It would be nice if there were a more "canonical" way to compute it directly from the lattice $L$ (i.e. without this rotate-and-pick-a-basis-vector story which breaks symmetry). I'm looking for a formula similar to that of the Eisenstein series where one sums over all points of the lattice:
$$
G\_n (L) = \sum\_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n}
$$
We have the theorem of Jacobi that the 24th power of $\eta$ computes as the discriminant of the lattice,
$$
(2\pi)^{12} \eta(\tau)^{24} = 20G\_4(\mathbb{Z} + \tau \mathbb{Z})^3 -49 G\_6 (\mathbb{Z} + \tau \mathbb{Z})^2,
$$
which is great, since it shows that the 24th power of $\eta$ can be defined canonically via a sum over the lattice points... but how about $\eta$ itself?
| https://mathoverflow.net/users/401 | How can one express the Dedekind eta function as a sum over the lattice? | The functions $G\_k$ and $\Delta = \eta^{24}$ can be regarded as functions on the set of lattices because they're modular of level 1. The $\eta$ function isn't modular of level 1 (its level is 24) so there's no natural way to regard it as a function on lattices -- it's a function on lattices with additional "level structure".
Similarly, in order to regularise $G\_2$ to get something with good convergence, you end up having to make it depend on some level structure as well.
| 8 | https://mathoverflow.net/users/2481 | 42478 | 27,036 |
https://mathoverflow.net/questions/42479 | 3 | Recently my work has led me to consider octonion algebras. Not having much of a background with non-associative anything, I decided to check out a basic text on the subject, R.D. Schafer's *Introduction to Nonassociative Algebras*.
I was reading it happily for a good while, until I got to the discussion of alternative algebras. A little background: let $A$ be an arbitrary $F$-algebra: i.e., just endowed with an $F$-bilinear product $A \times A \rightarrow A$, no further assumptions. In analogy to the more familiar commutator, it is also useful to define for any $x,y,z \in A$ the **associator**
$[x,y,z] = (xy)z - x(yz)$,
the obvious point being that an algebra is associative iff all of its associators identically vanish. But there is a more subtle merit to this: the associator is an $F$-trilinear map from $A^3$ to $A$. From this it follows that it is entirely determined by its values on any $F$-basis $\{e\_i\}\_{i \in I}$ of $A$. And from *that* it follows that associativity can be checked on basis elements and moreover that associativity is faithfully preserved by scalar extension: clearly any trilinear map on an $F$-vector space is identically zero iff its extension to some field extension $K/F$ is identically zero.
On to alternativity: an $F$-algebra $A$ is said to be **alternating** if for all $x,y \in A$,
$[x,x,y] = [x,y,y] = 0$.
(These two identities are easily seen to imply the **flexible law** $[x,y,x] = 0$.)
Note however that these identities are not multilinear any more: e.g. the left alternator $[x,x,y]$ is quadratic in $x$ and linear in $y$. Thus both of the above consequences of multilinearity are in question: is it sufficient to check left alternativity on basis elements, and is a scalar extension of an alternative algebra necessarily alternative?
Presumably the first question has a negative answer. Compare for instance the quadratic form $q(x,y) = xy$: it vanishes on the two standard basis elements of $F^2$ yet is nondegenerate. What we need to do is *linearize*, i.e., replace the quadratic form with the associated bilinear form. In the present context, this amounts to replacing the alternating condition with the skew-symmetric condition, i.e.,: for all $x,y,z \in A$,
$[x,y,z] = -[y,x,z] = [y,z,x]$.
The skew-symmetry condition looks much better: as a pair of equalities among trilinear maps, again it suffices to check it on basis elements and again it is faithfully preserved by scalar extension.
As is well-known, alternation implies skew-symmetry and the converse holds when $\frac{1}{2} \in F$.
>
> But what about when $F$ has characteristic $2$?
>
>
>
In this case, unfortunately (and somewhat embarrassingly) I am not even seeing why if $A$ is an alternating $F$-algebra and $K/F$ is a field extension, then $A\_K = A \otimes\_F K$ is an alternating $K$-algebra. Schafer does address this in his book: for $x \in A$, we have
the left and right multiplication operators $L\_x, R\_x$ as elements of $\operatorname{End}\_K(A)$.
Then equation (3.1) asserts that left and right alternating laws are equivalent to
$L\_{x^2} = (L\_x)^2$ and $R\_{x^2} = (R\_x)^2$.
He also says that the skew-symmetry of the associator is equivalent to (3.2), which is:
$R\_x R\_y - R\_{xy} = L\_{xy} - L\_y L\_x = L\_y R\_x - R\_x L\_y = L\_x L\_y - L\_{yx} = R\_y L\_x - L\_x R\_y = R\_{yx} - R\_y R\_x$.
(I have no problem with these identities.)
Then he says (on the top of p. 28) that "It follows from (3.1) and (3.2) that any scalar extension of an alternative algebra is alternative."
Unfortunately I don't follow. Can someone help me out?
| https://mathoverflow.net/users/1149 | Alternative algebras in characteristic 2, especially scalar extension | A good samaritan has delivered an answer directly to my email account. It is indeed almost obvious, as long as one ignores the bit about the multiplication operators!
Take for instance the left alternative property: for all $x,y \in A$, $[x,x,y] = 0$. As I said above, this condition is linear in $y$, so it is enough to check that for each element $e$ of a fixed $F$-basis of $A$, the bilinear form $B\_e(x,y) = [x,y,e]$ is alternating. Thus, the left alternation property itself will be preserved by base extension iff the alternating property of a bilinear map is preserved by base extension. I must have had a little mental block about this, but of course it is true: let $\{e\_i\}$ be an $F$-basis for $A$ hence a $K$-basis for $A\_K = A \otimes\_F K$. Then any $x \in A\_K$ may be uniquely written as $x = \sum\_i x\_i e\_i$ with $x\_i \in K$, and thus
$[x,x] = \sum\_{i=1}^n x\_i^2 [e\_i,e\_i] + \sum\_{i \neq j} x\_i x\_j [e\_i,e\_j]$
$= 0 + \sum\_{i < j} x\_i x\_j ([e\_i,e\_j] + [e\_j,e\_i]) = 0$,
since alternation implies skew-symmetry always.
| 3 | https://mathoverflow.net/users/1149 | 42487 | 27,040 |
https://mathoverflow.net/questions/42482 | 1 | Is there a name for the class of finite graphs $G$ with the following property?
* Every two graphs that can be created by removing one edge from $G$ are isomorphic.
(Edited to add the word "finite.")
| https://mathoverflow.net/users/10121 | Name the class of graphs G s.t. every two graphs that can be created by removing one edge from G are isomorphic. | If all the edge-deleted subgraphs of a finite graph are isomorphic then the graph is edge-transitive. [Here](http://www.newcastle.edu.au/school-old/math-physical-science/our_staff/downloads/macdougall_jim_onereductions.pdf) is a possible reference, though you might want to look at [this](http://staff.um.edu.mt/jlau/research/SURV_ARS.ps) survey on pseudosimilarity. (Pseudosimilar edges give isomorphic subgraphs after being deleted but are not part of the same orbit under the automorphism group of the graph). For finite graphs and small $k$, having $k$ edge orbits is the same as having $k$ isomorphism classes of edge deleted subgraphs, as is proved [here](http://www.springerlink.com/content/h321401n5046718w/).
| 7 | https://mathoverflow.net/users/2384 | 42494 | 27,046 |
https://mathoverflow.net/questions/42396 | 3 | I want to calculate the second homotopy group of the blow-up of the unit cotangent disk bundle of a closed surface $\Sigma$, i.e $\pi\_2\left(D^\*T\Sigma\#\overline{\mathbb{C}P^2}\right).$ Actually, I want to understand whether these symplectic manifolds are monotone. If yes, I want to find a closed monotone Lagrangian submanifold in it.
| https://mathoverflow.net/users/10104 | Homotopy groups of the blow-ups and monotone symplectic manifolds | Write $X$ for the manifold in question (i.e. the symplectic blowup of the unit disc bundle in the cotangent bundle of a surface $\Sigma$).
If you're interested just in monotonicity, what's relevant isn't so much $\pi\_2(X)$ as the image of $\pi\_2(X)$ under the Hurewicz map to $H\_2(X;\mathbb{Z})$, which is quite a bit simpler--it has rank either one or two (generated at most by the zero section and the exceptional sphere, depending on what $\Sigma$ is). Now $\omega$ evaluates as zero on the zero section and is positive on the exceptional sphere, while the first Chern class is zero on the zero section and $1$ on the exceptional sphere. Thus $[\omega]$ and $c\_1$ are positively proportional as elements of $H^2\left(X;R\right)$,
so $X$ is always monotone, regardless of $\Sigma$.
Regarding monotone Lagrangian submanifolds, the obvious candidate is the zero section $\Sigma$. Since $\pi\_1(\Sigma)\to \pi\_1(X)$ is an isomorphism, all classes in the relative $\pi\_2$ come from the absolute $\pi\_2$, so since the ambient manifold is monotone it directly follows that $\Sigma$ is a monotone Lagrangian submanifold.
When $\Sigma$ is $S^2$, there's another interesting example of a monotone Lagrangian $L\subset D^\*TS^2$ discussed in [this paper](https://arxiv.org/abs/math/0608356) of Albers and Frauenfelder. If you carefully choose the size of the blowup (so that the proportionality constant on the exceptional sphere is the same as the monotonicity constant for $L$), then it should be possible to arrange for $L$ to lift to a monotone Lagrangian torus in the blowup. (However for a generic size blowup this probably won't work.)
| 2 | https://mathoverflow.net/users/424 | 42503 | 27,054 |
https://mathoverflow.net/questions/42497 | 9 | Let $k$ be a field and let $X$ be a smooth irreducible variety over $k$.
Suppose that I know that the image of $X$ in the Grothendieck group
of varieties over $k$ is equal to that of
a) ${\mathbb A}^n$ for some $n$
b) ${\mathbb A}^{n\_1}\times {\mathbb G}\_m^{n\_2}$ for some $n\_1, n\_2$.
Does it follow that $X$ is isomorphic to the varieties appearing in a) and b)?
If the answer is "yes", then can one drop the smoothness assumption on $X$?
| https://mathoverflow.net/users/3891 | Motivic characterization of affine spaces | The answer is no: Let $X$ the projective plane $\mathbb P^2$ minus a smooth
quadric $Q$. Then $[X]=[\mathbb P^2]-[Q]=\mathbb L^2+\mathbb L+1-(\mathbb
L+1)=[\mathbb A^2]$ but $X$ is not isomorphic to $\mathbb A^2$ as its Picard group is
$\mathbb Z/2$. For b) just cross $X$ with $\mathbb G\_m^n$, the Picard group is still $\mathbb Z/2$.
