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https://mathoverflow.net/questions/37324 | 8 | Let $S$ be a scheme. Let $\mathcal X$ be a algebraic $S$-stack and be $Y$ a $S$-scheme. Let $f:\mathcal X\longrightarrow Y$ be a $S$-morphism of algebraic stacks which is an open embedding (resp. a closed embedding). Is $\mathcal X$ automatically a open(resp. closed) subscheme of $Y$?
| https://mathoverflow.net/users/8736 | Is a sub-stack of a scheme a scheme? | Let $f:\mathcal{X}\rightarrow Y$ be a morphism from an Artin stack to a scheme such that $f$ is an immersion. Then $\mathcal{X}$ is automatically an algebraic space, so we're done by Knutson, Algebraic spaces, II.6.16.
Additions prompted by Brian's comment
-------------------------------------
Assume that $f:\mathcal{X}\rightarrow Y$ is a schematic map, and that $Y$ is a scheme; then $f$ is the pullback of $f$ over the map of schemes $\mathrm{id}\_Y$, so $\mathcal{X}$ must be a scheme. Knutson needs lemma II.6.16 because he doesn't use the now-standard definition of *schematic*, but atlases instead.
When using *immersion*, I always mean $j\circ i$, where $i$ is a closed immersion and $j$ an open one, following EGA I. But I understand that this is not a better choice than the other way round, and that they are only equivalent when the morphism is quasicompact.
| 2 | https://mathoverflow.net/users/307 | 37328 | 24,011 |
https://mathoverflow.net/questions/37329 | 24 | (I've previously asked this question on the sister site [here](https://math.stackexchange.com/questions/3609/is-there-a-natural-model-of-peano-arithmetic-where-goodsteins-theorem-fails), but got no responses).
[Goodstein's Theorem](http://en.wikipedia.org/wiki/Goodstein%27s_theorem) is the statement that every Goodstein sequence eventually hits 0. It turns out that this theorem is unprovable in Peano Arithmetic ($PA$) but is provable in $ZFC$.
I'd like to "discuss" this (both the proof in $ZFC$ as well as the proof that it's impossible in $PA$) in an hour long lecture to a group of grad students (with no assumed background, handwaving is not only allowed, but encouraged). Because of previous talks I've given, I think it will not take *too* long to cover/remind them of the basics of a semester course in first order logic (e.g., the compactness and completeness theorem, etc).
The problem is that the proofs of unprovability I've found (the same as those linked in the Wikipedia article) are rather too difficult for this setting. In a nutshell, I'm looking for the easiest known proof.
For example, I would love a proof of unprovability which works by exhibiting a model of $PA$ in which Goodstein's theorem fails. Such models neccesarily exist by the completeness theorem, since "$PA$ + Goodstein's theorem is false" is consistent.
>
> Has anyone proven the independence of Goodstein's theorem by exhibiting a model of $PA$ where Goodstein's theorem has failed?
>
>
>
In the interest of getting as simple a proof as possible, I'd *love* to see a proof which uses the compactness and completeness ideas - something like showing there is a set $\Sigma = \{\phi\_n\}$ of explicit first order sentences(in a slightly larger language, say) such that
1) for any finite $\Sigma\_0\subseteq PA\cup \Sigma$, the standard model $\mathbb{N}$ models $\Sigma\_0$ and
2) The theory $PA\cup \Sigma$ proves Goodstein's theorem is false.
>
> Is such a proof known? More generally, is there a known proof of the unprovability of Goodstein's theorem which is accessible to someone with only a semester or 2 of logic classes?
>
>
>
Thank you and please feel free to retag as necessary.
| https://mathoverflow.net/users/1708 | Is there a known natural model of Peano Arithmetic where Goodstein's theorem fails? | Your proof strategy via (1) and (2) is impossible. If $PA\cup\Sigma$ proves that Goodstein's theorem is false, then the proof will have finite length, and so there will be some finite $\Sigma\_0\subset\Sigma$ such that $PA\cup\Sigma\_0$ proves that Goodstein's theorem is false. This would imply by (1) that Goodstein's theorem is false in the standard model. But Goodstein's theorem holds in the standard model, as Goodstein proved.
