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https://mathoverflow.net/questions/23624 | 4 | I am trying to find a reference for the following theorem:
Let $R$ be a complete DVR, and let $Y$ be a scheme projective and flat over $R$. Suppose that $X\_0 \longrightarrow Y\_0$ is a finite etale morphism, where $Y\_0$ is the fiber of $Y$ over the unique closed point of Spec $R$. Then there exists a scheme $X$, finite etale over $Y$, together with a closed immersion $X\_0 \longrightarrow X$ such that $X\_0 = X \times\_Y Y\_0$.
In a special case, this allows us to lift etale morphisms in characteristic $p$ to etale morphisms in characteristic zero, and can thus be important in studying the etale fundamental group of varieties in finite characteristic.
Note: While any reference would be appreciated, what I am really interested in is a reference that I can give for attribution purposes. My advisor thinks that this result was proved by Deligne about 40 years ago, but I would like something a little more solid.
| https://mathoverflow.net/users/5094 | Lifting Etale Morphisms | SGA 1, IX, 1.10: Let $Y$ be a scheme proper over a complete local noetherian ring $R$,
and let $Y\_0$ be the closed fibre of $Y/R$. Then the functor $X\mapsto X\_0$ from
finite etale coverings of $Y$ to finite etale coverings of $Y\_0$ is
an equivalence of categories.
| 14 | https://mathoverflow.net/users/930 | 23636 | 15,544 |
https://mathoverflow.net/questions/23573 | 6 | I have looked through all my standard algebraic geometry texts and tried many tricks using Zariski's main theorem and Noether normalization, but remain stuck by the following:
Let $\pi:X\to S$ be a morphism of finite type between integral, Noetherian schemes and let $x$ be a point of $X$. Does there exist an open neighbourhood of $X$ which admits a finite, surjective morphism onto a smooth $S$-scheme?
In this generality I think that the answer is 'no', though I do not have a counterexample. What if we impose additional assumptions such as $\pi$ being flat or proper (or even projective)?
A related question, an affirmative answer to which would imply the same for the previous question in the projective case, is the following: if $X$ is a projective scheme over a local ring $A$, then does $X$ admit a finite surjection to $\mathbb{P}\_A^d$ for some $d\ge 1$?
(I am imagining that $A=\mathbb{Z}\_p$, so please do not assume that the residue field of $A$ is infinite!)
Thank you!
---
Update
------
With Brian's help (thank you), the interesting remaining problem is the following: does every projective variety $V$ over a finite field $k$ admit a finite surjective morphism to $\mathbb{P}\_k^d$ for some $d$? I have a gnawing suspicion that the answer is 'no'.
It is useful to remember Noether normalisation in this case: if $I$ is a non-zero ideal of $k[X\_1,\dots,X\_n]$, then one can find a finite morphism $k[Y\_1,\dots,Y\_{n-1}]\to k[X\_1,\dots,X\_n]/I$ by sending $Y\_i$ to $X\_i-X\_n^e$ for some big enough $e\ge 1$. Unfortunately, projectivising this construction produces a morphism $\mathbb{P}\_k^n\setminus C\to\mathbb{P}\_k^{n-1}$ where $C$ is quite a large closed subscheme of $\mathbb{P}\_k^n$ (unless I have made a mistake); so if $V$ is our variety inside $\mathbb{P}\_k^n$, then it is difficult to ensure that $V$ doesn't meet $C$. Therefore we can't successively project down to smaller dimensional spaces. (In contrast with the case when $k$ is infinite, for then we use changes of variables looking like $Y\_i=X\_i-\alpha\_iX\_n$ and the resulting morphism between projective spaces is defined everywhere except for one point, which we can assume doesn't lie on $V$).
| https://mathoverflow.net/users/5830 | Do morphisms locally decompose into finite surjective followed by smooth? (update: Is every projective variety over a finite field a finite cover of $\mathbb{P}^d$ for some $d$?) | I claim that every projective scheme $V$, over a finite field $k$, all of whose components have dimension $\leq d$, admits a finite morphism to $\mathbb{P}^d\_k$. Let $q=|k|$.
**Key Lemma:** Let $V\_1$, $V\_2$, ... $V\_s$ be a finite collection of subvarieties of $\mathbb{P}\_k^N$. Then there is a homogenous polynomial $f$ in $k[x\_0, x\_1, \ldots, x\_N]$ such that $f|\_{V\_i}$ is nonzero for every $i$.
**Proof:** (improved thanks to comments below)
Choose a closed point $v\_i$ on each $V\_i$; we will force $f$ not to vanish at any of the $v\_i$. Choose $M$ large enough that every $V\_i$ is in $\mathbb{P}^N(\mathbb{F}\_{q^M})$. Let $a\_0$, $a\_1$, ... $a\_N$ be a basis for $\mathbb{F}\_{q^{(N+1)M}}$ over $\mathbb{F}\_{q^M}$. Then the linear form
$$a\_0 x\_0 + a\_1 x\_1 + \cdots a\_N x\_N$$
does not vanish on any $Vvi$. This, of course, does not have coefficients in $k$. But the product
$$\prod\_{i=0}^{(N+1)M-1} \left( a\_0^{q^i} x\_0 + a\_1^{q^i} x\_1 + \cdots a\_N^{q^i} x\_N \right)$$
is similarly nonzero, and does have coefficients in $k$. This concludes the proof of the key lemma.
---
Now, let $V$ be the projective variety, for which we want to prove the result. By the key lemma, there is some polynomial $f\_0$, of degree $r\_0$, which does not vanish on $V$. Every irreducible component of $V \cap \{ f\_0 =0 \}$ is of dimension $\leq d-1$. Apply the lemma again to these irreducible components to find a polynomial $f\_1$, of degree $r\_1$, so that every component of $V \cap \{ f\_0 = f\_1 = 0 \}$ has dimension $\leq d-2$. Continuing in this manner, we construct polynomials $f\_0$, $f\_1$, ..., $f\_d$ such that $V \cap \{ f\_0 = f\_1 = \cdots = f\_d = 0 \}$ is empty.
Set $R=\prod r\_i$. Our map $V \to \mathbb{P}^d\_k$ will be given by
$$(f\_0^{R/r\_0} : f\_1^{R/r\_1} : \cdots : f\_d^{R/r\_d})$$.
By our construction, this map has no base points, and is thus well defined.
---
By the "standard argument", this map is finite. I now recall the standard argument.
Since $V$ is proper, we just need to check that the map has finite fibers.
Suppose, for the sake of contradiction, that the the fiber above
$$(a\_0: a\_1 : \cdots a\_{i-1} : 1 : a\_{i+1} : \cdots : a\_d)$$
has positive dimension.
Let $C$ be a curve in this fiber, so $f\_j^{R/r\_j} - a\_j f\_i^{R/r\_i}$ vanishes on $C$ for all $j \neq i$.
The curve $C$ must meet the hypersurface $f\_i=0$, say at $z$. Then all the $f\_i$ vanish at $z$, contradicting our construction.
| 6 | https://mathoverflow.net/users/297 | 23650 | 15,556 |
https://mathoverflow.net/questions/23523 | 9 | [I want to think that this question has an answer, but it may be more a "community wiki" discussion. Feel free to re-tag.]
>
> What are trig classes like within a universe that's "noticeably"[\*] hyperbolic?
>
>
>
Using an appropriate model to peek into such a universe from our Euclidean one shows us that all the fundamental trig relations (Law of Sines, Law of Cosines, etc) involve transcendental functions in both angle measures and lengths. But, then, in *our* earliest mathematical days, similar triangles allowed us to build a theory of (circular) trig functions based on ratios of lengths of sides, and eventually our mathematics became sophisticated enough to include hyperbolic functions and to be comfortable enough with them and how they might apply to our non-Euclidean models. (We also had the benefit of being able to interpret products as areas of rectangles whose side-lengths correspond to factors.) What if you don't --*can't*-- draw from our experience base?
How do you even *get started* developing trigonometry (or explaining it to your hyperbolic trig students[\*\*]) without a concept of similar triangles?
Obviously(?), the Unit Circle is out.
It makes some sense that the Angle of Parallelism would be the fundamental bridge between angle-information and distance-information; one can imagine that hyperbolic people would be "aware" of the phenomenon on some level, and we know that it's a "universal" property. Even so, the most-concise representations of the AoP relationship are transcendental in both angle measure and length. How insightful would a hyperbolic mathematician have to be to discern the equations from tables of observed measurements? And is there a clear path from those equations to, say, (what we know as) the Law of Sines and the Laws of Cosines?
Or perhaps the fundamental figure in an "intrinsic" hyperbolic trig class is the Equilateral Triangle. This idea actually *exploits* non-similarity to set up a bridge between angle-information and distance-information (with area-information thrown in as a bonus); and it seems that it *might* be more likely to provide a path to the Laws of Sines and Cosines, since it already relates angles and sides of triangles. But, is it really a particularly helpful starting point? Can you get from there to the Angle of Parallelism relation?
Something else?
[\*] For instance, anyone can pull out a protractor and easily see that triangles have an angle-sum less than 180 degrees.
[\*\*] "hyperbolic" modifying both "trig" and "students". :)
| https://mathoverflow.net/users/5609 | What are trig classes like within a universe that's "noticeably" hyperbolic? | Chapter V of Harold E. Wolfe's book: Introduction to Non-Euclidean Geometry (Holt, Rinehart, and Winston), 1945 (and reprinted, 1966) is entitled: Hyperbolic Plane Trigonometry, and has a systematic treatment of this topic.
| 3 | https://mathoverflow.net/users/1618 | 23653 | 15,559 |
https://mathoverflow.net/questions/23654 | 4 | There's an exercise at the end of Ch. 2 of Mosher & Tangora's "Cohomology Operations and Applications in Homotopy Theory", which says:
Suppose the cocycle $u\in C^{2p}(X;Z)$ satisfies $\delta u=2a$ for some $a$.
i. Show that $u \cup\_0 u + u\cup\_1 u$ is a cocycle mod 4.
ii. Define a natural operation, the Pontrjagin square, $P\_2:H^{2p}(-;Z\_2)\rightarrow H^{4p}(-;Z\_4)$.
iii. Show that $\rho P\_2(u)=u\cup u$, where $\rho:H^\*(-;Z\_4)\rightarrow H^\*(-;Z\_2)$ denotes reduction mod 2.
iv. Show that $P\_2(u+v)=P\_2(u)+P\_2(v)+u\cup v$, where $u\cup v$ is computed with the non-trivial pairing $Z\_2 \otimes Z\_2\rightarrow Z\_4$.
---
First of all, I'm confused by the first sentence. Shouldn't it just say $u$ is a cochain, not a cocycle? Probably more importantly, I don't really see what's supposed to be going on in part (i), since those two addends aren't in the same cohomological dimensions. I'm getting that
$\delta (u\cup\_0 u+u\cup\_1 u) = 4(a\cup u)+2(a\cup\_1 u+u\cup\_1 a)$,
and I'm not sure how to cancel the last terms. The addends in the RHS came straight from the addends on the LHS, i.e. there's no interaction between the two terms as far as I can tell, which makes me doubt myself. This could be just a lot of silliness on my part, but I'd appreciate it if someone could clear this up for me. And of course I'd love to hear about fun or unexpected applications of this particular operation...
**EDIT**
Following Tyler's suggestion, I've found that changing the formula to $u\cup\_0 u+u\cup\_1 \delta u$ does wonders.
Now that I've reached it, I'm having difficulty with part (iv). I have that
$P\_2(u+v)=P\_2(u)+P\_2(v)+u\cup\_0 v+v\cup\_0 u+u\cup\_1 \delta v+v\cup\_1 \delta u$.
So somehow those last four terms are supposed to collapse into $u\cup v$ as computed using the non-trivial pairing. How exactly should this work? I have a vague sense that a representing cochain should only be spitting out the values 0(mod 4) and 2(mod 4), the latter only when the cup guys both spat out 1(mod 2)'s. So I'm slightly inclined to believe that the first two loose terms, which are equal since $\deg(u)=\deg(v)=2p$ and $\cup\_0=\cup$ (the usual cup product), might sum to the desired "$u\cup v$" thing. Assuming we represent $u$ and $v$ by the same cochains throughout the equation, which I'm pretty sure is a valid thing to do(???), then if they evaluate to an even number then that sum will be 0(mod 4) while if they evaluate to an odd number then that sum will be 2(mod 4). So I'm thinking that somehow those last two terms should vanish. But why? And is there a concise way of writing everything I've just said, e.g. using notation $\rho:Z\_2\otimes Z\_2\rightarrow Z\_4$ for the pairing, etc.? ***And probably more importantly, is there any significance to the fact that the cup product measures the failure of the Pontrjagin square to be a group homomorphism?***
| https://mathoverflow.net/users/303 | Pontrjagin square (and possible typo in Mosher & Tangora?) | I believe the formula shoud be
$$
u \cup\_0 u + u \cup\_1 \delta u
$$
instead.
EDIT: In order to answer your second question about part (iv), consider the expression
$$
u \cup\_0 v + v \cup\_0 u + u \cup\_1 \delta v + v \cup\_1 \delta u.
$$
Upon looking at this, you might say that the two cup-0 products should be "basically" the same, because they're multiplied in the opposite order. However, because the cup-product is not actually commutative, you can't just switch $v \cup\_0 u$ with $u \cup\_0 v$ - you need to subtract off a coboundary (involving the cup-1 product). So essentially you need to take the expression that you already have and introduce correction coboundary terms (which don't change the cohomology class) to reduce it to the form you're interested in.
So far as the significance of the additivity formula: The Pontrjagin square constructs a lift of the "squaring" operation from mod-2 cohomology to mod-4 cohomology. However, taking an element to its square isn't additive mod 4 like it is mod 2, because $(x+y)^2 = x^2 + 2xy + y^2$. The formula tells you the lift of the squaring operation to mod-4 cohomology satisfies this formula.
| 7 | https://mathoverflow.net/users/360 | 23656 | 15,561 |
https://mathoverflow.net/questions/16817 | 5 | What is the expected length of the longest consecutive subsequence of a random permutation of the integers 1 to N? To be more precise we are looking for the longest string of consecutive integers in the permutation (disregarding where this string occurs). I believe the answer should be ~ c ln(n), but I have been unable to prove this.
Update: We are looking for the longest increasing subsequence. This is to say 9,1,5,2,7,3 has an increasing subsequence 1,2,3.
| https://mathoverflow.net/users/630 | longest consecutive subsequence of a random permutation | The purpose of this answer is to use the [second moment method](http://en.wikipedia.org/wiki/Second_moment_method) to make rigorous the heuristic argument of Michael Lugo. (Here is why his argument is only heuristic: If $N$ is a nonnegative integer random variable, such as the number of length-$r$ increasing consecutive sequences in a random permutation of $\{1,2,\ldots,n\}$, knowing that $E[N] \gg 1$ does not imply that $N$ is positive with high probability, because the large expectation could be caused by a rare event in which $N$ is very large.)
**Theorem:** The expected length of the longest increasing block in a random permutation of $\{1,2,\ldots,n\}$ is $r\_0(n) + O(1)$ as $n \to \infty$, where $r\_0(n)$ is the smallest positive integer such that $r\_0(n)!>n$. ("Block" means consecutive subsequence $a\_{i+1},a\_{i+2},\ldots,a\_{i+r}$ for some $i$ and $r$, with no conditions on the relative sizes of the $a\_i$.)
**Note:** As Michael pointed out, $r\_0(n)$ is of order $(\log n)/(\log \log n)$.
**Proof of theorem:** Let $P\_r$ be the probability that there exists an increasing block of length at least $r$. The expected length of the longest increasing block is then $\sum\_{r \ge 0} r(P\_r-P\_{r+1}) = \sum\_{r \ge 1} P\_r$. We will bound the latter sum from above and below.
*Upper bound:* The probability $P\_r$ is at most the expected *number* of increasing blocks of length $r$, which is exactly $(n-r+1)/r!$, since for each of the $n-r+1$ values of $i$ in $\{0,\ldots,n-r\}$ the probability that $a\_{i+1},\ldots,a\_{i+r}$ are in increasing order is $1/r!$. Thus $P\_r \le n/r!$. By comparison with a geometric series with ratio $2$, we have $\sum\_{r > r\_0(n)} P\_r \le P\_{r\_0(n)} \le 1$. On the other hand $\sum\_{1 \le r \le r\_0(n)} P\_r \le \sum\_{1 \le r \le r\_0(n)} 1 \le r\_0(n)$, so $\sum\_{r \ge 1} P\_r \le r\_0(n) + 1$.
*Lower bound:* Here we need the second moment method. For $i \in \{1,\ldots,n-r\}$, let $Z\_i$ be $1$ if $a\_{i+1}<a\_{i+2}<\ldots<a\_{i+r}$ and $a\_i>a\_{i+1}$, and $0$ otherwise. (The added condition $a\_i>a\_{i+1}$ is a trick to reduce the positive correlation between nearby $Z\_i$.) The probability that $a\_{i+1}<a\_{i+2}<\ldots<a\_{i+r}$ is $1/r!$, and the probability that this holds while $a\_i>a\_{i+1}$ fails is $1/(r+1)!$, so $E[Z\_i]=1/r! - 1/(r+1)!$. Let $Z=\sum\_{i=1}^{n-r} Z\_i$, so
$$E[Z]=(n-r)(1/r! - 1/(r+1)!).$$
Next we compute the second moment $E[Z^2]$ by expanding $Z^2$. If $i=j$, then $E[Z\_i Z\_j] = E[Z\_i] = 1/r! - 1/(r+1)!$; summing this over $i$ gives less than or equal to $n/r!$. If $0<|i-j|<r$, then $E[Z\_i Z\_j]=0$ since the inequalities are incompatible. If $|i-j|=r$, then $E[Z\_i Z\_j] \le 1/r!^2$ (the latter is the probability if we drop the added condition in the definition of $Z\_i$ and $Z\_j$). If $|i-j|>r$, then $Z\_i$ and $Z\_j$ are independent, so $E[Z\_i Z\_j] = (1/r! - 1/(r+1)!)^2$. Summing over all $i$ and $j$ shows that
$$E[Z^2] \le \frac{n}{r!} + E[Z]^2 \le \left(1 + O(r!/n) \right) E[Z]^2.$$
The second moment method gives the second inequality in
$$P\_r \ge \operatorname{Prob}(Z \ge 1) \ge \frac{E[Z]^2}{E[Z^2]} \ge 1 - O(r!/n).$$
If $r \le r\_0(n)-2$, then $r! \le (r\_0(n)-1)!/(r\_0(n)-1)$, so $r!/n \le 1/(r\_0(n)-1)$, so $P\_r \ge 1 - O(1/r\_0(n))$. Thus
$$\sum\_{r \ge 1} P\_r \ge \sum\_{r=1}^{r\_0(n)-2} \left( 1 - O(1/r\_0(n)) \right) = r\_0(n) - O(1).$$
| 12 | https://mathoverflow.net/users/2757 | 23669 | 15,571 |
https://mathoverflow.net/questions/23676 | 0 | In Sato's theory, the following formal delta function is defined:
$\delta(\lambda,z)=\frac{1}{\lambda}\sum\_{n=-\infty}^\infty(\frac{z}{\lambda})^n=\frac{1}{z}\frac{1}{1-\lambda/z}+\frac{1}{\lambda}\frac{1}{1-z/\lambda}$
Given a function
$f(z)=\sum a\_iz^i$,
$f(\lambda)\delta(\lambda,z)=f(z)\delta(\lambda,z)$.
I want to know the properties as many as possible. Or useful references are welcome to be provided. Thanks!
| https://mathoverflow.net/users/5705 | Who can tell me the properties for the delta function in Sato's theory? | The formal delta function obeys the usual properties that the Dirac delta function does, but relative to the pairing defined by the residue. For instance,
$$ \mathrm{Res}\_z f(z)\delta(z,w) = f(w)$$
for any formal distribution $f(z)$.
This and more can be found in Kac's [*Vertex algebras for beginners*](http://books.google.com/books?id=PIhm9-37IlUC), particularly Proposition 2.1 in §2.1.
| 3 | https://mathoverflow.net/users/394 | 23688 | 15,581 |
https://mathoverflow.net/questions/23692 | 11 | Let $N$ be a normal subgroup of $G \times H$, and let $\pi\_1: G \times H \to G$ and $\pi\_2: G \times H \to H$ be the canonical projections. Then $\pi\_1(N)$ is normal in $G$ and $\pi\_2(N)$ is normal in $H$. What else can we say? I know that it is not true, in general, that $N \simeq \pi\_1(N) \times \pi\_2(N)$.
I'm particularly interested in the case where $G$ and $H$ are simple. In that case, $N \simeq \pi\_1(N) \times \pi\_2(N)$ except possibly in the case where $\pi\_1(N) = G$ and $\pi\_2(N) = H$. In that case, what do we know?
| https://mathoverflow.net/users/913 | What are the normal subgroups of a direct product? | See [Goursat's lemma](http://en.wikipedia.org/wiki/Goursat_lemma).
| 14 | https://mathoverflow.net/users/2757 | 23693 | 15,583 |
https://mathoverflow.net/questions/23680 | 1 | What is $I\_{0.5}(a,b)$ where I is the regularized incomplete beta function?
| https://mathoverflow.net/users/634 | What is the value of the regularized incomplete beta function at x=0.5? | You mean this?
<http://en.wikipedia.org/wiki/Beta_function>
$$
\frac{\int\_{0}^{\frac{1}{2}} t^{a - 1} (1 - t)^{b - 1} d t}{\int\_{0}^{1} t^{a - 1} (1 - t)^{b - 1} d t} = \\
\quad{}\quad{}\frac{\mathrm{hypergeom} \Bigl([a,-b + 1],[1 + a],\frac{1}{2}\Bigr) \Gamma (a + b)}{2^{a} a \Gamma (b) \Gamma (a)}
$$
Why do you think there is anything simpler in general?
| 0 | https://mathoverflow.net/users/454 | 23696 | 15,585 |
https://mathoverflow.net/questions/23684 | 7 | **Update:** The question is completely answered. I had overlooked a reduction to the self-adjoint case, and the latter can be proved using a Hahn-Banach separation theorem. Thanks to Matthew Daws for first pointing this out to me. Thanks also to Willie Wong for pointing out that I should have asked the author; I later did so, and he confirmed that a Hahn-Banach separation theorem was intended (and generously provided another argument). I want to point out, because this post may highlight a minor error in the book, that it is [a great book](https://books.google.com/books?id=VtSFHDABxMIC&printsec=frontcover&dq=paulsen+completely&hl=en&ei=WhnjS7rPLYXCNaOP5PAC&sa=X&oi=book_result&ct=result#v=onepage&q&f=false).
---
When I hear "[Krein-Milman](https://planetmath.org/KreinMilmanTheorem)", I think of the result at the link, a standard result in functional analysis. But recently I came across an invocation of "the Krein-Milman theorem (for cones)" which looks more like a Hahn-Banach type result, and I am having trouble tracking down what precisely the author is referring to (and why the result is true).
To make my question precise, I'll have to give the basic setup. Let $V$ be a complex vector space with a conjugate linear involution $\*$, a norm that makes $\*$ an isometry, and a norm-closed cone $C\subset\{v\in V:v=v^\*\}$, meaning $C$ is closed under addition and nonnegative real scalar multiplication, and satisfies $C\cap(-C)=\{0\}$. (Such a $V$ is called an ordered $\*$-vector space.)
To paraphrase part of Paulsen's [exposition](https://books.google.com/books?id=4WgnHzOdHt4C&lpg=PP1&dq=paulsen%25252520completely%25252520bounded&pg=PA179#v=onepage&q&f=false) of the Choi-Effros abstract [characterization](https://mathscinet.ams.org/mathscinet-getitem?mr=430809) of operator systems1:
>
> If $x$ is in $V\setminus C$, then by the Krein-Milman theorem (for cones) there exists a linear functional $s:V\to\mathbb{C}$ with $s(C)\subseteq[0,\infty)$ and $s(x)<0$.
>
>
>
Because $C$ is a closed convex subset of a normed (and thus locally convex) space, my first inclination was to adapt some version of Hahn-Banach. However, I can't see how to do this while keeping complex linearity and the conclusion of the above claim. (There would be no problem in the real case, e.g. by Theorem 3.4 of Rudin's *Functional Analysis*, 2nd edition, page 59.) I've looked across the internet and my library's bookshelves to try to find what is meant, with no luck so far.
>
> Question 1: Does anyone have a reference for (or statement and explanation of) the "Krein-Milman theorem for cones"?
>
>
> Question 2: Am I missing a straightforward argument proving the above claim, for example by an application of Hahn-Banach?
>
>
>
I believe I have faithfully presented enough to cover everything relevant to my question, but the spaces in Paulsen's proof actually have a lot more structure, which you can find at the link above if you think it will help. Hopefully I didn't lose something essential in an effort to not get bogged down--this is clearly a risk because I don't know what theorem is being cited.
1 An operator system is a self-adjoint unital subspace of the algebra of bounded operators on a Hilbert space (or an abstract space that is completely order isomorphic to one of these).
| https://mathoverflow.net/users/1119 | What is the "Krein-Milman theorem for cones"? | Something is wrong with the question, as here's a counter-example. Let $V=c\_0$ with the pointwise involution (so this is a commutative C\*-algebra). Let $C$ be the obvious cone: the collection of vectors all of whose coordinates are positive. Let $x=(i,0,0,\cdots)$. Then $V^\* = \ell^1$, so if $s=(s\_n)\in\ell^1$ satisfies $s(C)\subseteq[0,\infty)$, we need that $s\_n\geq 0$ for all $n$. But then $s(x)$ is purely imaginary!
So, maybe you also need $x^\*=x$. Under this assumption, here's a proof, but it has nothing to do with "Krein-Milman"...
As C is closed, $V\setminus C$ is open, so let A be an open ball about x which doesn't intersect C. Then A and C are disjoint, non-empty, convex, so by Hahn-Banach, as A is open, we can find a bounded linear map $\phi:V\rightarrow\mathbb C$ and $t\in\mathbb R$ with
$$ \Re \phi(a) < t \leq \Re \phi(c) $$
for $a\in A$ and $c\in C$. This is e.g. from Rudin's book. As $0\in C$, we see that $t\leq 0$.
Now, we can lift the involution \* from V to the dual of V. In particular, define
$$ \phi^\*(x) = \overline{ \phi(x^\*) } \qquad (x\in V)$$
So let $\psi = (\phi+\phi^\*)/2$. For $c\in C$, as $c^\*=c$, notice that $\psi(c) = \Re \phi(c)$. Hence $0 \leq \psi(c)$ for all $c\in C$. Similarly, as $x^\*=x$, we have that $\psi(x) = \Re\phi(x)<t\leq 0$, as $x\in A$.
| 4 | https://mathoverflow.net/users/406 | 23697 | 15,586 |
https://mathoverflow.net/questions/23690 | 14 | I have two related questions on the representability of integers
by quadratic forms in two variables :
(1) Let $f: {\mathbb Z} \times {\mathbb Z} \to {\mathbb Z} $ be such a quadratic
form, i.e. we have $f(x,y)=ax^2+bxy+cy^2+dx+ey+g$ for some integer constants
$a,b,c,d,e,g$. Suppose that $f$ is not surjective, i.e. some integer is not represented
by $f$. Is it true that there is an integer constant $C$ such that in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
(2) If the answer to (1) is yes, is there a uniform bound ? In other words,
is there a uniform constant $C$ such that for any non-surjective $f$, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
**Update** : Good answers to my original questions appeared quickly. It seems the only
interesting subquestion left is the one asked by fedja :
(2') is there a universal $C$ such that for any $f$ with positive definite quadratic part, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
One may also ask,
(3) is there a universal $C$ such that for any non-surjective and irreducible $f$, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
| https://mathoverflow.net/users/2389 | representability of consecutive integers by a binary quadratic form | Such a $C$ does not exist, if I got it right.
My example is the function $f(x,y)=(2x+1)(5y+1)$. An integer is represented by $f$ iff it can be written as the product of an odd integer and a number which is 1 mod 5. So for example it is not hard to check that 2 cannot be written in this way (the odd number had better be $\pm1$, and neither of $\pm2$ are 1 mod 5). However a tedious construction using the Chinese Remainder Theorem gives arbitrarily large sequences of consecutive integers which are representable by $f$.
Let me explain a bit more about the CRT argument. Clearly any positive integer $N$ can be written as the product $M2^r$ of an odd integer and a power of two. If furthermore this odd integer $M$ is divisible by three distinct primes, one of which is 2 mod 5, one of which is 3 mod 5 and one of which is 4 mod 5, then moving one of these factors $p$ from $M$ to the power of two gives $N=(M/p)(p2^r)$, and if $2^r$ isn't 1 mod 5 then there will be some $p$ such that $p2^r$ will be 1 mod 5, so we have our decomposition of $N$ which is hence representable by $f$.