For the singular case the following is an interesting example: Let the symmetric group
$\Sigma\_n$ act on $(\mathbb A^m)^n$ by permuting the factors. Then it is a fact
(proved by Totaro, see Lemma 4.4 of L. Göttsche, On the motive of the Hilbert
scheme of points on a surface, Math. Res. Lett. 8 (2001), no.~5-6, 613--627.)
that the class of $(\mathbb A^m)^n/\Sigma\_n$ in the Grothendieck group of
varieties is equal to that of $\mathbb A^{mn}$. However, when $m>1$ then the
quotient is always singular and hence not isomorphic to $\mathbb A^{mn}$.
**Addendum**: In response to the further question by Alexander, I can't think
of any strengthening that would make it true. For a moment I thought that if one
instead asked that $X$ be smooth and proper and have the same class as $\mathbb
P^n$ would imply that $X$ is isomorphic to $\mathbb P^n$. That however is
counterexampled by a smooth quadric of odd dimension greater than $1$. We can
project from a point to make the blowing up of one point isomorphic to the
blowing up of a smooth quadric of $\mathbb P^n$ which gives that the class of
$X$ is equal to $\mathbb P^n$. On the other hand $X$ is not isomorphic to
$\mathbb P^n$ as its cohomology ring is not generated by its degree $2$ part.
**Addendum 1**: Some comments related to Alexander's latest question. It is known that the stable birational class of $X$ can be recovered from the class $[X]$. Let us assume that we actually get that it determines the birational class (this may very well always be true and is true in small dimensions).
* For curves (we shall always assume that $X$ is smooth and irreducible) we get that we can recover the projective model of $X$ but as $[X]$ is equal to the class of the projective model minus a number of points we see that if $[X]$ is not proper we can not recover which points we have removed from the projective model to get $X$. Hence the reconstruction problem is reasonable only when $X$ is also proper which we shall assume.
* Let us consider a ruled surface $X\to C$ with fibres $\mathbb P^1$'s. If we blow up a point on a fibre and then blow down the strict transform of the fibre containing the point, we get a new ruled surface which in general (ever?) is not isomorphic to $X$ but has the same class.
* On the other hand in the case of non-ruled surfaces we have a minimal model such that all other surfaces birational to it is constructed by a succession of blowing ups. Hence we can distinguish the minimal model from all other birational surfaces. However, we can not tell which points we have blown up and by blowing up different points we get different surfaces.
* In higher dimensions things seem worse as there almost never is a unique minimal model. For instance the two small resolutions of an ordinary double point in three dimensions have the same class but are in general non-isomorphic.
* There are a few cases where a minimal model is unique. The one that comes to mind is when $X$ is an abelian variety in which case $X$ can actually be recovered from its class.
Hence I think that there will only be a very limited number of situations when one can hope to recover $X$ from its class.
| 19 | https://mathoverflow.net/users/4008 | 42504 | 27,055 |
https://mathoverflow.net/questions/42510 | 17 | Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?
| https://mathoverflow.net/users/1916 | How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? | I think that the answer is **yes**.
Indeed, there are examples of integral domains $D$ such that every non-zero prime ideal of $D$ has infinite height.
Look at the paper
"Anti-archimedean rings and power series rings"
D.D. Anderson; B.G. Kang; M H. Park
Communications in Algebra, 1532-4125, Volume 26, Issue 10, 1998, Pages 3223 – 3238.
| 14 | https://mathoverflow.net/users/7460 | 42515 | 27,062 |
https://mathoverflow.net/questions/42516 | 3 | I would like to know if the the following exist or are defined
1. The Fourier transform $\mathcal{F}\left(\frac{d^{\frac{1}{2}}y}{dx^\frac{1}{2}}\right)$ of a fractional differential operator such as $\frac{d^{\frac{1}{2}}y}{dx^\frac{1}{2}}$. (I'm aware that the fractional Fourier transform exists, but this isn't quite the same thing.)
2. An equivalent of the Plancherel theorem for fractional Lebesgue norms. I'm aware that the Plancherel theorem is defined for $\mathcal{L}\_n$ where $n = 2$. What I'd like to know is if a similar theorem exists for positive non-integer values of $n \ne 2$.
Plancherel theorem where $n = 2$
$\int\_{\mathbb{R}^m} \mid f({x}) \mid^n \; d{x} \; = \; \int\_{\mathbb{R}^m} \mid \tilde{f}({\omega}) \mid^n d{\omega}$
Where $m$ is the number of dimensions. The answer to these questions would rule out or extend certain possibilities.
As always I'm sorry if these questions are total nonsense. I'm just a computer scientist teaching myself mathematics while writing my thesis.
| https://mathoverflow.net/users/7486 | Fourier transform of fractional differential operator and Plancherel formula equivalent for fractional norms | Regarding your first question, the fractional derivative operators are in fact DEFINED by how they act on the Fourier transform side. If the Fourier Transform of $f(x)$ is $\hat f(\xi)$ then the Fourier transform of its derivative $f'(x) = Df(x)$ is $2\pi i \xi \hat f(\xi)$ and hence for a positive integer $k$, the Fourier Transform of its $k$-th derivative $D^kf(x)$ is $(2\pi i \xi)^k \hat f(\xi)$. To define the $k$-th derivative operator when $k$ is a fraction or even an arbitrary real number, that same formula is used---i.e., Fourier Transform, multiply the resulting function of $\xi$ by $(2\pi i \xi)^k$, and then inverse Fourier Transform.
As for your second question, I do not know of anything that is in the nature of a generalization of the Plancherel Theorem from $L^2$ to other $L^p$ spaces, and certainly the obvious generalization is false.
| 6 | https://mathoverflow.net/users/7311 | 42526 | 27,072 |
https://mathoverflow.net/questions/42538 | 3 | I am talking about the principle that is to DC what the global choice is to the usual axiom of choice. Global choice involves existential quantification over classes, but global DC can be stated as a schema in first-order set theory.
$(\forall x (\phi(x) \to \exists y (\phi(y) \wedge \psi(x,y)))) \to \forall x (\phi(x) \to \exists f (f(0)=x \wedge \forall n \in \omega (\phi(f(n)) \wedge \psi(f(n),f(n+1)))))$
Searching for this I come up with nothing, other than in constructive set theory, where it is called "relativised" DC, and has a simple computational interpretation. I guess that classically, it just follows from ZF+DC. Is that right? I can't figure out how.
| https://mathoverflow.net/users/6787 | Global or Relativised Dependent Choices | Given any $x$, let $\alpha\_x$ be 0 if $\phi(x)$ fails. If it holds, for each $x'$ with $\phi(x')$ and $x'$ of set-theoretic rank less than or equal to $x$, let $\alpha^{x'}$ be the least ordinal $\alpha$ such that there is a $y$ of rank $\alpha$ with $\phi(y)\land\psi(x',y)$. Now set $\alpha\_x$ as the supremum of the $\alpha^{x'}$.
Then $\alpha\_x$ is an upper bound for the rank of a `witness' $y$ when $\phi(x)$ holds. By replacement, there is a sequence $(g(n)\mid n<\omega)$ with $g(0)$ the rank of $x$, $g(1)$ an upper bound for the rank of a $y$ with $\phi(y)\land\psi(x,y)$, $g(2)$ an upper bound for the rank of a $z$ such that there is a $y$ with $\phi(y)\land\psi(x,y)\land\phi(z)\land\psi(y,z)$, etc.
Taking $\beta$ as the supremum of the $g(n)$, we see that $\beta$ is such that if we restrict $\phi$, $\psi$ to sets of rank at most $\beta$, then DC ensures that there is an $f$ as wanted, starting with $f(0)=x$.
[By the way, this is a standard approach, you replace your problem with one about ranks, and things become easier.]
| 3 | https://mathoverflow.net/users/6085 | 42542 | 27,082 |
https://mathoverflow.net/questions/42557 | 5 | In exercise 15 of Milnor's *Topology from a Differentiable Viewpoint*, one is asked to compute the Hopf invariant of the Hopf map. The way one is supposed to do this is to compute the linking number of two of the fibres, but Milnor doesn't define the linking number in terms of an integral. He says to compute it as the degree of the map $\frac{x-y}{||x-y||}$ from the product of two compact oriented boundaryless manifolds embedded in $\mathbf{R}^{k+1}$ to the sphere of dimension $k$ where the sum of the dimension of the manifolds is $k$.
I'm aware of other ways to compute the Hopf invariant by using deRham cohomology (see Bott and Tu, for instance), but I'm curious how one is actually supposed to do it by hand. Is there a particularly concrete way to compute the linking number without using this other machinery? Most of the other exercises in the book have cute little solutions, but is that true of this problem?
(Not homework!!)
| https://mathoverflow.net/users/1353 | Computing the hopf invariant (without integration or homology, as in Milnor) of the hopf map | If you have the Hopf link embedded in some standard way in $\mathbb{R}^3$, you can see the linking number as given by the degree of a map $S^1 \times S^1 \to S^2$ in a number of ways. For instance, the pre-image of the north pole in $S^2$ consists of pairs of points stacked vertically above each other, i.e., crossings between the two components in the knot diagram given by projection to the $xy$ plane. (Crossings will correspond to preimages of the north pole or south pole, depending on your conventions.) For the standard diagram for the Hopf link, there's only one crossing that counts. The hard part from this point of view is getting the orientation right (is the Hopf invariant $-1$ or $+1$?), but that can be done with care and attention.
| 3 | https://mathoverflow.net/users/5010 | 42559 | 27,088 |
https://mathoverflow.net/questions/42562 | 4 | Is there an intuitive way to define what the p-quotient of an integer partition is?
In response to the comments:
my vague understanding is that the p-quotient is somehow division with remainder generalized to partitions. They turn up in the classic by Littlewood that I was reading:
<http://rspa.royalsocietypublishing.org/content/209/1098/333.abstract>
See also Macdonalds book on symmetric functions and Hall polynomials.
| https://mathoverflow.net/users/692 | p-quotient of an integer partition | I learned the following perspective in part from Bill Doran.
Redefine a partition through its boundary, extended along the axes in both directions:
A *partition* is a bi-infinite sequence of vertical $(0,1)$ and horizontal $(1,0)$ steps $(... S\_{-2},S\_{-1},S\_0,S\_1,S\_2,...)$ whose prefix is all vertical steps, and whose suffix is all horizontal steps. A *square* of the partition is a pair of steps so that the first step is horizontal and the second step is vertical. The *hook length* of that square is the distance between the steps.
Then the $p$-*quotient* of a partition is the $p$-tuple of partitions you get by the steps in each congruence class mod $p$: $P\_i = (...S\_{-2p+i},S\_{-p+i},S\_i,S\_{p+i},S\_{2p+i},...)$ for $i \in \{0,..., p-1\}$.
Each square of a partition of the $p$-quotient corresponds to a square of the original partition whose hook length is divisible by $p$.
$p$-*cores* are partitions whose $p$-quotient is a $p$-tuple of empty partitions. They can be parametrized by an equivalence class of $p$-tuples of values $k\_i$ so that $S\_{p k + i}$ is vertical iff $k \le k\_i$, i.e., the offsets of the empty partitions.