A second point is that you may find that there are no specific "natural" models of PA at all other than the standard model. For example, Tennenbaum proved that there are no computable nonstandard models of PA; that is, one cannot exhibit a nonstandard model of PA so explicitly that the addition and multiplication of the model are computable functions. (See [this related MO question](https://mathoverflow.net/questions/12426/is-there-a-computable-model-of-zfc).) But I do not rule out that there could be natural nonstandard models in other senses.
| 21 | https://mathoverflow.net/users/1946 | 37333 | 24,014 |
https://mathoverflow.net/questions/37305 | 3 | Giving some motivation is hard here, so I'll just ask the question. I want an element $a=(a\_n)\in\ell^1(\mathbb Z)$ such that:
* $\|a\|>1$
* a is power bounded (turn $\ell^1(\mathbb Z)$ into a Banach algebra for the convolution product)
* we have also that $\|a^m\|\_\infty \rightarrow 0$.
I'm sure a clever use of the Fourier transform would work. For example, the third condition is ensured if, letting $f\in C(\mathbb T)$ be the Fourier transform of $a$, we have that $|f|<1$ almost everywhere. The 2nd condition implies that $|f|\leq1$, but of course this isn't if and only if.
| https://mathoverflow.net/users/406 | Odd element of L^1 group algebra of the integers | I'm a bit uncertain what is meant in the third condition: is this the supremum norm of the Gelfand/Fourier transform of $a^m$, or the norm of $a\*a\*\dots\*a$ in $\ell^\infty(\mathbb Z)$?
In the first interpretation, it would clearly suffice to find an element satisfying the first two conditions, and then multiply it by an appropriate scalar betwen $0$ and $1$.
In the second interpretation, as Matthew says, it would suffice to find $a$ satisfying conditions 1) and 2) with the additional property that the Fourier transform of $a$ has modulus $<1$ at all but finitely many points of the unit circle.
Anyway, it turns out that we can find (trigonometric) polynomials satisfying conditions 1) and 2), which should by the previous remarks be enough. I don't know where the first examples are in the literature, but you can find explicit *quadratic* examples in this paper of D. J. Newman:
MR0241980 (39 #3315)
Newman, D. J.
*Homomorphisms of $l\_{+}$.*
Amer. J. Math. 91 1969 37--46
which I personally think is a gem.
| 5 | https://mathoverflow.net/users/763 | 37336 | 24,017 |
https://mathoverflow.net/questions/37356 | 27 | Frucht showed that every finite group is the automorphism group of a finite graph. The paper is [here](http://www.numdam.org/item?id=CM_1939__6__239_0).
The argument basically is that a group is the automorphism group of its (colored) Cayley graph
and that the colors of edge in the Cayley graph can be coded into an uncolored graph that has the same automorphism group.
This argument seems to carry over to the countably infinite case.
Does anybody know a reference for this?
In the uncountable, is it true that every group is the automorphism group of a graph?
(Reference?)
It seems like one has to code ordinals into rigid graphs in order to code the uncountably many colors of the Cayley graph.
| https://mathoverflow.net/users/7743 | Realizing groups as automorphism groups of graphs. | According to the [wikipedia page](http://en.wikipedia.org/wiki/Frucht%27s%5Ftheorem), *every* group is indeed the automorphism group of some graph. This was proven independently in
de Groot, J. (1959), [*Groups represented by homeomorphism groups*](https://eudml.org/doc/160702), Mathematische Annalen 138
and
Sabidussi, Gert (1960), [*Graphs with given infinite group*](https://eudml.org/doc/177088), Monatshefte für Mathematik 64: 64–67.
| 22 | https://mathoverflow.net/users/2233 | 37357 | 24,029 |
https://mathoverflow.net/questions/37349 | 10 | My trouble is best described by the following diagram:
$$ \begin{array}{ccccc} \mathrm{Alt}^k V &\stackrel{\sim}{\rightarrow}& (\Lambda^k V)^\* &\stackrel{\sim}{\rightarrow}& \Lambda^k V^\* \cr
i \downarrow &&&& \downarrow \mathrm{Sk}\cr
\mathrm{Mult}^k V &\stackrel{\sim}{\leftarrow} & (\otimes^k V)^\* & \stackrel{\sim}{\leftarrow} & \otimes^k V^\*
\end{array} $$
The problem is that this diagram is not commutative but let me explain the terminology first.