So it suffices to find arbitrarily large strings of integers each of which has a prime factor which is 2 mod 5, a prime factor which is 3 mod 5 and a prime factor which is 4 mod 5 (in fact by using sign changes one can even get away with primes which are 2 mod 5). But now this is easy: enumerate the primes which are 2 mod 5, 3 mod 5 and 4 mod 5, let C\_n be the product of the n'th 2-mod-5, the n'th 3-mod-5 and the n'th 4-mod-5, and now choose a big positive $N$ with $N=0$ mod C\_1, $N+1=0$ mod $C\_2$,... $N+10^6=0$ mod $C\_{10^6+1}$ (such an $N$ exists by CRT) and there you have it.
| 11 | https://mathoverflow.net/users/1384 | 23700 | 15,588 |
https://mathoverflow.net/questions/23706 | 2 | In the answer of my question:
[On the full reducibility of representations of reductive Lie algebras](https://mathoverflow.net/questions/23085/on-the-full-reducibility-of-representations-of-reductive-lie-algebras)
James E. Humphreys replied to me saying that:"the notion of "reductive" for a Lie algebra in characteristic 0 has no intrinsic interest, unless you study the Lie algebra of a Lie (or algebraic) group and relate their representations carefully."
Can please someone explain that to me or give to me any reference?
thank you!
| https://mathoverflow.net/users/4821 | Reductive Lie algebra of a Lie group | What Jim means is that one naive definition of reductive Lie algebra
* $\mathfrak{g}$ is **reductive** if all its finite-dimensional representations are semi-simple.
already has a name: **semi-simple.**
Another one
* $\mathfrak{g}$ is **reductive** if all its representations are semi-simple.
is actually trivial; there are no (**EDIT:** nonzero) Lie algebras that satisfy it.
Of course, there actually is a pretty good definition that matches better with reductive for groups:
* $\mathfrak{g}$ is **reductive** if its adjoint representation is semi-simple.
but it's important to keep in mind that the properties above don't follow from that.
| 6 | https://mathoverflow.net/users/66 | 23714 | 15,596 |
https://mathoverflow.net/questions/23679 | 13 | Forgive me if this question does not meet the bar for this forum. But i would really appreciated some help.
I'm trying to construct a function according to some conditions in the frequency domain of the Fourier transformation. I want the function to be analytic and real when I transform it back to the time domain. The Fourier transformation of $f$ has of course some symmetry criteria to make $f$ real. But what about the Analytic property. As an analytic function imply some convergent power series expansion, and the Fourier transform of a polynomial is a sum of derivatives of Delta functions, I assume that there is a corresponding criteria of the Fourier transformation.
So the question is:
If a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is assumed to be analytic, what is the corresponding criteria for the Fourier transform of the function $\mathcal{F}[f] (k)$?
*Edit*: what I am trying to construct is probability distribution with the following condition
$f(x/\mu)/\mu=\frac{2}{3} f(x) + \frac{1}{3} (f\ast f)(x)\quad$
where $\ast$ mark the convolution, and $\mu=\frac{4}{3}$. $f$ is positive and real for $x\in [0,\infty)$
Taking the fourier transformation make the condition simpler:
$\tilde f(\mu k) = \frac{2}{3}\tilde f(k) + \frac{1}{3}\tilde f^2(k)$
So my problem is to construct $f$ (I am in particular interested in the tail behavior) and I try to use the properties of $\tilde f$. I posted a similar problem a while ago ([see here](https://mathoverflow.net/questions/19186/finding-functional-form-for-a-given-scaling-condition)). Julián Aguirre answered how to construct $\tilde f$ if it is analytic. But the inverse transformation of the power expansion is an infinite sum of derivatives of Delta functions, and is of little help.
| https://mathoverflow.net/users/4626 | Fourier transform of Analytic Functions | What is sufficient (though not necessary) is that the Fourier transform decays exponentially at $\infty$ (if you want just analyticity on the line) or faster than any exponent (if you want your original function to be entire). In particular, anything with compact support will do. If this is too restrictive for your construction, you'd better just tell what exactly you are trying to construct.
| 18 | https://mathoverflow.net/users/1131 | 23721 | 15,599 |
https://mathoverflow.net/questions/23717 | 2 | This is quite likely to be a solved problem, perhaps even a standard exercise. However, being a non-[number theorist], I don't know where to look. A quick perusal of the basic starting references of Google, CLRS, and Bach+Shallit does not seem to help.
*Problem.* I have an integer *N*, and a divisor *d*. What is a good upper bound on the time required to compute coprime integers *n*1 and *n*2 , such that *N* = *n*1*n*2 , and such that *d* divides *n*1?
**Actual Question.** What is a good reference for the solution / time requirements for this problem?
*Solution to problem.* As I'm aware that this may also be an exercise in some number-theory class, I'll outline a very reasonable iterative approach as a good-faith gesture.
Define sequences *xj* , *yj* , and *gj* by the recurrences
$\begin{align}
\quad x\_1 =& d & \quad && x\_{j+1} =& x\_j g\_j \\\\
y\_1 =& N/d &&& y\_{j+1} =& y\_j / g\_j \\\\
g\_1 =& \gcd(x\_1, y\_1) &&& g\_{j+1} =& \gcd(x\_j, y\_j)
\end{align}$
which eventually converge. When this occurs (i.e. for *j* sufficiently large that *gj* = 1), we may let *n*1 = *xj* and *n*2 = *yj* .
Note that for any *j* such that *xj* ≥ *yj* , we may show without too much difficulty that *g**j*+1 = 1; so the last few iterations take time O( log(*N*)2 ), and the time required for the preceding iterations increases monotonically with *xj* . Considering the prime-power decompositions of *xj* and of *N*, we may note that the exponent of the maximal power of each prime *p* dividing *xj* doubles with each succesive iteration, until it saturates the exponent of the maximal power of *p* which divides *N*. Thus, the number of iterations required will be bounded above by something like log log(*N*). The cumulative run-time of all but the last few iterations depends exponentially on the number of iterations; one can then bound the time required for all but the last few iterations by something like O( log(*N*)2 ) again.
This is then an upper bound for the whole procedure.
**Remark.** I doubt that one can do better than the upper bound of O( log(*N*)2 ) above. I also doubt that I'm the first person to solve this problem, and I'd rather not clutter up a paper describing this solution if I can cite another paper instead.
| https://mathoverflow.net/users/3723 | Reference request: given a divisor d of N, how quickly can I obtain the largest factor of N coprime to d? | Take a look at these [papers](http://cr.yp.to/coprimes.html) from Dan Bernstein. It's not quite what you are looking for, but he does even more than you need in time $n(\lg n)^{2+o(1)}$ where $n$ = number of bits of $N\cdot d$ (one of the elements of the coprime base will be $n\_2$). Maybe your problem can be solved even faster than that.
| 5 | https://mathoverflow.net/users/5542 | 23728 | 15,604 |
https://mathoverflow.net/questions/17216 | 7 | Do any treatises on real analysis take the following as the basic completeness axiom for the reals?
"Let $A$ and $B$ be set of real numbers such that
(a) every real number is either in $A$ or in $B$;
(b) no real number is in $A$ and in $B$;
(c) neither $A$ nor $B$ is empty;
(d) if $\alpha \in A$, and $\beta \in B$, then $\alpha < \beta$.
Then there is one (and only one) real number $\gamma$ such that
$\alpha \leq \gamma$ for all $\alpha \in A$, and $\gamma \leq \beta$
for all $\beta \in B$."
This appears as Theorem 1.32 in Walter Rudin's "Principles of Mathematical Analysis", and can be traced back to Dedekind's "Continuity and Irrational Numbers" (section V, subsection IV). Both Rudin and Dedekind derive this result from the construction of the reals via cuts of the rationals.
Authors who prefer to axiomatize the reals directly (instead of constructing them from the rationals) might be expected to take the above property as an axiom, but I haven't found anyone who does this. Instead, they all assume the least upper bound property as an axiom, or the nested interval property, or the convergence of Cauchy sequences.
I personally think the way to go is to take Rudin's Theorem 1.32 as an axiom (because it is simple and compelling) and then derive the least upper bound property (since it is more useful in practice than 1.32) and then get to work building up the apparatus of real analysis. But leaving aside the issue of whether this is the right way to go: have any authors taken this approach?
I should remark that the geometrical analogue of Theorem 1.32, characterizing the completeness of the line, appears to be well known to geometers (especially those interested in the foundations of geometry; see for instance Marvin Jay Greenberg's very nice article in the March 2010 issue of the Monthly).
| https://mathoverflow.net/users/3621 | completeness axiom for the real numbers | As Akhil says, yes. Another somewhat standard name for this axiom is the "Dedekind cut axiom". If you Google that with quotes, you will find some references.
| 4 | https://mathoverflow.net/users/1450 | 23750 | 15,620 |
https://mathoverflow.net/questions/23719 | 4 | Is the following true? What's a nice proof?
>
> Let $M$ and $N$ be von Neumann algebras, and let $\phi:M\rightarrow N$ be a normal, surjective, \*-homomorphism. Is there a normal \*-homomorphism $\theta:N\rightarrow M$ with $\phi\circ\theta$ being the identity? If I cannot choose $\theta$ as a \*-homomorphism, can I at least get a normal complete contraction?
>
>
>
This is, well, hinted at in the proof of Lemma 3.2 of <http://pjm.math.berkeley.edu/pjm/2002/205-1/p09.xhtml> but I don't follow the details (and they are proving it for weak\* TROs: I think surely the von Neumann algebra case should be easier). Normal \*-homomorphisms between von Neumann algebras have a very nice structure theorem, and maybe if I stared at that long enough I'd see an answer, but I thought I'd ask on MO...
| https://mathoverflow.net/users/406 | Lifting surjective von Neumann algebra homomorphisms | Yes you can get a $\phi$ that is a homomorphism. Here is a quick sketch.
First let $p=sup$ {$p\_\alpha,$ projections in $Ker \theta$}. So $p\in Ker \theta$. Furthermore $p\in Z(M)$, the center of M.
To see this note that if this were not true then we could find a unitary $u\in M$ with $p\neq upu^\star$. So then $p\wedge upu^\star$ would be a projection in $Ker \theta$ bigger than $p$.
From here you can get that $Ker \theta=pMp$, and so we can decompose $M=Ker \theta \oplus M\_1$ and $\theta|\_{M\_1}$ is injective and thus an isomorphism, thus $\phi$ can be chosen to just be the inverse of $\theta$ on $M\_1$.
Note that if we demand that $\phi$ be unital, this doesn't work and I don't think it can be done in general. I will have to give it more thought.
| 4 | https://mathoverflow.net/users/5732 | 23751 | 15,621 |
https://mathoverflow.net/questions/23748 | 17 | I am studying graph algorithms.
I need a database of graphs on which I can test my algorithms.
Where can I find a reliable database of graphs of all kinds?
Thanks!
| https://mathoverflow.net/users/5360 | Where on the internet I can find a database of graphs? | You might want to look at Donald Knuth's *Stanford GraphBase: A Platform for Combinatorial Computing* (1994, 2009) and the accompanying [website](https://www-cs-faculty.stanford.edu/~knuth/sgb.html).
See also [The Stony Brook Algorithm Repository](http://www.cs.sunysb.edu/~algorith/).
| 15 | https://mathoverflow.net/users/965 | 23761 | 15,629 |
https://mathoverflow.net/questions/23386 | 2 | Snippet portion:
From Iwaniec and Kowalski's Analytic Number Theory:
If the class number $h=h(D)$ is small, then there are only few
prime ideals $\bf{p}$ of degree one with small norm. Indeed, if
$p=\bf{p \bar{p}}$ with $(\bf{p},\bf{\bar{p}})=1$, then $\bf{p}^h$
is a principal ideal generated by $\frac{1}{2}(m+n\sqrt{D})$ with
$n \ne 0$, when $p^h = \frac{1}{4}(m^2-n^2 D) \ge \frac{|D|}{4}$.
Therefore the least prime $p\_1 = p\_1(D)$ with $\chi\_D(p\_1)=1$
satisfies $p\_1 \ge {(\frac{|D|}{4})}^{1/h}$.
Hence $\chi\_D(n)$ agrees with $\mu(n)$ on all squarefree numbers
$n \le {(\frac{|D|}{4})}^{1/h}$ with
$(n,{(\frac{|D|}{4})}^{1/h})=1$. This property is not likely to
hold in long segments (because $\chi\_D$ is periodic while $\mu$ is
not), therefore this suggests that h is rather large.
Question portion:
Although the above argument would not work in a Real quadratic field ($D > 0$ so the last inequality in the first paragraph does not hold), it seems that if we replace the class number h with h times the regulator this should work.
Any ideas on how to actually show this?
| https://mathoverflow.net/users/695 | Lower bounds for split primes in Real quadratic fields | The following is not a full answer, but perhaps gives you an idea of how to approach the result.
Let us consider the claim
$$ p^{hR} \ge \Big(\frac{D}4\Big) $$
for the smallest noninert prime. I first show that the inequality holds whenever $p \ge 11$.
In fact, we have $h \ge 1$ and $R \ge \frac12 \log D + O(1)$: the latter claim is clear since the fundamental unit $(t+u\sqrt{D})/2$ satisfies $u \ge 1$ and $t \approx u \sqrt{D}$. Since $\log p > 2$, the claimed inequality follows.
Thus we only have to look at small primes. If $R$ is large enough, the claim holds.
If $R$ is as small as possible, the following happens. If $u = 1$, the discriminant must have a very special form. One possibility is $D = 4n^2 +1$. Here the fundamental unit is $\varepsilon = 2n+\sqrt{D}$ (unless $n = 1$), so $R \approx \frac12 \log D$. On the other hand, "Davenport's Lemma" shows that the minimal nontrivial norm is $n = N((2n+1+\sqrt{D})/2)$; see
* Ankeny, Chowla & Hasse, *On the class-number of the maximal real subfield of a cyclotomic field* J. Reine Angew. Math. 217, 217-220 (1965);
a different idea for a proof of this and similar results can be found in
* Lemmermeyer & Pethö, *Simplest cubic fields*, Manuscr. Math. 88, No.1, 53-58 (1995).
Observe that we only have to check the claimed inequality for $n \le 7$.
Since the result holds for $R$ large and very small, I would guess that its proof is within reach. Good luck!
| 4 | https://mathoverflow.net/users/3503 | 23763 | 15,631 |
https://mathoverflow.net/questions/23770 | 20 | This question is going to be extremely vague.
It seems that wherever I go (especially about Grothendieck's circle of ideas) the higher-dimensional analogue of a curve minus a finite number of points is a scheme minus a normal crossing divisor.
Why is that? What's so special about a normal crossing divisor that it simulates a curve minus a finite number of points better?
| https://mathoverflow.net/users/5309 | Why are normal crossing divisors nice? | It mostly has to do with finding nice compactifications. Compactifications of
varieties are a good thing as they allow us to control what happens at
"infinity". If the variety itself is smooth it seems a good idea (and it is!) to
demand that the compactification also be smooth. However, you need the situation
to be nice at infinity in order to make the study of asymptotic behaviour at
infinity to be as easy as possible. The best behaviour at infinity would be if
the complement were smooth but that is in general not possible. What is always possible
is to demand that the complement be a divisor with normal crossings. In practice
it works essentially as well as having a smooth complement: You have a bunch of
smooth varieties intersecting in as nice a manner as possible.
| 27 | https://mathoverflow.net/users/4008 | 23776 | 15,641 |
https://mathoverflow.net/questions/23759 | 10 | Schubert's Theorem in Knot Theory says that any knot can be uniquely decomposed as the connected sum of prime knots.
Unfortunately the original paper is in German.
Does anyone know a good english reference for this. Or just the special case of the unknot. (i.e. that the unknot can't be written as the connected sum of two knots which aren't the unknot.)
| https://mathoverflow.net/users/5732 | A Reference for Schubert's Theorem | A fairly standard reference would be "Knot theory" by G.Burde and H. Zieschang,
Chapter 7.
<http://books.google.nl/books?id=DJHI7DpgIbIC&pg=PR1&dq=Burde+Zieschang&cd=1#v=onepage&q&f=false>
Roland
| 4 | https://mathoverflow.net/users/5914 | 23778 | 15,642 |
https://mathoverflow.net/questions/23710 | 21 | Several months ago a paper was posted at
<http://arxiv.org/abs/1001.4164>
called "Another way of answering Henri Poincaré's fundamental question." The author gave a talk on it today at my institution. If it's correct, it is a major breakthrough in terms of proof length (~10 pages). However, it is very outside my specialty. There's apparently been very little feedback, but the author is ok with public discussion. Therefore, can anyone say whether they have read the paper? Whether it is correct, missing details, clearly flawed, or what have you?
| https://mathoverflow.net/users/4020 | Does this approach for the Poincaré conjecture work? | I had a quick look. Although I haven't found a specific error, as far as I can tell, he's not using the hypothesis of simple-connectivity anywhere in an essential way. Even though he posits this as a hypothesis in Prop. 5.6, the proof of this proposition works for any manifold with 2-sphere boundary (also, Prop. 4.4 works for any homology ball). Thus, in Step (4) of his proof of Prop. 5.8, when he refers to Prop. 5.6, this step is not making essential use of simple-connectivity.
| 21 | https://mathoverflow.net/users/1345 | 23783 | 15,644 |
https://mathoverflow.net/questions/23781 | 5 | In Rudyak's *On Thom Spectra, Orientability, and Cobordism*, two variants of Brown's representability theorem are presented: given a natural transformation $f^\*: E^\* \to F^\*$ of cohomology theories, Brown's representability theorem asserts that we can lift $f^\*$ uniquely to a map $f: E \to F$ of spectra, and we can also lift $f^\*$ to map $f^i: \Omega^\infty \Sigma^i E \to \Omega^\infty \Sigma^i F$ of representing spaces -- but he does not assert that this second kind of lift is unique.
My feeling is that this second map becomes unique after stabilization. Is there an accessible example of such a natural transformation with two nonhomotopic representing maps on classifying spaces, illustrating nonuniqueness?
| https://mathoverflow.net/users/4701 | Nonuniqueness of maps of representing spaces | The actual work being done here is by the Yoneda lemma. Brown's representability theorem tells you that these spaces represent the cohomology theory, turning natural transformations into morphisms is done by Yoneda.
That said, uniqueness holds.
Of course, one has to be a little bit more careful about what one means by "natural transformation" in this case. Natural transformation of *what*? If your natural transformation is of the *whole* cohomology theory in one go (aka a stable operation) then, indeed, you get a morphism of spectra. If your natural transformation acts on one particular level (aka an unstable operation) then you get a morphism of spaces.
Where you do **not** get uniqueness is if you have a family of unstable operations (aka a family of morphisms of *spaces*) which look as if they fit together to give a stable operation. You can get "phantom" morphisms, and there's a "$\lim^1$" term that controls this. A nice place to read about all of this is the papers by Boardman and Boardman, Johnson, and Wilson on stable and unstable cohomology operations (Handbook of algebraic topology, also available via Steve Wilson's homepage).
---
**Update**: To expand on that last point (I was deliberately vague because I didn't have Boardman's paper in front of me and couldn't remember which way the maps went): a *stable* operation (morphism of spectra) defines a compatible (under suspension) family of *unstable* operations (morphisms of spaces, indeed, morphisms of infinite loop spaces) but this assignment may not be injective (it is always surjective). There is a short exact sequence (attributed to Milnor, and (9.7) in Boardman's paper):
$$
0 \to \lim\_n{}^1 E^{k-1}(\underline{E}\_n,o) \to E^k(E,o) \to \lim\_n E^k(\underline{E}\_n,o) \to 0
$$
The "o" means "pointed" and $\underline{E}\_n$ is the $n$th component in the spectrum $E$. So if the $\lim^1$ term vanishes, you get an isomorphism (whence injectivity) but if not, then there may be "phantom" stable operations that are non-trivial but can't be detected in the unstable realm.
---
**Update**: And now I've been shown a copy of the book references in the question and have read what may well be the statements themselves. The second one is subtly different to the summary above. In the book, the statement is that a natural transformation of cohomology theories of spaces can be lifted to a morphism of the representing spectra. *This* is not necessarily unique. The basic reason being that the Yoneda lemma does not apply to this case because the representing object **is in a different category**. So cohomology of *spaces* represented by a *spectrum* is the composition of $\Sigma^\infty$ and the cohomology theory and thus not an honest representable functor. In general, there's no way (that I can think of) to pull-back or push-forward natural transformations along a functor.
Nonetheless, because spectra and spaces are closely related, it is possible in this case to construct a natural-transformation-of-cohomology-of-spaces from one of spectra. This works because the *pieces* of the cohomology theory are individually representable as spaces, and so again the Yoneda lemma comes into play. Then one can ask about this map, and the $\lim{}^1$ sequence says that it is always surjective but not necessary injective.
So, in summary:
* $E^\ast(R)$: both spectram: Yoneda => $Nat(E^\ast,F^\ast) \cong \{E,F\}$
* $E^k(X)$: both spaces: Yoneda => $Nat(E^k,F^k) \cong [\underline{E}\_k, \underline{F}\_k]$
* $E^\ast(X)$: spectrum and space: Milnor => $\{E,F\} \to Nat(E^\ast,F^\ast)$ surjective
| 6 | https://mathoverflow.net/users/45 | 23784 | 15,645 |
https://mathoverflow.net/questions/23791 | 4 | Suppose $M$ and $N$ are two Stein manifolds of dimension at least $3$ with compact subsets $U$ and $V$ such that $M\setminus U$ is biholomorphic to $N \setminus V$. It it true that $M$ is biholomorphic to $N$?
It this is not true, what is the simplest example? And if this is true, what would be the reference for such a statement?
| https://mathoverflow.net/users/943 | Stein manifolds isomorphic at infinity | I believe the answer is positive in dimension at least two. Stein manifolds admit proper embeddings on vector spaces. An isomorphism from $M$ to $N$ can be represented by a collection of holomorphic functions from $M$ to $\mathbb C$. Each one of these extends to the whole $M$ according to Hartogs. Thus the isomorphism at infinity extends to a holomorphic map. Arguing in the same way with the inverse of the isomorphism at infinity yields an affirmative answer.
| 4 | https://mathoverflow.net/users/605 | 23797 | 15,651 |
https://mathoverflow.net/questions/23794 | 6 | [Edit: For a category $C$ let $C^\wedge$ denote the category of presheaves on $C$.]
Let $f:C\to D$ be a functor. Precomposition with $f^{op}$ induces a functor
$f^\wedge:D^\wedge \to C^\wedge$. This functor has both a left- and a right adjoint, called *left-* and *right kan extension*:
$f\_\wedge \dashv f^\wedge \dashv f\_+$.
Now for $c\in C$ we get $D^\wedge(D(-,fc),Y)=Y(fc)=f^\wedge Y(c)=C^\wedge(C(-,c),f^\wedge Y)$. This gives us the restriction of $f\_\wedge$ to $C$ along the yoneda embedding: It is $f$ (composed with the yoneda embedding).
Now here's my question:
>
> What is the restriction of $f\_+$ to $C$ along the yoneda embedding?
>
>
>
It seems not to agree with $f$ but:
>
> Is there a nice connection between $f\_+C(-,c)$ and $D(-,fc)$?
>
>
>
| https://mathoverflow.net/users/1261 | Kan extensions and the yoneda embedding. | For all $Z \in C^\wedge, Y \in D^\wedge$, we have $C^\wedge(f^\wedge Y,Z)=D^\wedge(Y,f\_+ Z)$. If we put $Y = D(-,d), Z = C(-,c)$, we get
$(f\_+ C(-,c))(d) = C^\wedge(f^\wedge D(-,d),C(-,c)) = C^\wedge(D(f-,d),C(-,c))$
There seems to be no connection between $f\_+ C(-,c)$ and $D(-,fc)$ (only when $f$ is an equivalence). For example,
$D^\wedge(D(-,fc),f\_+ C(-,c)) = C^\wedge(f^\wedge D(-,fc),C(-,c)) = C^\wedge(D(f-,fc),C(-,c))$
and it is possible to construct an example where there is no natural transformation $D(f-,fc) \to C(-,c)$ at all. For example if $D(fx,fc)$ is nonempty, but $C(x,c)$ is empty. Take $C^{op}=D=Set, f = Hom(-,2), x = 0, c = 1$.
| 1 | https://mathoverflow.net/users/2841 | 23802 | 15,655 |
https://mathoverflow.net/questions/23713 | 7 | A *finite abstract simplicial complex* is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.
For $n\geq 0$ the topological space $\Delta^n=\{(x\_0,...,x\_n)\in\mathbb{R}^{n+1}\mid x\_i\geq 0, \sum x\_i =1\}$ is called the *standard $n$-simplex*. A topological space homeomorphic to the standard $n$-simplex is called an *$n$-simplex*. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.
A *finite topological simplicial complex* is a pair $(X,F)$ where $X$ is a topological space and $F=(F\_1,...,F\_m)$ is a finite sequence of embeddings $F\_k:\Delta^{i\_k}\to X$ such that
* $X=\cup\_k F\_k(\Delta^{i\_k})$
* $F\_k\neq F\_l$ if $k\neq l$
* for every $1\leq k\leq m$ with $i\_k\geq 1$ and for every face $A$ of the $i\_k$-simplex $F\_k(\Delta^{i\_k})$ there is a $1\leq l\leq m$ with $F\_l(\Delta^{i\_l})=A$
* for every $1\leq k\neq k'\leq m$ the simplex $F\_k(\Delta^{i\_k})\cap F\_{k'}(\Delta^{i\_{k'}})$ is a face of each of them.
I hope, the definitions are correct.
There is the notation of a *geometric realization* of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.
~~If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.~~
~~My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?~~
I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:
A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the **same** directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in **opposite** directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?
| https://mathoverflow.net/users/4676 | Simplicial complexes vs. geometric realization of abstract simplicial complexes | At least in the realm of topology an abstract simplicial complex is equivalent to a topological (or geometric) simplicial complex, and neither of these two notions involves anything about orienting the simplices or ordering the vertices. If one has a simplicial complex of either type, one can choose a partial ordering of the vertices that restricts to a linear ordering of the vertices of each simplex, and this gives the notion of an ordered simplicial complex. For a simplicial complex that is a manifold (or more generally a pseudo-manifold) one can define an orientation to consist of a choice of orientations of the top-dimensional simplices such that the two induced orientations on each codimension one face are opposite. For a general simplicial complex the only definition of an orientation that I have seen is the trivial one where an orientation is chosen for each simplex with no compatibility assumptions on these orientations.
| 11 | https://mathoverflow.net/users/23571 | 23807 | 15,658 |
https://mathoverflow.net/questions/23798 | 4 | Is there a nice fundamental domain for the symmetric group $S\_n$ acting on the Grassmannian of $k$-planes in $\mathbb{R}^n$? (The action of $S\_n$ is by permuting the coordinates, of course.)
I'm looking for a way to efficiently test whether two subspaces of $\mathbb{R}^n$ are related by permuting and/or negating the coordinates. (So I really care not about $S\_n$, but about $(\mathbb{Z}/2)^n \rtimes S\_n$, the Weyl group of the B/C series.) I'm somewhat skeptical this is easy, but has anyone thought about it?
| https://mathoverflow.net/users/5010 | Fundamental domain for symmetric group $S_n$ acting on $\mathop{Gr}(k,n)$? | Here is a suggestion that ought to work generically.
A general point on $\operatorname{Gr}(k,n)$ corresponds to the graph of a linear transformation $\mathbb{R}^k \to \mathbb{R}^{n-k}$, and hence to an $(n-k) \times k$ matrix. In its $S\_n$-orbit, there is one point that is distinguished by the following requirements (assuming that ties do not occur):
1) The maximum of the absolute values of the entries of the matrix is as large as possible (this determines a $k$-element subset of distinguished coordinates).
2) If $a\_i$ is the maximum of the absolute values of the entries in the $i$-th row, then $a\_1<a\_2<\cdots<a\_{n-k}$.
3) If $b\_j$ is the maximum of the absolute values of the entries in the $j$-th column, then $b\_1<b\_2<\cdots<b\_k$.
You can compute it by first trying all $k$-element subsets, and then permuting the rows and columns of the $(n-k) \times k$ matrix as needed. (Whether this is efficient enough for you will depend on the size of $n$ and $k$.)