If you strip $p$-rimhook after $p$-rimhook off of a partition, this always results in the same $p$-core, and the choices don't matter. The choices correspond to taking one corner square (hook length 1) at a time off of a partition in the $p$-quotient.
Although in some applications, $p$ is prime, that is not necessary.
| 8 | https://mathoverflow.net/users/2954 | 42566 | 27,091 |
https://mathoverflow.net/questions/42565 | 1 | This question is motivated by the Euclidean Traveling Salesman Problem, i.e. finding the shortest Hamiltonian path of a complete graph of N randomly placed vertices. To eliminate boundary effects I consider the problem on the unit square with periodic boundary conditions. The idea is to find a "direction" and connect points in an ordered manner along this direction to find a short path (Wheeler, J.A., On recognizing law without law, Am. J. Phys, 51(5), pp. 398, 1983).
The question is:
given N points with real coordinates on the unit square, is there a neat way to find the least-square fit line $y=rx+b \mod 1$, where $r=p/q$ is rational? The line also needs to be as short as possible. The length of the line in the square is $\sqrt{p^{2}+q^{2}}$, so minimizing the following quantity should yield a short line that passes near the points:
$\sum d\_{i}^{2}+\alpha \left( p^{2}+q^{2}\right) $.
Here $d\_{i}$ is the "distance" (vertical or perpendicular) of point i from the line and $\alpha$ is a positive weight.
By "neat" I mean something like the regular least-square line fit solution (<http://mathworld.wolfram.com/LeastSquaresFitting.html>).
| https://mathoverflow.net/users/9766 | Least-square fit of line with rational slope to points on a square with periodic boundary conditions | I take it that the points themselves do not have to have rational coordinates. I don't think that there would have to be a best approximation. Consider the two point set set $\{(0,0),(1,\frac{1}{\sqrt{2}})\}$. Then lines such as $y=\frac{12}{17}x$ and $y=\frac{408}{577}x$ based on convergents to $\frac{1}{\sqrt{2}}$ provide better and better approximations. Nothing changes if we use more points $(x\_i,\frac{x\_i}{\sqrt{2}})$ or use some other irrational slope.
Also, at the eight x values $\frac{\sqrt{3}-\sqrt{2}+j}{8}$ for $0 \le j \le 7$, the two functions $3x+\sqrt{2}$ and $-5x+\sqrt{3}$ are equal $\mod 1$ so either is a perfect fit.
**later** Based on some of the comments, here is one idea for a question: We are given some points $(x\_i,y\_i)$ in the unit square. For each integer $k \ge 0 $ consider all the lines $y=rx+b$ (with $r$ rational if desired, although it might be better to not make this restriction) such that $0 \le b <1$ and $k \le r <k+1$. For each such line find the squared distance to the points $(x\_i,y\_i+h\_i$ where the $h\_i$ are integers chosen to minimize each distance. QUESTION: Is there an elegant way to find the best approximating line(s) for each $k$ ? One should then also consider negative slopes.
| 0 | https://mathoverflow.net/users/8008 | 42576 | 27,093 |
https://mathoverflow.net/questions/33095 | 5 | In the [**paper**](http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.3592v1.pdf), by János Kollár there is *problem 19* (page 8).
It is one more strict resolution. A resolution that leaves untouched the semi-simple-normal-crossings singularities of pairs.
**My question is:** *How/where is that kind of resolution used/needed?*
**Quick definitions:**
*Pair:* $(X,D)$ with $X$ algebraic variety and $D$ a Weil divisor on it.
*Semi-simple-normal-crossings:* A point in $X$ where $X$ is (locally) a union of coordinates hyperplanes and $D$ is given by intersecting $X$ with some of the other coordinate hyperplanes not contained in $X$.
| https://mathoverflow.net/users/5506 | How/where are semi-log resolutions used? | Perhaps a little more explanation would be this:
One of the first things we learn in algebraic geometry is *normalization* and we are told that it is "harmless" to assume that something is normal since the normalization exists and canonical and all that jazz. This is fine as long as one studies a stand alone object, but once they come in a family, it is no longer true.
Consider a family of curves. Recall that for curves normal=smooth so if not all members of the family are smooth, which is the likely scenario, then they are also not all normal. Now the problem is, there is no way to normalize the family members that they stay in a family. For instance, a family of smooth cubic plane curves degenerates to a singular one, but the normalization of the singular cubic has genus $0$, while the cubic curves have genus $1$ so they can't be members of the same family. Also, if you try to resolve the singularities by blowing up you'll see that you can resolve all singularities to be normal crossings, but you cannot do better and you also add new irreducible components to the singular fibers. This leads one to do semi-stable reduction, which is actually another story, so I won't get into that.
Anyway, for curves, we can actually make do with handling only smooth and simple normal crossing points. In higher dimensions if one tries to do the same, then there are other singularities that one must allow and these are the semi-log canonical (a.k.a. slc) singularities Zsolt mentioned.
OK, so maybe the above convinces you that if you want to do moduli theory and you want to study compact moduli spaces, that is, you actually would like to understand degenerations as well and not just the nice part, then you have to deal with non-normal, in particular with slc singularities. (Actually, "s-something" is usually the non-normal version of "something", accordingly *slc* is the non-normal version of *lc*).
Well, now how do you define a non-normal version of a singularity that is otherwise defined via some properties of exceptional divisors (or saying it in a more enlightened way: exceptional set)? You cannot take a full fledged resolution of singularities, because it will resolve the non-normality of the singularity as well. This would not be a huge problem from the point of view of making it simple, but it sabotages the entire operation. The issue is, that if you look at the definition of lc (and klt, dlt, etc) more closely, then it becomes clear that it kind of needs that the resolution used in the definition is an isomorphism in codimension $1$ on the target, that is, the singular guy. It is also important that the exceptional set is a divisor. These will fail for non-normal but $S\_2$ singularities, for instance for slc but not lc singularities.
So, you need a partial resolution that resolves the singularities to something that is close to being smooth but has the above properties. The "close to being smooth" is called "semi-smooth", these are double normal crossings and pinch points, exactly the singularities that cannot be made better by only changing something in codimension $2$. (This last statement is left to the reader. If you have difficulty with it, ask).
OK, I better wrap this up. So the point of a semi-resolution is that it has those properties that make it possible to define discrepancies but it does not go "too far". However, it produces varieties with very mild singularities that are almost as good as smooth, at least from the point of view of this definition.
| 19 | https://mathoverflow.net/users/10076 | 42578 | 27,094 |
https://mathoverflow.net/questions/42580 | 1 | Let $X\_{p} = \mathbb{Q}\_{p} / \sim $, where $\sim$ is defined by:
$x\sim 0 \Leftrightarrow x\in \mathbb{Q}$
$X\_{p}$ is path-connected, because (unless I'm making some horrible mistake,) for any $x\in X\_{p} \backslash \mathbb{Q}$, we have that $\lbrace x,0\rbrace$ under the subspace topology is path-connected.
Is $X\_p$ contractible?
| https://mathoverflow.net/users/9455 | Is this quotient space of Q_p contractible? | The answer is yes. Whenever you crunch a dense subspace $Y$ of a topological space $Z$ to a point $q$ in the quotient $Z/Y$, you have the following contracting homotopy: For any $x \in Z/Y$ and any $t \in [0,1]$, set $f(x,0) = x$ and $f(x,t) = q$ for $t > 0$. Here, $q$ is the equivalence class of zero, $Z = \mathbb{Q}\_p$, and $Y = \mathbb{Q}$.
| 7 | https://mathoverflow.net/users/121 | 42597 | 27,104 |
https://mathoverflow.net/questions/42598 | 7 | It is known, from the works of G.Margulis, etc. that lattices in semi-simple real (algebraic) groups are "often" arithmetic subgroups, as long as the split rank is high enough. Here by a lattice in a Lie group $G$ is understood a discrete subgroup $\Gamma$ in $G$ such that the measure on $\Gamma\backslash G$ induced from the Haar measure on $G$ is of finite volume.
Typical example: for $G$ a simple $\mathbb{Q}$-group, with $\mathbb{Q}$-rank at least 2, then a lattice in $G(\mathbb{R})$ is arithmetic, in the sense that it is commensurable with a congruence subgroup, the latter being the intersection $G(\mathbb{Q})\cap K$ inside $G(\mathbb{A}^f\_\mathbb{Q})$ for some compact open subgroup $K$ in the group of adelic points of $G$.
And what should be said about discrete subgroups in $G(\mathbb{Q}\_p)$? Write for simplicity $G\_p=G(\mathbb{Q}\_p)$. At least one knows that for a discrete subgroup $\Gamma$ in $G\_p$, the quotient $\Gamma\backslash G\_p$ is compact if and only if it is of finite volume with respect to the Haar-induced measure. If $G$ is defined over a global field, say $\mathbb{Q}$, then a conguence subgroup of $G$ "should" be also a lattice in $G\_p$. Moreover, for general semi-simple $\mathbb{Q}\_p$-group $G$, how to classify the co-compact discrete subgroups in $G\_p$? Do they admit explicit constructions like those coming from global fields?
Any comments and references are welcome.
Thanks!
| https://mathoverflow.net/users/9246 | discrete subgroups in p-adic Lie groups? | The arithmeticity theorem of Margulis also characterizes lattices in product of simple groups over $R$ as well as $p$-adic fields. All such lattices in the higher rank case are S-arithmetic. When all but one of the factors is compact, the projection onto the non-compact factor (that you can take to be a p-adic field) gives you a lattice there. For instance, take a quadratic form $q$ that is anisotropic over $R$, but isotropic over ${\mathbb Q}\_p$ for some $p$. Then $SO({\mathbb Z}[1/p])$ is a lattic in the product $SO(q, {\mathbb R}) \times SO(q, {\mathbb Q}\_p)$. The projection onto the first component is dense, and onto the second component is a lattice there. For the precise statement of the (S-)arithmeticity theorem, the best place to look is Margulis's book "Discrete Subgroups of Semisimple Lie Group".
| 10 | https://mathoverflow.net/users/3635 | 42610 | 27,114 |
https://mathoverflow.net/questions/42609 | 4 | It is well known that every stable sheaf on a K3 surface $S$ is simple but the contrary is not true. Moreover, if $M$ denotes the (coarse) moduli space of stable sheaves on $S$ with fixed Chern classes and $Spl$ the moduli space of simple sheaves on $S$ with the same Chern classes, $M$ is an open subscheme of $Spl$. Can there exist a component of $Spl$ not intersecting $M$ or is $M$ dense in $Spl$?
What about the relation between semistable and simple sheaves? Is a simple sheaf on a K3 surface always semistable? If not, can you give me a counterexample?
| https://mathoverflow.net/users/33841 | Stable, semistable and simple sheaves on a $K3$ | $M(c\_1, c\_2)$ is **not** always dense in $Spl(c\_1, c\_2)$, in other words irreducible components of $Spl(c\_1, c\_2)$ not intersecting $M(c\_1, c\_2)$ may actually exist.