Here $\mathrm{Alt}^k V$ and $\mathrm{Mult}^k V$ are the spaces of alternating and multilinear $k$-forms, respectively, on the vector space $V$. All the horizontal isomorphisms are canonical. The left vertical arrow is the inclusion of the alternating forms in the multilinear ones. The only "questionable" arrow is the right-hand vertical one ($\mathrm{Sk}$, following the notation in Birkhoff-MacLane "Algebra", Section XVI.10). It is given as an extension of the following alternating map
$$ (\*)\quad (v\_1^\*, \ldots, v\_k^\*) \mapsto {1\over k!} \sum\_{\sigma\in S\_k} (-1)^{\sigma} v\_{\sigma(1)}^\* \otimes \cdots \otimes v\_{\sigma(k)}^\*. $$
If the characteristic is zero (which I assume) then Sk is an inclusion. There are two good things about this inclusion:
First, it is a linear inversion of the canonical projection modulo the graded ideal generated by squares of elements of grade 1, i.e. of $\otimes^k V^\* \rightarrow \Lambda^k V^\*$.
Second, if $\mathrm{Sk'}$ is a map $\otimes^k V^\* \rightarrow \otimes^k V^\* $ which is again obtained by extending a multilinear map (\*), then we have
$$ Sk(a \wedge b) = Sk'(Sk(a)\otimes Sk(b)) $$
(i.e. to learn what $a\wedge b$ is you map both to tensors via $Sk$ and then antisymmetrize their tensor product in the tensor algebra).
So the above argument suggests that $Sk$ is somewhat canonical as well. However, here is a strange situation. Suppose that $e\_1, \ldots, e\_n$ are the basis of $V$. Then consider the alternating 2-form that operates on $V\times V$ as follows:
$$ f(v\_1, v\_2) = e\_1^\*(v\_1) e\_2^\*(v\_2) - e\_1^\*(v\_2) e\_2^\*(v\_1) $$
Its image in $\Lambda^k V^\*$ is $e\_1^\* \wedge e\_2^\*$, and thus under $Sk$ it gets mapped to
$$ {1\over 2} (e\_1^\* \otimes e\_2^\* - e\_2^\* \otimes e\_1^\*) $$
Therefore applying the other two bottom isomorphisms we arrive at a multilinear form that operates on $V\times V$ as follows:
$$ g(v\_1, v\_2) = {1\over 2} (e\_1^\*(v\_1) e\_2^\*(v\_2) - e\_1^\*(v\_2) e\_2^\*(v\_1)) $$
Clearly $g\neq f$ and this is precisely the non-commutativity I was talking about.
Can somebody explain if I made a mistake somewhere? And if not, why then so many physicists happily use "skew-symmetric tensors" and refuse to use "differential forms" and still arrive to the very same answers never loosing coefficient like $1\over 2$?
Thanks in advance! This looks really puzzling to me and I know this is too easy for MO, but I am in a situation much different from the rest of MO having zero mathematicians around to ask such silly questions to. Again, thanks for reading!
**Added later:** As Andrew and Georges point out it is easy to make the diagram commute by either redefining the $\mathrm{Sk}$ without the ${1\over k!}$ or by changing $(\Lambda^k V)^\* \rightarrow \Lambda^k V^\* $ from the $\mathrm{det}$-map to ${1\over k!}\mathrm{det}$. Let me explain why I think why either approach is confusing.
First, redefining the $\mathrm{Sk}$ map as Georges suggests revokes its first property: namely it is no longer a right inverse of the projection $\otimes^k V \to \Lambda^k V$. On the other hand, map $(\Lambda^k V)^\* \rightarrow \Lambda^k V^\* $ determines what we call a wedge-product in the graded algebra $\mathrm{Alt}^\* V$ (since the wedge-product in $\Lambda^\* V^\* $ canonically comes from $\otimes^\* V^\* $ via projection). Therefore, if we are to redefine the meaning of $(\Lambda^k V)^\* \rightarrow \Lambda^k V^\* $ as Andrew proposes, then we have to agree that now
$$ (\* \*)\quad (dx \wedge dy) (\partial\_x, \partial\_y) = {1\over 2},$$
which I think many will find somewhat weird. (Although, it seems things like Stokes theorem do not depend on the agreement (\*\*), right?)
To sum up: if we agree what $\wedge$ means in $\mathrm{Alt}^k$ then this determines the definition of $\mathrm{Sk}$. And thus with the usual definition of $\wedge$ in $\mathrm{Alt}^k$ we unfortunatelly obtain the $\mathrm{Sk}$ which is not the right-inverse of the projection. Am I correct in this?
| https://mathoverflow.net/users/3637 | Alternating forms as skew-symmetric tensors: some inconsistency? | I can't speak to what is actually *used*, particularly what is used by physicists! However, I can try to shed some light on the diagram and the maps in question. In actual fact, there are two diagrams here and you are conflating them. This, simply put, is the source of the confusion. Let me expand (at a bit more length than I intended!) on that.