If you also want to allow negating the coordinates, then you can choose the sign changes in a unique way (up to an overall sign change) so that all entries in the first row and first column are positive (assuming that the original point is general enough that these entries are nonzero).
| 3 | https://mathoverflow.net/users/2757 | 23808 | 15,659 |
https://mathoverflow.net/questions/23805 | 9 | Let E be an elliptic curve over the rationals, and let $TE = \lim\_\leftarrow E[n]$ be the Tate module of the elliptic curve. The action of the Galois group of $\bf Q$ gives rise to a representation $\rho\_E : G\_{\bf Q} \rightarrow GL\_2(\widehat{\bf Z})$.
My first question is why is the image of this representation not surjective? I seem to recall there is a very easy argument for it, but I can't remember what that is.
My second question is can one use this to give a level structure to all rational elliptic curves? Specifically, can I find a collection of modular curves with some level structure $\{X\_i/{\bf Q}\}$ such that $\cup j(X\_i({\bf Q})) = X(1)({\bf Q})$, where $j:X\_i \rightarrow X(1)$ is the natural forgetful map? (I'm sure the answer to this is no, but it never hurts to be overly optimistic.)
| https://mathoverflow.net/users/92 | Images of action of Galois on the Tate module of Elliptic Curve, | Let $\Delta$ be the discriminant of $E$. Then the action of $G\_{\mathbf{Q}}$ on $E[2]$ determines the action on $\sqrt{\Delta}$. On the other hand, the action of $G\_{\mathbf{Q}}$ on $E[n]$ determines the action on a primitive $n$-th root of unity $\zeta\_n$, via the Weil pairing. The Kronecker-Weber theorem implies that $\sqrt{\Delta} \in \mathbf{Q}(\zeta\_n)$ for some $n$, and this forces a compatibility between the actions on $E[2]$ and $E[n]$, which forces the image of $G\_{\mathbf{Q}}$ into an index-2 subgroup of $\operatorname{GL}\_2(\hat{\mathbf{Z}})$. But this index-2 subgroup varies with $E$, so you do not get a rational point on a nontrivial modular curve corresponding to any *one* kind of level structure.
**Remarks:**
1) Nathan Jones in his Ph.D. thesis at UCLA, building on earlier work of William Duke, proved that in a precise sense, asymptotically `100%` of elliptic curves are such that the image of Galois is of index 2.
2) Over higher number fields $K$, quadratic extensions are not necessarily contained in cyclotomic ones, and in fact there exist elliptic curves over certain number fields $K$ other than $\mathbf{Q}$ for which $G\_K \to \operatorname{GL}\_2(\hat{\mathbf{Z}})$ is surjective. The first example was given by Aaron Greicius in his Ph.D. thesis.
3) David Zywina then proved that under mild necessary conditions on a number field $K$, asymptotically `100%` of elliptic curves are such that $G\_K \to \operatorname{GL}\_2(\hat{\mathbf{Z}})$ is surjective.
| 22 | https://mathoverflow.net/users/2757 | 23809 | 15,660 |
https://mathoverflow.net/questions/23817 | 0 | If [1:0:....:0] is an s-fold singularity of a degree $r$ hypersurface $F$ in $\mathbb{P}^n$ then the hypersurface can be written as $F=x\_0^{r-s}g\_s(x\_1,...,x\_n) + x\_0^{r-s-1}g\_{s+1}(x\_1,...,x\_n) + ... +g\_r(x\_1,...x\_n)$. After we dehomogenize it is known that the initial term $g\_s(x\_1,...x\_n)$ is the tangent cone at the singularity in $\mathbb{C}^n$. My question is known about the higher order terms such as $g\_{s+1}(x\_1,..,x\_n)$? Do they admit a some geometric interpretation?
I know that the common locus of $g\_s=g\_{s+1}=...=g\_{s+h}=0$ give the set of points whose line through the origin has intersection multiplicity s+h+1 with the hypersurface. I would like to find a geometric interpretation of just the hypersurface $g\_{s+1}=0$.
| https://mathoverflow.net/users/2565 | What is known beyond the tangent cone for hypersurface singularities? | It seems unlikely that there is something nice. An interpretation should preferably be invariant under linear coordinate transformations and a homogeneous component itself isn't, it is only invariant modulo the ones of lower order.
| 3 | https://mathoverflow.net/users/4008 | 23819 | 15,665 |
https://mathoverflow.net/questions/23834 | 12 | The existence of non-measurable subsets and functions on $\mathbb{R}$ require the use of the axiom of choice. That is, there exist models of ZF in which all subsets of (and hence all functions defined on) $\mathbb{R}$ are measurable. Does this mean that if I can define a subset or function on $\mathbb{R}$ without invoking the axiom of choice, it must be measurable?
Let's say I fully accept AC and am just looking for a quick and dirty way of proving a function is measurable.
| https://mathoverflow.net/users/5513 | Drawing conclusions by NOT using AC. | With some difficulty, you can define a set of reals which is measurable under
one extra-ZF assumption (the axiom of determinacy) and nonmeasurable under
another (V=L). Under V=L you have a definable well-ordering of the reals, and
this enables you to *define* any of the nonmeasurable functions you normally get using the axiom of choice. On the other hand, under the axiom of determinacy, *every* real function is measurable.
That's the strict answer. However, I think that any subset or function you are *likely*
to define will be Borel, and hence provably measurable in ZF.
| 6 | https://mathoverflow.net/users/1587 | 23844 | 15,678 |
https://mathoverflow.net/questions/23849 | 4 | I've been reading about the rotating calipers algorithm for solving the minimum-area enclosing rectangle problem. It relies on a theorem: The rectangle of minimum area enclosing a convex polygon has a side collinear with one of the edges of the polygon.
Can someone explain why is this true?
| https://mathoverflow.net/users/5938 | Minimum enclosing rectangle of a convex polygon has a collinear side | Suppose you've got your polygon *P*, embedded in the Euclidean plane with some standard coordinate system. Then there's some rectangle *R* with horizontal and vertical sides which encloses *P* and is as small as possible among all such rectangles (just take horizontal lines through the topmost and bottommost points, and vertical lines through the rightmost and leftmost points). Translating the coordinate system doesn't change the rectangle or its area, but rotating it through an angle *θ* will; so the rectangle *R**θ* and its area *A**θ* are really functions of *θ* (and (*π*/2)-periodic functions at that). As such, *A* definitely has a minimum at some angle *θ*min.
The meat of the theorem is that, for any angle *φ* where *R**φ* does not meet the conclusion, the function *A* does not even have a local minimum. To see this, note that *R**φ* intersects *P* at a finite set of points *S* — usually |*S*| = 4, but 3 or 2 are also possible. Now consider the rectangle *R'**θ* which is the smallest rectangle with sides parallel to the *θ*-axes and which contains *S*, and let *A'**θ* be its area. Note two things:
* For *θ* near *φ*, *R'**θ* = *R**θ*, and hence *A'**θ* = *A**θ*.
* Near *φ*, the function *A'* is *concave downward*. Why? Divide the rectangle *R'θ* along the lines between the points of *S*. We get a convex polygon determined by *S* (and hence constant), plus |*S*|-many right triangles. Each triangle has a hypotenuse which stays constant as *θ* changes, and an interior angle which changes as *θ*/2, hence an area which is concave downward.
Under our assumption, *φ* can't possibly be a local minimum of *A*, let alone a global one; so wherever the global minimum is, it must have a side coincident with one of the sides of *P*.
| 11 | https://mathoverflow.net/users/4133 | 23854 | 15,684 |
https://mathoverflow.net/questions/23848 | 19 | The Torelli theorem states that the map $\mathcal{M}\_g(\mathbb{C})\to \mathcal{A}\_g(\mathbb{C})$ taking a curve to its Jacobian is injective. I've seen a couple of proofs, but all seem to rely on the ground field being $\mathbb{C}$ in some way. So:
>
> Under what conditions does the Torelli Theorem hold?
>
>
>
Is algebraically closed necessary? Characteristic zero? It is known if there are any other base rings/schemes over which it is true?
| https://mathoverflow.net/users/622 | When does the Torelli Theorem hold? | The Torelli theorem holds for curves over an arbitrary ground field $k$ (in particular, $k$ need not be perfect). A very nice treatment of the "strong" Torelli theorem may be found in the appendix by J.-P. Serre to Kristin Lauter's 2001 Journal of Algebraic Geometry paper *Geometric methods for improving the upper bounds on the number of rational points on algebraic curves over finite fields*. It is available on the arxiv:
<http://arxiv.org/abs/math/0104247>
Here are the statements (translated into English):
Let $k$ be a field, and let $X\_{/k}$ be a nice (= smooth, projective and geometrically integral) curve over $k$ of genus $g > 1$. Let $(\operatorname{Jac}(X),\theta\_X)$ denote the Jacobian of $X$ together with its canonical principal polarization. Let $X'\_{/k}$ be another nice curve.
Theorem 1: Suppose $X$ is hyperelliptic. Then for every isomorphism of polarized abelian varieties $(\operatorname{Jax}(X),\theta\_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta\_{X'})$, there exists a *unique* isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ such that $F = \operatorname{Jac} f$.
Theorem 2: Suppose $X$ is not hyperelliptic. Then, for every isomorphism $F: (\operatorname{Jax}(X),\theta\_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta\_{X'})$ there exists an isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ and $e \in \{ \pm 1\}$ such that $F = e \cdot \operatorname{Jac} f$. Moreover, the pair $(f,e)$ is uniquely determined by $F$.
| 23 | https://mathoverflow.net/users/1149 | 23856 | 15,686 |
https://mathoverflow.net/questions/23859 | 7 | Suppose $S$ and $S'$ are two compact Riemann surfaces of genus $g$. Does there exist a sequence of genera $g\_i \to \infty$ and covers $S\_i, S\_{i}'$ of $S,S'$, both of genus $g\_i$, such that $d(S\_i,S\_{i}')\to 0$? Here $d$ a "natural" distance function on Teichmuller space, of which I suppose there are many, but for definiteness let's take it to be induced by the Teichmuller metric.
This question was asked to me by Rick Kenyon last year, and some brief thought on it got me nowhere.
| https://mathoverflow.net/users/1464 | Covers of Riemann surfaces which become arbitrary close in Teichmuller space | This is the Ehrenpreis Conjecture, and is still open.
Jeremy Kahn and Vlad Markovic have made some progress recently.
UPDATE: Kahn and Markovic have now announced a proof of the entire conjecture. See <http://arxiv.org/abs/1101.1330>
| 10 | https://mathoverflow.net/users/1335 | 23861 | 15,688 |
https://mathoverflow.net/questions/23857 | 31 | A finite group $G$ can be considered as a category with one object. Taking its nerve $NG$, and then geometrically realizing we get $BG$ the classifying space of $G$, which classifies principal $G$ bundles.
Instead starting with any category $C$, what does $NC$ classify? (Either before or after taking realization.) Does it classify something reasonable?
| https://mathoverflow.net/users/3557 | What does the classifying space of a category classify? | Ieke Moerdijk has written a small Springer Lecture Notes tome addressing this question:["Classifying Spaces and Classifying Topoi" SLNM 1616](http://openlibrary.org/books/OL800793M/Classifying_spaces_and_classifying_topoi).
Roughly the answer is: A $G$-bundle is a map whose fibers have a $G$-action, i.e. are $G$-sets (if they are discrete), i.e. they are functors from $G$ seen as a category to $\mathsf{Sets}$. Likewise a $\mathcal C$-bundle for a category $\mathcal C$ is a map whose fibers are functors from $\mathcal C$ to $\mathsf{Sets}$, or, if you want, a disjoint union of sets (one for each object of $\mathcal C$) and an action by the morphisms of $\mathcal C$ — a morphism $A \to B$ in $\mathcal C$ takes elements of the set corresponding to $A$ to elements of the set corresponding to $B$.
There is a completely analogous version for topological categories also.
| 26 | https://mathoverflow.net/users/733 | 23865 | 15,691 |
https://mathoverflow.net/questions/23868 | 3 | Let $(M,\mathcal F)$ be a smooth foliated manifold. An automorphism of $(M,\mathcal F)$ is a diffeomorphism of $M$ that takes leaves of $\mathcal F$ onto leaves. Let now $L$ be a leaf of $\mathcal F$. It may happen that $L$ has an open neighborhood $U$ which is a sum of leaves and such that for every leaf $L'\subset U$ there exists an automorphism $\phi$ taking $L$ onto $L'$.
My questions are:
>
> Does every $(M,\mathcal F)$ have a leaf $L$ with the described property?
>
>
> If the answer is no, how much can we hope for instead? And is there a simple counterexample? Are there some natural classes of foliations which still have this property?
>
>
>
Context:
Ideally, I would like to consider a smooth foliated manifold $(M,\mathcal F)$ such that in some flat chart $\psi\colon U\to \mathbb R^n$ there exists an open set $V\subset U$ and a family $\tau\_x$ of automorphisms of $(M,\mathcal F)$ indexed by a neighborhood of 0 in $\mathbb R^n$, where $\tau\_x$ acts on $V$ like a translation by $x$ in the chart $\psi$. This condition is satisfied for instance when there is a neighborhood $U$ of a leaf $L$ and a foliation-preserving diffeomorphism $U\to L\times W$, where $L\times W$ has the corresponding product foliation.
| https://mathoverflow.net/users/4698 | A local transitivity property of the automorphism group of a foliated manifold | No. There exist foliations $F$ of the plane that are rigid in the sense that no automorphism of $F$ can permute the leaves. See the following paper:
MR2407104 (2009e:54055)
Gartside, Paul(1-PITT); Gauld, David(NZ-AUCK); Greenwood, Sina(NZ-AUCK)
Homogeneous and inhomogeneous manifolds. (English summary)
Proc. Amer. Math. Soc. 136 (2008), no. 9, 3363--3373.
| 4 | https://mathoverflow.net/users/317 | 23870 | 15,692 |
https://mathoverflow.net/questions/23869 | 1 | I'm fixing a software defect that occurs 1 in *n* test runs. If I want to know that the probability of it being fixed is *>= p* for some *0 <= p < 1*, how many times, *m*, do I need to run the test successfully (without the defect occurring)?
| https://mathoverflow.net/users/5945 | Chance of something being fixed | According to my statistics final which I took yesterday, the answer should be
$m=\lceil 2\left(1-\frac{1}{n}\right)\text{InverseErf}^2[1-p]\rceil$ where InverseErf[*x*] is the [Inverse Error Function](http://en.wikipedia.org/wiki/Error_function#Inverse_function).
| 0 | https://mathoverflow.net/users/1982 | 23872 | 15,694 |
https://mathoverflow.net/questions/23864 | 0 | I'm trying to work through calculating the order of orthogonal groups in characteristic $\neq 2$. However there is one proof by induction used that i can't quite follow. Could someone help me understand where the formula for $z\_{m+1}$ comes from and how we know $U$ must contain $2q-1$ vectors with norm $0$ and $q-1$ vectors of every non-zero norm in the following extract:
>
> Let $z\_m$ denote the number of non-zero isotropic vectors in an orthogonal space with dimension $2m$ or $2m+1$. We claim that:
>
>
> $z\_m = q^{2m}-1$ for dimension $2m+1$
>
>
> $z\_m = (q^m-1)(q^{m-1}+1)$ for plus type with dimension $2m$
>
>
> $z\_m = (q^m-1)(q^{m-1}-1)$ for minus type with dimension $2m$
>
>
> For our inductive step we look at a $n+2$ dimensional space $V$ to ensure all spaces remain of the same type. Split V into the direct sum of $U$ and $W$ where $U$ is a $2$-dimensional space of plus type and $W$ is an $n$-dimensional space with the same type as $V$. Any isotopic vector in $V$ can be written as $u+w$ for isotropic vectors $u\in U$ and $w\in W$. Either $u$ and $w$ both have norm $0$ (with one being non-zero) or $u$ has norm $\lambda \neq 0$ and $w$ has norm $-\lambda$. Since $U$ contains $2q-1$ vectors of norm $0$ (including the zero vector) and $q-1$ vectors of every non-zero norm we count:
>
>
> $z\_{m+1}=(2q-1)(1+z\_m)+(q-1)(q^n-1-z\_m)-1$
>
>
> The other three cases are similar.
>
>
>
Thanks
| https://mathoverflow.net/users/5943 | Calculating norms over a finite field (orthogonal groups). | It is the case that each isotropic vector in $V$ has the form
$u+w$ where $u\in U$ and $w\in W$ but $u$ and $w$ need not be isotropic.
To see where $2q-1$ and $q-1$ come from, the quadratic form on $U$ has
norm given by $(x,y)\mapsto xy$. Now count how many pairs of elements of
$\mathbb{F}\_q$ give zero, and any given nonzero element.
| 1 | https://mathoverflow.net/users/4213 | 23875 | 15,696 |
https://mathoverflow.net/questions/23878 | 1 | I have a set of items, for example: {1,1,1,2,2,3,3,3}, and a restricting set of sets, for example {{3},{1,2},{1,2,3},{1,2,3},{1,2,3},{1,2,3},{2,3},{2,3}. I am looking for permutations of items, but the first element must be 3, and the second must be 1 or 2, etc.
One such permutation that fits is: {3,1,1,1,2,2,3}
Is there an algorithm to count all permutations for this problem in general? Is there a name for this type of problem?
For illustration, I know how to solve this problem for certain types of "restricting sets". Set of items: {1,1,2,2,3}, Restrictions {{1,2},{1,2,3},{1,2,3},{1,2},{1,2}}. This is equal to 2!/(2-1)!/1! \* 4!/2!/2!. Effectively permuting the 3 first, since it is the most restrictive and then permuting the remaining items where there is room.
| https://mathoverflow.net/users/5948 | Permutations with extra restrictions | In the case where all the elements in your set are different, this problem is known as "rook theory" and there's a substantial literature on it.
What you're trying to do, then, is rook theory on multisets ("multiset" is the usual name for a set with repeated elements); I can't tell if this exists or not.
| 1 | https://mathoverflow.net/users/143 | 23879 | 15,698 |
https://mathoverflow.net/questions/23269 | 44 | An often-cited principle of good mathematical exposition is that a definition should always come with a few examples and a few non-examples to help the learner get an intuition for where the concept's limits lie, especially in cases where that's not immediately obvious.
**Quillen model categories** are a classic such case. There are some easy rough intuitions—“something like topological spaces”, “somewhere one can talk about homotopy”, and so on—but various [surprising examples](https://mathoverflow.net/questions/18336/what-are-surprising-examples-of-model-categories) show quite how crude those intuitions are, and persuade one that model categories cover a much wider range of situations than one might think at first.
However, I haven't seen any non-examples of model structures written up, or even discussed—that is, **categories and classes of maps which one might think would be model structures, but which fail for subtle/surprising reasons**. Presumably this is because, given the amount of work it typically takes to construct an interesting model structure, no-one wants to write (or read) three-quarters of that work without the payoff of an actual example at the end. **Has anyone encountered any interesting non-examples of this sort?**
---
Background on my motivations: I'm currently working with Batanin/Leinster style weak higher categories, and have a problem which seems amenable to model-theoretic techniques, so I'm trying to see if I can transfer/adapt/generalise the model structures defined by Cisinski et al, Lafont/Métayer/Worytkiewicz, etc. in this area. So I have some candidate (cofibrantly generated) classes of maps, and am trying to prove that they work; and there are lots of good examples around of how to prove that something *is* a model structure, but it would also be helpful to know what kinds of subtleties I should be looking out for that might make it *fail* to be.
| https://mathoverflow.net/users/2273 | Non-examples of model structures, that fail for subtle/surprising reasons? | Here is a classical example.
Let CDGA be the category of commutative differential graded algebras over a fixed ground field k of characteristic $p$. Weak equivalences are quasi-isomorphisms, fibrations are levelwise surjections. These would determine the others, but cofibrations are essentially generated by maps $A \rightarrow B$ such that on the level of the underlying DGA, $B$ is a polynomial algebra over $A$ on a generator $x$ whose boundary is in $A$.
CDGA is complete and cocomplete, satisfies the $2$-out-of-$3$ axiom, the retract axiom, satisfies lifting, and a general map can be factored into a cofibration followed by an acyclic fibration by the small object argument.
However, you don't have factorizations into acyclic cofibrations followed by fibrations, because of the following.
Suppose $A \rightarrow B$ is a map of commutative DGAs which is a fibration in the above sense. Then for any element $[x]$ in the (co)homology of $B$ in even degree, the $p$-th power $[x]^p$ is in the image of the cohomology of $A$. In fact, pick any representing cycle $x \in B$ and choose a lift $y \in A$. Then the boundary of $y^p$ is $py^{p-1} = 0$ by the Leibniz rule, so $[y^p]$ is a lift of $[x]^p$ to the (co)homology of $A$.
(As a result, there are a lot of other "homotopical" constructions, such as homotopy pullbacks, that are forced to throw you out of the category of commutative DGAs into the category of $E\_\infty$ DGAs.)
Nothing goes wrong in characteristic zero.
| 50 | https://mathoverflow.net/users/360 | 23885 | 15,702 |
https://mathoverflow.net/questions/23873 | 11 | This is most probably widely known and discussed here many times, so I am preliminay sorry.
Does Riemann conjecture imply some lower estimates on values, say $|\zeta(3/4+it)|$ for real $t$, when $|t|$ tends to infinity?
Are any such results known without assuming Riemann conjecture (many doubts here)?
Thanks!
| https://mathoverflow.net/users/4312 | Lower bounds on zeta(s+it) for fixed s | Yes, such conditional results are covered in Chapter 14 of the standard reference - the second edition of The Theory of the Riemann Zeta-Function by E. C. Titchmarsh. This edition has end-of-chapter notes by D. R. Heath-Brown bringing it up to date as of 1986.
In particular a lower bound
$$
|\zeta(3/4 + it)| \gg e^{-c\sqrt{\log(t)}/\log\log(t)}
$$
holds with some $c > 0$, conditionally on RH. See page 384 of the cited reference.
Such results are only known unconditionally for a region to the left of the line $\sigma = 1$ that
narrows to zero width as $t \rightarrow {\pm}\infty$. Not coincidentally, the best zero-free region known is also of this form. See page 135 of the cited reference for the best result of this kind.
| 12 | https://mathoverflow.net/users/3304 | 23892 | 15,707 |
https://mathoverflow.net/questions/23813 | 3 | The fact that acyclicity corresponds to having unique paths powers a lot of useful arguments in various areas of mathematics. What is the most fundamental reason you can come up with to explain the correspondence?
Also, what are more sophisticated generalizations of this correspondence? By this, I mean connections between negative and positive structural properties of an object, which in some way, perhaps only informally, generalize the fact about acyclicity. I'm looking to collect examples.
Please feel free to close this question if it's deemed too vague or philosophical for MathOverflow.
| https://mathoverflow.net/users/5926 | Acyclicity equivalent to unique paths | One possible generalization comes from the graph minors project of Robertson and Seymour. In particular the notion of [tree-width](http://en.wikipedia.org/wiki/Tree_decomposition) is in some sense dual to the notion of a *bramble* (I will define this in a second). Note that a connected graph has tree-width 1 if and only if it is a tree.
Now, let $G$ be a graph. Two subsets of $V(G)$ *touch* if they have a vertex in common or $G$ contains an edge between them. A set of pairwise touching
connected vertex sets in $G$ is a *bramble*. A subset of vertices *covers* a bramble $\mathcal{B}$ if it intersects every set in $\mathcal{B}$. The least number of vertices covering $\mathcal{B}$ is the *order* of $\mathcal{B}$.
Here is the duality relation that I alluded to earlier.
**Theorem.** A graph has tree-width $< k$ if and only if it does not contain a bramble of order $>k$.
So, loosely think of tree-width as a generalized unique paths property and brambles as generalized cycles.
| 2 | https://mathoverflow.net/users/2233 | 23894 | 15,709 |
https://mathoverflow.net/questions/23891 | 11 | Let $f:X\longrightarrow Y$ be a morphism of noetherian schemes. Under what conditions is $f\_\ast \mathcal{O}\_X$ a locally free $\mathcal{O}\_Y$-module?
**Example 1**. Suppose that $f$ is affine. Then $f\_\ast\mathcal{O}\_X$ is a quasi-coherent $\mathcal{O}\_Y$-module.
**Example 2**. Suppose that $f$ is finite. Then $f\_\ast \mathcal{O}\_X$ is even coherent.
**Example 3**. Suppose that $f:X\longrightarrow Y$ is a finite morphism of regular integral 1-dimensional schemes. Then $f\_\ast \mathcal{O}\_X$ is coherent and locally free. (The local rings $\mathcal{O}\_{Y,y}$ are discrete valuation rings.)
In view of the above examples, I'm basically looking for a higher-dimensional analogue of Example 3. But, I don't require quasi-coherence in my question. (Although this is quite unnatural.)
**Idea**. For any finite morphism $f:X\longrightarrow Y$, we have that $f\_\ast \mathcal{O}\_X$ is locally free. Only what are the precise conditions on $X$ and $Y$?
| https://mathoverflow.net/users/4333 | When is the push-forward of the structure sheaf locally free | It is of course not true that for any finite morphism $f:X\to Y$ we have $f\_\*\mathcal{O}\_X$ locally free : think about a closed immersion.
In fact, your question is about the important topic of "base change and cohomology of sheaves" for proper morphisms, which is treated by Grothendieck in EGA3. The simplest answer one can give, I think, is that if $f$ is proper [EDIT : and flat, as t3suji points out] and for all $y\in Y$ we have $H^1(X\_y,\mathcal{O}\_{X\_y})=0$ then $f\_\*\mathcal{O}\_X$ is locally free.
You may want to avoid going to find the exact reference in EGA3, since as Mumford says, that result is "unfortunately buried there in a mass of generalizations". In that case, go to chapter 0, section 5 of *Geometric Invariant Theory* (3rd ed.) by Mumford, Fogarty and Kirwan. This is where Mumford's comment is taken from.
| 8 | https://mathoverflow.net/users/17988 | 23895 | 15,710 |
https://mathoverflow.net/questions/23898 | 5 | It is well known that a non-abelian free group is residually a finite simple group. Katz and Magnus proved, in fact, that non-abelian free groups are residually alternating and residually $PSL\_{2}$. S. J. Pride has some nice results along these lines as well. The best result that I know of is the theorem of Weigel that can be formulated as follows. If $\mathfrak{X}$ is a group-theoretic class containing an infinite set of pairwise non-isomorphic finite non-abelian simple groups, then every non-abelian free group is residually an $\mathfrak{X}$-group.
---
My question is this:
>
> Is a non-abelian free group fully residually a finite non-abelian simple group?
>
>
>
It seems likely that the answer to such an obvious question is known, but I have not been able to find it in the literature.
I should probably add that I suspect we can probably replace "finite non-abelian simple" with "alternating", but I haven't yet given any thought to the other infinite series. I'd like to learn whether anything is known before spending more time on this.
| https://mathoverflow.net/users/1392 | Is a non-abelian free group fully residually a finite non-abelian simple group? | Yes! See my recent preprint [Alternating quotients of free groups](http://arxiv.org/abs/1005.0015).
I expect that what you want is well known, but I too couldn't find it in the literature. In fact, I prove the much stronger result that free groups are something like 'locally extended fully residually alternating'. Specifically:
>
> Let $F$ be any free group of rank at least two, let $H$ be a finitely generated subgroup of infinite index in $F$ and let $\{g\_1,\ldots,g\_n\}$ be a finite subset of $F\smallsetminus H$. Then there is a surjection $f$ from $F$ to a finite alternating group such that $f(g\_i)$ is not in $f(H)$ for any $i$.
>
>
>
| 6 | https://mathoverflow.net/users/1463 | 23899 | 15,713 |
https://mathoverflow.net/questions/23593 | 67 | OK so let's see if I can use MO to explicitly compute an example of something, by getting other people to join in. Sort of "one level up"---often people answer questions here but I'm going to see if I can make people do a more substantial project. Before I start, note
(1) the computation might have been done already [I'd love to hear of a reference, if it has]
(2) the computation may or may not be worth publishing
(3) If it is worth publishing, there may or may not be a debate as to who the authors are.
I personally don't give a hang about (2) or (3) at this point, but others might. Let me get on to the mathematics. Oh---just a couple more things before I start---this project is related to the mathematics at [this question](https://mathoverflow.net/questions/22908/does-anyone-want-a-pretty-maass-form), but perhaps pushes it a bit further (if we can get it to work). I had initially thought about these issues because I was going to give them to an undergraduate, but the undergraduate tells me today that he's decided to do his project on the holomorphic case, and it seemed a bit daft to let my initial investment in the problem go to waste, so I thought I'd tell anyone who was interested. If no-one takes the bait here, I'll probably just make this another UG project.