In fact, in his paper
"Moduli of simple rank-$2$ sheaves on $K3$-surfaces",
Manuscripta Math. 79 (1993), no. 3-4, 253–265,
Z. Qin constructs irreducible components of $Spl(c\_1, c\_2)$, for suitable values of $c\_1$, $c\_2$, in which **no** sheaf is stable. Moreover, he proves the following
**Theorem.** Let $X$ be a $K3$-surface. Assume that $(4c\_2-c^2\_1)>16$, and that $S$ is an irreducible component in $Spl(c\_1,c\_2)$ such that no sheaf in $S$ is stable. Then $S$ is birational to either ${\rm Hilb}^d(X)$ or $X\times{\rm Hilb}^{d-1}(X)$.
I did not check whether these components contain some strictly semistable sheaves.
| 2 | https://mathoverflow.net/users/7460 | 42614 | 27,117 |
https://mathoverflow.net/questions/42615 | 5 | *[Disclaimer: this may be a very trivial question; it certainly looks like it ought to have been studied and understood. I started thinking about it this morning when writing some notes for Rellich-Kondrachov, but cannot find a simple counterexample.]*
For the time being, let us just work on $\mathbb{R}^d$ with the Lebesgue measure. It is well-known that for an open, bounded domain (hence with finite measure) $\Omega$, the inclusion
$$ L^p(\Omega) \to L^q(\Omega) $$
is continuous for $\infty \geq p \geq q \geq 1$.
**Question**: Is the inclusion completely continuous (i.e. compact)? If not, what is a simple counterexample?
I dug around a bit and cannot find any references to a proof (or even the statement). The usual non-compactness mechanisms, of course, do not work. Let $f\_i$ be a sequence of functions with $\|f\_i\|\_{L^p(\Omega)} \leq 1$, so by continuous inclusion it is a bounded sequence in $L^q(\Omega)$. Because $\Omega$ is compact, we cannot have the problem fixed-scale translations: if $f\_i(x) = f(x + y\_i)$ for a sequence of points $y\_i$, since $y\_i$ must have a converging subsequence, then so must $f\_i$, even in $L^p$. The other usual non-compactness mechanism is dilations. WLOG assume $0\in supp(f) \subset\subset \Omega$ and that the $supp(f)$ is convex. Then the sequence $f\_j = 2^{jd/p} f(2^j x)$ is a bounded but non-compact sequence in $L^p(\Omega)$. But in $L^q(\Omega)$ for $q < p$, the sequence converges strongly to 0.
| https://mathoverflow.net/users/3948 | Is the inclusion of Lebesgue spaces compact? | Take $f\_j(x)=\sin(jx\_1)$. This is a bounded sequence in every $L^p(\Omega)$ when the domain $\Omega$ is bounded. Yet, it is non-compact in every $L^q(\Omega)$. It happens to converge weakly-star in these spaces towards $f=0$, but not strongly.
| 5 | https://mathoverflow.net/users/8799 | 42616 | 27,118 |
https://mathoverflow.net/questions/42472 | 3 | Let $X=\{x\_{1}, \cdots , x\_{n}\}$ be a set of $n$ positive integers and integer $i \ge 1$. Let’s say that the set $X$ is $i$-sum-avoiding if for any nonnegative integers $c\_{1}, \cdots, c\_{n}$ such that $\sum\_{j=1}^{n}c\_{j} = n$ and $(c\_{1},\cdots, c\_{n}) \ne (1,\cdots, 1)$, it holds that
$\displaystyle \sum\_{j=1}^{n}c\_{j}x\_{j}^{i} \ne \displaystyle \sum\_{j=1}^{n}x\_{j}^{i}$
Let $f(n,i)$ be the minimum value for a given $i \ge 1$ such that there exists an $i$-sum-avoiding set $X$ consisting of $n$ positive integers at most $f(n,i)$. Does there exist a constant $k\_{i}$ for every $i$ such that $\forall n \in \mathbb N$, it holds $f(n,i) \le n^{k\_{i}}$? If it does, what is the minimum of such $k\_{i}$ for every $i$?
Showing such a set would help solve hard problems in computer science given some space relaxations. It seems that the hardness of such problems is directly related to non-existence of such sets. I could only show such sets when $k=n$ that is $k$ is not a constant. My example for $X$ is $X = \{n^{1}, n^{2},\cdots, n^{n}\}$.
$\underline{Conjecture}$: $k\_{i} = \infty$ $\forall i \ge 1$.
| https://mathoverflow.net/users/10035 | A conjecture on a Subset Power Sum Problem motivated by Computer Science | Here is what I think proves that for any *i*, there is no constant *k**i* satisfying *f*(*n*,*i*)≤*n**k**i*. That is:
**Claim**. Let *i* be a positive integer. Then the function *f*(*n*,*i*) is not polynomially bounded in *n*.
**Proof**. First consider the case of *i*=1. A key observation is that if *X*={*x*1,…,*x**n*} contains two subsets *A* and *B* such that |*A*|=|*B*|, *A*≠*B*, and the sum of *A* is equal to the sum of *B*, then *X* cannot be 1-sum-avoiding since assigning *c**j* as follows violates the condition: *c**j*=2 if *x**j* belongs to *A* but not to *B*, *c**j*=0 if *x**j* belongs to *B* but not to *A*, and *c**j*=1 if neither holds.
Let *m* be a positive integer and *X* be a 1-sum-avoiding set of size 2*m*. By the above observation, all *m*-element subsets of *X* must have distinct sums, and therefore the largest sum must be at least $\binom{2m}{m}$. Therefore, the largest element in *X* must be at least $\binom{2m}{m}/m>2^{m-1}$, which implies that *f*(2*m*,1) > 2*m*−1. This establishes the claim for *i*=1.
Now observe that if a set *X*={*x*1,…,*x**n*} is *i*-sum-avoiding, then the set {*x*1*i*,…,*x**n**i*} is 1-sum-avoiding. This means that *f*(*n*,1) ≤ *f*(*n*,*i*)*i*. Since we already know that *f*(*n*,1) is not polynomially bounded, *f*(*n*,*i*) is not polynomially bounded in *n*, either. **QED**.
| 4 | https://mathoverflow.net/users/7982 | 42620 | 27,121 |
https://mathoverflow.net/questions/42463 | 17 | I am interested in learning about the derived categories of coherent sheaves, the work of Bondal/Orlov and T. Bridgeland. Can someone suggest a reference for this, very introductory one with least prerequisites. As I was looking through the papers of Bridgeland, I realized that much of the theorems are stated for Projective varieties (not schemes), I've just started learning Scheme theory in my Algebraic Geometry course, my background in schemes is not very good but I am fine with Sheaves. It would be better if you suggest some reference where everything is developed in terms of Projective varieties.
| https://mathoverflow.net/users/9534 | Derived categories of coherent sheaves: suggested references? | Kapustin-Orlov'a *survey* of derived categories of coherent sheaves is pretty good,
* A. N. Kapustin, D. O. Orlov, *Lectures on mirror symmetry, derived categories, and D-branes*, Uspehi Mat. Nauk **59** (2004), no. 5(359), 101--134; translation in Russian Math. Surveys **59** (2004), no. 5, 907--940, [math.AG/0308173](https://arxiv.org/abs/math/0308173)
but more slow/elementary exposition starting with fundamentals of derived categories is in an earlier survey of Orlov
* D. O. Orlov, *Derived categories of coherent sheaves and equivalences between them*, Uspekhi Mat. Nauk, 2003, Vol. **58**, issue 3(351), pp. 89–172, [Russian pdf](http://www.mathnet.ru/php/getFT.phtml?jrnid=rm&paperid=629&volume=58&year=2003&issue=3&fpage=89&what=fullt&option_lang=eng), English transl. in Russian Mathematical Surveys (2003),58(3):511, [doi link](http://dx.doi.org/10.1070/RM2003v058n03ABEH000629), [pdf](http://www.mi.ras.ru/%7Eorlov/papers/Uspekhi2003.pdf) at Orlov's webpage (not on arXiv!)
There are also Orlov's handwritten slides in djvu from a 5-lecture course in Bonn
* [djvu](http://www.irb.hr/korisnici/zskoda/orlovMPIslides.djvu), but the link is temporary
For derived categories per se, apart from Gelfand-Manin methods book and Weibel's homological algebra remember that a really good expositor is Bernhard Keller. E.g. his text
* Bernhard Keller, *Introduction to abelian and derived categories*, [pdf](http://www.math.jussieu.fr/%7Ekeller/publ/cam.pdf)
...and also his Handbook of Algebra entry on derived categories:
[pdf](http://www.math.jussieu.fr/%7Ekeller/publ/dcu.pdf)
| 16 | https://mathoverflow.net/users/35833 | 42621 | 27,122 |
https://mathoverflow.net/questions/42590 | 5 | I'm not very experienced with respect to Category Theory. So if this question makes no sense I'm sorry. At any rate here is my question: If the existence or non-existence of specific sets can be independent of set theory, then how can it be that the category Set is complete under small limits?
For example, suppose you have a small category A, with objects that are linearly ordered spaces X such that: X is without smallest or largest element, X has CCC, X is complete, and X is dense in itself. And as morphisms for A, you take order preserving bijections. Now, let F be the forgetful functor from A to the underlying set.
How exactly can you define the limit over F inside of Set?
Another example, would be considering some set sized collection of Whitehead groups, call it X. Now, consider X as a category equipped with homomorphisms as morphisms. And let G be the forgetful functor from X into Set.
How exactly can one define the limit over G inside Set?
Or even better, suppose that G is the identity functor from X into Grps. Then, what happens?
Am I correct in saying that the answer depends drastically on the set theoretic universe you pick? If so, how is this not a problem with category theory?
| https://mathoverflow.net/users/8843 | Independence and Category Theory | Like all fields of mathematics, Category Theory is not immune to foundational questions. Although different foundations have been presented, the most common foundation for Category Theory is within Set Theory: Category Theory starts with a given universe of sets and then develops its theory. The computation of limits, colimits, and whatnot will be affected by the overlying structure of sets. However, this overlying structure of sets is not variable. When a computation depends on certain principles of Set Theory, it must be labeled as such. For example, [Grothendieck Universes](http://en.wikipedia.org/wiki/Grothendieck_universe) and [Vopenka's Principle](http://en.wikipedia.org/wiki/Vop%C4%9Bnka%27s_principle) are large cardinal axioms which have direct applications in Category Theory.