Firstly, there are too many maps flying around and some are more canonical than others. The most canonical is the identification of $(\bigotimes^k V)^\*$ with $\operatorname{Mult}^k(V)$ since this is by (one of the) definition(s) of the tensor product. So let us start with that. The inclusion $\operatorname{Alt}^k(V) \to \operatorname{Mult}^k(V)$ is probably next in line since it is the inclusion of a subspace. After that, I'd put the map $\bigotimes^k V^\* \to (\bigotimes^k V)^\*$. So, so far we have a diagram:
$$
\begin{array}{ccccc} \operatorname{Alt}^k V \\
i \downarrow \\
\operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
\end{array}
$$
That the horizontal maps are isomorphisms is nice, but only holds for finite dimensional vector spaces so I'm not going to write in the fact that they are isomorphisms. I want to emphasise what's really canonical and what's not.
Now let us consider $(\Lambda^k V)^\*$. We appear to have a canonical map from this to $\operatorname{Alt}^k(V)$ but in fact, we don't. We have a canonical map from this to $(\bigotimes^k V)^\*$ given by:
$$
f(v\_1 \otimes \cdots \otimes v\_k) = f(v\_1 \wedge \cdots \wedge v\_k)
$$
This is dual to the projection map $\bigotimes^k V \to \Lambda^k V$. That projection map is pretty canonical as we usually define $\Lambda^k V$ as a quotient of $\bigotimes^k V$. Taking its dual is a natural thing to do, so this also appears on my list of "canonical maps". Now when we go "down" and "across" we happen to end up in the subspace $\operatorname{Alt}^k(V)$ so we can add a horizontal arrow $(\Lambda^k V)^\* \to \operatorname{Alt}^k(V)$ if we like, but the new map that we add by doing this is one step removed from the really canonical maps so I'm going to leave it out at this stage.
Now we come to $\Lambda^k V^\*$. This is, as for $\Lambda^k V$, defined as a quotient of the tensor product. So we have a projection $\bigotimes^k V^\* \to \Lambda^k V^\*$. This, again, is pretty canonical. So our "canonical" diagram looks like this:
$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* && \Lambda^k V^\* \cr
i \downarrow &&{p\_V}^\* \downarrow&& \uparrow p\_{V^\*}\cr
\operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
\end{array}
$$
At this point, an obvious question is as to whether or not we can fill in the gaps. I've already said that we can in the top-left. Can we in the top-right? That is, is there a map $\Lambda^k V^\* \to (\Lambda^k V)^\*$ making the diagram commute? (Thinking about infinite dimensions says that this is the correct direction.) The answer is: (drum roll) No. And the reason is quite simply that we start in $\bigotimes^k V^\*$ and can choose any element there as our starting point, but would want to end up in the *alternating* part of $(\bigotimes^k V)^\*$.
Okay, now we throw in the Alternator (probably time for another drum roll). The Alternator does what it says on the tin: it alternates stuff. But we have to be careful and ensure that we only apply it to stuff that can genuinely be alternated. So we have an alternator: $\operatorname{Alt} \colon \operatorname{Mult}^k(V) \to \operatorname{Alt}^k(V)$ given by
$$
\operatorname{Alt}(f)(v\_1,\dotsc,v\_k) = \frac{1}{k!} \sum (-1)^{\sigma} f(v\_{1\sigma}, \dotsc, v\_{k\sigma})
$$
The $1/k!$ is to make this a left inverse of the inclusion $\operatorname{Alt}^k(V) \to \operatorname{Mult}^k(V)$.