OK so here's the deal. Say $K/\mathbf{Q}$ is a finite Galois extension, and $\rho:Gal(K/\mathbf{Q})\to GL(2,\mathbf{C})$ is an irreducible 2-dimensional representation. General conjectures in the Langlands philosophy predict that $\rho$ comes from an automorphic form on $GL(2)$ over $\mathbf{Q}$. The idea is that we are going to "see" this form in an explicit example where general theory does not yet prove that it exists.
Now the determinant of $\rho$ is a 1-dimensional Galois representation, and it makes sense to ask whether $det(\rho(c))$ is $+1$ or $-1$, where $c$ is complex conjugation. It has to be one of these, because $c^2=1$. The nature of the automorphic form predicted to exist depends on the sign.
If the determinant is $-1$ then the form should be holomorphic, and a classical weight 1 cusp form. In this case the existence of the form is known, because it is implied by Serre's conjecture, which is now a theorem of Khare and Wintenberger.
If the determinant is $+1$ then the conjectural form should be a real-analytic function on the upper half plane, invariant under a congruence subgroup, and satisfying a certain differential equation (which is not the Cauchy-Riemann equations in this case). If the image of $\rho$ is a solvable group then the existence of this form is known by old work of Langlands and Tunnell.
So in summary then, the one case where the form is not known to exist is when the determinant of complex conjugation is $+1$ and the image of Galois is not solvable.
Here is an explicit example. The polynomial
```
g5=344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5
```
has splitting field $L$, an $A\_5$-extension of $\mathbf{Q}$ ramified only at the prime 1951. Now $A\_5$ is isomorphic to the quotient of $SL(2,\mathbf{F}\_5)$ by its centre $\pm 1$, and $L$ has a degree two extension $K\_0$, also unramified outside 1951, with $Gal(K\_0/\mathbf{Q})$ being $SL(2,\mathbf{F}\_5)$. Turns out that $K\_0$ can be taken to be the splitting field of the rather messier polynomial
```
g24 = 14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 - 1381768039105642956*x^6 + 4291028045077743465*x^8 - 2050038614413776542*x^10 + 287094814384960835*x^12 - 9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 + 152929135*x^20 - 50726*x^22 + x^24
```
I should perhaps say that David Roberts told me these polynomials in Jan 2008; they're in a paper by him and John Jones---but they learnt about them from a paper of Doud and Moore.
Now $SL(2,\mathbf{F}\_5)$ has two faithful 2-dimensional complex representations; the traces of each representation take values in $\mathbf{Q}(\sqrt{5})$ and one is of course the conjugate of the other. The determinant of both representations is trivial so in fact they are $SL(2,\mathbf{C})$-valued. Oh---also, all the roots of g24 are real---and hence $K\_0$ is totally real. So what we have here is a representation
$$\rho\_0:Gal(K\_0/\mathbf{Q})\to GL(2,\mathbf{C})$$
which is conjectured to come from automorphic forms, but, as far as I know, the conjecture is not known in this case.
Unfortunately the conductor of $\rho\_0$ is $1951^2$, which is a bit big. In fact let me say something more about what is going on at 1951. In the $A\_5$ extension the decomposition and inertia groups at 1951 are both cyclic of order 5. If my understanding of what David Roberts told me is correct, in the $SL(2,\mathbf{F}\_5)$ extension the decomposition and inertia groups are both cyclic of order 10 (in fact I just got magma to check this). But the upshot is that $\rho\_0$ restricted to a decomposition group at 1951 is of the form $\psi+\psi^{-1}$ with $\psi$ of order 10. Which character of order 10? Well there are four characters $(\mathbf{Z}/1951\mathbf{Z})^\times\to\mathbf{C}^\times$ of order 10, and two of them will do, and two won't, and which ones will do depends on which 2-dimensional representation of $SL(2,\mathbf{F}\_5)$ you chose.
The key point though is that if you get $\psi$ right, then the twist $\rho:=\rho\_0\otimes\psi$ will have conductor 1951, which is *tiny* for these purposes.
Now, as Marty did in an $A\_4$ example and as I did and Junkie did in a dihedral example
in the Maass form question cited above, it is possible to figure out explicitly numbers $b\_1$, $b\_2$, $b\_3$,..., with the property that
$$L(\rho,s)=\sum\_{n\geq1}b\_n/n^s.$$
If one had a computer program that could calculate $b\_n$ for $n\geq1$, then there are not one but two ways that one could attempt to give computational evidence for the predictions given by the Langlands philosophy:
(A) one could use techniques that Fernando Rodriguez-Villegas explained to me a few months ago to try and get computational evidence that $L(\rho,s)$ had analytic continuation to the complex plane and satisfied the correct functional equation, and
(B) one could compute the corresponding real analytic function on the upper half plane, evaluate it at various places to 30 decimal places, and see if the function was invariant under the group $\Gamma\_1(1951)$.
I don't know much about (A) but I once tried, and failed, to do (B), and my gut feeling is that my mistake is in the computer program I wrote to compute the $b\_n$. But as junkie's response in the previous question indicates, there now seem to be several ways to compute the $b\_n$ and one thing I am wondering is whether we can use the methods he/she indicated in this question.
Let me speak more about how I tried to compute the $b\_n$. The character of the Maass form is $\psi^2$, the determinant of $\rho$. General theory tells us that $b\_n$ is a multiplicative function of $n$ so we only need compute $b\_n$ for $n$ a prime power. Again general theory (consider the local $L$-functions) says that for $p\not=1951$ one can compute $b\_{p^n}$ from $b\_p$. If $p=1951$ then $b\_{p^n}=1$ for all $n$, because the decomposition and inertia groups coincide for 1951 in $K\_0$ [EDIT: This part of the argument is wrong, and it explains why my programs didn't work. I finally discovered my mistake after comparing the output of mine and Junkie's programs and seeing where they differed. It's true that decomposition and inertia coincide in $K\_0$ but when one twists by the order 10 character this stops being true. In fact $b\_{1951}$ is a primitive 5th root of unity that I don't know how to work out using my method other than by trial and error.]. Finally, if the $L$-function of $\rho\_0$ is $\sum\_n a\_n/n^s$ then $b\_n=\psi(n)a\_n$ for all $n$ prime to 1951, so it suffices to compute $a\_p$ for $p\not=1951$.
To compute $a\_p$ I am going to compute the trace of $\rho\_0(Frob\_p)$. I first compute the GCD of the degrees of the irreducible factors of g24 mod $p$. If $p$ doesn't divide the discriminant of g24 then this GCD is the order of $Frob\_p$ in $SL(2,\mathbf{F}\_5)$. If $p$ does divide the discriminant of g24 then rotten luck, I need to factorize $p$ in the ring of integers of the number field generated by a root of g24. I did this using magma. Here are the results:
```
prime order
2 6
3 10
5 4
163 5
16061 1
889289 10
451400586583 2
1188493301983785760551727 2
120450513180827412314298160097013390669723824832697847 1
```
Now unfortunately just computing the order of the conj class of Frobenius is not enough to determine the trace of the Galois representation, because $SL(2,\mathbf{F}\_5)$ contains two conj classes of elements of order 5, and two of order 10. However, in both cases, the conj classes remain distinct in $A\_5$ and so it suffices to have an algorithm which can distinguish between the two conjugacy classes in $A\_5$. More precisely, we have to solve the following problem: we label the conj classes of elements of order 5 in $A\_5$ as C1 and C2, and we want an algorithm which, given a prime $p$ for which g5 is irreducible mod $p$, we want the algorithm to return "C1" or "C2" depending on which class $Frob\_p$ is in. Here is a beautiful way of doing it, explained to me by Bjorn Poonen: if g5 is irreducible mod $p$ then its roots in an alg closure of $\mathbf{F}\_p$ are $x,x^p,x^{p^2},...$. Set $x\_i=x^{p^i}$ and compute $\prod\_{i<j}(x\_i-x\_j)$. This product is a square root of the discriminant of g5. Choose once and for all a square root of the discriminant of g5 in the integers; if the product is congruent to this mod $p$ return "C1", else return "C2".
That's it! I implemented this. I had a program which returned a bunch of $a\_n$'s, and hence a bunch of $b\_n$'s. I built the function on the upper half plane as the usual sum involving Bessel functions and so on, but it computationally did not come out to be invariant under $\Gamma\_1(1951)$.
If anyone wants to take up the challenge of computing the $b\_n$ that would be great. I have explained one way to do it above, but I am well aware that there might be other ways to compute the $b\_n$ analogous to junkie's approach from the previous question.
| https://mathoverflow.net/users/1384 | Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. | Here is Magma code that gets you the answer in a few seconds. I made a special case for the bad primes, and did them by hand.
```
_<x> := PolynomialRing(Rationals());
f5 := 344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5;
g24 := 14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 -
1381768039105642956*x^6 + 4291028045077743465*x^8 -
2050038614413776542*x^10 + 287094814384960835*x^12 -
9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 +
152929135*x^20 - 50726*x^22 + x^24;
K := NumberField(f5);
_,D := IsSquare(Integers()!Discriminant(f5));
prec := 30;
CHAR_TABLE := CharacterTable(GaloisGroup(g24));
chi := CHAR_TABLE[2];
BAD_FACTORS :=
[ <2,Polynomial([1,-1,1])>,
<3,Polynomial([1,-ComplexField(prec)!chi[9],1])>,
<5,Polynomial([1,0,1])>,
<7,Polynomial([1,0,1])>,
<71,Polynomial([1,0,1])>,
<137,Polynomial([1,1,1])>,
<163,Polynomial([1,-ComplexField(prec)!chi[5],1])>,
<1951,Polynomial([1])>,
<16061,Polynomial([1,-2,1])>,
<889289,Polynomial([1,-ComplexField(prec)!chi[8],1])> ];
BAD := [bf[1] : bf in BAD_FACTORS];
FACTORS := [bf[2] : bf in BAD_FACTORS];
function LOCAL(p,d : Precision:=prec)
if p in BAD then return FACTORS[Position(BAD,p)]; end if;
R := Roots(ChangeRing(f5,GF(p)));
if #R eq 1 then return Polynomial([1,0,1]); end if;
if #R eq 2 then
ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]);
return Polynomial([1,ord eq 3 select 1 else -1,1]); end if;
if #R eq 5 then
ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]);
return Polynomial([1,ord eq 1 select -2 else 2,1]); end if;
r := Roots(ChangeRing(f5,GF(p^5)));
x := r[1][1];
prod := GF(p)!&*[x^(p^i)-x^(p^j) : j in [(i+1)..4], i in [0..4]];
wh := prod eq GF(p)!D;
ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]);
if ord eq 10 then class := wh select 8 else 9; // compatible with FACTORS
else class := wh select 6 else 5; end if;
return Polynomial([1,-ComplexField(prec)!chi[class],1]);
end function;
L := LSeries(1, [0,0], 1951^2, LOCAL : Precision:=prec);
// s->1-s, Gamma(s/2)^2
psi := DirichletGroup(1951, CyclotomicField(10)).1;
p1951 := Polynomial([1,-ComplexField(prec)!CyclotomicField(5).1]);
TP := TensorProduct(L, LSeries(psi : Precision:=prec), [<1951, 1, p1951>]);
CheckFunctionalEquation(TP);
```
Here is the special values:
```
ev := Evaluate(TP,0); // 2-1.453085056...
rel := PowerRelation(ev,4 : Al:="LLL");
NF := NumberField(rel);
Q5<zeta5> := CyclotomicField(5);
assert IsIsomorphic(NF,Q5);
Q5!NF.1;
```
So $L(\rho,0)=-4\zeta\_5(1+\zeta\_5)$ for Marty. I get $L(\rho\_0,-1)=32(48723\sqrt{5} - 778741)$ as an algebraic. I get $L(\rho,-2)=8800\zeta\_5^3 - 14444\zeta\_5^2 + 35604\zeta\_5 + 17412$ with more precision. I determined the TensorProduct factor at 1951 via trial and error, making the obvious guesses until one worked (the failure is at 100-110 digits). With this, I take it to 240 digits and I can even get $$L(\rho,-4)=-18475535360\zeta\_5^3 - 11142861380\zeta\_5^2 - 12091894020\zeta\_5 - 7107607296$$ and $$L(\rho,-6)=25255057273186244\zeta\_5^3 - 1015274469604000\zeta\_5^2 - 15695788409197884\zeta\_5 +
9459547822189412$$ The precision can go higher if you want more.
Finally, the Maass form:
```
function MaassEval(L,z)
x:=Real(z); y:=Imaginary(z);
printf "Using %o coefficients\n", Ceiling(11/y);
C := LGetCoefficients(L,Ceiling(11/y));
pi := Pi(RealField());
a := Sqrt(y)*&+[C[n]*KBessel(0,2*pi*n*y)*Sin(2*pi*n*x) : n in [1..#C]];
return a;
end function;
zz:=0.0001+0.0001*ComplexField().1;
MaassEval(TP,zz);
// Using 110000 coefficients
// -1.71477211817772949974178783985E-8 + 9.01673609747756708674470686948E-9*i
MaassEval(TP,zz/(1951*zz+1));
// Using 161297 coefficients
// -1.71477211817772949974179078240E-8 + 9.01673609747756708674496293450E-9*i
```
| 27 | https://mathoverflow.net/users/5267 | 23913 | 15,723 |
https://mathoverflow.net/questions/23887 | 10 | Apparently it's 'well known' that if $P$ is a presheaf on $C$ then there is an equivalence $\widehat{C}/P \simeq \widehat{\int P}$, where $\int P$ is the usual category of elements and $\widehat{C} = [C^{\rm op},{\rm Set}]$. (I've seen a reference to Johnstone's *Topos Theory* for this, but I don't have easy access to the book. It's also Exercise III.8(a) in Mac Lane--Moerdijk)
Now, I've come across comma categories $\widehat{C}/H$ for functors $H \colon D \to \widehat{C}$ (mainly when $H = D(F-,-)$ for $F \colon C \to D$), and I'd like to have a similarly useful/interesting result for them. The equivalence doesn't generalize to $\widehat{C}/H \simeq \widehat{\int H}$, or at least I can't see any way to make that work. So I want to understand the first equivalence from a more abstract-nonsensical point of view, to figure out what's really going on.
Is there a way to see the above equivalence as living in a fibrational cosmos or something similar? If not, is there any other kind of machinery that might help me understand categories of the form $\widehat{C}/H$?
| https://mathoverflow.net/users/4262 | Slices of presheaf categories | Yes, you can see this as happening in a "fibrational cosmos." I'll describe how it goes, but then we'll see that the description of $\widehat{C}/H$ that comes out could also be deduced pretty naively.
The universal property of $\widehat{C}$ is that the category of functors $A\to \widehat{C}$ is equivalent to the category of discrete fibrations from $A$ to $C$, via pullback of the universal such. (A discrete fibration from $A$ to $C$ is a span $A\leftarrow E \to C$ such that $E\to C$ is a fibration, $E\to A$ is an opfibration, the two structures are compatible, and $E$ is discrete in the slice 2-category over $A\times C$.)
Now the slice category $\widehat{C}/P$, for $P\in\widehat{C}$, also has a universal property: it is the comma object of $\mathrm{Id}\_{\widehat{C}}$ over $P:1\to \widehat{C}$. Thus a functor $A\to \widehat{C}/P$ is equivalent to a functor $A\to \widehat{C}$ and a natural transformation from it to the composite $A \to 1 \overset{P}{\to} \widehat{C}$. By the universal property of $\widehat{C}$, this is the same as giving a discrete fibration $A\leftarrow E \rightarrow C$ together with a map to the discrete fibration classified by $P:1\to \widehat{C}$, which is just $1\leftarrow \int P \rightarrow C$.
Next, (discrete) fibrations have the special property that not only is the composite of two (discrete) fibrations again a (discrete) fibration, but if $g$ is a discrete fibration and $g\circ f$ is a fibration, then $f$ is a fibration. Therefore, given a discrete fibration from $A$ to $C$ together with a map $E\to \int P$ over $C$, the map $E\to \int P$ is itself a fibration, and we can actually show that $A\leftarrow E \to \int P$ is itself a discrete fibration from $A$ to $\int P$, and that this is an equivalence. Hence, the slice category $\widehat{C}/P$ has the same universal property as $\widehat{\int P}$, so they are equivalent.
Now we can ask about replacing $P:1\to \widehat{C}$ with a more general functor $H:D\to \widehat{C}$. Here the fibrational-cosmos argument fails, because in the discrete fibration $D \leftarrow \int H \to C$ it is no longer true that the solitary map $\int H \to C$ is itself a *discrete* fibration, so cancellability no longer holds and $E\to \int H$ is no longer necessarily a fibration. However, at least in the 2-category $Cat$, we can still use it to figure out what $\widehat{C}/H$ should look like, by looking at the case $A=1$. In this case, to give a map $1 \to \widehat{C}/H$ means to give a map $d:1\to D$ (i.e. an object of $D$) along with a discrete fibration $E\to C$ and a map from $E$ to $D \leftarrow \int H \to C$ over $d$ and $\mathrm{Id}\_C$. This is the same as a map from $E$ to $d^\*(\int H) = \int H(d)$ over $C$, and we can then apply the previous argument here since both are discrete fibrations over $C$.
Thus, an object of $\widehat{C}/H$ consists of an object $d\in D$ together with a presheaf on $\int H(d)$. This is not a surprise, though, since we're just applying the original fact $\widehat{C}/P \simeq \widehat{\int P}$ objectwise. Since $d$ can vary between objects, what we're saying is really that $\widehat{C}/H$ is the category of elements of the functor $D\to Cat$ defined by $d\mapsto \widehat{\int H(d)}$, or more evocatively
$$ \widehat{C}/H = \int\_d \widehat{\int H(d)} $$
as a "double integral".
| 14 | https://mathoverflow.net/users/49 | 23919 | 15,726 |
https://mathoverflow.net/questions/23917 | 8 | From wikipedia: <http://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem>
"Shoenfield's theorem shows that if there is a model ZF in which a given $\Pi^1\_3$ statement $\phi$ is false, then $\phi$ is also false in the constructible universe of that model."
The problem is that the only proofs I can find seem to use "a model of ZF+DC" or "a model of ZFC".
Is wikipedia right?
If so, is there a proof from just "a model of ZF" online or short enough to sketch here?
| https://mathoverflow.net/users/nan | Shoenfield's Absoluteness Theorem | Wikipedia is correct in that the Shoenfield Absoluteness Theorem holds for plain ZF.
Since the proof of the theorem relies heavily on the absoluteness of well-foundedness, it is tempting to assume DC. However, since the trees that occur in the usual proof of the theorem are canonically well-ordered, DC is not necessary to prove that the well-foundedness of these trees is absolute. For a different approach, see the proof given by Barwise and Fisher in *The Shoenfield Absoluteness Lemma*. [Israel J. Math. 8 1970, 329-339, [MR278934](http://www.ams.org/mathscinet-getitem?mr=278934)]
| 14 | https://mathoverflow.net/users/2000 | 23920 | 15,727 |
https://mathoverflow.net/questions/23788 | 13 | According to the Wikipedia ACA0 is a conservative extension of First Order logic + PA.
<http://en.wikipedia.org/wiki/Reverse_Mathematics>
First of all I have a few questions about the proof:
a - What is the general sketch of this proof, is it based on models?
b - Consider the theorem that ACA0 is a conservative extension of First Order + PA, and the proof of that theorem is proven in a formal system, what kind of logic is needed? If the proof is based on models, then it requires second order logic. However, the theorem itself is a ∏02 question as far as I understand, and can be expressed in First Order logic + PA. Is there also a proof in First Order logic + PA?
Then I am interested in the following:
c - Given an ACA0 formal proof that ends in a theorem that is part of First Order logic + PA, is there an algorithm that reduces the ACA0 proof to First Order + PA proof?
One could just do a breath first search on First Order logic + PA and given the fact that ACA0 is a conservative extension, it is guaranteed to end. So, the answer to question c is definitely "yes", but I am looking for something more clever.
I am struggling with this algorithm for months. In general an ACA0 proof, with a First Order + PA end theorem reduces rather easier. However, there are some non-trivial cases. If the answer to question b is "yes", then that proof might give hints for constructing the algorithm.
I want to use this algorithm to reduce proofs of full second order, such that the reduced proof is First Order logic + PA, or contains the use of the induction scheme with a second order induction hypothesis.
In many cases the use of second order induction hypothesis, can be reduced by using the "Constructive Omega Rule". I want to understand the limitations of this (if any).
Thanks in advance,
Lucas
| https://mathoverflow.net/users/5917 | Reducing ACA₀ proof to First Order PA | Chapter nine of Simpson (1999) *Subsystems of Second-Order Arithmetic* proves (a) by showing how to construct a second-order model for ACA0 from a first-order model of PA.
(b) The "second-order" we are talking about is really first-order multi-sorted logic, i.e., the second-order quantifiers have Henkin semantics. So it's all first order, all the way down.
(c) Yes, you are right in your thoughts about getting PA proofs from ACA0. Why do you want to do this? Proofs in PA of a given theorem may be much longer than in ACA0, to the extent that they may useless as witness objects. Paulo Oliva, a student of Kohlenbach's, has studied the application of Kohlenbach's "proof mining" to subsystems of second-order arithmetic in his PhD dissertation, [Proof Mining in Subsystems of Analysis](http://www.brics.dk/DS/03/12/BRICS-DS-03-12.pdf); maybe you will find this of use? Kohlenbach's works in general are relevant to this kind of question, see [his publications page](http://www.mathematik.tu-darmstadt.de/~kohlenbach/) and his book, *Applied proof theory: proof interpretations and their use in mathematics* (2009) Springer Verlag.
| 8 | https://mathoverflow.net/users/3154 | 23922 | 15,729 |
https://mathoverflow.net/questions/23911 | 43 | I am teaching a course on Riemann Surfaces next term, and would **like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties** (over, say, $\mathbb{C}$). I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.
Here are some examples that I thought of:
**1.** Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$. By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$. Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.
**2.** Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$. Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that
agrees with $f$ on $V$ and is identically zero outside of $U$.
By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function. It may be impossible find a global regular function $g$ that agrees with $f$ on $V$. When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.
**3.** If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian. The local ring of a variety at a point is always noetherian.
***What are some more examples?***
Answers illustrating the difference between real manifolds and complex manifolds are also welcome.
| https://mathoverflow.net/users/5337 | What's the difference between a real manifold and a smooth variety? | Here is a list biased towards what is remarkable in the complex case. (To the potential peeved real manifold: I love you too.) By "complex" I mean holomorphic manifolds and holomorphic maps; by "real" I mean $\mathcal{C}^{\infty}$ manifolds and $\mathcal{C}^{\infty}$ maps.
* Consider a map $f$ between manifolds of *equal* dimension.
In the complex case: if $f$ is injective then it is an isomorphism onto its image. In the real case, $x\mapsto x^3$ is not invertible.
* Consider a holomorphic $f: U-K \rightarrow \mathbb{C}$, where $U\subset \mathbb{C}^n$ is open and $K$ is a compact s.t. $U-K$ is connected. When $n\geq 2$, $f$ extends to $U$. This so-called Hartogs phenomenon has no counterpart in the real case.
* If a complex manifold is compact or is a bounded open subset of $\mathbb{C}^n$, then its group of automorphisms is a Lie group. In the smooth case it is always infinite dimensional.
* The space of sections of a vector bundle over a compact complex manifold is finite dimensional. In the real case it is always infinite dimensional.
* To expand on Charles Staats's excellent answer: few smooth atlases happen to be holomorphic, but even fewer diffeomorphisms happen to be holomorphic. Considering manifolds up to isomorphism, the net result is that many complex manifolds come in continuous families, whereas real manifolds rarely do (in dimension other than $4$: a compact topological manifold has at most finitely many smooth structures; $\mathbb{R}^n$ has exactly one).
On the theme of *zero subsets* (i.e., subsets defined locally by the vanishing of one or several functions):
* *One* equation always defines a *codimension one* subset in the complex case, but
{$x\_1^2+\dots+x\_n^2=0$} is reduced to one point in $\mathbb{R}^n$.
* In the complex case, a zero subset isn't necessarily a submanifold, but
is amenable to manifold theory by Hironaka desingularization. In the real case, *any* closed subset is a zero set.
* The image of a proper map between two complex manifolds is a zero subset, so isn't too bad by the previous point. Such a direct image is hard to deal with in the real case.
| 24 | https://mathoverflow.net/users/2109 | 23927 | 15,733 |
https://mathoverflow.net/questions/23437 | 20 | The generating function $f(z)$ of the Catalan numbers which is characterized by $f(z)=1+zf(z)^2$ is D-finite, or holonomic, i.e. it satisfies a linear differential equation with polynomial coefficients.
The generating function $F(z)$ of the $q$-Catalan numbers is analogously characterized by the functional equation $F(z)=1+z F(z) F(qz)$. I suspect that $F(z)$ is not $q$-holonomic, i.e. does not satisfy a linear $q$-differential equation with polynomial coefficients. But I have no proof. Is there a proof in the literature or references which may lead to a proof?
Since there were some misunderstandings I want to clarify the situation.
A power series $F(z)$ is called $q - $holonomic if there exist polynomials $p\_i (z)$ such that $\sum\limits\_{i = 0}^r {p\_i (z)D\_q^i } F(z) = 0$ where $D\_q $ denotes the $q - $differentiation operator defined by $D\_q F(z) = \frac{{F(z) - F(qz)}}{{z - qz}}.$ Equivalently if there exist (other) polynomials such that $\sum\limits\_{i = 0}^r {p\_i (z)F(q^i } z) = 0.$
Let $f(z)$ be the generating function of the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . Then $f(z) = 1 + zf(z)^2 $ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$
There are 3 simple $q - $analogues of the Catalan numbers:
a) The polynomials $C\_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. My question is about these polynomials. Their generating function satisfies a simple equation, but there is no known formula for the polynomials themselves.
b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function.
c) The $q - $Catalan numbers $c\_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q - $analogue of $ \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q - $analogue of $sqrt {(1 + z)}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function. But they are not polynomials in $q.$
Both b) and c) are $q$-holonomic. My question is a proof that a) is not $q$-holonomic.
| https://mathoverflow.net/users/5585 | Are the q-Catalan numbers q-holonomic? | The problem is to show that the function $F(z)=1+z+(1+q)z^2+O(z^3)$
satisfying the functional equation $F(z)=1+zF(z)F(qz)$ does not satisfy
$\sum\_{j=0}^{n-1}P\_j(z)F(q^jz)+Q(z)=0$ identically in $z$ for some $n$;
here $P\_j$ and $Q$ are polynomials in both $z$ and $q$.
(Although the original question assumes the homogeneous equation,
$Q(z)=0$, the limiting case $q\to1$ suggests to consider $Q(z)$ more
generally.) In what follows we show that such a functional equation
implies the algebraicity of $F(z)$; this is known to be false.
First of all, switch to the function $G(z)=zF(z)$ which satisfies
$$
G(qz)G(z)=q(G(z)-z).
$$
The problem is then to show that the newer function does not satisfy
$\sum\_{j=0}^{n-1}\tilde P\_j(z)G(q^jz)+\tilde Q(z)=0$ for some $n$.
By applying $z\mapsto q^{-k}z$ we can assume that $\tilde P\_0(z)\ne0$
in this relation. The substitution $z\mapsto qz$ results in the relation
$\sum\_{j=1}^n\hat P\_j(z)G(q^jz)+\hat Q(z)=0$ where $\hat P\_1(z)\ne0$.
The next step is to show, by iterating the functional equation for $G(z)$, that
$$
G(q^nz)\dots G(qz)G(z)=X\_n(z)G(z)-Y\_n(z).
$$
Indeed, we have $X\_0=1$, $Y\_0=0$, and
$$
X\_n(z)G(z)-Y\_n(z)=\bigl(X\_{n-1}(qz)G(qz)-Y\_{n-1}(qz)\bigr)G(z)
$$
implying
$$
X\_n(z)=qX\_{n-1}(qz)-Y\_{n-1}(qz), \quad Y\_n(z)=qzX\_{n-1}(qz)
\qquad\text{for}\quad n\ge1.
$$
By means of the formula we see that $\deg Y\_n$ does not decrease with $n$,
so that $Y\_n\ne0$ for $n\ge1$. Note that our computation implies
$$
G(q^nz)=\frac{X\_n(z)G(z)-Y\_n(z)}{X\_{n-1}(z)G(z)-Y\_{n-1}(z)}
\qquad\text{for}\quad n\ge1.
$$
Substitute the above finding into the equation
$\sum\_{j=1}^n\hat P\_j(z)G(q^jz)+\hat Q(z)=0$ with $\hat P\_1(z)\ne0$.