That said, Category Theory is in a unique position to deal with independence results that arise from Set Theory. Indeed, every forcing construction in Set Theory has an analogue in Category Theory via sheaves over an appropriate site. More precisely, forcing poset can be viewed as a small category and when endowed with the double-negation topology, the Grothendieck topos of sheaves over this site is equivalent to the Boolean-valued model that one obtains in via forcing. Thus independence results from Set Theory are directly visible to Category Theory by doing the computations inside a Boolean Grothendieck topos instead of the topos of sets.
| 10 | https://mathoverflow.net/users/2000 | 42628 | 27,125 |
https://mathoverflow.net/questions/39300 | 2 | Assume $C$ is a $1$-category and is $M : C \to Cat$ a $1$-functor, such that for every morphism $f : x \to y$, the functor $M(f) : M(x) \to M(y)$, which is denoted by $f\_\\*$, has a left-adjoint functor, which is denoted by $f^\* : M(y) \to M(x)$. Now I expect that $x \to M(x), f \mapsto f^\*$ defines a contravariant pseudo-functor $M \to Cat$. For example, it should be possible to conclude by this method that $X \to \text{Mod}\_X, f \mapsto f^\*$ is a pseudo-functor on the category of ringed spaces.
Now here is what I have tried: Using the units and counits of the adjunction, we get morphisms
$\epsilon\_x : (id\_x)^\* = (id\_x)^\* (id\_x)\_\* \to id\_{M(x)}$,
$\alpha\_{f,g} : f^\* g^\* \to f^\* g^\* (gf)\_\* (gf)^\* = f^\* g^\* g\_\* f\_\* (gf)^\* \to f^\* f\_\* (gf)^\* \to (gf)^\*$,
and these turn out to be isomorphisms. I have already done some calculations resp. diagram chases (which you probably don't want to see), but the necessary compatibility conditions ([Vistoli's Descent notes](http://homepage.sns.it/vistoli/descent.pdf), Def. 3.10) don't seem to be true automatically. Or am I missing something?
If not: Where can I find a nice exposition when these adjoint functors $f^\*$ may be chosen in such a way that we get a pseudo-functor? Or at least, why does everything works out fine in the example with modules, without tedious calculations?
| https://mathoverflow.net/users/2841 | Pointwise left-adjoint yields a pseudo-functor | Here is a way of proving the following (condition b) of Vistoli in your question). I'm assuming condition a) to be strict (the $id$ is the $id$) in here for the moment.
Let us state condition b)
>
> Consider a pseudo-functor $(-)\_\*$ with structural morphisms in the direction $(fg)\_\* \to f\_\* g\_\*$, >that is associative, meaning that
>
>
> $$\begin{matrix} (fgh)\_\* & \to & f\_\* (gh)\_\* \\
> > \downarrow & & \downarrow \\
> > (fg)\_\* h\_\* & \to & f\_\* g\_\* h\_\* \end{matrix}$$
>
>
> commutes. Then if $f\_\*$ has a left adjoint $f^\*$ for any $f$, there is a unique way of defining a >pseudo-functor structure $(-)^\*$ such that $(f)^\*=f^\*$ and such that the two following diagrams >commute:
>
>
> $$\begin{matrix}
> >Id & \to & (gf)\_\*(gf)^\* \\
> >\downarrow & & \downarrow \\
> >g\_\* f\_\* f^\* g^\* & \to & g\_\*f\_\* (gf)^\*
> >\end{matrix}
> >$$ and
>
>
> $$\begin{matrix}
> >f^\*g^\*(gf)\_\* & \to & f^\*g^\* g\_\* f\_\* \\
> >\downarrow & & \downarrow \\
> >(gf)^\* (gf)\_\* & \to & Id
> >\end{matrix}
> >$$
>
>
> Furthermore, if $(-)\_\*$ is associative, meaning the following diagram is commutative:
>
>
> $$\begin{matrix}
> >(fgh)\_\* & \to & (fg)\_\*h\_\* \\
> >\downarrow & & \downarrow \\
> >f\_\*(gh)\_\* & \to & f\_\*g\_\*h\_\*
> >\end{matrix}
> >$$
>
>
> Then $(-)^\*$ will be associative too, meaning the following diagram is commutative:
>
>
> $$\begin{matrix}
> >h^\*g^\*f^\* & \to & (gh)^\*f^\* \\
> >\downarrow & & \downarrow \\
> >h^\*(fg)^\* & \to & (fgh)^\*
> >\end{matrix}
> >$$
>
>
>
Ok, so now, let's prove this statement.
Now here is a way of looking at this type of diagrams obtained from adjunctions.
From now on, some diagonal arrows will appear in the diagrams. They are the natural transformations you are asking about such as $(fg)\_\* \to f\_\*g\_\*$ (or 2-maps, if you want), whereas the other arrows (vertical and horizontal) in the diagrams will simply be usual functors (or 1-maps), such as $f\_\*$.
First of all, you have to convince yourself that when you put together a bunch of diagrams to make a bigger one, the order in which you apply the 2-maps (when you have a choice) is irrelevant, because they act on 1-maps that are in different parts of the formulas giving objects. For example, in the following diagram,
$$\begin{matrix}
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet
\end{matrix}
$$
you have to apply first the lower left 2-map, but then you can do either the lower right or the middle left, and so on. It does not matter, in the end, you get the same 2-map from the lower left part of the perimeter to the upper right one.
All this had nothing to do with adjunctions, so far. Now let's start using them.
You have to convince yourself of the following lemma (see <http://arxiv.org/abs/0806.0569>, lemma 1.2.6).
Given a diagram
$$\begin{matrix}
a & \to & a' \\
\downarrow & \nearrow & \downarrow \\
b & \to & b'
\end{matrix}
$$
such that the vertical maps have left adjoints. Then, there is a diagram
$$\begin{matrix}
a & \to & a' \\
\uparrow & \nwarrow & \uparrow \\
b & \to & b'
\end{matrix}
$$
such that two diagrams commute (see loc. cit.). The point is the existence of the 2-map given by the diagonal arrow, and it is defined exactly as you do it in your question, in the particular example you are considering.
**Whenever this lemma will be applied in what follows, the maps with adjoints will be vertical. The horizontal maps will never be changed.**
Now convince yourself of the following fact: if I put two such diagrams beside each other
$$\begin{matrix}
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet
\end{matrix}
$$
you can compose the 2-maps. Now assuming all vertical maps have left adjoints, you can apply the lemma to both squares and then compose the resulting 2-maps. You can also apply the lemma to the rectangle (forgetting the middle vertical arrow) with the composition as diagonal 2-map. The point is that you get the same result.
You can prove a similar lemma, but with two squares stacked one above each other instead of besides each other.
Finally, here is the relevant point. It helps you getting some commutative diagrams involving left adjoints from commutative diagrams involving right adjoints. I'm not going to draw this here, because it is too painful, but look at Lemma 1.2.7 p. 8 of [loc. cit.](http://arxiv.org/pdf/0806.0569v1). It is stated in the opposite way of what you want: it starts with left adjoints, and says something about right ones. But you can reverse the statement (so start with the second cube and get the first one). The proof of this lemma is left as an exercise in the reference, but the point is that you have no choice as to how to prove it, because it is very general. Write both compositions, and try to complete with smaller commutative diagrams, you'll have no choice as to which diagram to add at each step.
Now, at last, let's apply all this to your own problem. Here is a file with a description of the cube you have to use.
<http://www.math.uni-bielefeld.de/~bcalmes/fichiers/diagram.pdf>
The 2-maps appear as double arrows. I've called $\xi$ and $\zeta$ the pseudo-functor morphisms you are interested in. All unlabeled single arrows are identities. Unlabeled double arrows are either identity or some obvious adjunction morphism (unit or counit).
You have to check that the 2-maps you are getting by adjunction are what you want, but it is completely straightforward.
Now, for condition a) of Vistoli on identities, you can use the same kind of tricks.
| 2 | https://mathoverflow.net/users/4763 | 42638 | 27,132 |
https://mathoverflow.net/questions/42623 | 21 | Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.
Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even ... [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]
So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.
Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is
$w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.
This leads me to ask the question:
**If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?**
| https://mathoverflow.net/users/450 | Are the Stiefel-Whitney classes of a vector bundle the only obstructions to its being invertible? | The answer is no. Let $G$ be a cyclic group of order $n$ not divisible by $2$, let $V$ be an irreducible $2$-dimensional representation of $G$, and consider the associated vector bundle $EG\times\_G V\to BG$ (which I'll call $V$ again). Then $V$ has trivial Stiefel-Whitney classes since $H^q(BG;\mathbb{Z}/2)=0$ if $q>0$, but $V$ can have non-vanishing Pontryagin class (we have $H^\*(BG;\mathbb{Z})=\mathbb{Z}[[x]]/(nx)$ where $x\in H^2$, and $p(V)=1-a^2x^2$ with $a\in \mathbb{Z}/n$ depending on the original representation.) Since the Pontryagin class satisfies Whitney sum up to $2$-torsion, this gives a counterexample: the virtual bundle $-V$ is does not come from a vector bundle, since $p(-V)=(1-a^2x^2)^{-1}$.
Now you ask, what if we also require that the Pontryagin classes are (finitely) invertible? There's probably a counterexample here too, though I don't have one at hand.
**Added later.** Here's one. (I hope: I keep needing to fix it.) If $G$ is a finite *$p$-group*, then the "Borel construction" defines a bijection between of: the set $\mathrm{Rep}\_n(G)$ of isomorphism classes of real $n$-dimensional representations of $G$ into the set $\mathrm{Bun}\_n(G)$ of isomorphism classes of real $n$-dimensional vector bundles over the classifying space $BG$. In representation theory, there are no additive inverses, so non-trivial bundles over $BG$ which come from representations cannot have inverses. So it's enough to find a non-trivial representation $V$ whose characteristic classes all vanish.
Let $G$ be a cyclic group of order $p^2$, where $p$ is an odd prime, generated by an element $\sigma$. Let $L$ be the $1$-dimensional complex representation given by $\sigma|L=e^{2\pi i/p}$, and write $V$ for the *real* $2$-dimensional vector bundle underlying the complex line bundle $EG\times\_G L\to BG$. It appears that the Pontryagin classes vanish: up to signs, the total Pontryagin class is the total Chern class of $V\otimes \mathbb{C}\approx L\oplus \overline{L}$, and we compute $c(V)=c(L)c(\overline{L}) = (1+px)(1-px) = 1-p^2x^2 = 0$.
(The fact about the bijection $\mathrm{Rep}(G)$ into $\mathrm{Bun}(G)$ is non-trivial; it follows from a theorem of Dwyer and Zabrodsky ("Maps between classifying spaces", LNM 1298). I don't know a more elementary proof, but a condition such as "$G$ is a $p$-group" is probably necessary.)
**End addition.**
There are obstructions in $K$-theory to "inverting" bundles. There are exterior power operators $\lambda^k:KO(X)\to KO(X)$, such that if $[V]$ is the $K$-class of an actual bundle $V$, we have $\lambda^k([V])=[\Lambda^k V]$, the class of the exterior power bundle. The formal sum $\Lambda\_t(x)\in KO(X)[[t]]$ given by
$$
\Lambda\_t(x)= 1+\lambda^1(x)t+\lambda^2(x)t^2+\dots
$$
has a Whitney formula ($\Lambda\_t(x+y)=\Lambda\_t(x)\Lambda\_t(y)$), coming from the usual decomposition $\Lambda^nV=\sum \Lambda^iV\otimes \Lambda^{n-i}V$. Furthermore, if $x=[V]$ is the class of an honest $n$-dimensional bundle, we must have $\lambda^i([V])=[\Lambda^iV]=0$ for $i>n$, so that $\Lambda\_t(x)$ is polynomial.