We obtain
$$
\sum\_{j=1}^n\hat P\_j(z)\frac{X\_j(z)G(z)-Y\_j(z)}{X\_{j-1}(z)G(z)-Y\_{j-1}(z)}
+\hat Q(z)=0.
$$
The term corresponding to $j=1$ is equal to
$$
\hat P\_1(z)\frac{X\_1(z)G(z)-Y\_1(z)}{X\_0(z)G(z)-Y\_0(z)}
=\hat P\_1(z)\frac{qG(z)-qz}{G(z)}.
$$
Note that the denominator $G(z)$ in this expression is not
canceled by the other denominators in the former relation
because $X\_{j-1}(z)G(z)-Y\_{j-1}(z)$ is never divisible by $G(z)$
as $Y\_{j-1}(z)\ne0$ for $j\ge2$. In other words, after multiplying
$$
\sum\_{j=1}^n\hat P\_j(z)\frac{X\_j(z)G(z)-Y\_j(z)}{X\_{j-1}(z)G(z)-Y\_{j-1}(z)}
+\hat Q(z)=0
$$
by $G(z)$ we get an algebraic relation
$$
-qz\hat P\_1(z)+G(z)\cdot\text{rational function in }z,q,G(z)
=0
$$
where the rational function does not involve $G(z)$ as a multiple in the denominator.
Since $\hat P\_1(z)\ne0$, this gives us a nontrivial algebraic relation for $G(z)$.
Thus the function $G(z)$, hence $F(z)$ as well, are algebraic.
**Addition.**
I have never seen Carlitz's function $F(z)$ before, so I was pretty sure that
things like its transcendence had been already established. Especially, since
it is not hard (although required some time from me). Any way, this explains
my reasons for not having a reference to this fact.
If we write
$$
F(z)=F\_q(z)=\sum\_{n=0}^\infty a\_n(q)z^n=\sum\_{n=0}^\infty a\_nz^n=1+z+(1+q)z^2+\dots,
$$
the functional equation $F(z)=1+zF(z)F(qz)$ implies
$$
a\_{n+1}=\sum\_{k=0}^nq^ka\_ka\_{n-k} \quad\text{for}\; n\ge0,
\qquad a\_0=1.
$$
The clear induction on $n$ shows that $a\_{n+1}(q)$ is a polynomial from $\mathbb Z[q]$
with leading term $q^{n(n-1)/2}$. In particular, the denominator of $a\_{n+1}(1/2)$ is
exactly $2^{n(n-1)/2}$.
If the function $F\_q(z)$ were algebraic then its specialization $F\_{1/2}(z)$ should be
algebraic. This would imply that for a certain integer $A$ the $z$-expansion of
$F\_{1/2}(Az)$ has *integral* coefficients. But no $A\in\mathbb Z$ with the property
$A^{n+1}/2^{n(n-1)/2}\in\mathbb Z$ for all $n$ could be given. Therefore, $F\_{1/2}(z)$ and
$F\_q(z)$ in general are transcendental.
| 14 | https://mathoverflow.net/users/4953 | 23930 | 15,736 |
https://mathoverflow.net/questions/23924 | 1 | Given a fibered surface $X \rightarrow C$, with generic fiber $Y$ and a vector bundle $E$ on $X$.
Then the first Chern class $c\_1(E)$ is a divisor on $X$, so one can restrict this divisor to the generic fiber $c\_1(E)\_{|Y}$.
On the other hand one can restrict $E$ to the generic fiber and look at its first Chern class as a divisor on $Y$, i.e. $c\_1(E\_{|Y})$.
Do we always have the equality $c\_1(E)\_{|Y}=c\_1(E\_{|Y})$ as divisors on $Y$, if $X$, $C$ and $Y$ are "nice" enough?
| https://mathoverflow.net/users/3233 | Restriction of divisors to the generic fiber | Yes, even when $X$, $C$ and $Y$ are as nasty as they can be. Chern classes are functorial.
| 2 | https://mathoverflow.net/users/4790 | 23932 | 15,738 |
https://mathoverflow.net/questions/23935 | 1 | ~~Hi,
I need to know if this relation is correct for a metric:~~
$g\_{a[b}g\_{c]d}=\frac{1}{2}\epsilon\_{ace}\epsilon\_{bdf}gg^{ef}$
I know that :
$\frac{1}{2}\epsilon\_{ace}\epsilon\_{bdf}g^{ef}=g\_{b[a}g\_{c]d}$
but I don't see how the determinant $g$ of the metric could appear.
Edit:
Ok so the previous relation emerged when computing the area of a surface $S$ in terms of the "densitized" triad $E\_{i}^{a}=ee\_{i}^{a}$ where $a,b,c,...$ are the spatial coordinates and $i,j,k,...$ are $SU(2)$ coordinates, `e` the determinant of the triad matrix defined by $g\_{ab}=e\_{a}^{i}e\_{b}^{j}\delta\_{ij}$ where $g\_{ab}$ is the spatiale metric. So, since the computation of the area uses the determinant of the the metric $h\_{\alpha\beta}$ induced by $g\_{ab}$ on $S$: ($\alpha,\beta,... =1,2\;and\; a,b,..=1,2,3$)
$h\_{\alpha\beta}=g\_{ab}\frac{\partial x^{a}}{\partial\sigma^{\alpha}}\frac{\partial x^{b}}{\partial\sigma^{\beta}}$
So in computing the determinant $h$ explecitely on finds the term
$g\_{a[b}g\_{c]d}$ which needs to equal to $\frac{1}{2}\epsilon\_{ace}\epsilon\_{bdf}gg^{ef}$
in order to obtain the final result:
$h=E\_{i}^{a}E\_{j}^{b}\delta^{ij}n\_{a}n\_{b}$ where $n$ are normal vectors $n\_{a}=\epsilon\_{abc}\frac{\partial x^{b}}{\partial\sigma^{1}}\frac{\partial x^{c}}{\partial\sigma^{2}}$
EDIT2:
After the notification of Willie Wong, I decided to put my original problem as a question, i.e: deriving the expression of the determinant of the induced metric on $S$ in terms of the densitized triad.
| https://mathoverflow.net/users/2597 | Area of a surface in terms of the densitized triad | Edit 2:
Contrary to your description in your question, I am going to assume that your E's are tensor densities of weight 1, not 2 (otherwise the units don't work out right). Then a sketch of the argument goes something like this:
Let $\sigma^\alpha$ be a coordinate system on $S$, extend this with the normal vector $n$ so that we get a local basis in the tangent space. In this basis $f\_0 = n, f\_1 = \partial/\partial \sigma^1, f\_2 = \partial/\partial\sigma^2$ the metric tensor $g$ looks like
$$ (g) = \begin{pmatrix} g\_{00} & 0 \\\\ 0 & (h) \end{pmatrix} $$
So in this coordinate system $|h| g\_{00} = |g|$, or $|h| = (g^{-1})\_{00} |g|$ since $(g)$ is block diagonal.
Now, using that $E\_i^a = e e\_i^a$ where $e\_i^a$ are an orthonormal basis, we have that the inverse metric $g^{-1}$ is equal to $\sum\_i e\_i^a\otimes e\_i^b$. So the right hand side of your formula reads
$$ E\_i^aE\_j^b \delta^{ij}n\_a n\_b = e^2 g^{ab}n\_an\_b$$
So if $e$ is chosen to be the square root of the metric determinant, then the expression agrees with what is shown above.
---
Original answer:
Take a look at the Wikipedia entry for Levi-Civita symbols <http://en.wikipedia.org/wiki/Levi-Civita_symbol>
Now, I assume that when you say metric you mean a (pseudo-)Riemannian metric tensor. Note that the Levi-Civita symbol by itself is not a tensor. Nor is the metric determinant a scalar (they are both frame/coordinate dependent). But with a choice of an orientation, the object which sometimes is written as $\sqrt{|g|}\epsilon\_{abc}$ and sometimes just $\epsilon\_{abc}$ is a covariant object: it is a way of writing the volume form.
So I suspect you are just confused about the "frame" expression of the volume form (in an orthonormal frame the metric determinant is just 1, and the volume form agrees with the Levi-Civita symbol) and the coordinate expression of the volume form...
EDIT (after OP's edit):
Something is wrong with the relation you are trying to prove: Take trace relative to $g^{ac}$, The left hand side gives $-2 g\_{bd}$ and the right hand side vanishes. So I would start by checking your computation for the determinant of $h$.
| 2 | https://mathoverflow.net/users/3948 | 23938 | 15,742 |
https://mathoverflow.net/questions/15942 | 19 | The title is a little tongue-in-cheek, since I have a very particular question, but I don't know how to condense it into a pithy title. If you have suggestions, let me know.
Suppose I have a **Lie groupoid** $G \rightrightarrows G\_0$, by which I mean the following data:
* two finite-dimensional (everything is smooth) manifolds $G,G\_0$,
* two surjective submersions $l,r: G \to G\_0$,
* an embedding $e: G\_0 \hookrightarrow G$ that is a section of both the maps $l,r$,
* a **composition law** $m: G \times\_{G\_0} G \to G$, where the fiber product is the pull back of $G \overset{r}\to G\_0 \overset{l}\leftarrow G$, intertwining the projections $l,r$ to $G\_0$.
* Such that $m$ is associative, by which I mean the two obvious maps $G \times\_{G\_0} G \times\_{G\_0} G \to G$ agree,
* $m(e(l(g)),g) = g = m(g,e(r(g)))$ for all $g\in G$,
* and there is a map $i: G \to G$, with $i\circ l = r$ and $i\circ i = \text{id}$ and $m(i(g),g) = e(r(g))$ and $m(g,i(g)) = e(l(g))$.
Then it makes sense to talk about smooth functors of Lie groupoids, smooth natural transformations of functors, etc. In particular, we can talk about whether two Lie groupoids are "equivalent", and I believe that a warm-up notion for "smooth stack" is "Lie groupoid up to equivalence". Actually, I believe that the experts prefer some generalizations of this — (certain) bibundles rather than functors, for example. But I digress.
Other than that we know what equivalences of Lie groupoids are, I'd like to point out that we can work also in small neighborhoods. Indeed, if $U\_0$ is an open neighborhood in $G\_0$, then I think I can let $U = l^{-1}(U\_0) \cap r^{-1}(U\_0)$, and then $U \rightrightarrows U\_0$ is another Lie groupoid.
Oh, let me also recall the notion of **tangent Lie algebroid $A \to G\\_0$ to a Lie groupoid**. The definition I'll write down doesn't look very symmetric in $l\leftrightarrow r$, but the final object is. The fibers of the vector bundle $A \to G\\_0$ are $A\\_y = {\rm T}\\_{e(y)}(r^{-1}(y))$, the tangent space along $e(G\\_0)$ to the $r$-fibers, and $l: r^{-1}(y) \to G\\_0$ determines a God-given **anchor** map $\alpha = dl: A \to {\rm T}G\\_0$, and because $e$ is a section of both $l,r$, this map intertwines the projections, and so is a vector bundle map. In fact, the composition $m$ determines a Lie bracket on sections of $A$, and $\alpha$ is a Lie algebra homomorphism to vector fields on $G\_0$.
Suppose that I have a smooth function $f: G\_0 \to \mathbb R$ that is constant on $G$-orbits of $G\_0$, i.e. $f(l(g)) = f(r(g))$ for all $g\in G$. I'd like to think of $f$ as a Morse function on "the stack $G\_0 // G$". So, suppose $[y] \subseteq G\_0$ is a critical orbit, by which I mean: it is an orbit of the $G$ action on $G\_0$, and each $y \in [y]$ is a critical point of $f$. (Since $f$ is $G$-invariant, critical points necessarily come in orbits.) If $y$ is a critical point of $f$, then it makes sense to talk about the **Hessian**, which is a symmetric pairing $({\rm T}\\_yG\\_0)^{\otimes 2} \to \mathbb R$, but I'll think of it as a map $f^{(2)}\\_y : {\rm T}\\_yG\\_0 \to ({\rm T}\\_yG\\_0)^\*$. In general, this map will not be injective, but rather the kernel will include $\alpha\\_y(A\\_y) \subseteq {\rm T}\\_yG\\_0$. Let's say that the critical orbit $[y]$ is **nondegenerate** if $\ker f^{(2)}\_y = \alpha\\_y(A\\_y)$, i.e. if the Hessian is nondegenerate as a pairing on ${\rm T}\\_yG\\_0 / \alpha\\_y(A\\_y)$. I'm pretty sure that this is a condition of the orbit, not of the individual point.
Nondegeneracy rules out some singular behavior of $[y]$, like the irrational line in the torus.
Anyway, my question is as follows:
>
> Suppose I have a Lie groupoid $G \rightrightarrows G\_0$ and a $G$-invariant smooth function $f: G\_0 \to \mathbb R$ and a nondegenerate critical orbit $[y]$ of $f$. Can I find a $G$-invariant neighborhood $U\_0 \supseteq [y]$ so that the corresponding Lie groupoid $U \rightrightarrows U\_0$ is equivalent to a groupoid $V \rightrightarrows V\_0$ in which $[y]$ corresponds to a single point $\bar y \in V\_0$? I.e. push/pull the function $f$ over to $V\_0$ along the equivalence; then can I make $[y]$ into an honestly-nondegenerate critical *point* $\bar y \in V\_0$?
>
>
>
I'm assuming, in the second phrasing of the question, that $f$ push/pulls along the equivalence to a $V$-invariant function $\bar f$ on $V\_0$. I'm also assuming, so if I'm wrong I hope I'm set right, that ${\rm T}\\_{\bar y}V\\_0 \cong {\rm T}\\_yG\\_0 / \alpha\\_y(A\\_y)$ canonically, so that e.g. $\bar f^{(2)}\_{\bar y} = f^{(2)}\_y$.
| https://mathoverflow.net/users/78 | How difficult is Morse theory on stacks? | Yes, it's possible to find such a neighbourhood $U\_0$ of $[y]$.
Here's how you do it.
Pick a submanifold $M\subset G\_0$, $y\in M$, transverse to $[y]$.
By your Morse-ness assumption, the restriction $f|\_M$ is Morse,
with critical point $y$. Pick a neighborhood $V\_0\subset M$ of $y$, such that
$y$ is the *only* critical point of $f$ in $V\_0$.
Let $U\_0$ be the orbit of $V\_0$ under $G$. Clearly, $U\_0$ is a neighborhood of $[y]$.
Now consider the restriction of $G$ to $U\_0$. This is defined to be the groupoid with object space $U\_0$, and morphism space $U\_0\times\_{G\_0}G\times\_{G\_0}U\_0$. That's your groupoid $U\rightrightarrows U\_0$. Similarly, you can consider the restriction $V\rightrightarrows V\_0$
of $G$ to $V\_0$.
The inclusion $(V\rightrightarrows V\_0) \hookrightarrow (U\rightrightarrows U\_0)$ is a Morita equivalence because it's essentially surjective and fully faithfull.
---
Note: *The notions "essentially surjective" and "fully faithfull" for Lie groupoids are somewhat stronger than what you might initially guess. The first one also requires the existence of locally defined smooth maps $U\_0\to V\_0$, while "fully faithfull" means that $V$ is the pullback of $V\_0\times V\_0 \to G\_0\times G\_0\leftarrow G$.*
| 10 | https://mathoverflow.net/users/5690 | 23945 | 15,747 |
https://mathoverflow.net/questions/23940 | 10 | This is a question of the motivation for a common assumption found in the literature.
The free topological group $F(X)$ on a space $X$ exists for all spaces $X$ (It seems this was first shown by Katutani and Samuel). I mean "free topological group" in the sense that $F:Top\rightarrow TG$ is left adjoint to the forgetful functor $U:TG\rightarrow Top$ from the category of topological groups to the category of topological spaces.
$F(X)$ is well studied when $X$ is a Tychonoff space. This permits the application of pseudometrics which seems to be a powerful tool for describing the complicated topological structure of $F(X)$. Also, it seems to be a useful fact that the canonical map $\sigma:X\rightarrow F(X)$ is an embedding when $X$ is Tychonoff.
These two conveniences do seem to make it convenient to study $F(X)$ when $X$ is Tychonoff but it seems almost no one is interested in $F(X)$ when $X$ is not Tychonoff. Why is this? Are these uninteresting for some reason?
| https://mathoverflow.net/users/5801 | Why free topological groups on Tychonoff spaces? | Let $X$ be a topological space. If $F(X)$ is $T\_0$ then I think $F(X)$ is isomorphic (as topological groups) to $F(Y)$, where $Y$ is the Tychonofficiation (see below) of $X$. So it is enough to study topological free groups on a Tychonoff space.
### Explanation:
First let me remind myself about some notation. *Completely regular* means that any point can be separated from a closed set not containing it by a continuous real-valued function. *Tychonoff* then means completely regular and $T\_2$(=Hausdorff). Any topological group is completely regular; and for topological groups $T\_0$ is equivalent to $T\_2$.
Suppose $X$ is an arbitrary topological space, and let $Y$ be its "Tychonoffication" (!). That is, set theoretically $Y$ is the quotient of $X$ by the equivalence relation $x\sim x'$ if and only $f(x)=f(x')$ for all continuous $f:X\to\mathbb{R}$; each such $f$ descends to $Y$ and we give $Y$ the weak topology induced by all these real-valued maps. This makes $Y$ into a Tychonoff space which I think satisfies the following universal property: any continuous map from $X$ to a Tychonoff space factors uniquely through $Y$. Also, the natural map $X\to Y$ induces an isomorphism $C(Y)\stackrel{\cong}{\to} C(X)$, where $C(-)$ denotes the ring of real-valued continuous functions.
Assuming $F(X)$ is $T\_0$, then the natural map $F(X)\to F(Y)$ seems to be an isomorphism of topological groups. It is enough to construct an inverse. Since $F(X)$ is $T\_0$, it is even $T\_2$, and therefore it is Tychonoff. So the natural map $X\to F(X)$ factors through $Y$ and induces $F(Y)\to F(X)$, which surely does the trick?
### What about that $T\_0$ assumption?
Lots of people are only interested in Hausdorff topological groups, so it seems reasonable to only study spaces $X$ for which $F(X)$ is $T\_0$ (hence $T\_2$). Otherwise you could replace $Y$ by the "complete-regularization" of $X$ (i.e. $X$ equipped with the weak topology induced by $C(X)$) and repeat the argument, but it doesn't work so nicely.
### Edit:
While I was typing my answer, you asked about this Tychonoffication business!
| 9 | https://mathoverflow.net/users/5830 | 23951 | 15,750 |
https://mathoverflow.net/questions/23950 | 11 | Let $f:X\longrightarrow Y$ be a finite morphism of smooth projective varieties over a field $k$ of characteristic zero, where $\dim X=\dim Y$. Then $f$ is flat. Hence $f\_\ast \mathcal{O}\_X$ is a coherent locally free sheaf on $Y$.
Now, my question is based on the following example. (For simplicity, take $k=\overline{k}$.)
**Example**. Suppose that $\dim X =\dim Y = 1$ (i.e., curves). Apply Grothendieck-Riemann-Roch to $f$ and $\mathcal{O}\_X$. In degree 0, we get the fact that $\textrm{deg} \ f = \textrm{rk} \ f$. In degree 1 we get a "Hurwitz theorem". In fact, with little effort the formula reads $$2c\_1(f\_\ast \mathcal{O}\_X) =\deg f \cdot K\_Y - f\_\ast(K\_X) = f\_\ast(-R), $$ where $R$ is the ramification divisor on $X$.
Now for my two questions that are based on this formula.
**Q1**. The divisor $R$ is not called the ramification divisor for nothing. Its support is the set of ramification points and the multiplicity of $R$ at a point $P$ is precisely $e\_P-1$. So in my opinion, it "measures" the ramification. What about $c\_1(f\_\ast \mathcal{O}\_X) = c\_1(\det f\_\ast \mathcal{O}\_X)$? How does he "measure" the ramification? (I'm probably missing something really elementary here.)
**Q2**. In higher-dimensions, if I understand correctly, one should get a "higher-dimensional" Hurwitz formula: $$2c\_1(f\_\ast \mathcal{O}\_X) =f\_\ast(\textrm{td}(X/Y)).$$ I doubt that this "measures" all the ramification. And, to be frank, I don't really know what it "measures". Can anyone provide some insight?
| https://mathoverflow.net/users/4333 | How does $f_* O_X$ measure ramification and Grothendieck-Riemann-Roch | The following does not exactly answer your question, but you may find it interesting. It is the Riemann-Hurwitz formula for surfaces.
Let $\phi:S\_1\to S\_2$ be a finite morphism between smooth, projective surfaces (over an algebraically closed field of characteristic zero) of degree $n$, and let $B\subseteq S\_2$ be the set of $y\in S\_2$ such that $\phi^{-1}(y)$ does not contain $n$ points (i.e. $B$ is the ramification locus). Zariski's purity theorem states that $B$ is pure of dimension one; let $B\_1,\dots,B\_r$ be its irreducible components, and let $n\_i$ be the degree of the morphism $\phi|\_{\phi^{-1}(B\_i)}:\phi^{-1}(B\_i)\to B\_i$. Then
$$\chi(S\_1)=\chi(S\_2)\deg \phi-\sum\_{i=1}^r(n-n\_i)\chi(B\_i)+\sum\_{y\in B}\left(|\phi^{-1}(y)|-n+\sum\_{i=1}^r(n-n\_i)m\_i(y)\right)$$
where $m\_i(y)$ denotes the number of local branches of $B\_i$ at $y$. Here $\chi$ is the $\ell$-adic Euler characteristic of the surface ( topological Euler characteristic if $k=\mathbb{C}$), which can be translated into a Chern class if you prefer.
The proof is B. Iversen, 'Numerical invariants and multiple planes', Amer. J. Math. 92 (1970), 968-996. When $k=\mathbb{C}$, you can prove it by thinking of the topological Euler characteristic as a measure on constructible sets (e.g. O. Ya. Viro, Some integral calculus based on Euler characteristic); then the formula is equivalent to Fubini's theorem ($\int\int dxdy=\int\int dydx$) for the graph of $\phi$.
| 9 | https://mathoverflow.net/users/5830 | 23955 | 15,751 |
https://mathoverflow.net/questions/23811 | 8 | Hi, I was looking to traverse a planar graph and report all the faces in the graph (vertices in either clockwise or counterclockwise order). I have build a random planar graph generator that creates a connected graph with iterative edge addition and needed a solution to report all the faces that were created in the final graph. I was contemplating several strategies such as doing a sweep line algorithm and tracking the areas between lines or tracking the faces as I generate the graph however I have not been able to find much material regarding this matter and was wondering where I could find some assistance/ideas how to do this.
So far I found a Boost algorithm here <http://www.boost.org/doc/libs/1_36_0/boost/graph/planar_face_traversal.hpp> however I am having a lot of trouble decrypting the boost library to determine how it is done.
Any thoughts, ideas or existing algorithms would be welcome.
| https://mathoverflow.net/users/5924 | Reporting all faces in a planar graph | I'll assume the graph is connected, and that you have the clockwise or counterclockwise ordering of the edges around each vertex. Then it's easy, given a directed edge e, to walk around the face whose counterclockwise boundary contains e. So make a list of all directed edges (i. e., two copies of each undirected edge). Pick one directed edge, walk counterclockwise around its face, and cross off all the directed edges you traverse. That's one face. Pick a directed edge you haven't crossed off yet and walk around its face the same way. Keep doing that until you've crossed off all of the edges. (Note that the "counterclockwise" boundary of the exterior unbounded face actually goes clockwise around the outside of the graph.)
If the graph isn't assumed to be connected, then things could be more complicated, since the boundary of a face could have multiple connected components. In that case, you might as well use a standard general-purpose algorithm for computing planar subdivisions. I don't think you lose anything (in asymptotic complexity, anyway) by doing this.
I don't have it in front of me, so I can't tell you with certainty, but I *think* this is covered in "The Dutch Book" (*Computational Geometry: Algorithms and Applications*, by de Berg et al.). In any case, that's one of the main references for these types of computations.
| 9 | https://mathoverflow.net/users/302 | 23958 | 15,752 |
https://mathoverflow.net/questions/23956 | 9 | In Quelques proprietes globales des varietes differentiables, Thom classifies unoriented manifolds up to cobordism. I've been struggling a bit to understand this paper, and while Stong's cobordism notes have helped a bit, I was wondering if an English translation (of the entire paper or just parts) exists. Thank you in advance.
| https://mathoverflow.net/users/5969 | Thom's seminal cobordism paper in English? | An English translation of this paper is included in the first volume of the "Topological Library", edited by Novikov and Taimanov. See the following website : <http://www.worldscibooks.com/mathematics/6379.html>
By the way, Thom's paper is rather hard to read. There are alternative expositions (often with somewhat easier proofs) of various pieces of it in various places. What portion in particular is giving you trouble?
| 13 | https://mathoverflow.net/users/317 | 23959 | 15,753 |
https://mathoverflow.net/questions/9961 | 53 | This is related to [another question](https://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes).
I've found many remarks that the category of schemes is not cocomplete. The category of locally ringed spaces is cocomplete, and in some special cases this turns out to be the colimit of schemes, but in other cases not (which is, of course, no evidence that the colimit does not exist). However, I want to understand in detail a counterexample where the colimit does not exist, but I hardly found one. In FGA explained I've found the reference, that Example 3.4.1 in Hartshorne, Appendix B is a smooth proper scheme over $\mathbb{C}$ with a free $\mathbb{Z}/2$-action, but the quotient does not exist (without proof). To be honest, this is too complicated to me. Are there easy examples? You won't help me just giving the example, because there are lots of them, but the hard part is to prove that the colimit really does not exist.
| https://mathoverflow.net/users/2841 | Colimits of schemes | **Edit:** [BCnrd](https://mathoverflow.net/users/3927/bcnrd) gave a proof in the comments that this example works, so I've edited in that proof.
A ~~possible~~ proven example
-----------------------------
~~I suspect~~ There is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.
~~I don't have a proof, but I thought other people might have ideas if I posted this here.~~
If a coequalizer $P$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $P$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.
On the other hand, let $U$ be an affine open around the image of the generic point in $P$. $U$ has dense open preimages $V$ and $V'$ in both affine lines. Let $W=V\cap V'$ inside the affine line, so we have two maps from $W$ to the affine $U$ which coincide at the generic point of $W$, and hence are equal (as $U$ is affine). In particular, the two maps from affine line to categorical pushout $P$ coincide at each "common pair" of closed points of the two copies of $W$, contradicting the previous paragraph.
---
**Edit:** The questions below are no longer relevant, but I'd like to leave them there for some reason.
Here are some questions that might be helpful to answer:
>
> If the coequalizer above *does* exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?
>
>
>
(see the related question [Can a coequalizer of schemes fail to be surjective?](https://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective))
>
> Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of **separated** schemes equal to $\mathbb A^1$? (probably)
>
>
>
>
> What are some ways to determine that a functor $Sch\to Set$ *is not* corepresented by a scheme?
>
>
>
| 37 | https://mathoverflow.net/users/1 | 23966 | 15,759 |
https://mathoverflow.net/questions/23957 | 2 | Let $m,n$ be positive integers, and $\displaystyle \Phi\_{m,n}~:~ {\mathbb{R}\_+^\*}^m \to \mathbb{R}\_+^\*, \ \ \ (x\_1,x\_2, \ldots , x\_m) \mapsto \sum\_{k=1}^m \sqrt[n]{x\_k}$.
Clearly for $m=1$ if for all positive integer $n$, we have $\Phi\_{1,n}(x) \in \mathbb Q$, then $x=1$.
It seems that the same conclusion holds for $m>1$ (or at least the subset of ${\mathbb{R}\_+^\*}^m$ for which $\Phi\_{m,n}(x) \in \mathbb Q$ is finite).
Is it true (or even obvious and I missed it)?
| https://mathoverflow.net/users/3958 | Sum of n-th roots is rarely rational | The following conclusion is true: If $\Phi\_{m,n}(x)\in\mathbb{Q}$ for all positive integers n, then x1=x2=...=xn=1.
It follows in what I believe is a fairly routine, or at least not too difficult, manner from the following Claim:
Let K be the extension field of ℚ generated by all n-th roots of all xi. Then K is a finite extension of ℚ.
Proof. Let yi be the N!-th root of xi. Then the power sum symmetric functions of the yi are all rational, hence the elementary symmetric functions are all rational, so the y's lie in a field extension of ℚ of degree at most m. Take N as large as you like. Voila!
| 3 | https://mathoverflow.net/users/425 | 23967 | 15,760 |
https://mathoverflow.net/questions/23952 | 5 | Suppose $X$ is an orientable surface with non-empty boundary and $f:X\to X$ is a pseudo-Anosov automorphism that acts identically on $H\_1(X,\mathbf{Z})$. Let $x$ be a fixed point of $f$.
For any $\gamma\in\pi\_1(X,x)$ we have $\gamma^{-1}f(\gamma)\in [\pi\_1(X,x),\pi\_1(X,x)]$, the commutant of $\pi\_1(X,x)$. More generally, we have $\gamma\cdot g^{-1}f g(\gamma)\in [\pi\_1(X,x),\pi\_1(X,x)]$ where $g$ is an automorphism of $X$ that fixes $x$.
I would like to ask what one can say about the normal closure in $\pi\_1(X,x)$ of the set of all elements $\gamma\cdot g^{-1}f g(\gamma)$ where $\gamma$ runs through $\pi\_1(X,x)$ and $g$ runs through the set of all diffeomorphisms $X\to X$ that fix $x$. In particular, does this closure coincide with the commutant of $\pi\_1(X,x)$?
| https://mathoverflow.net/users/2349 | Automorphisms of $\pi_1$ induced by pseudo-Anosov maps | No. Let $\Gamma\_i$ be the lower central series defined by $\Gamma\_1=\pi\_1(X,x)$, $\Gamma\_{i+1}=[\Gamma\_1,\Gamma\_i]$. The *Johnson filtration* $\text{Mod}\_g(k)$ is the descending filtration of the mapping class group relative to $x$ defined by:
$f\in \text{Mod}\_g(k)\iff f$ acts trivially on $\Gamma\_1/\Gamma\_k$
The first term $\text{Mod}\_g(2)$ is the *Torelli group*, consisting of diffeomorphisms acting trivially on homology. The next term $\text{Mod}\_g(3)$ is the *Johnson kernel*. By a beautiful theorem of Johnson, this is the subgroup generated by Dehn twists around separating curves.
By residual nilpotence of surface groups, we have $\bigcap \text{Mod}\_g(k)=\{1\}$, but every individual term in the filtration is nontrivial. It is not hard to see that every term of the Johnson filtration contains pseudo-Anosovs. Indeed every normal subgroup of the mapping class group contains pseudo-Anosovs (see Lemma 2.5 of Long, "A note on the normal subgroups of mapping class groups") from which Long concluded that any two normal subgroups intersect nontrivially!
Thus since $\text{Mod}\_g(k)$ is normal, if we take $f\in\text{Mod}\_g(k)$ we have $\gamma^{-1}\cdot g^{-1}fg(\gamma)\in \Gamma\_k$ for all $g$ and all $\gamma$.
| 8 | https://mathoverflow.net/users/250 | 23968 | 15,761 |
https://mathoverflow.net/questions/23960 | 7 | This question is inspired by [How kinky can a Jordan curve get?](https://mathoverflow.net/questions/23954/how-kinky-can-a-jordan-curve-get)
What is the least upper bound for the Hausdorff dimension of the graph of a real-valued, continuous function on an interval? Is the least upper bound attained by some function?
It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. However, this does not rule out the possible existence of a function graph of dimension two.
| https://mathoverflow.net/users/802 | How big can the Hausdorff dimension of a function graph get? | The answer is 2.
Besicovitch and Ursell, Sets of fractional dimensions (V): On dimensional
numbers of some continuous curves. J. London Math. Soc. 12 (1937) 18–25. [doi:10.1112/jlms/s1-12.45.18](http://dx.doi.org/10.1112/jlms/s1-12.45.18)
| 14 | https://mathoverflow.net/users/454 | 23972 | 15,763 |
https://mathoverflow.net/questions/23977 | 6 | What is a good book/article explaining the mathematics behind perspective painting? I have already looked at the Wikipedia article on the topic, so I am looking for something more advanced than this.
I am a research mathematician of limited artistic ability and knowledge.
| https://mathoverflow.net/users/5337 | Reference Request: Perspective Painting | The geometry of an art by Kirsti Andersen ([amazon](http://rads.stackoverflow.com/amzn/click/0387259619))
Mathematics for the non-mathematician by Morris Kline (See Chapter 10- math and painting in the renaissance)
Mathematics and its history by John Stillwell (See chapter 8 on Projective Geometry)
| 4 | https://mathoverflow.net/users/5372 | 23979 | 15,766 |
https://mathoverflow.net/questions/23989 | 33 | It is known that there is a gap between 2 and the next largest norm of a graph.
Is there an interval of the real line in which norms of graphs are dense?
| https://mathoverflow.net/users/5973 | Are the norms of graphs dense in any interval? | I found a reference that seems to answer your question:
Shearer, James B.
[On the distribution of the maximum eigenvalue of graphs](https://mathscinet.ams.org/mathscinet-getitem?mr=986863), 1989.
The theorem in this paper is that the set of largest eigenvalues of adjacency matrices of graphs is dense in the interval $\left[\sqrt{2+\sqrt{5}},\infty\right)$. [Here's an online version](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0R-45W3BTD-1V&_user=440026&_coverDate=04%252F30%252F1989&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000020939&_version=1&_urlVersion=0&_userid=440026&md5=99474a7c10cf69e8e86314cf27d6507a).
---
Here's a related paper:
Hoffman, Alan J.
[On limit points of spectral radii of non-negative symmetric integral matrices](https://mathscinet.ams.org/mathscinet-getitem?mr=347860), 1972.
In this paper limit points less than $\sqrt{2+\sqrt{5}}$ are described. In particular, they form an increasing sequence starting at 2 and converging to $\sqrt{2+\sqrt{5}}$. [Here's an online version](https://doi.org/10.1007/BFb0067367). The author also posed the problem that led to Shearer's paper.
| 38 | https://mathoverflow.net/users/1119 | 23993 | 15,771 |
https://mathoverflow.net/questions/23992 | 6 | In Milne's notes on Class Field Theory (<http://www.jmilne.org/math/CourseNotes/CFT.pdf>), he initially defines group cohomology in terms of injective resolutions, then he talks about computing cohomology using cochains. I don't see him mention anywhere that the group has to be finite in order for cochains to work, but this seems to be the case?
Later, he discusses profinite groups, in which he says that cohomology of profinite groups can be computed using continuous cochains. What isn't clear is the following: is the cohomology using continuous cochains a modified cohomology theory, different from the one using injective resolutions? In this case, then, do we get the cohomology theory using injective resolutions if we use all cochains, not just continuous ones? Or do the continuous cochains give the same cohomology as injective resolutions, and cochains which are not necessarily continuous only give cohomology in the case of finite groups? I.e., is there only one such cohomology theory? At the very least, using general cochains versus continuous cochains in the case of infinite profinite groups is different, for in one case $H^1$ is $\mathrm{Hom}(G,M)$ when M is trivial, and in the other case $H^1$ is $\mathrm{Hom}\_{\mathrm{cts}}(G,M)$.
Assuming that set-theoretic cochains only work for finite groups, why is it the case? It seems that the proof that cochains compute cohomology (i.e. by looking at a projective resolution of $\mathbb{Z}$) fails because the modules used in the case of finite groups, i.e. tensor powers of $\mathbb{Z}[G]$, aren't necessarily projective when $G$ is infinite (in the case when $G$ is finite, they are free). Is this correct?
| https://mathoverflow.net/users/1355 | Question about computing group cohomology using cochains | You should take a look at the beginning of chapter 2 of Serre's Galois Cohomology.
He explains there that if G is a profinite group, then the category of discrete abelian groups with a continuous action of G has enough injectives (but not enough projectives in general), and that cohomology can be "computed" as a direct limit of cohomology of a finite group.
To sum up:
-if G is discrete (i.e. no topology), then the category has enough injectives and projectives (it is the category of left $\mathbb{Z}[G]$-modules), and using a projective resolution for $\mathbb{Z}$ gives you the equivalence between the derived functor definition and the cochains definition (using the fact that Ext can be computed two ways). You can find this in Serre's Local Fields.
-if G is profinite, and we consider the category of discrete modules with a continuous action of G, then there are enough injectives (this can be seen quite easily from the discrete case), but not enough projectives. Luckily though, the two definitions agree (thanks to the "direct limit computability"). You can find this in Serre's Galois Cohomology.
-if G is an arbitrary topological group, there is not much left. There aren't enough injectives nor projectives in general, and if you define cohomology with cochains, you don't get an homological functor (only the beginning of the long exact sequence exists). However, see the end of J.-M. Fontaine and Yi Ouyang's book (it's a pdf, I found it on Fontaine's web page) about p-adic representations, they mention that if you have a continuous set-theoretic section in your short exact sequence, you get a long exact sequence. I haven't read the reference they provide, though.
| 9 | https://mathoverflow.net/users/5735 | 24007 | 15,781 |
https://mathoverflow.net/questions/24028 | 4 | Hello,
I've been used to writing logical transformations using equality, but the other day it struck me that perhaps I should be using the biconditional $\iff$?
So my question is:
What is the difference between the biconditional iff. $\iff$ and equality = ? Can they be used interchangeably? And when should one be used instead of the other? Is this matter of style? meaning? or correctness?
Simple Examples to illustrate the point: here p,q are propositions, ^ is "and", v is "or", and ~ is "not":
Using equality:
(1) p ^ (q v r) = (p ^ q) v (p ^ r)
and
(2) p $\Rightarrow$ q = ~ p v q
But is this more correctly written using the biconditional iff.?
(1') p ^ (q v r) $\iff$ (p ^ q) v (p ^ r)
and
(2') p $\Rightarrow$ q $\iff$ ~ p v q
In rendering into English, are these logically equivalent statements? Equal statements (equality being considered as an equivalence relation that then establishes an equivalence class)? Or is it that the very meaning of equality in the context of logical propositions is iff.?
(Can't find the Community Wiki tag! someone please add it, thanks.)
| https://mathoverflow.net/users/39723 | What is the difference between the biconditional iff. and equality = ? | Usually the biconditional is denoted by $\leftrightarrow$ and logical equivalence is represented by $\Leftrightarrow$.
Given two compound propositions $P$ and $Q$, the proposition $P \Leftrightarrow Q$ means that $P$ and $Q$ have the same truth value for each possible combination of truth values of the variables of which they are composed. That is, $P \Leftrightarrow Q$ means that $P \leftrightarrow Q$ is a tautology (i.e., a proposition that is always true). It's important to note that $\leftrightarrow$ is a connective, whereas $\Leftrightarrow$ is like an "equals sign" for propositions. More formally, the expression $P \Leftrightarrow Q$ is really a proposition about a proposition, namely that "$P \leftrightarrow Q$ is a tautology", whereas the expression $P \leftrightarrow Q$ is just a compound proposition that may or may not be a tautology.
Likewise, the conditional is usually denoted by $\rightarrow$ and logical implication is represented by $\Rightarrow$. Given two compound proposition $P$ and $Q$, the proposition $P \Rightarrow Q$ means $Q$ is true whenever $P$ is true, i.e., $P \Rightarrow Q$ means that $P \rightarrow Q$ is a tautology.
It seems that you may have confused logical equivalence ($\Leftrightarrow$) with the biconditional connective ($\leftrightarrow$) and logical implication ($\Rightarrow$) with the conditional connective ($\rightarrow$).
| 18 | https://mathoverflow.net/users/3029 | 24030 | 15,796 |
https://mathoverflow.net/questions/23825 | 4 | In the beginning Shelah classifies all $\aleph\_1$-free Abelian groups into 3 possibilities each of which is satisfied by some $\aleph\_1$-free Abelian group and the classification depends on the group G up to isomorphism (in page 250 is the Israel Journal of Mathematics for those who have the article around). He also defines pure subgroups.
What I want to ask is : why are pure subgroups so important for the proof? It seems like decomposition of pure subgroups of groups is central in the V=L chunk of the proof and groups of possibility 1 and 2 are defined in terms of how well can they split as a direct sum of pure subgroups. Is it just because they are of prime importance for Abelian groups in general.
According to V=L, possibility 1 and 2 contradict Whiteheadness but with MA possibility 2 does not contradict Whiteheadness.
I hope my question is precise.
Thx
| https://mathoverflow.net/users/3859 | Shelah's proof of the independence of the Whitehead Problem | A good source for understanding Shelah’s proof of the undecidability of Whitehead’s problem is Paul Eklof’s “Whitehead’s Problem is undecidable” article in the American Mathematical Monthly. The key property about pure subgroups relevant to Shelah’s proof is that a countable torsion-free group is free iff every finitely generated subgroup is contained in a finitely generated pure subgroup. This gives a characterization of countable free groups is terms of objects that are “almost finite”. This also gives a decomposition of a countable free group into an increasing chain of finitely generated pure subgroups. Recall, that a group is torsion-free iff every finitely generated subgroup is free. This is generalized by the notion of $\aleph\_1$-free which states that every countable subgroup is free. Pure subgroups are generalized to $\aleph\_1$-pure subgroups where the quotient is correspondingly $\aleph\_1$-free. The characterization of free groups of cardinality $\aleph\_1$ with the generalized notion is not so simple. For a group of size $\aleph\_1$ to be free it is not sufficient that every countable subgroup be contained in a countable $\aleph\_1$-pure subgroup. However any group having the above property can be decomposed into an $\aleph\_1$ chain of free subgroups with the subgroups indexed by the successor ordinals being $\aleph\_1$-pure. If furthermore “enough” of the limit stages are also $\aleph\_1$-pure, more precisely if the set of ordinals indexing $\aleph\_1$-pure subgroups is stationary, then the group is free. Again this gives a characterization of free groups of size $\aleph\_1$ in terms of smaller, namely countable subgroups. The $V=L$ case uses the fact that the $\Diamond$-principle holds in $L$. The $\Diamond$-principle allows you to anticipate properties of an object of size $\aleph\_1$ as it is constructed from countable objects indexed by countable ordinals. This is where the characterization of a free group of size $\aleph\_1$ in terms of its countable subgroups is crucial. I believe the fact that pure subgroups are “almost finite” allows Shelah to prove that a certain poset has the countable chain condition in the Martin’s axiom part of the proof.
| 7 | https://mathoverflow.net/users/5984 | 24035 | 15,798 |
https://mathoverflow.net/questions/24034 | 39 | More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?
Some days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.
| https://mathoverflow.net/users/1508 | Can Cantor set be the zero set of a continuous function? | Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from $I\_0 = [0,1]$ by repeatedly removing the middle third of any ensuing interval. So let's denote by $I\_n$ the
$n$-th set in this process.
Now let's make a smooth function $f\_n$ on $[0,1]$ such that its zero set is exactly $I\_n$.
Starting with $f\_0 = 0$ we obtain $f\_{n+1}$ from $f\_n$ as follows:
Set $f\_{n+1} = f\_n$ on $I\_{n+1}$ and
on an interval that is removed from $I\_n$ make $f\_{n+1}$ equal to a bump function
that is 0 only at the boundary of the interval. We can choose the bump function to be of height $2^{-2^n}$.
This choice of heights of the bump functions will ensure that
the derivatives of $f$ all converge uniformly to their pointwise limits.
Hence the limit function $f\_n$ is again smooth. By construction its zero set is exactly
the Cantor set.
| 34 | https://mathoverflow.net/users/692 | 24037 | 15,800 |
https://mathoverflow.net/questions/24043 | 2 | If I have a square and want to place four equally large circles within this square, how large can the maximum radius be (compared to the lenght of the side of the square)?
Just an answer would be ok, but answer and explanation would be better.
| https://mathoverflow.net/users/5986 | fit 4 circles within a square | See the links below. The solutions have been show to be optimal up to 20 circles into a square.
<http://mathworld.wolfram.com/CirclePacking.html>
A fun site about all kinds of packing results is:
<https://erich-friedman.github.io/packing/cirinsqu/>
| 6 | https://mathoverflow.net/users/692 | 24045 | 15,805 |
https://mathoverflow.net/questions/24039 | 1 | we know $u$ (if it is a solution to the wave equation in $\mathbb{R}^3$) decays as $1/t$ as $t$ goes to $\infty$. this comes easily from spherical means. but how do we know this is the maximum possible rate of decay?
| https://mathoverflow.net/users/5985 | maximum decay rate | Let $\bar u(t,r)$ be the spherical mean over the sphere of radius $r$, centered at (say) the origin. Then $r\bar u(t,r)$ satisfies the one-dimensional wave equation. If $u$ has compact support initially, then eventually $r\bar u(t,r)$ is an outward traveling wave. This wave has constant shape, and so $\bar u(t,r)$ decays as $1/r$.
| 0 | https://mathoverflow.net/users/802 | 24048 | 15,807 |
https://mathoverflow.net/questions/23908 | 2 | Hello, this is my first post here. I am no mathematician and English is not my first language, so please excuse me if my question is too stupid, it is poorly phrased, or both.
I am developing a program that creates timetables. My timetable-creating algorithm, besides creating the timetable, also creates a graph whose nodes represent each class I have already programmed, and whose arcs represent which pairs of classes should not be programmed at the same time, even if they have to be reprogrammed. The more "heavily linked" a node is, the more inflexible its associated class is with respect to being reprogrammed.
Sometimes, in the middle of the process, there will be no option but to reprogram a class that has already been programmed. I want my program to be able to choose a class that, if reprogrammed, affects the least possible number of other already-programmed classes. That would mean choosing a node in the graph that is "not very heavily linked", subject to some constraints with respect to which nodes can be chosen.
Do you know any algorithms that solve this problem?
| https://mathoverflow.net/users/1871 | Measuring how "heavily linked" a node is in a graph | I suggest you calculate the shortest path between every two nodes. Several algorithm exist for that:
<http://en.wikipedia.org/wiki/Shortest_path>
Then you can give each node a value, based on the shortest path to the other nodes. If the shortest paths are longer, it is less heavily connected.
Of course, you can do other solutions as suggested, but I think an important factor in your calculation is whether you may return to your own node or note. If you don't want take into account going back to your own node, you should base it on shortest paths. If you do want to return, then matrix multiplications is the solution, I think.
| 1 | https://mathoverflow.net/users/5917 | 24051 | 15,808 |
https://mathoverflow.net/questions/24056 | 1 | a friend and i are working on a research problem and we just hit a wall:
if $u$ is a solution of the 3-dimensional helmholtz operator $\Delta + k^2$ in the ball $B\_r(x)$ of radius $r > 0$, then does $u$ satisfy anything like the mean value for harmonic functions? basically, we are wondering what an analogous "mean value property" would be for a nonzero $k$ term. we can't find anything obvious like a harmonic function, where $k = 0$.
| https://mathoverflow.net/users/5985 | helmholtz zero in R^3 | If $u$ is a solution to the equation $\triangle u +k^2 u=0$ in a 3D domain $\Omega$, then
for any $x\in\Omega$ and any $r>0$ such that $\{y\in\mathbb R^3:\ |x-y|\leq r \}\subset\Omega$, we have
$$u(x)=\frac {p(r)}{4\pi r^2}\int\_{|x-y|=r} u(y)dS\_y,\qquad\qquad\qquad(1)$$
where
$$p(r)=\frac{rk}{\sin rk}.$$
Formula (1) is an analogue of the mean value theorem for harmonic functions (in the case of spherical means).
**Edit added:** relation (1) is valid for all $r\_1\leq r$. If we multiply it by $4\pi r^2/p(r)$ and integrate between $0$ and $r$ we will obtain that
$$u(x)=\frac{k^3}{4\pi(\sin rk-rk\cos rk)}\int\_{|x-y|\leq r} u(y)dy.$$
The latter formula generalizes the property that the value of a harmonic function at $x\in\Omega$ is equal to function's average value over a ball with the center at $x$.
A short derivation of formula (1) can be found in chapter IV of *Methods of Mathematical Physics* (Vol. 2)
by Courant and Hilbert (or see Harald's comment below).
| 2 | https://mathoverflow.net/users/5371 | 24057 | 15,810 |
https://mathoverflow.net/questions/24059 | 7 | Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an *[absolutely continuous](http://en.wikipedia.org/wiki/Absolute_continuity)* map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?
Remarks:
1. The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the [Cantor function](http://en.wikipedia.org/wiki/Cantor_function) .
2. If $f$ is assumed to be $C^1$, then the answer is yes - a nice proof can be found at [this page.](http://sci.tech-archive.net/Archive/sci.math.research/2005-02/0171.html) Essentially the same proof works if it is assumed that $f$ is not differentiable at at most countably many points.
**Edit:** I would like to retract the previous sentence. Now I don't see why it should be true.
| https://mathoverflow.net/users/1508 | Is the absolutely continuous image of a nowhere dense set is also nowhere dense? | I think this is a counter-example.
Let $C$ be a cantor set of positive measure, so $C$ is nowhere dense, perfect and is the countable decreasing intersection of sets $C\_{n}$ each of which are a finite union of closed disjoint intervals in $[0,1]$.
Let $f(x)=\int\_{0}^{x}\chi\_{C}(t)\mbox{ }dt,$ so $f$ is certainly absolutely continuous.
Assuming I did this right $f(C)$ contains $[0,m(c)).$ Consider each $C\_{n},$ since $f$ is increasing and continuous $f([0,1])$ contains $[0,m(C)].$ Also $[0,1]\setminus C\_{n}$ is a union of open intervals each of which are in the compliment of $C$ so it follows that $f$ is constant on each such interval. Since $f([0,1])=[0,m(C)]$ it follows that for any $x\in [0,m(C))$ we can find $t\in C\_{n}$ so that $f(t)=x.$ Indeed we already know we can do this with $t\in [0,1]$ but since $f$ is constant on the intervals in the complement of $C\_{n}$ we can force $t\in C\_{n}$ (for instance if $x$ is in some interval $I$ in $[0,1]\setminus C\_{n}$ then its left endpoint is in $C\_{n}$ and since $f$ is constant on $I$ we have that $f$ has the same value at the left-endpoint of $I$ as on $I$.)
Now fix $y\in [0,m(C)).$ Since $\lbrace x\in C\_{n}:f(x)=y\rbrace$ is non-empty and these sets are decreasing, (since the $C\_{n}$ are decreasing) by compactness we can find $x\in C$ so that $f(x)=y.$ Thus $f(C)=[0,m(C))$ and we have found a nowhere dense set which is mapped to a set which is not nowhere dense.
| 5 | https://mathoverflow.net/users/5994 | 24063 | 15,813 |
https://mathoverflow.net/questions/24083 | 12 | Hello,
Let T be a Turing machine such that
1) it operates on the alphabet {0,1},
2) its set of states is A
3) the language it accepts is $L$ .
Does there exists a Turing machine S which also operates on the alphabet {0,1} and such that the language it accepts is L (the set of states might be different though) and such that, crucially, S is reversible?
By reversible I mean "the computational paths of S are disjoint". More precisely, the transition table of S gives rise to a map $K\_S: \text{Tapes}\times B \to \text{Tapes} \times B$, where Tapes is the subset of the infinite product $\{0,1\}^Z$ consisting of those sequences which have a finite number of 1's, and B is the set of states of S. S is reversible iff, by definition, $K\_S$ is injective on the set
$$
\bigcup\_{i=0}^\infty K\_S^{i}(\text{Tapes}\times \{Initial \} ),
$$
where $Initial\in B$ is the initial state of S.
If the answer to the above question is "no" then what if we allow S to operate on an alphabet which is larger then {0,1}?
| https://mathoverflow.net/users/2631 | reversible Turing machines | Using the method of the universal Turing machine, consider the Turing machine $S$ that on input $x$ simulates the computation of $T$ on $x$, and keeps track of the entire computation history. That is, $S$ writes down on the tape complete descriptions (tape contents, state, head position) of each successive step of the computation of $T$ on $x$. If eventually $S$ finds that $T$ accepts or rejects $x$, then $S$ should also accept or reject $x$.
This computation is reversible, since at any stage of the computation of $S$ on $x$, we can easily read off $x$ from the description of the first configuration, and so we know how $S$ started and therefore how we got to where we are.
Another way to do it would be the following: using a pairing function, regard the tape as two tapes, and then on input $x$, make a copy of $x$ on one of the tapes, and then with each step of simulated computation increment a counter on this tape (by adding one more $1$). This process leads to a reversible computation, since from any stage in the computation we can tell what the input was, and how many simulated steps have been performed. So we know where the computation came from.
These computation are reversible in the weaker sense that they can be reversed from the configurations that actually arise in computation, whereas your definition seems to require reversibility on all possible finite confgurations, even if these would not arise during an actual computation. Is that really what you want?
| 5 | https://mathoverflow.net/users/1946 | 24086 | 15,824 |
https://mathoverflow.net/questions/24082 | 16 | Let $k$ be a field. Then $k[[x,y]]$ is a complete local noetherian regular domain of dimension $2$. What are the prime ideals?
I've browsed through the paper "Prime ideals in power series rings" (Jimmy T. Arnold), but it does not give a satisfactory answer. Perhaps there is none. Of course you might think it is more natural to consider only certain prime ideals (for example open/closed ones w.r.t. the adic topology), but I'm interested in the whole spectrum.
A first approximation is the subring $k[[x]] \otimes\_k k[[y]]$. If we know its spectrum, perhaps we can compute the fibers of $\text{Spec } k[[x,y]] \to \text{Spec } k[[x]] \otimes\_k k[[y]]$. Now the spectrum of the tensor product consists of $(x),(y),(x,y)$ and $\text{Spec } k((x)) \otimes\_k k((y))$. The latter one is still very complicated, I think. For example we have the kernel of $k((x)) \otimes\_k k((y)) \to k((x))$. Also, for every $p \in k[[x]]$, we have the prime ideal $(y - p)$.
| https://mathoverflow.net/users/2841 | What are the prime ideals of k[[x,y]]? | The ring $k[[x,y]]$ is a local UFD of dimension 2; so its prime ideals are the zero ideal, the maximal ideal, and all the ideals generated by an irreducible element. They are all closed (all ideals in a noetherian local ring are closed). Classifying them is an extremely complicated business, already when $k = \mathbb C$.
The ring $k[[x]] \otimes\_k k[[y]]$ is truly nasty, it is not even noetherian, and I doubt it would help.
| 24 | https://mathoverflow.net/users/4790 | 24088 | 15,826 |
https://mathoverflow.net/questions/24081 | 7 | Is the following true ?
>
> Every solvable transitive subgroup
> $G\subset\mathfrak{S}\_p$ (the symmetric group on
> $p$ letters, where $p$ is a prime)
> contains a unique subgroup $C$ of
> order $p$ and is contained in the
> normaliser $N$ of $C$ in $\mathfrak{S}\_p$. The
> quotient $G/C$ is cyclic of order
> dividing $p-1$. If $G$ is not cyclic,
> then it has exactly $p$ subgroups of
> index $p$.
>
>
>
I need such a result for an arithmetic application. A reference or a short argument will be appreciated.
**Addendum.** For those interested in the arithmetic application, see <http://arxiv.org/abs/1005.2016>
| https://mathoverflow.net/users/2821 | Solvable transitive groups of prime degree | A transitive subgroup $G$ of $S\_p$ contains a Sylow $p$-subgroup $P$
having order $p$. If it has only the one, then $P$ is normal in $G$
and so $G$ lies in the normalizer $N$ of $P$ in $S\_p$. This is the affine
linear group $\mathrm{AGL}(1,p)$ which is soluble. Thus $G$ is soluble.
Otherwise $G$ has more than one Sylow $p$-subgroup. By Sylow's theorems,
these $p$-subgroups are conjugate in $G$.
If $H$ is a nontrivial normal subgroup of $G$ then $H$ must be transitive
since $G$ is primitive (the orbits of $H$ form a partition invariant under the
action of $G$). So $H$ contains a Sylow $p$-subgroup $P$ of $G$. So
$H$ contains all the Sylow $p$-subgroups of $G$ (as they are conjugate under $G$).
Therefore $G$ cannot be soluble, as by repeatedly taking nontrivial normal subgroups
we always get groups with more than one Sylow $p$-subgroup.
| 13 | https://mathoverflow.net/users/4213 | 24091 | 15,828 |
https://mathoverflow.net/questions/24069 | 4 | I need to reference the following result. Do you know a good source?
The following conditions on an $n$-dimensional simplicial complex $S$ are equivalent:
a) $S$ is an $n$ manifold;
b) The link of every vertex of $S$ is homeomorphic to the $(n - 1)$-sphere;
c) The link of every $k$-simplex is homeomorphic to the $(n - k - 1)$-sphere.
| https://mathoverflow.net/users/5498 | Characterization of combinatorial manifolds in terms of links | The usual term for objects like this is "combinatorial manifolds".