In this case, trivial bundles don't have trivial $\Lambda$-class; instead, writing "$n$" for the trivial $n$-plane bundle, we have $\Lambda\_t(n)=(1+t)^n$. Thus, an isomorphism $V\oplus W= n$ implies an identity $\Lambda\_t([V])\Lambda\_t([W])=(1+t)^n$.
So an even stronger form of your question is: if $V$ is a vector bundle (over a nice space) such that $\Lambda\_t([V])/(1+t)^n \in KO(X)[[t]]$ is a polynomial, must it follow that there exists a bundle $W$ such that $V\oplus W\approx n$? Again, I don't have a counterexample here.
| 21 | https://mathoverflow.net/users/437 | 42639 | 27,133 |
https://mathoverflow.net/questions/42624 | 14 | In the <http://arxiv.org/abs/math/0606464v1> I read
"If you want to prove existence of exotic smooth structure on $\mathbb R^4$ you can do this if you are in
possession of a knot which is topologically slice but not smoothly slice (slice means zero slice genus).
Freedman has a result stating that a knot with Alexander polynomial 1 is topologically slice. We now
have an obstruction (s being non-zero) to being smoothly slice." (p.28)
What does it mean? Does anyone know the construction of exotic $\mathbb R^4$ using slice knots? Please, give me a references, if there are several constructions.
Added: <http://arxiv.org/abs/math/0408379>. Does there exist some other construction?
| https://mathoverflow.net/users/4298 | Slice knots and exotic $\mathbb R^4$ | From Jacob Rasmussen's paper "Knot polynomials and knot homologies", arXiv:math/0504045, p.13 of ArXiv version:
>
> Bob Gompf kindly pointed out another such application [of Rasmussen's $s$-invariant, a concordance invariant of knots extracted from Khovanov homology]. Namely, $s$ can also be used to give a gauge-theory free proof of the existence of an exotic $\mathbb{R}^4$. Indeed, Gompf has shown that to construct such a manifold, it suffices to exhibit a knot $K$ which is topologically but not smoothly slice (see Gompf and Stipsicz, "4-manifolds and Kirby calculus", p. 522 for a proof). By a theorem of Freedman, any knot with Alexander polynomial 1 is topologically slice, so we need only find a knot $K$ with $\Delta\_K(t)=1$ and $s(K) \neq 1$. It is not difficult to provide such a knot - for example, the $(-3,5,7)$ pretzel knot will do.
>
| 14 | https://mathoverflow.net/users/2356 | 42649 | 27,138 |
https://mathoverflow.net/questions/42647 | 45 | Is there a special name for the class of (commutative) rings in which every non-unit is a zero divisor? The main example is $\mathbf{Z}/(n)$. Are there other natural or interesting examples?
| https://mathoverflow.net/users/532 | Rings in which every non-unit is a zero divisor | A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. Also, every $T(A)$ has this property. Thus probably there will be no special terminology except "total rings of fractions".
Artinian rings provide examples: If $x \in A$, the chain $... \subseteq (x^2) \subseteq (x) \subseteq A$ is stationary, say $x^k = y x^{k+1}$ for some minimal $k \geq 0$. If $k=0$, $x$ is a unit. If $k \geq 1$, $x (x^k y - x^{k-1})=0$ and $x^{k-1} \neq y x^k$, i.e. $x$ is a zero divisor.
The class of total rings of fractions is closed under (infinite) products and directed unions. Is it the smallest such class containing the artinian rings?
| 46 | https://mathoverflow.net/users/2841 | 42657 | 27,144 |
https://mathoverflow.net/questions/42660 | 6 | Let $K$ be a field of characteristic zero. Let $G/K$ be a group scheme of finite type.
Assume that $G$ is commutative and connected. For a natural number $n$ denote by $n\_G: G\to G$ the multiplication by $n$ morphism. Is it true that $n\_G$ is surjective with finite kernel?
(I know that the answer is yes provided $G$ is an abelian variety. But the proof of this fact makes use of a very ample invertible sheaf on $G$, so it does not carry over to the general case directly.)
| https://mathoverflow.net/users/8680 | Multiplication by $n$ on commutative algebraic groups | Yes, this is true. The derivative of $n\_G$ at the identity element $e$ is multiplication by $n$ on the Lie algebra (tangent space at $e$) which gives that the kernel is finite and the image of $n\_G$ has the same dimension as $G$ which implies that it is equal to $G$ as $G$ is connected.
**Addendum**: Jim may be right: The statements may be checked over an algebraic closure of $K$ so we may assume $K$ algebraically closed. As $dn\_G$ is an isomorphism the kernel is finite and the dimension of the image is equal to that of $G$. As $G$ is of finite type the image is constructible by Chevalley's theorem and hence contains an open subset of $G$. As it is also a subgroup it is open and as $G$ is connected (and the cosets are also open) we get that the image equals $G$.
| 10 | https://mathoverflow.net/users/4008 | 42664 | 27,148 |
https://mathoverflow.net/questions/42051 | 12 | Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too?
| https://mathoverflow.net/users/9871 | Negative holomorphic sectional curvature | Here is the answer.
Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x\_0\in X$ an choose local holomorphic coordinates $(z\_1,\dots,z\_n)$ centered at $x\_0$ and such that $(\partial/\partial z\_1,\dots,\partial/\partial z\_n)$ is unitary at $x\_0$. Let
$$
\Theta\_{x\_0}(T\_X,\omega)=\sum\_{j,k,l,m=1}^nc\_{jklm}\hspace{0.3mm}dz\_j\wedge d\bar z\_k\otimes\left(\frac\partial{\partial z\_l}\right)^\*\otimes\frac\partial{\partial z\_m}
$$
be the Chern curvature at the point $x\_0$. Consider the induced hermitian form on rank one tensors of $T\_X\otimes T\_X$ given by
$$
\theta\_{T\_{X,x\_o}}(v\otimes w)=\sum\_{j,k,l,m}^nc\_{jklm}\hspace{0.3mm}v\_j\bar v\_k w\_l\bar w\_m,
$$
where
$$
v,w\in T\_{X,x\_0},\quad v=\sum v\_j\hspace{0.3mm}\frac\partial{\partial z\_j},\quad w=\sum w\_j\hspace{0.3mm}\frac\partial{\partial z\_j}.
$$
With this notation, the holomorphic sectional curvature in the direction of $v\in T\_{X,x\_0}\setminus\{0\}$ is given by
$$
\frac{1}{||v||\_\omega^4}\theta\_{T\_{X,x\_o}}(v\otimes v).
$$
The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x\_0$ which is given by
$$
s(x\_0)=2\sum\_{j,k=1}^nc\_{jjkk}.
$$
So, let's compute the integral
$$
\int\_{S^{2n-1}}\sum\_{j,k,l,m}^nc\_{jklm}\hspace{0.3mm}\xi\_j\bar \xi\_k \xi\_l\bar \xi\_m\hspace{0.3mm}d\sigma(\xi),
$$
where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral
$$
\int\_{S^{2n-1}}\xi\_j\bar \xi\_k \xi\_l\bar \xi\_m\hspace{0.3mm}d\sigma(\xi)
$$
vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute
$$
\int\_{S^{2n-1}}|\xi\_j|^2|\xi\_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n.
$$
It is classically known that
$$
\int\_{S^{2n-1}}|\xi\_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n,
$$
and
$$
\int\_{S^{2n-1}}|\xi\_j|^2|\xi\_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n.
$$
Then, we get
$$
\begin{aligned}
\int\_{S^{2n-1}}\sum\_{j,k,l,m}^nc\_{jklm}\hspace{0.3mm}\xi\_j\bar \xi\_k \xi\_l\bar \xi\_m\hspace{0.3mm}d\sigma(\xi) & =\sum\_{j,k=1}^nc\_{jjkk}\left(\delta\_{jk}\frac 2{n(n+1)}+(1-\delta\_{jk})\frac 2{n(n+1)}\right) \\
& = \frac 2{n(n+1)}\sum\_{j,k=1}^nc\_{jjkk}=\frac 1{n(n+1)}s(x\_0),
\end{aligned}
$$
where we have used the Kähler identity $c\_{jklm}=c\_{jmlk}$.
Thus, if $\frac{1}{||v||\_\omega^4}\theta\_{T\_{X,x\_o}}(v\otimes v)$ is negative, so is its average and we are done.
| 9 | https://mathoverflow.net/users/9871 | 42695 | 27,170 |
https://mathoverflow.net/questions/42703 | 0 | Let $k$ be a number field, and $F/k$ a finite extension. I would like to find a countable family of extensions $k\_i/k$ of degree 2 and a place $v\_i$ of $k\_i$ such that if $v$ is the place of $k$ lying below $v\_i$, then $[k\_{v\_i}:k\_v]$=2, where $k\_{v\_i}$ and $k\_v$ are the completions of $k\_i$ and $k$ at $v\_i$ and $v$. Furthermore, for some place $w$ of $F$ lying above $v$, I want the $F\_w = k\_v$. Is this possible?
| https://mathoverflow.net/users/4561 | extensions of number fields | There are infinitely many places $v$ of $k$ such that $F\_w=k\_v$ for some $w$ above $k$. For each such place, take $a\_v \in k, v(a\_v)=1$ and consider the extension $k(\sqrt{a\_v})/k$. It has the property you want at $v$ and you will get infinitely many such extensions as you vary $v$, so you are done.
| 4 | https://mathoverflow.net/users/2290 | 42705 | 27,178 |
https://mathoverflow.net/questions/42706 | 7 | Is there an algorithm which takes as input two lists of words $v\_1,...,v\_n$ and $w\_1,...,w\_n$ over an alphabet $X$ and decides if there is an infinite sequence $(k\_i)$ where $1 \leq k\_i \leq n$ for all $i$ such that $v\_{k\_1}v\_{k\_2}...=w\_{k\_1}w\_{k\_2}...$? It seems that undecidability of the original Post Correspondence problem should imply there is no such algorithm. Is there a reference that shows undecidability of this variation of Post? Thanks.
| https://mathoverflow.net/users/10159 | post correspondence problem variant | See Halava, Vesa, Harju, Tero, Karhumäki, Juhani
Decidability of the binary infinite Post correspondence problem. If the alphabet consists of $\le 2$ letters, then the problem is decidable, if the number of letters is at least 7, then the problem is undecidable. The latter result is proved in Y. Matiyasevich, G. Sénizergues, Decision problems for semi-Thue systems with a few rules, in: Proceedings, 11th Annual IEEE Symposium on Logic in Computer Science, New Brunswick, NJ, 27–30 July 1996, IEEE Computer Society, Silver Spring, MD, pp. 523–531.
| 6 | https://mathoverflow.net/users/nan | 42721 | 27,187 |
https://mathoverflow.net/questions/42652 | 10 | In an earlier post (Use Lie Sub-Groups of GL(3, R) for elastic deformation ? [here](https://mathoverflow.net/questions/40470/use-lie-sub-groups-of-gl3-r-for-elastic-deformation?)), I mentioned polar decompositions as in F = RU where R in SO(3) & U in symmetric positive-semidefinite matrices. In response, I received the following comment: "The decompositions you mention are well-known in the theory of Lie groups, e.g., F=RU is called Iwasawa decomposition."
I have been trying to understand how a polar decomposition & Iwasawa decomposition are related. I have found the following definitions of both.
Iwasawa: G=KAN for G in GL(n, R) where K = orthogonal matrices (rot'ns ?), A = positive diag. matrices & N = upper triag. matrices with diag. entries = 1.
Polar: G=KAK where G = semi-simple Lie group, K = maximal compact subgroup of G, A = abelian subgroup of G. For GL(n, R), I think K in SO(n) but unsure about A. The only abelian subgroup of GL(n, R) appears to be the set of all nonzero scalar matrices (i.e. scalar multiples of identity matrix). I'm also puzzled by G=KAK having 3 terms but F=RU only 2.
Given all of this, I'm still unsure how these two decompositions are related.
| https://mathoverflow.net/users/9624 | Iwasawa Decomposition & Polar Decomposition related how ? | You can obtain the $G=KAK$ decomposition from a decomposition of the type $F=UR$. To avoid unnecessary complications, let's assume that our reductive group $G$ is a selfadjoint subgroup of $\operatorname{GL}(n,\mathbb{R})$. Then the map $g \mapsto g^{-t}$ is an involution of $G$, which is called the Cartan involution and is typically denoted by $\theta$. The first observation to make is that the fixed-point set $K = \{ g \in G \colon \theta(g)=g \}$ of $\theta$ is a maximal compact subgroup of $G$. For example, if $G=\operatorname{GL}(n,\mathbb{R})$, then $K=\operatorname{O}(n)$.
Next we observe that $\theta$ induces an involution (also denoted by $\theta$) at the Lie algebra level: explicitly, this is the map $X \mapsto -X^t$. If $\mathfrak{p}$ denotes the $-1$-eigenspace of this latter involution, then one has the following result.
>
> The map $K \times \mathfrak{p} \to G$ given by $(k, X) \mapsto k e^X$ is a diffeomorphism.
>
>
>
In particular, every $g \in G$ can be expressed as $k e^X$ for some $k \in K$ and $X \in \mathfrak{p}$. This decomposition is known as the Cartan decomposition; it is a generalization of the polar decomposition to $G$ (and is, I presume, the $F=UR$ decomposition stated in the OP). Indeed, if $G = \operatorname{GL}(n,\mathbb{R})$, then $\mathfrak{p}$ is just the set of symmetric matrices, and thus the set $\exp \mathfrak{p}$ consists of symmetric, positive semidefinite matrices.
Now let $\mathfrak{a}$ denote a maximal abelian subspace of $\mathfrak{p}$. Then it can be shown that $A = \exp \mathfrak{a}$ is a closed abelian subgroup of $G$ with Lie algebra $\mathfrak{a}$. It can also be shown that $\mathfrak{a}$ is unique up to conjugacy via an element of $K$. That is to say, if $\mathfrak{a}'$ is another maximal abelian subspace of $\mathfrak{p}$, then there is a $k \in K$ such that $\text{Ad}(k) \mathfrak{a} = \mathfrak{a}'$. With this information we can obtain the decomposition $G=KAK$: given $g \in G$, one observes that $p=gg^t \in \exp \mathfrak{p}$, say $p=e^X$. Thus there is a $k \in K$ such that $\text{Ad}(k)X \in \mathfrak{a}$, and then $e^{-\text{Ad}(k)X/2}kg \in K$ (because it is fixed by $\theta$), whence $g \in KAK$.
This hopefully alleviates your 3-terms-vs-2-terms issue.
I'm not aware of any relationship between the Iwasawa decomposition and the $KAK$ (polar) decomposition.
| 14 | https://mathoverflow.net/users/430 | 42724 | 27,189 |
https://mathoverflow.net/questions/42744 | 30 | [This](https://mathoverflow.net/questions/42709/question-about-hodge-number/42735#42735) question made me wonder about the following:
Are there orientedly diffeomorphic Kähler manifolds with different Hodge numbers?
It seems that this would require that those manifolds are not deformation equivalent.
However, there are examples by Catanese and Manetti that that happens already for smooth projective surfaces.
| https://mathoverflow.net/users/10076 | Diffeomorphic Kähler manifolds with different Hodge numbers | This question was debated in another forum a few years ago. The result was [a note by Frédéric Campana](http://iecl.univ-lorraine.fr/%7EPierre-Yves.Gaillard/DIVERS/hodgenumbers.pdf) in which he describes a counterexample as a corollary of another construction. In 1986 Gang Xiao (*An example of hyperelliptic surfaces with positive index Northeast. Math. J.* **2** (1986), no. 3, 255–257.) found two simply connected complex surfaces $S$ and $S'$ (that is, complex dimension 2), with different Hodge numbers, that are *homeomorphic* by Freedman's classification. The homeomorphism has to be orientation-reversing, but $S \times S$ and $S' \times S'$ are orientedly diffeomorphic and of course still have different Hodge numbers. Freedman's difficult classification is not essential to the argument, because in 8 real dimensions you can use standard surgery theory to establish the diffeomorphism.
Campana also explains that Borel and Hirzebruch found the first counterexample in 1959, in 5 complex dimensions.
| 38 | https://mathoverflow.net/users/1450 | 42746 | 27,197 |
https://mathoverflow.net/questions/42751 | 4 | I am reading [this paper of Teleman and Woodward](http://arxiv.org/abs/math/0312154).
On page 4, they say that $H^4(BG;\mathbb{R})$ can be identified with the space of invariant symmetric bilinear forms on $\mathfrak{g}\_k$. Why is this true? Is there an easy way to see this?
$G$ is a complex reductive Lie group and $\mathfrak{g}\_k$ is the Lie algebra of the compact form of $G$.
It seems like this must be a "standard fact", but I haven't seen it before.
| https://mathoverflow.net/users/83 | Invariant symmetric bilinear forms and H^4 of BG | You need to combine two classical theorems:
1. if $K$ denotes the compact form of $G$, then $K \to G$ is a homotopy equivalence; this is discussed in Bröcker-tom Diecks book "Representations of compact Lie groups", in the section with "complexification" in the title. So $H^{\ast}(BG)=H^{\ast}(BK)$.
2. The Chern-Weil homomorphism $CW : Sym^{\ast} (k^{\ast})^K \to H^{\ast} (BK;\mathbb{R})$ (which exists for an arbitrary Lie group $K$ with Lie algebra $k$) is an isomorphism for compact groups, this goes back to Borel or Cartan (?).
This theorem is discussed and proven in Dupont's beautiful book "Curvature and characteristic classes" (there you find the construction of $CW$ as well, in case you do not now it already).
| 6 | https://mathoverflow.net/users/9928 | 42752 | 27,201 |
https://mathoverflow.net/questions/42749 | 2 | I find in my books it is given by Bott periodicity, but this is not direct and Bott periodicity is not easy. Is there an easy and direct way to define $K^n(X)$, like $K^{-n}(X)$? I just start to learn this stuff...
| https://mathoverflow.net/users/7341 | Easy way to define positive higher K groups? | There is a definition of $K^n$ for positive $n$, without Bott periodicity. This approach goes back to Karoubi, and you can find it in his book "K-theory". The definition (for both, positive and negative $n$) uses Clifford algebras, but no Bott periodicity (Karoubi uses his new, more algebraic and direct definition to prove Bott periodicity). On the other hand, without Bott periodicity, topological $K$-theory is not very interesting. Also, the identification of Karoubi's negative $K$-theory with the ordinary theory is not a triviality - this is more or less equivalent to Bott periodicity.
Another way to phrase the problem: To define $K^{-n}(X)$, you can take the n-fold suspension of $X$ or the n-fold loop space of $Z \times BU$. If you wish to define $K^n (X)$, you need a space $Y\_n$ whose $n$-fold loop space is $Z \times BU$. So what you need to know is that $Z \times BU$ is what topologists call an "infinite loop space", see Adams nice book with the same title. This is what Bott periodicity does, and it is absolutely crucial to define $K$-theory as a USEFUL cohomology theory.
As pointed in other answers to this question, there is another way to construct K-theory as a cohomology theory - the "infinite loop space machines". You can read in Adams book "Infinite loop spaces" about it - this is one of the most pleasant math books I know, not least because he leaves out all the technicalities. If you go into the details of the infinite loop space machines (May or Segal), it becomes WAY more technical than most the proofs of Bott periodicity, at least if you take the details seriously - it takes
May about 80 pages to write everything down. Segals approach looks substantially easier, but that might be due to his very condensed writing.
Using this theory of infinite loop spaces, you get a cohomology theory $k^n$. This theory is closely related to, but by no means equal to the ordinary K-theory (it is connective, as is any spectrum coming from infinite loop space machines).
If you want to study K-theory seriously, there is absolutely no way around the Bott periodicity theorem: I am not aware of any application of K-theory that can be done without it.
The reason is that you cannot compute anything without Bott periodicity.
This starts with the first computations (spheres and projective spaces) and goes on to applications like
Hopf-invariant one problem, vector fields on spheres, Adams work on Im (J), the Adams conjecture a la Becker-Gottlieb, not to mention Index theory...
Last but not least, there are several proofs of B.P. which I found very pleasant to read, for example Atiyah's "Bott periodicity and elliptic operators".
| 10 | https://mathoverflow.net/users/9928 | 42756 | 27,203 |
https://mathoverflow.net/questions/42726 | 12 | Given a modular form, what is the precise formulation of BSD (in particular, the residue formula for the $L$-function at special values)? And what about the special values if the $L$-function is twisted by some character? Does there exist a good reference?
| https://mathoverflow.net/users/5730 | BSD for modular forms | My comments are getting too long, so here is a tentative answer.
First, a general statement: conjectures predicting special values of $L$-functions are formulated for all motives over number fields. Because normalized eigenforms (even twisted by finite order characters $\chi$) are attached to motives (or vice-versa), there exists conjectures predicting the special values of $L(f,s,\chi)$.