However, the result is not quite true as you have stated. Definitely b is true if and only if c is true, and b implies a. However, a does not imply b. There definitely exist simplicial complexes which do not satisfy b or c but which are topological manifolds. For example, the famous [double suspension theorem](http://en.wikipedia.org/wiki/Double_suspension_theorem) of Cannon (weaker versions were proved by Edwards) says that if $X$ is a homology $n$-sphere, then the space $Y$ obtained by suspending $X$ twice is homeomorphic to the $(n+2)$-sphere. If neither $X$ nor the suspension of $X$ is an actual sphere (for instance, this holds if $X$ is the [Poincare homology sphere](http://en.wikipedia.org/wiki/Poincar%C3%A9_homology_sphere#Poincar.C3.A9_homology_sphere)), the vertices of $Y$ corresponding to the suspension points will then not satisfy b.
| 5 | https://mathoverflow.net/users/317 | 24094 | 15,831 |
https://mathoverflow.net/questions/21643 | 1 | In determining the monotonicity of coefficients in a series expansion (which appeared in one of my study), I come across the following problem.
Let $p\ge 2$ be an integer, and $$6p^3(i+3)d\_{i+3}=6p^2(i+2+p)d\_{i+2}+3p(p-1)(i+1+2p)d\_{i+1}+(p-1)(2p-1)(i+3p)d\_i,~~i\ge0$$ with $d\_0=d\_1=d\_2=1$. How to show $d\_i>d\_{i+1}$ for all $i\ge3$?
By easy calculation, $d\_3=1, d\_4=\frac{18p^3+11p^2-6p+1}{24p^3}$. For a recursion of 2 or 3 terms, it is easy to proceed with induction, but what about 4 terms in this case?
| https://mathoverflow.net/users/3818 | monotonicity from 4 term-recursion. | I am a little bit upset that my previous answer to the question
was downvoted by somebody. As far as I understand the idea of MO is
not necessarily to produce final answers/solutions/responses but
mostly to provide the ideas of approaching a given problem.
It's a question of time to solve this particular problem (I still
wonder about where it comes from). Denote
$$
F(x)=\sum\_{i=0}^\infty d\_ix^i=1+x+x^2+x^3+\dots
$$
the generating function of the sequence. The given recursion can
be then written as the differential equation of order 1,
$$
\frac{F'}{F}=\frac{3p(2p^2+2p(p-1)x+(p-1)(2p-1)x^2)}{6p^3-6p^2x-3p(p-1)x^2-(p-1)(2p-1)x^3}
$$
which can be explicitly solved:
$$
F(x)=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
(one checks that the expansion indeed starts $1+x+x^2+x^3+\dots$).
The desired property $d\_i>d\_{i+1}$ for $i\ge3$ is equivalent to showing that
the expansion
$$
1-(1-x)F(x)=c\_0x^4+c\_1x^5+c\_2x^6+\dots
$$
has all coefficients $c\_0,c\_1,c\_2,\dots$ positive. In turn, the latter is
equivalent to the positivity of all coefficients in the expansion
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)=4c\_0x^3+5c\_1x^4+6c\_2x^5+\dots.
$$
We have
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)
=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
$$
-p(1-x)\biggl(\frac1p+\frac{(p-1)x}p+\frac{(p-1)(2p-1)x^2}{2p^3}\biggr)
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\Biggl(1+\sum\_{n=1}^\infty\frac{(p+1)(p+2)\dots(p+n)}{n!}
\biggl(\frac xp+\frac{(p-1)x^2}{2p^2}+\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^n\Biggr),
$$
and the desired positivity follows.
| 2 | https://mathoverflow.net/users/4953 | 24096 | 15,832 |
https://mathoverflow.net/questions/24062 | 2 | I've been struggling with this question for a while.
In theorem 4.1 of ["Smoothing and extending cosmic time functions"](https://doi.org/10.1007/BF00759586 "H.-J. Seifert, Gen. Relat. Gravit. 8, 815–831 (1977). https://zbmath.org/?q=an:0425.53032") Seifert proves that a time function defined on a compact subset of a stably causal Lorentzian manifold is extendible to a global time function. To do this he constructs a countable collection of non-intersecting stable spacelike boundaries, $C\_{\tau\_k}$. The boundaries are inductively defined so that
$$ C\_{\tau\_k}=\tilde{J}^-\_{\theta\_k}(Q\_k)=\bigcup\_{\eta>\theta\_k}J^-(Q\_k;g\_{\eta}),$$
where $Q\_k$ is some compact spacelike set, $\theta\_k,\eta\in[a,b]$ for some $a,b\in\mathbb{R}$,
$J^-(Q\_k;g\_{\eta})$ is the causal past of $Q\_k$ with respect to the metric $g\_{\eta}$ and for all
$\alpha,\beta\in [a,b]$ we have that $g\lt g\_\alpha$ and $g\_\alpha\lt g\_\beta$ if and only if $\alpha$ is less than $\beta$.
Seifert defines a boundary as a set $A$ so that there exists some $W\subset M$ so that $\partial W=A$. It seems to me that $I^-(C\_{\tau\_k},g)\subset C\_{\tau\_k}$ implying that $C\_{\tau\_k}$ has non-empty interior and therefore isn't a boundary. The set $\partial C\_{\tau\_k}$ seems to fit what Seifert wants, did he just forget a boundary symbol?
Since this paper was published in 1977 and has been cited roughly 15 times I would expect other researchers to comment on this problem. I've looked through the literature, however, and can't see anything that comments on this.
So my question is, "Is $C\_{\tau\_k}$ a boundary?"
| https://mathoverflow.net/users/5993 | What are the spacelike boundaries referred to in theorem 4.1 of "Smoothing and extending cosmic time functions" by Seifert | Looking at Figure 3 and Step B of the proof of the theorem, it looks like the $C\_{\tau\_k}$ should be of the form $\partial \tilde{J}^-\_{\theta\_k}(Q\_k)$. I am also pretty sure that he chose the symbol $C$ for "cone". Essentially the idea is to construct the surface as a union of a bunch of truncated cones that are wider than the causal cone for the given metric (so that the surface defined is automatically space-like).
| 1 | https://mathoverflow.net/users/3948 | 24097 | 15,833 |
https://mathoverflow.net/questions/24098 | 12 | In functional analysis, there is the term "integral kernel". Examples are Possion kernel, Dirichlet kernel etc.
In algebra, the term kernel of a homomorphism refers to the inverse image of the zero element.
Are these two terms related? If not, where did the word "kernel" in the term "integral kernel" come from?
| https://mathoverflow.net/users/nan | What does "kernel" mean in integral kernel? | I think it simply denotes the inner part.
According to dictionary, kernel is "the important, central part of anything".
(This is the third meaning in Chambers Concise Dictionary). From O.E. cyrnel=corn,grain + dimin. suffix -el).
I also know the kernel of an operating system.
| 6 | https://mathoverflow.net/users/5864 | 24104 | 15,836 |
https://mathoverflow.net/questions/24079 | 9 | A standard homework in measure theory textbooks asks the student to prove that there are not countably infinite $\sigma$-algebras. The only proof that I know is via a contradiction argument which yields no estimate on the minimum cardinality of an infinite $\sigma$-algebra.
Given an a set $X$ of infinite cardinality $\kappa$, the $\sigma$-algebra of all co-countable subsets of $X$ is of cardinality ~~$2^\kappa$~~ $\kappa^{\aleph\_0}$. This example doesn't tell me whether there are $\sigma$-algebras of cardinality below $2^{\aleph\_0}$, if I don't assume the Continuum Hypothesis.
My question is as the title says: Are there $\sigma$-algebras of every uncountable cardinality?
Edit: The combined answer with Stephen, Matthew proves that the cardinality of a $\sigma$-algebra is necessarily at least $2^{\aleph\_0}$. Further, for each cardinality $\kappa\ge 2^{\aleph\_0}$ with uncountable cofinality, the $\sigma$-algebra of countable (or cocountable) subsets of a set $X$ with cardinality $\kappa$, is of cardinality $\kappa$.
What is left is whether for $\kappa\ge 2^{\aleph\_0}$ with $cf(\kappa)=\aleph\_0$ are there $\sigma$-algebras of cardinality $\kappa$. (I changed the title to reflect this.)
Thanks Stephen, Matthew, Apollo, for the combined work!
| https://mathoverflow.net/users/nan | Are there sigma-algebras of cardinality $\kappa>2^{\aleph_0}$ with countable cofinality? | I'm really not used to thinking about this sort of question (which is why I'm giving it a shot...) but here goes.
Given a $\sigma$-algebra $A$ of subsets of a set $X$, assume that $A$ is infinite. Then $A$ is a poset, with inclusion as the order relation. Apply Zorn's lemma to the poset $P$ of all linearly ordered subsets of $A$ to conclude that there is a maximal linearly ordered subset of $A$; since $A$ is infinite this implies there is an infinite chain $S\_1 \subset S\_2 \subset \cdots$ of elements of $A$. The pairwise differences $S\_{i+1}-S\_{i}$ are pairwise disjoint and generate a $\sigma$-subalgebra of $A$ of size at least the size of the power set of $\mathbb{Z}\_{\geq 0}$. So any infinite $\sigma$-algebra is at least that big.
Edit: Actually, this combined Matthew's argument below *almost* finishes the problem: if the cardinality of X is at least that of the power set of $\mathbb{Z}\_{\geq 0}$, then the $\sigma$-algebra of countable or cocountable sets in $X$ has cardinality $X$ *if $X$ is not of countable cofinality*. So a cardinal number is the cardinality of a sigma algebra exactly if it is at least the cardinality of the continuum (no CH needed) *and not of countable cofinality*.
| 5 | https://mathoverflow.net/users/nan | 24105 | 15,837 |
https://mathoverflow.net/questions/24090 | 26 | What is the background of the terminology of [spectra](http://en.wikipedia.org/wiki/Spectrum_(homotopy_theory)) in homotopy theory? In what extend does the name "spectrum" fit to the definition and the properties? Also, are there relations to other spectra in mathematics (algebraic geometry, operator theory)?
PS: The title is an allusion to [this question](https://mathoverflow.net/questions/17357/what-is-so-spectral-about-spectral-sequences) ;-)
| https://mathoverflow.net/users/2841 | What is so "spectral" about spectra? | It seems reasonable to me that in operator theory the term "spectrum" comes from the Latin verb *spectare* (paradigm: *specto, -as, -avi, -atum, -are*), which means "to observe". After all in quantum mechanics the spectrum of an observable, i.e. the eigenvalues of a self adjoint operator, is what you can actually see (measure) experimentally.
**Edit:** after having a look to an online etymological dictionary, it seems the relevant Latin verb is another: *spècere* (or interchangeably *spicere*)= "to see", from which comes the root *spec-* of the latin word *spectrum*= "something that appears, that manifests itself, vision". Furthermore, *spec-* = "to see", *-trum* = "instrument" (like in *spec-trum*). Also the term "spectrum" in astronomy and optics has the same origin.
In algebraic geometry, I believe the term "spectrum", and the corresponding concept, has been introduced after the development of quantum mechanics became well known. In this context, the concept of spectrum as a space made of ideals is perfectly analogous of that in operator theory (think of Gelfand-Naimark theory, and that the Gelfand spectrum of the abelian C-star algebra generated by one operator is nothing but the spectrum of that operator).
I wouldn't be surprised if the term "spectral sequence" had something to do with "inspecting" [b.t.w. also "to inspect" comes from *in + spècere*...] step by step the deep properties of some cohomological constructions.
Maybe the term "spectrum" in homotopy theory and generalized (co)homology -but I don't know almost anything about these- has to do with "probing", "testing", a space via maps from (or to?) certain standard spaces such as the Eilenberg-MacLane spaces or the spheres. Does it sound reasonable?
**Edit:** The following paragraph from the wikipedia article on "primon gas" seems to support my guess:
>
> "The connections between number theory and quantum field theory can be somewhat further extended into connections between topological field theory and K-theory, where, corresponding to the example above, **the spectrum of a ring takes the role of the spectrum of energy eigenvalues**, the prime ideals take the role of the prime numbers, the group representations take the role of integers, group characters taking the place the Dirichlet characters, and so on"
>
>
>
| 7 | https://mathoverflow.net/users/4721 | 24107 | 15,839 |
https://mathoverflow.net/questions/24108 | 9 | Suppose we have a sequence $d\_i<2n$ for $i=1,\ldots,n$ and we want to select $n$ disjoint pairs from $Z\_p$, $x\_i,y\_i$ such that $x\_i-y\_i=d\_i \mod p$. Then how big $p$ has to be compared to $n$ to do this? I am primary interested on an upper bound on $p$. Is it true that there is always a $p\le (1+\epsilon)2n+O(1)$?
My comments. It is trivial that $p\ge 2n$ because all the numbers $x\_i,y\_i$ must be different and $d\_1=1, d\_2=2$ shows that this is not always enough. I also guess that it helps if $p$ is a prime, maybe the smallest prime bigger than $2n$ works which would answer the question.
| https://mathoverflow.net/users/955 | Can select many disjoint pairs with prescribed differences from Z_n? | Your last guess is correct. The smallest prime number $>2n$ works, see [Preissmann, Emmanuel; Mischler, Maurice
Seating couples around the King's table and a new characterization of prime numbers.
Amer. Math. Monthly 116 (2009), no. 3, 268--272.]
| 9 | https://mathoverflow.net/users/4556 | 24113 | 15,841 |
https://mathoverflow.net/questions/24122 | 7 | I am interested in the following question on products of finite groups. Let $\Gamma$ be a subgroup of $U\_1\times U\_2$ such that the compositions with the canonical projections $\Gamma \subset U\_1\times U\_2 \rightarrow U\_1$ and $\Gamma \subset U\_1\times U\_2 \rightarrow U\_2$ are both surjective.
Does it follow that there is a group $G$ such that $\Gamma$ is isomorphic to the fiber product $U\_1 \times\_G U\_2$? This means that there are surjections $\pi\_1:U\_1\rightarrow G$ and $\pi\_2:U\_2\rightarrow G$ such that $\Gamma$ is the set of pairs $(u\_1,u\_2)$ with $\pi\_1(u\_1)=\pi\_2(u\_2)$.
[Goursat's](http://en.wikipedia.org/wiki/Goursat_lemma) Lemma mentioned [in this question](https://mathoverflow.net/questions/23692/what-are-the-normal-subgroups-of-a-direct-product) proves the statement in the case $\Gamma$ is a normal subgroup of $U\_1\times U\_2$.
If the statement is not true without the normality assumption, then what would be a general characterization of these subgroups $\Gamma$?
| https://mathoverflow.net/users/2805 | Subgroups of direct product of groups | Goursat's Lemma provides a complete characterization of subgroups of a direct product of two groups as fiber products. In the language I am used to: subgroups correspond to the graphs of isomorphisms between isomorphic sections of the two factors. Some subgroup embedding properties can be read from the embedding of the sections in the factors (for instance being normal in the factors, or being central in the factors), but there are no embedding properties required to use the lemma.
Goursat's lemma appears in Roland Schmidt's Lattice of Subgroups book in chapter 1.6 [(google books)](http://books.google.com/books?id=EuVadOnix5MC&pg=PA34), and as an exercise in several textbooks.
| 11 | https://mathoverflow.net/users/3710 | 24124 | 15,847 |
https://mathoverflow.net/questions/24120 | 4 | Weak convergence can be tricky when dealing with infinite dimensional spaces. For example, the usual Levy's continuity theorem does not extend readily to separable Banach spaces.
Consider a (separable) Hilbert space $H$: we know that the sequence of $H$-valued random variables $Z^N$ converges in laws towards the random variable $Z$. We also know that the sequence of random processes
$W^N \in C([0,T],H)$ (continuous functions from $[0,T]$ to $H$) converges in laws to a Brownian motion $W$.
**Question:** If we can show that for any $k,h\_1, \ldots, h\_n \in H$ and time $t\_j$
$$E[e^{i(k,Z^N)}e^{\sum\_{j=1}^N i(h\_j,W\_{t\_j})}] \to E[e^{i(k,Z^N)}] \ E[e^{\sum\_{j=1}^N i(h\_j,W\_{t\_j})}],$$
is it enough to conclude that the sequence of couples $(Z^N,W^N)$ also converge in laws to $(Z,W)$, with $Z$ and $W$ independent ?
| https://mathoverflow.net/users/1590 | weak convergence in infinite dimensional spaces | This seems to be a typical case where one can apply Prokhorov's theorem.
Since both sequences $(Z^N)$ and $(W^N)$ converge in distribution, both families of distributions are tight due to Prokhorov theorem. It easily follows that the sequence of couples $(Z^N,W^N)$ is tight, and again due to Prokhorov theorem, it is relatively compact, and we have only to see that there is a unique limiting point in distribution for any subsequence of $(Z^N,W^N)$. But each limiting point has to have the characteristic functional that you give in the r.h.s., and this characterizes the limiting distribution uniquely.
| 7 | https://mathoverflow.net/users/2968 | 24127 | 15,850 |
https://mathoverflow.net/questions/24137 | 8 | Hello,
Are there any examples of varieties which are not Shimura varieties or abelian varieties
and whose L-functions have been shown to be a product of automorphic L-functions?
Thanks.
N
| https://mathoverflow.net/users/36285 | modularity of algebraic varieties | Too long for a comment:
Yes. One family of examples is the singular K3 surfaces - a recent paper generalizing this is
<http://arxiv.org/pdf/0904.1922>
This is a consequence of a result of Livné about the modularity of 2-dimensional orthogonal Galois representations.
Rigid Calabi-Yau 3-folds also give examples, after Serre's conjecture, cf. the following paper:
<http://arxiv.org/pdf/0902.1466>
(Although this implication was already in Serre's original paper: you can deduce a similar result for any motive with the right Hodge numbers).
These examples are however "close" to abelian varieties in some sense, so you might not find them very satisfying. I don't know of any others though.
Edit: I want also to mention information that potential automorphy theorems can give you. For example, in his thesis Barnet-Lamb showed that the zeta function of the Dwork hypersurface in $\mathbb{P}^4$ has meromorphic continuation, by showing that the cohomology is automorphic after possibly restricting to a totally real field extension of $\mathbb{Q}$.
| 10 | https://mathoverflow.net/users/1594 | 24140 | 15,856 |
https://mathoverflow.net/questions/24143 | 17 | I would appreciate either an explanation or a reference for what is going on here.
Motivation:
Let $f : X \rightarrow Y$ be a morphism of algebraic varieties. The derived projection formula implies that for a sheaf $\mathcal{F}$ on $Y$, we have
$$Rf\_\ast Lf^\ast \mathcal{F} \cong Rf\_\ast \mathcal{O}\_X \otimes^L \mathcal{F}.$$
Suppose for example that $R^i f\_\ast \mathcal{O}\_X = 0$ when $i>0$, and is $\mathcal{O}\_Y$ when $i=0$. It's natural to guess that the cohomology of the right hand side is just $\mathcal{F}$ in degree 0 and 0 otherwise.
Question:
Is this true? More generally, is there a spectral sequence calculating the cohomology of the composite of a right derived- with a left derived functor?
Is there an exact sequence of terms of low degree, as there is for the composite of two right derived functors?
| https://mathoverflow.net/users/1594 | Composing left and right derived functors | The answer is yes. Now, concerning the general question - it is not important here that one functor is right derived and the other is left derived. You just have triangulated functors between triangulated categories. Also you have t-structures on all categories which give you filtrations on all objects. And the spectral sequence controls their interaction.
To be more precise, let $F:T\_1 \to T\_2$ and $G:T\_2 \to T\_3$ be two triangulated functors. You want to apply $G\circ F$ to a pure (with only one nontrivial cohomology) object $X \in T\_1$ and compute the cohomology of $G(F(X))$. First, let us denote by $Y\_i$ the $i$-th cohomology of $F(X)$. In other words, there is a filtration on $F(X)$ with factors $Y\_i$. Now we apply the functor $G$. We obtain an object $G(F(X))$ with a filtration with factors $G(Y\_i)$. Now we want to compute the cohomology of this object. These are computed by the spectral sequence. So, you see, there is nothing special about the functors, the question is about filtered objects in triangulated categories.
| 29 | https://mathoverflow.net/users/4428 | 24151 | 15,862 |
https://mathoverflow.net/questions/24054 | 44 | First a little background. Microwaves do not heat uniformly. To help overcome this, your food is rotated, however this is not usually sufficient to produce totally uniform heating. Informally, this is the question: Is there a way of moving our food in order to heat it uniformly throughout?
Let $f : \mathbb{R}^n \to R$ be our heat function. Let $I^n = [-0.5,0.5] \times \cdots \times [-0.5,0.5]$ be the unit n-dimensional cube centered at the origin; it will be our food. Let $\gamma : [0,1] \to \mathbb{R}^n \times SO(n)$ be a map specifying a path along which to translate and rotate $I^n$. If $x \in I^n$ then let $h(x)$ denote the total heat absorbed by $x$ as it travels along $\gamma$.
Note that if $\gamma(t) = (\gamma\_1(t), \gamma\_2(t))$ then $h(x) = \int\_0^1 f(\gamma\_2(t)(x) + \gamma\_1(t)) dt$.
We will call a curve $\gamma$ 'uniformly heated' iff $\forall x,y \in I^n$, $h(x) = h(y)$.
How sufficiently nice must our heat function $f$ be in order to guarantee that there exists a uniformly heated curve? Do these requirements change if we consider a different food to heat, for example, if we heat $I^m \times 0^{n-m}$ in $\mathbb{R}^n$?
Note that in $\mathbb{R}^1$, as $SO(1) = 1$, if $f$ is a strictly monotonic function then there cannot exist any uniformly heated curves as (assuming wlog $f$ is increasing) $h(-0.5) < h(0.5)$.
| https://mathoverflow.net/users/3121 | Microwaving Cubes | You can uniformly cook the cube if $f$ is harmonic, i.e. $\Delta f=0$. Note that, if e.g. $\Delta f>0$ everywhere, then the center of the cube will always receive less heat than the average over a sphere with the same center. Thus if $\Delta f$ happens to have constant sign, it must be zero.
To achieve uniform cooking in the harmonic case, let the center of the cube stay at the origin and rotate the cube using a Peano-like curve $\gamma:[0,1]\to SO(n)$ such that the push-forward measure $\gamma\_\*m$ (where $m$ is the Lebesgue measure on $[0,1]$) is the normalized Haar measure on $SO(n)$. Then the heat received by a point $x\in I^n$ is the average of $f$ over the sphere of radius $r=|x|$ centered at the origin. Since $f$ is harmonic, this average value equals $f(0)$.
To construct such a curve $\gamma$, follow the standard procedure for the Peano curve: partition $SO(n)$ into reasonable sets (connected and with piecewise smooth boundaries) and visit all of them by a continuous path; this path is the first approximation. Then subdivide the partition and change the path so that it visits all sub-parts but do not make new intersections with boundaries of old parts. And so on. At each step, choose the parametrization so that the time spent in each piece equals its Haar volume. Let the diameters go to zero, then the paths will converge to a Peano-like curve $\gamma$. The push-forward measure $\gamma\_\*m$ coincides with the Haar measure on all elements of the partitions and hence on all Borel sets.
**Remark.**
A similar trick works if $f$ has compact support (more precisely, its support should be separated away from the microwave walls by distance at least 1). Just move around so as to realize a Haar measure on the relevant subset of the translation group rather than $SO(n)$.
| 23 | https://mathoverflow.net/users/4354 | 24162 | 15,871 |
https://mathoverflow.net/questions/24161 | 2 | Given two normally distributed variables `x_1, x_2`, is there a non-simulation method of calculating the probability that `x_1 > x_2`?
Generalizing a bit, what is the probability that given a list of normally distributed variables `x_i`, the probability that `x_a = max x_i`?
| https://mathoverflow.net/users/6012 | Comparing normally distributed variables | Yes. For normal random variables, the probability P(X > Y) can be calculated in closed form. See this post on [random inequalities](http://www.johndcook.com/blog/2008/07/26/random-inequalities-ii-analytical-results/).
Regarding your more general question, see this article on random inequalities with [three or more random variables](http://www.johndcook.com/blog/2008/09/06/random-inequalities-vii-three-or-more-variables/).
| 3 | https://mathoverflow.net/users/136 | 24168 | 15,876 |
https://mathoverflow.net/questions/24175 | 5 | In [this link](http://www.math.leidenuniv.nl/scripties/Trevisan.pdf), Corollary 3.2.2, page 59 the author claims that: The Euler characteristic of the toric variety $X\_K$ associated to a convex polytope $K$ is the number of vertices of $K$.
I want to see how it works. Could someone please illustrate this for me by using this method to compute the Euler characteristic of $\mathbb{P}^{2}$ and $\mathbb{P}^{1}\times \mathbb{P}^{1}$? Thanks so much.
| https://mathoverflow.net/users/5136 | Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope? | Merely observe that a toric variety is the union of torus orbits $(\mathbb C^\\*)^r$ for various dimensions $r$, and that the Euler characteristic of $(\mathbb C^\\*)^r$ is zero if $r>0$ and $1$ if $r=0$.
Vertices of a polytope correspond to 0-dimensional orbits, $r$-dimensional faces -- to $r$-dimensional orbits.
$\mathbb P^2$ corresponds to a triangle, $\mathbb P^1\times\mathbb P^1$ to a square. It is not very hard to count their vertices.
| 16 | https://mathoverflow.net/users/1784 | 24179 | 15,884 |
https://mathoverflow.net/questions/24102 | 10 | I'm interested in find out what were some of the first uses of mathematical induction in the literature.
I am aware that in order to define addition and multiplication axiomatically, mathematical induction in required. However, I am certain that the ancients did their arithmetic happily without a tad of concern about induction.
When did induction get mentioned explicitly in the mathematical literature? Definitely this places before about 1800 when the early logicians started formulating axioms for arithmetic.
| https://mathoverflow.net/users/nan | Historically first uses of mathematical induction | There are several questions here, so my answer overlaps with some of the
others.
1. First use of induction in some form. I would nominate the "infinite
descent" proof that $\sqrt{2}$ is irrational -- suppose that $\sqrt{2}=m/n$,
then show that $\sqrt{2}=m'/n'$ for smaller numbers $m',n'$
-- which probably goes back to around 500 BC.
2. First published use of induction in some form. Euclid's infinite descent
proof that every natural number has a prime divisor, in the *Elements*.
3. First use of induction in the "base step, induction step" form. I
suggest Levi ben Gershon and (more definitely) Pascal, as mentioned in
danseetea's answer.
4. First *mention* of "induction". The one suggested by Gerald Edgar is
the earliest I know of.
5. First realization that induction is fundamental to arithmetic:
Grassmann's *Lehrbuch der Arithmetik* of 1861, where he defines addition
and multiplication by induction, and proves their ring properties by
induction. This idea was rediscovered, and built into an axiom system by
Dedekind, in his *Was sind und was sollen die Zahlen?* of 1888. It
became better known as the Peano axiom system, when Peano redeveloped
it a couple of years later.
| 19 | https://mathoverflow.net/users/1587 | 24182 | 15,886 |
https://mathoverflow.net/questions/24123 | 4 | Is the category of topological spaces locally presentable? n-lab claims that it is not locally FINITELY presentable, but how about for some larger cardinal? Here I really mean the 1-category of topological spaces and am not willing to identify it with simplicial sets. Essentially, I want to know if (after I fix appropriate Grothendieck universes) representable presheaves on Top are characterized by those presheaves which send colimits in Top to limits in Set, which would follow from local presentablility.
| https://mathoverflow.net/users/4528 | Local presentability and representable presheaves over the category of topological spaces | The category of topological spaces is not locally $\lambda$-presentable for any $\lambda$. The reason for this is the existence of spaces which aren't $\lambda$-presentable (a.k.a. $\lambda$-small) for any $\lambda$ (in a locally presentable category every object is $\lambda$-presentable for some $\lambda$). An example of such a space is the Sierpinski space; a proof of this can be found in Mark Hovey's book on model categories, on page 49.
There is a convenient category of topological spaces which *is* locally presentable, the category of $\Delta$-generated spaces. This category contains most of the spaces usually studied by algebraic topologists (for example, the geometric realization of any simplicial set is a $\Delta$-generated space). Daniel Dugger has some expository notes on this [here](https://pages.uoregon.edu/ddugger/delta.html). A proof that the category of $\Delta$-generated spaces is locally presentable can be found [this paper](http://www.tac.mta.ca/tac/volumes/21/1/21-01abs.html) of L. Fajstrup and J. Rosický.
The second question was already answered in the comments: if $G\colon \mathbf{Top}^{\mathrm{op}} \rightarrow \mathbf{Set}$ is continuous, then it has a left adjoint $F$ by the special adjoint functor theorem. Therefore we have natural isomorphisms
$G(X) \cong \mathbf{Set}(\ast,GX) \cong \mathbf{Top}^{\mathrm{op}}(F(\ast),X)=\mathbf{Top}(X,F(\ast))$,
which shows that $G$ is represented by $F(\ast)$.
Edit: added the missing op's mentioned in the comment.
| 13 | https://mathoverflow.net/users/1649 | 24185 | 15,888 |
https://mathoverflow.net/questions/24174 | 3 | Suppose $f:X \to Y$ is a map of sets and $F$ a filter on $X$ such that its image filter is contained in an ultrafilter $G$ on $Y$. Can I find an ultrafilter $H$ on $X$ whose image is $G$?
If this question is too elementary, I apologize. I have not worked much with ultrafilters, so sometimes basic properties escape me.
EDIT: I messed up the formulation of the question earlier, sorry!
(The previous formulation said that $F$ was on $Y$ etc., to make sense of the responses already posted).
| https://mathoverflow.net/users/4528 | Ultrafilters containing the image of a filter | Note that the image of a filter on $X$ will be $G$ if and only if it contains the filter consisting of all preimages of ''big'' sets in $Y$. Note also that we can combine two filters $F$ and $F'$ (that is, find a filter containing both of them) if and only if the intersection of any pair of sets $S\in F$, $S'\in F'$, is nonempty. Simply take the filter consisting of all sets which contain such an intersection.
So it suffices to show that if $S\in F$ and $T\in G$, then $S\cap f^{-1}(T)$ is nonempty. Suppose it were empty. Then $f(S)$ lies in the complement of $T$, hence is not in $G$. But $S\subset f^{-1}(f(S))$ implies that $f^{-1}(f(S))$ is in $F$, which contradicts that the image filter of $F$ was contained in $G$.
| 3 | https://mathoverflow.net/users/5513 | 24188 | 15,890 |
https://mathoverflow.net/questions/24145 | 3 | I'm not sure if this is standard, but we'll call the property that every weak homotopy equivalence is an honest homotopy equivalence the *Whitehead property* (from Whitehead's theorem for CW complexes).
Then the question: Is there any (nontrivial) category of spaces that is cartesian-closed and has the Whitehead property? If not, is there some counterexample we can construct to show that these properties are mutually exclusive?
I am familiar with the category of m-cofibrant spaces, but it is not cartesian closed (even though there is the weaker result allowing us to take loop spaces etc.).
| https://mathoverflow.net/users/1353 | Cartesian-closed category of spaces with the Whitehead property? | This is only a partial answer Harry, but, to my knowledge, all Cartesian-closed categories of topological spaces that I know of arise as the monocoreflective hull (in either Top or Hausdorff spaces etc.) of a productive class $C$ of generating exponentiable spaces. The requirement to be a productive class is that binary products of generators lie in the monocoreflective hull. Also, it is a well known theorem that a space is exponentiable if and only if it is core-compact. So, for example if $C$ is compact Hausdorff space, we get the category of compactly generated spaces as its monocoreflective hull. If you tried to use this machinery to produce a Cartesian-closed category of "whitehead spaces", I think it would fail: a natural choice would be compact Hausdorff m-cofibrant spaces for C- this should be a productive class (if I am not making a silly mistake), BUT, I don't think that being homotopy equivalent to a CW-complex is stable under colimits, as CW-complexes aren't. So, my guess of an answer would be "No." However, this is hardly a proof. By the way, what did you want to do with these spaces anyhow?
| 3 | https://mathoverflow.net/users/4528 | 24189 | 15,891 |
https://mathoverflow.net/questions/24190 | 8 | The formula for 1 + a + a^2 + .... where 0 < a < 1 is $\frac{1}{1-a}$, but what if you wanted to sum only those where the exponent is a power of 2? That is,
$S = a + a^2 + a^4 + a^8 + \cdots$
I feel like this is an easy one but I just can't seem to find a closed expression for it, nor search for it on Google.
| https://mathoverflow.net/users/5534 | Sum of subset of geometric series: a^2^n | Mahler proved in the 1930s that the values of $f(z)=\sum\_{n=0}^\infty z^{d^n}$, $d>1$ is an integer, are transcendental for any algebraic $z$ satisfying $0<|z|<1$. A related problem of transcendence of the function $f(z)$ was discussed in [this question](https://mathoverflow.net/questions/21290/whats-an-example-of-a-transcendental-power-series/21554#21554). This motivates nonexistence of simple formula like $1/(1-z)$ for $f(z)$.
| 16 | https://mathoverflow.net/users/4953 | 24197 | 15,895 |
https://mathoverflow.net/questions/24205 | 7 | The set up is $C$ is a curve and $J$ is its Jacobian. On the $C \times J$ there is the Poincare bundle $P$ which is the universal family of degree zero line bundles on $C$. For every integer $d$ there is also a line bundle $P(d)$ on $C \times J$ which is a family of line bundles of degree $d$ on $C$.
I've seen a construction of $P$ and given $P$ an example of a $P(d)$ would be $P^L:= q\_1^\*L\otimes P$ where $q\_1 \colon C \times J \to C$ is the projection and $L$ is a line bundle of degree $d$ on $C$. If $L,L'$ both have degree $d$ you can form either $P^L$ or $P^{L'}$. I've seen $P(d)$ defined as the inverse limit of $P^L$ as $L$ ranges over all degree $d$ line bundles. I don't really know how to think of such an inverse limit. Is there a more concrete way to describe $P(d)$? It seems up to isomorphism, $P(d) = P^L$, but this is "very" non canonical which seems bad. For example does $P^L$ have a universal property like $P$ does?
I have a more specific question related to this. This question is coming from Prop. 21.6 of Polishchuk's book on Abelian varieties and the Fourier Mukai tranform, if anyone is curious. If $q\_2 \colon C \times J \to J$ is the projection, then apparently $q\_{2\*}P(g-1) = 0$. Why is this so?
One argument for this (that I don't understand) is the following.
1) Embed $P(g-1) \to F$ with $F$ flat over $J$ and $R^1q\_{2\*}F = 0$.
2) From $0 \to P(g-1) \to F \to F/P=: E \to 0$ and cohomology we get
$0 \to q\_{2\*}P(g-1) \to q\_{2\*}F \to q\_{2\*}F \to R^1q\_{2\*}P(g-1) \to 0$
3)The restriction of $R^1q\_{2\*}P(g-1)$ to $a \in J$ is $H^1(C \times a, P(g-1)|\_{C \times a})$ which is $H^1$ of a line bundle of degree $g-1$. This is zero outside the theta divisor and Riemann-Roch says $h^1(L) = h^0(L)$ for degree $g-1$ line bundles, so
$q\_{2\*}F \to q\_{2\*}F$
is an isomorphism outside a divisor. And hence
4) $q\_{2\*}F \to q\_{2\*}F$ is injective.
Part 4) is the part I don't see.
I don't necessarily want to understand this line of reasoning. But I don't see why $q\_{2\*}P(g-1) = 0$ since it seems to be supported on the theta divisor.
| https://mathoverflow.net/users/7 | The Poincare Bundle(s) on C \times J | For your first question, if you want a proper universal property it is defined
by a variety $J^{(d)}$ and a line bundle $L^{(d)}$ on $C\times J^{(d)}$ which is
of degree in the $C$-direction, i.e., of degree $d$ on each fibre $C\times
x$. The universality then says that for every $X$ and every line bundle $M$ on
$C\times X$ of degree $d$ in the $C$-direction there is a unique morphism $f\colon
X\rightarrow J^{(d)}$ such that $(\mathrm{id}\times f)^(L^{(d)})$ and $M$ differ
by a line bundle from $X$. Picking any line bundle of degree $d$ on $C$ (which
may not exist if the base field is not algebraically closed) allows you to
construct such a line bundle on $J$ but different choices will give different
line bundles on $J$ and hence an automorphism of $J$. This automorphism is then
explicitly given by a translation.
As for your second question $q\_{2\ast}P(g-1)$ is a torsion free sheaf on $J$ whose
restriction to the complement of the theta divisor is $0$ and must therefore be
$0$. What happens is that the base change formula makes $H^0(C\times
a,P(g-1)\_{|C\times a})$ come from $R^1q\_{2\ast}P(g-1)$. An algebraic model is the
exact sequence $0\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$
where $k[t]\rightarrow k[t]$ is multiplication by $t$ (you should think of the
first $k[t]$ as $q\_{2\ast}F$, the second as $q\_{2\ast}E$ and $k$ as
$R^1q\_{2\ast}P(g-1)$). When you tensor this with $k=k[t]/(t)$ the map
$k[t]\rightarrow k[t]$ grows a kernel, i.e., we have an exact sequence
$0\rightarrow k\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$. This
new kernel is $\mathrm{Tor}^1(k,k)$ which models $H^0(C\times
a,P(g-1)\_{|C\times a})=\mathrm{Tor}^1(k(a),R^1q\_{2\ast}P(g-1))$ (which comes from
the base change formula).
| 4 | https://mathoverflow.net/users/4008 | 24208 | 15,901 |
https://mathoverflow.net/questions/23670 | 3 | I am actually not reading the original paper "Diophantine problems over local fields, I (1965)" by Ax and Kochen but the revised version " The model theory of local fields (1975) " by Kochen which is simpler.
>
> Side question: Are the ideas in this two papers significantly different from one another?
>
>
>
The central result of the paper is the isomorphism theorem:
Let $V$ and $V'$ be unramified $\omega$-pseudo-complete Hensel fields of cardinality $\aleph\_1$ with normalized cross section ($x$-section). Then there is a x-analytic isomorphis $\varphi: V \rightarrow V'$ if and only if $ \bar{V} \simeq \bar{V'}$ and $\text{ord}\ V \simeq \text{ord}\ V'$.
Here unramified means the characteristic of the residue field is 0 or the characteristic of the residue field is $p$, and the absolute ramification index is 1 i.e. p is the element of least positive value. $\bar{V}$ denotes the residue field of $V$ and $\text{ord}\ V$ denotes the valuation group of $V$. An x-isomorphism is and isomorphism that send the cross section in $V$ to the cross section in $V'$.
I have read that in "One Definable subsets of p-adic fields ", MacIntyre proved that in the case of p-adic fields, the x-section can be replaced with the quantifier $P\_n$.
>
> Is it possible to get rid of the x-section in the above isomorphism theorem in general?
>
>
>
The motivation is that the x-section is not among the usual functions. ( There is certain discussion on the same paper by Kochen about removing the x-section, but it appears not what I want).
| https://mathoverflow.net/users/2701 | Can the cross-section (x-section) be removed from the Ax-Kochen proof in "Diophantine problems over local fields"? | It is not necessary to assume the existence of a cross-section to conclude the relative completeness part of the Ax-Kochen-Ershov theorem, but for the more refined relative quantifier elimination is not true in general without the cross section.
Let me explain.
First of all, we should make a slight correction to your statement of the Ax-Kochen theorem. The existence of an isomorphism between V and V' requires the continuum hypothesis. [See Shelah, Saharon, Vive la différence. II. The Ax-Kochen isomorphism theorem,
Israel J. Math. 85 (1994), no. 1-3, 351--390, in which Shelah shows that there models of ZFC in which there are nonprincipal ultrafilters ${\mathcal U}$ on the primes for which $\prod\_{\mathcal U} {\mathbb F}\_p((t))$ and $\prod\_{\mathcal U} {\mathbb Q}\_p$ are
not isomorphic.]
What Matthew Morrow says about completeness relative to $RV(F)$ is correct (mathematically, though not historically --- these structures have been studied under different names, by for example, Franz-Viktor Kuhlmann and Serban Basarab, in connection with quantifier elimination and relative completeness and by the founders of the theory of valued fields already in the 1930s. I discuss these structures in detail in my paper published in the proceedings of the 1999 workshop on valuation theory in Saskatoon). There are many suitable languages for understanding RV, but the following might be the most straight forward.
We give ourselves four sorts: VF for the valued field, RV for the residue-valuation sort, $\Gamma$ for the value group, and $k$ for the residue field. On VF and k we have the language of rings, on RV the language of multiplicative groups and on $\Gamma$ the language of ordered abelian groups. These sorts are connected by the natural quotient map $r:K^\times \to RV(K) = K^\times/(1 + {\mathfrak m})$, the valuation map $v:K^\times \to \Gamma$, the induced valuation map (still denoted by $v$) $v:RV \to \Gamma$, the reduction map $\pi:{\mathcal O} \to k$, and the natural inclusion $\iota:k^\times \to RV$.
It should be clear that $k$ and $\Gamma$ are interpretable in RV with its full induced structure. That is, if we would prefer to work with a two-sorted language, then we should treat RV as a structure with multiplication, a predicate for the image of $\iota$, a three place predicate for addition on the image of $\iota$ and a two-place relation $V(x,y)$ to be interpreted as $v(x) \leq v(y)$.
In this language it is now the case that if $K$ and $L$ are unramified henselian fields (so either $k(K)$ and $k(L)$ have residue characteristic zero or the residue characteristic is $p > 0$ and $v(p)$ is the least positive element of the value group), then $K$ and $L$ are elementarily equivalent (in the two-sorted language described above) if and only if $RV(K)$ and $RV(L)$ are elementarily equivalent.
Moreover, the theory of unramified henselian fields eliminates quantifiers relative to RV in the sense that for any formula $\phi$ there is another formula $\psi$ for which the quantifiers only range over RV for which the theory of unramified henselian fields proves that $\phi$ is equivalent to $\psi$.
Now, one may deduce the relative completeness theorem for the residue field and value group from the corresponding relative completeness theorem for RV. The point is that $RV(K)$ and $RV(L)$ are elementarily equivalent if and only if $k(K)$ and $k(L)$ are elementarily equivalent as fields and $\Gamma(K)$ and $\Gamma(L)$ are elementarily equivalent as ordered groups. Why? The question is absolute, so we may prove the result under the assumption of the continuum hypothesis. We can replace $K$ and $L$ with elementarily equivalent saturated models of size $\aleph\_1$, $K^\ast$ and $L^\ast$, respectively. From the saturation hypothesis, cross-sections $\chi\_K:\Gamma(K^\ast) \to RV(K^\ast)$ and $\chi\_L:\Gamma(L^\ast) \to RV(L^\ast)$ exist as do isomorphisms $k(K^\ast) \cong k(L^\ast)$ and $\Gamma(K^\ast) \cong \Gamma(L^\ast)$. Since with $\chi$, RV splits as a product, we conclude that $RV(K^\ast) \cong RV(L^\ast)$. In particular, $RV(K) \equiv RV(K^\ast) \equiv RV(L^\ast) \equiv RV(L)$ so that $K$ and $L$ are elementarily equivalent by the weak form AKE for RV.
Quantifier elimination simply relative to the residue field and value group is not true. For example, in ${\mathbb Q}((t))$ the elements $t^2$ and $5 t^2$ have the same [relative to the residue field and value group] quantifier-free type (ie they agree on all formulas in which only quantification over $k$ and $\Gamma$ are allowed) but they do not have the same type as the first is a square and the second is not.
Generalizing the RV construction to include sorts $RV\_n$ to be interpreted as $RV\_n(K) = K^\times/(1 + n {\mathfrak m})$ one may obtain a relative quantifier elimination and completeness theorem for all henselian fields of characteristic zero. That is, without any restriction on ramification.
| 8 | https://mathoverflow.net/users/5147 | 24212 | 15,904 |
https://mathoverflow.net/questions/24181 | 12 | This may be total nonsense. But I need to know the answer quickly and I am too tired to think about it thoroughly. Let $k$ be a positive integer. Roe's "Elliptic Operators" claims that there is a 1-to-1 correspondence between:
* representations of the Clifford algebra $\operatorname{Cl}\mathbb R^k$ of the vector space $\mathbb R^k$ with the standard inner product;
* representations of the Pin group of this vector space (i. e., of the subgroup of the multiplicative group of $\operatorname{Cl}\mathbb R^k$ generated by vectors from $\mathbb R^k$) on which the element $-1$ of the Pin group acts as $-\operatorname{id}$;
* representations of the subgroup $\left\lbrace \pm e\_1^{i\_1}e\_2^{i\_2}...e\_k^{i\_k} \mid 0\leq i\_1,i\_2,...,i\_k\leq 1 \right\rbrace$ of the Pin group (where $\left(e\_1,e\_2,...,e\_n\right)$ is the standard orthogonal basis of $\mathbb R^k$) on which the group element $-1$ acts as $-\operatorname{id}$.
I do see how representations restrict from the above to the below, and also how there is a 1-to-1 correspondence between the first and the third kind of representations. But is it really that obvious that there are no "strange" representations of the second kind? I mean, why is a representation of the Pin group uniquely given by how it behaves on $-1$, $e\_1$, $e\_2$, ..., $e\_k$ ?
Any help welcome, I'd already be glad to know whether it's really that obvious or not.
---
EDIT: This seems to have caused some confusion. Here is the core of the question:
Assume that we have a representation $\rho$ of the Pin group $\operatorname{Pin}\mathbb R^k$ such that $\rho\left(-1\right)=-\operatorname{id}$. This, in particular, means an action of each unit vector. By linearity, we can extend this to an action of every vector. Is this always a representation (i.e., does the sum of two vectors always act as the sum of their respective actions)?
| https://mathoverflow.net/users/2530 | Representations of Pin vs. Representations of Clifford | I'm not sure whether a representation of an algebra $A$ means a representation of the unit group of $A$, or an $A$-module. With the second interpretation, the statement is false.
Let's take $k=2$ and use the negative definite inner product. (This example will occur inside any larger example.) So the Clifford algebra is generated by $e\_1$ and $e\_2$, subject to $e\_1^2=e\_2^2=1$ and $e\_1 e\_2 = - e\_2 e\_1$. Let $S$ be the $2$-dimensional representation
$$\rho\_S(e\_1) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \rho\_S(e\_2) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$
Let $V=S^{\otimes 3}$. I claim that $V$ is a $\operatorname{Pin}$ representation where $-1$ acts by $- \operatorname{Id}$, but $V$ is not a module for the Clifford algebra.
For all $\theta$, the vector $v(\theta) := \cos \theta e\_1 + \sin \theta e\_2$ is in the Pin group. Clearly,
$$\rho\_S( v(\theta)) = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}.$$
Then $\rho\_V(v(\theta))$ is an $8 \times 8$ matrix I don't care to write down, whose entries are degree $3$ polynomials in $\sin \theta$ and $\cos \theta$. The point is,
$$ \rho\_V( v(\theta) ) \neq \cos \theta \rho\_V(e\_1) + \sin \theta \rho\_V(e\_2).$$
So $V$ is not an $A$-module. It is easy to build similar examples for the other signatures.
I'm not sure what happens if we read "representation of $A$" as "representation of the unit group of $A$."
| 12 | https://mathoverflow.net/users/297 | 24238 | 15,916 |
https://mathoverflow.net/questions/24221 | 10 | I am interested in clean algorithms for approximating solutions and so I am interested in numerical analysis, but most of the books I have seen get bogged down in error analysis or they spend a lot of time and effort in squeezing an additional 2% efficiency from a classical algorithm. What are some works which just have the fundamental ideas and methods?
| https://mathoverflow.net/users/4692 | Reference request for conceptual numerical analysis | Are you looking for a reference that links the field of numerical analysis to mathematical concepts moreso than algorithmic concepts? Matrix Computations by Golub and Van Loan is a fairly important book that studies the algebraic structures of matrices and derives algorithms from those properties. If you're looking for an entry-level work, I keep a copy of Michael Heath's book Scientific Computing on my desk. It covers fundamental concepts and algorithms fairly well, in my opinion.
Do you have a specific problem domain in mind?
| 5 | https://mathoverflow.net/users/5640 | 24247 | 15,919 |
https://mathoverflow.net/questions/24255 | 2 | Every group G is a subgroup of Isometry group of its Cayley graph.
What is essential property of being an Isometry group?
Lie group?
| https://mathoverflow.net/users/5980 | What kind group can be realized as a Isometry group of some space? | Every group is the full group of isometries of a connected, locally connected, complete metric space:
de Groot, J. "Groups represented by homeomorphism groups."
Math. Ann. 138 (1959) 80–102.
[MR119193](http://www.ams.org/mathscinet-getitem?mr=119193)
[doi:10.1007/BF01369667](http://dx.doi.org/10.1007/BF01369667)
Being a group of symmetries is the same thing as being a group.
You may also be interested to know that every group is the full automorphism group of a graph, not just a subgroup. References for this and various refinements are given at [the wikipedia page for Frucht's theorem](http://en.wikipedia.org/wiki/Frucht%27s_theorem).
| 14 | https://mathoverflow.net/users/3710 | 24257 | 15,924 |
https://mathoverflow.net/questions/24258 | 1 | Hi there,
I was wondering if you guys could be able to find the sum of the following series:
$ S = 1/((1\cdot2)^2) + 1/((3\cdot4)^2) + 1/((5\cdot6)^2) + ... + 1/(((2n-1)\cdot2n)^2) $, in which $\{n\to\infty}$ .
This question came to mind when I was looking at this (<http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf>) paper by Professor Anirban DasGupta. In the last section, a couple of specific examples of his 'unified' method to find the sums of infinite series is pressented. In equation (34), he states that the following series:
$ 1/(1\cdot2) + 1/(3\cdot4) + 1/(5\cdot6) + ... 1/(2n\cdot(2n-1)) = log(2) $ (Note that $\{n\to\infty}$ again). I was wondering If it's possible to find the sum if the values of the denominators of the terms are squared.
Thanks in advance,
Max Muller
| https://mathoverflow.net/users/93724 | Evaluation of the following Series | Yes, it equals $\frac{\pi^2}{3}-3$. This follows from applying partial fractions and using $\zeta(2)=\frac{\pi^2}{6}$.
| 6 | https://mathoverflow.net/users/1464 | 24260 | 15,926 |
https://mathoverflow.net/questions/24233 | 9 | It seems that there are two notions of strongly equivaraint $D\_X$- Modules and I would like to know if they are equivalent, or at least how they are related.
Let $\rho: G\times X \rightarrow X$ be an action of an algebraic group on a smooth variety over the complex numbers.
The first definition goes like this:
An equivariant $D\_X$ Module is just a $D\_X$ module $M$ together with an isomorphism
$$\rho^\* M\rightarrow \pi^\* M$$ of $D\_{G\times X}$ -modules. That isomorphism has to satisfy some cocycle condition.
The other definition is a bit more cumbersome to write down. First it requires just an isomorphism of $O\_{G\times X}$ modules, not necessarily of $D\_{G\times X}$-modules $$\rho^\* M\rightarrow \pi^\* M$$ modules, which again satisfies the cocycle conditon.
In addition it requires the action map $$D\_X\otimes M \rightarrow M$$ to be equivariant.
Finally there is another condition to be satisfied:
Observe that we get two operations of the liealgera on $M$:
One operation, by directly differentiating the action of $G$ on $M$.
Another operation in the following way: First we differentiate the action of $G$ on $X$, and get a map
$$Lie(G)\rightarrow Der\_X$$
from the liealgebra into vectorfields on $X$. Because $M$ is a $D\_X$ module we can compose this map with the action of vectorfields on $M$ and get our second operation.
We require these operations to coincide.
A more precise definition of the second kind is given here on pages 48-49:
<http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf>
So the question is, are these two notions equivalent?
Edit: If anybody else needs these facts, I found a reference which gives a proof:
<http://alpha.uhasselt.be/Research/Algebra/Publications/Geq.ps>
| https://mathoverflow.net/users/2837 | Are these notions of strongly equivariant D-modules equivalent? | Yes. They are both discussed in Chapter 7 (Hecke Patterns) of Beilinson-Drinfeld's Quantization of Hitchin Hamiltonians, accessible (like most things in the area) off Dennis Gaitsgory's page you quote above.
More precisely, there are two ways to think of D-modules on a stack $X/G$ (aka $G$-equivariant D-modules on $X$). First, you can just describe them by descent from $X$ -- the descent data is exactly your first description - the two pullbacks have to agree, in a way that's associative. The same description works for the (dg) derived category of D-modules on the stack $X/G$ -- you make the simplicial scheme with simplices $G\times G\times\cdots\times G\times X$, and take the (homotopy) limit of the dg categories of $D$-modules on the simplices. Again this is explicitly in BD chapter 7, as well as in the appendix to the long [paper](http://arxiv.org/abs/math/0508382) by Frenkel-Gaitsgory and in my Character Theory paper with Nadler.
The second description is quantum hamiltonian reduction (aka BRST). You can write the cotangent of $X/G$ as the quotient by $G$ of the zero fiber of the moment map on $T^\* X$. Now quantize, you can write the quantum version of $T^\*(X/G)$ --- ie D-modules on $X/G$ - by first taking G-equivariants as O-modules (this gives the quantization of $(T^\*X)/G$) and then imposing the zero value of the moment map, which means the compatibility of Lie algebra actions you wrote. This picture is also in BD..and again if you take invariants and coinvariants (ie impose 0 moment value) in a derived way the statement holds on the derived level.
One way Nadler and I like to say this is by thinking about D-modules as O-modules on the de Rham space $X\_{dR}$ (quotient of X by formal neighborhood of the diagonal). Then D-modules on $X/G$ means O-modules on the de Rham space of $X/G$, which is $X\_{dR}/G\_{dR}$ -- spelling that out gives the first description you wrote. To get the second picture, you first impose $G$-equivariance as an $O$-module, which means looking at $X\_{dR}/G$. But $G\_{dR}=G/\widehat{G}$ (formal group of $G$) is a very useful realization -- ie Lie groups DO have canonical normal "subgroups"!).. the second step, fixing the action of the Lie algebra (or equivalently the formal group) tells you which sheaves on $X\_{dR}/G$ actually come from $X\_{dR}/G\_{dR}$, ie are strongly equivariant..
| 14 | https://mathoverflow.net/users/582 | 24274 | 15,935 |
https://mathoverflow.net/questions/24275 | 3 | This is a question based on the heuristics that most things in algebraic/differential topology has an analogue in algebraic geometry.
The fundamental group classifies the covering spaces of a (pointed, connected, path connected, semilocally simply connected topological) space. First we construct the universal cover as a space of paths and equip it with compact-open topology. And then we use the action of the fundamental group on the cover to define the space as a quotient. And by actions of the subgroups, we construct all the covering spaces, which are in one-one correspondence with the conjugacy classes of subgroups of fundamental group of the pointed space.
The question is, up to what extent can we carry over this setup to algebraic geometry? There are the appropriate notions of etale coverings and etale fundamental groups already, and by GAGA for smooth algebraic varieties one can already see some hope.
So, are there algebraic geometric analogues of the theorem on existence of universal cover and the theorem classification of covering spaces in one-one correspondence with conjugacy classes of subgroups of the fundamental groups?
| https://mathoverflow.net/users/6031 | Does there exist a classification of covering spaces in algebraic geometry? | Yes, (and of course): The very definition of the étale fundamental group is that it classifies finite étale covers.
Precisely: Let $X$ be a connected scheme and $x$ be a geometric point of $X$. There is by construction an equivalence of categories between finite $\pi\_1^{\mathrm{ét}}(X,x)$--sets and finite étale coverings of $X$, connected coverings corresponding to transitive sets. This property characterises $\pi\_1^{\mathrm{ét}}(X,x)$ up to unique isomorphism.
A universal cover as in topology also exists, although only as a profinite cover.
A fine source for this is T. Szamuely's book "Fundamental groups and Galois groups" (Cambridge 2009).
| 4 | https://mathoverflow.net/users/5952 | 24278 | 15,938 |
https://mathoverflow.net/questions/24270 | 10 | This may be a soft question, but it's just something I thought of one night before sleeping. It's not my field at all, so I am just asking out of curiosity. Has anyone studied the number which is the sum over primes $\sum{ 2^{-p}}$? Its binary expansion (clearly) has a 1 in each prime^th "decimal place", and a zero everywhere else, so, it should be important in number theory I would guess.
| https://mathoverflow.net/users/4528 | A number encoding all primes | Here is Hardy & Wright's answer from "An Introduction to the Theory of Numbers", (5th ed, p344), where they discuss a similar number:
>
> "Although ... gives a 'formula' for the nth prime, it is not a very useful one. To calculate $p\_n$ from this formula, it is necessary to know the value of $a$ correct to $2^n$ decimal places; and to do this, it is necessary to know the values of $p\_1$, $p\_2$, ..., $p\_n$ ... There are a number of similar formulae which suffer from the same defect ... Any one of these formulae (or any similar one) would attain a different status if the exact value of the number $a$ which occurs in it could be expressed independently of the primes. There seems no likelihood of this, but it cannot be ruled out as entirely impossible."
>
>
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| 36 | https://mathoverflow.net/users/5860 | 24283 | 15,940 |
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