Now what do they look like? For simplicity and because you are especially interested in the value of the residue, I'll restrict to the case of a critical value (so $s=0,\cdots,k-1$ if $f$ is of weight $k$ EDIT: $s=1,\cdots,k-1$ Thanks to David Loeffler for pointing this out). The general conjectures then imply that there exists a special element which is a basis of the determinant of the motivic cohomology which is sent to $L(f,s,\chi)$ by the realization morphism from motivic cohomology to Betti cohomology and to a specific basis of the determinant of étale cohomology by the realization morphism to $p$-adic étale cohomology (for any $p$). But forget about this, because when $s≠k/2$ (which I assume henceforth), then $L(f,s,\chi)$ is non-zero (by Jacquet) and in this case, K.Kato has constructed a candidate $z$ for this conjectural element in his article $p$-adic Hodge theory and values of zeta functions of modular forms. I can't really say that this $z$ lives in the right space, simply because I am unsure whether motivic cohomology is properly defined in this case, but at least it lives in something that has all the property you would wish for motivic cohomology (namely the second $K$-theory group $K\_{2}$ of the modular curves) and it is sent to the right value through the realization morphism to Betti cohomology.
So now the Tamagawa Number Conjecture predicts that it should be sent to a specific basis of the determinant of the $p$-adic étale cohmology for any $p$. Unraveling what it means in this case, you get that $H^2(\textrm{Spec}\ \mathbb Z[1/p],T)$ is a finite group and the following conjectural equality:
\begin{equation}
\sharp H^2\_{et}(\textrm{Spec}\ \mathbb Z[1/p],T)=[H^1\_{et}(\textrm{Spec}\ \mathbb Z[1/p],T):z]
\end{equation}
Here $T$ is any lattice in the Galois representation of your modular form and $[-:-]$ denotes generalized index (so $[\mathbb Z\_{p}:1/p]=p^{-1}$).
Now you have a perfectly valid expression of a generalized BSD conjecture, and this is how I think of conjectures about special values in this case. However, you might want to express this conjecture in a way that recovers usual BSD when $f$ comes from an elliptic curve $E$. This again is a doable exercise (albeit one I find non-trivial) which is done for instance in Burns-Flach Math. Ann. 305 (section 1.7) or O.Venjakob London Math. Soc. Lecture Note Ser., 320 (section 3.1). I generally recommend the latter article because I learnt a lot from it myself but beware that, on this very specific question, there is a typo in the definition of one of the crucial objects, if memory serves well, so I rather recommend using both articles in parallel.
| 8 | https://mathoverflow.net/users/2284 | 42757 | 27,204 |
https://mathoverflow.net/questions/42753 | 1 | I have a random variable $X$ and I want to find the probability density function from transforming it through the Heaviside step function. So
$Y = H(X)$
where the $H$ is the Heaviside step function with
$H(0) = 1$
Given the cumulative distribution function of $X$ as $F\_X$, then I have the pdf of $Y$ as
$f\_Y(y) = (1-F\_X(0))\delta(y-1) + F\_X(0) \delta(y)$
However, I think this pdf is for the case when the step function is defined at $0$ as $H(0) = 0$. Do I need to incorporate the point probability
$F\_X(0) - \lim\_{(x \rightarrow 0^-)}F\_X(x)$
into $f\_Y$ to account for the case where $H(0)=1$, or is there a better/nicer way to do it?
| https://mathoverflow.net/users/10009 | Heaviside Step Function of a Random Variable | I don't think there's a much nicer way. You could either say
>
> Let $G\_X(x)=\mathbb{P}(X <x)=\lim\_{(t\rightarrow x^-)}F\_X(t)$; then
> $$
> f\_Y(y)=(1-G\_X(0))\delta(y-1)+G\_X(0)\delta(y)
> $$
>
>
>
or change the random variable. Since
$$\mathbb P(X<x)=1-\mathbb P(X\ge x)=1-\mathbb P(-X\le -x)=1-F\_{-X}(-x),$$
we have
>
> $$
> f\_Y(y)=F\_{-X}(0)\delta(y-1)+(1-F\_{-X}(0))\delta(y).
> $$
>
>
>
| 2 | https://mathoverflow.net/users/4600 | 42758 | 27,205 |
https://mathoverflow.net/questions/42755 | 11 | Let $\cal{A}$ be an abelian category with enough projectives and $\mathbf{C}\_+ (\cal{A})$ the category of bounded below chain complexes.
Since Quillen (Homotopical algebra, 1.2, examples B), there is a well-known "standard" model category structure on $\mathbf{C}\_+ (\cal{A})$ taking as weak equivalences the maps inducing isomorphisms on homology, as fibrations the degree-wise epimorphisms in $\cal{A}$ and with cofibrations maps $i$ which are injective and such that $\mathrm{cok}\ i$ is a complex having a projective object of $\cal{A}$ in each degree.
More recently, Hovey (Model categories), proved an analogous result for the category of *unbounded* chain complexes, but with ${\cal A} = R$-modules, $R$ a ring (but cofibrations are not so easy to characterise). Finally, it's folklore (at least, I don't know if it is published somewhere) that the same holds for $\cal{A}$ an abelian category with a projective generator -the fact that allows the small object argument to work, as Eric Wofsey points out in his answer to [this MO question](https://mathoverflow.net/questions/141/model-category-structures-on-categories-of-complexes-in-abelian-categories).
>
> I'm interested in the following variant of this problem: is there a model structure on $\mathbf{C}\_+ (R)$ taking as weak equivalences the **homotopy** equivalences?
>
>
>
If it's true, I think this should be easy to verify: just taking a look to the classical proof and seing if you can change "**homology** equivalences" everywhere by "**homotopy** equivalences". I'm willingly going to do it, but, prior to start, I would like to know if it is already done, much as in the case of topological spaces where, together with the "standard" (Quillen too) model structure with **weak homotopy** equivalences as weak equivalences, there is the Strom model structure (The homotopy category is a homotopy category), with **homotopy** equivalences as weak equivalences.
| https://mathoverflow.net/users/1246 | Non standard (?) model category structure on (co)chain complexes. | This is well known, but formulated in a slightly different way.
Recall that a Frobenius category is an exact category which has enough injectives as well as enough projectives, and such that an object is projective if and only if it is injective (injectivity (resp. projectivity) is defined with respect to inflations (resp. deflations)).
If you forget about the existence of (finite) limits and colimits, any Frobenius category satisfies all the axioms of a Quillen model category: the cofibrations (resp. fibrations) are the inflations (resp. deflations), while the weak equivalences are the maps which factor through some projective-injective object. Therefore, any Frobenius category which has finite limits as well as finite colimits is a (stable) closed model category in the sense of Quillen.
Now, given an additive category $A$, the category of chain complexes $C^\sharp(A)$ (where $\sharp=\varnothing$ for unbounded chain complexes, $\sharp=b$ for bounded chain complexes, etc) is a Frobenius category: inflations (resp. deflations) are the degreewise split monomorphisms (resp. split epimorphisms), and projective-injective objects are the contractible chain complexes. If $A$ has finite limits as well as finite colimits, this shows that the category $C^\sharp(A)$ is a stable closed model category whose cofibrations (fibrations) are the degreewise split monomorphisms (resp. epimorphisms), and whose weak equivalences are the chain homotopy equivalences.
| 20 | https://mathoverflow.net/users/1017 | 42761 | 27,207 |
https://mathoverflow.net/questions/42759 | 6 | Let $F\_1,...,F\_m$ be a partition of the 3-element subsets of $[n]$ into families such that no three subsets in any one family $F\_i$ are all contained in one 4-element subset of $[n]$. What is the minimum value of $m$?
| https://mathoverflow.net/users/10171 | partitioning the 3-sets of [n]={1,...,n} into families | Tony Huynh's update can be easily generalized to show that
$$n\geq k\left(R(\underbrace{3,3,\dots,3}\_{k-1})-1\right)+3\implies m\geq k$$ and so we get a very weak lower bound on $m$ which at least shows that $m \_{min}\to \infty$.
For an easy upper bound $m\_{min}\le \lfloor\frac{n+1}{2}\rfloor$, which you can see by partitioning the triples $(a,b,c)$ in classes according to $a+b+c\pmod{\lfloor\frac{n+1}{2}\rfloor}$.
| 3 | https://mathoverflow.net/users/2384 | 42768 | 27,212 |
https://mathoverflow.net/questions/42548 | 16 | Let $G$ be a compact Lie group, $X$ be a (compact, oriented) smooth manifold, with $G$ acts on $X$ smoothly. Then we can talk about the $G$-equivariant homology and cohomology.
My question: In what sense can we have a duality between the equivariant homology and cohomology, in analogue with the Poincare duality between the ordinary homology and cohomology of $X$? In particular, the degree of a nontrivial equivariant (co)homology class could exceed the dimension of the manifold $X$. Then in such case, what does the dual mean, geometrically?
| https://mathoverflow.net/users/2555 | Poincare dual in equivariant (co)homology? | There is a paper by Brion: "Poincaré duality and equivariant (co)homology," *Michigan Math. J.* 48 (2000). As mentioned in one of the comments, he uses Borel-Moore as the homology theory.
That said, one should always be careful about the word "duality" in this context -- often people just mean there's a canonical *isomorphism* $H\_G^k X \rightarrow H^G\_{n-k}X$. (Which does exist using equivariant Borel-Moore homology, when $X$ is smooth.) The geometric meaning is the same as in the ordinary case: if $V$ is a $G$-invariant sub(manifold/variety/cycle/etc) of codimension $k$, it defines a fundamental class in $H^G\_{n-k} X$, which is identified with $H\_G^kX$ by means of the isomorphism.
Duality could also refer to a pairing $H\_G^k X \otimes H\_G^\ell X \rightarrow H\_G^{k+\ell-n}(point)$, given by the equivariant pushforward (integral): $$\alpha\otimes\beta \mapsto \int \alpha\cdot\beta .$$ Using this pairing, and assuming that $H\_G^\*X$ is free as a module over $H\_G^\*(point)$, any $H\_G^\*(point)$-module basis for $H\_G^\*X$ has a dual basis, in the usual sense of linear algebra. (Arguably, this duality pairing is what should really be called "Poincaré duality".) Again, the geometric meaning is similar to the ordinary case: if you're lucky enough to have invariant subvarieties whose classes form a basis for $H\_G^\*X$, the Poincaré dual basis is given by classes of subvarieties intersecting the original ones transversally (if they exist). However, the existence of geometrically defined dual bases is a stronger statement, because the equivariant integral is generally nonzero on classes of degree greater than $\dim X$.
(I should say all this is prejudiced toward the equivariant cohomology rings that usually show up in algebraic geometry, e.g., $H\_G^\*X$ is a free module over $H\_G^\*(point)$, so that it makes sense to talk about dual bases.)
PS: Defining equivariant Borel-Moore homology requires a little more care, since the spaces $EG\times^G X \rightarrow BG$ are infinite-dimensional fiber bundles. But they have finite-dimensional approximations $EG\_m \times^G X \to BG\_m$, so it makes sense to define $$H^G\_k X = H\_{k+\dim BG\_m}(EG\_m\times^G X)$$
for $m\gg0$. The equivariant Poincaré isomorphism is just the ordinary one for these approximation spaces.
| 15 | https://mathoverflow.net/users/5081 | 42789 | 27,223 |
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