We also have an alternator $\Lambda^k V \to \bigotimes^k V$ given by:
$$
v\_1 \wedge \dotsb \wedge v\_k \mapsto \frac{1}{k!} \sum (-1)^{\sigma} v\_{1\sigma} \otimes \dotsb v\_{k\sigma}
$$
Again, the multiplier is chosen to ensure that this is a right inverse of the projection map. This is your $Sk$ map. Putting these into a diagram, we get:
$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* && \Lambda^k V^\* \cr
\operatorname{Alt} \uparrow &&{Sk\_V}^\* \uparrow&& \downarrow Sk\_{V^\*}\cr
\operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
\end{array}
$$
Again, the obvious question is: can we fill in the gaps? We can fill in the first one. Indeed, the same filler map works in this diagram as in the last. That was the map $\alpha \colon (\Lambda^k V)^\* \to \operatorname{Alt}^k(V)$ with the property that $i \alpha = \eta {p\_V}^\*$ (where $\eta \colon (\bigotimes^k V)^\* \to \operatorname{Mult}^k(V)$ is the isomorphism). So:
$$
i \alpha (Sk\_V)^\* = \eta {p\_V}^\*(Sk\_V)^\* = \eta (Sk\_V p\_V)^\* = \eta\;\text{and}\; i \operatorname{Alt} \eta = \eta
$$
Thus, as $i$ is an injection, $\alpha (Sk\_V)^\* = \operatorname{Alt} \eta$.
But it's the other gap that's more interesting. Now we *can* fill it in. And the "filler" map is laid out for us already: it's simply follow-the-arrows. If we work it out in detail, it's the following map:
$$
\begin{aligned}
f\_1 \wedge \dotsb \wedge f\_k \mapsto \Big((v\_1 \wedge \dotsb \wedge v\_k) \mapsto & Sk\_{V^\*}(f\_1 \wedge \dotsb \wedge f\_k) \big( {Sk\_V}^\*(v\_1 \wedge \dotsb \wedge v\_k)\big)\Big) \\
&= \frac{1}{k!} \frac{1}{k!} \sum\_\sigma \sum\_\tau (-1)^{\sigma} (-1)^{\tau} f\_{1\sigma}(v\_{1\tau}) \dotsb f\_{k\sigma}(v\_{k\tau})
\end{aligned}
$$
This simplifies considerably by rewriting $f\_{j\sigma}(v\_{j\tau})$ as $f\_{j\rho}(v\_j)$. Then we end up with $k!$ of each term, so we get:
$$
(f\_1 \wedge \dotsb \wedge f\_k)(v\_1 \wedge \dotsb \wedge v\_k) = \frac{1}{k!} \operatorname{det}(f\_i(v\_j))
$$
But notice the factor of $1/k!$ in this!
So to make that right-hand rectangle commute, one of the maps has to have a factor of $1/k!$ in it. It doesn't have to be the top one, but that's the most obvious one since if you modify one of the $Sk$s then you ought to modify the other one - though there's no reason to do so, and in fact this might be what's going on: the physicists are keeping one of the $Sk$s as it is and defining the other one to be suitably scaled so that the upper map is the determinant map. But that's speculation, returning to reality we have a diagram:
$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* &\stackrel{\frac{1}{k!}\operatorname{det}}{\leftarrow} & \Lambda^k V^\* \cr
\operatorname{Alt} \uparrow &&{Sk\_V}^\* \uparrow&& \downarrow Sk\_{V^\*}\cr
\operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
\end{array}
$$
Finally, let's compare this to your original diagram. The key thing to notice is that in my diagrams, I have two vertical maps in one direction and one in the other. In your diagram, you have two vertical maps in the same direction (and are missing the third). But whichever of my diagrams you prefer, one of your maps is going in the wrong direction.
So, in conclusion, the mistake is that your diagram isn't supposed to commute. Rather, there's two commuting diagrams there with some maps from one diagram and some from another.
(I have a feeling that I haven't really answered the question. This was what I wrote out when trying to make sense of the question rather than towards an answer. But I hope that it helps clarify the issue for you.)
| 14 | https://mathoverflow.net/users/45 | 37361 | 24,031 |
https://mathoverflow.net/questions/37360 | 2 | Let $p(n,m)$ be the number of partitions of an integer $n$
into integers $\le m$, we have a well-known asymptotic expression:
For a fixed $m$ and $n\to\infty$,
$$p(n,m)=\frac{n^{m-1}}{m!(m-1)!} (1+O(1/n)) $$
My question is: why the error $O(1/n)$ is independent of $m$?
Or how can it be extended for $m$ growing slowly with $n$?
Please help me to find the answer or the references. Thanks.
| https://mathoverflow.net/users/8933 | complete estimates of the error for a well-known asymptotic expression of partition p(n,m) | I'm not entirely sure of what you are asking, but note that Erdos and Lehner proved [here](https://projecteuclid.org/journals/duke-mathematical-journal/volume-8/issue-2/The-distribution-of-the-number-of-summands-in-the-partitions/10.1215/S0012-7094-41-00826-8.short) that
$$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has
$$p(n,A)=\frac{1}{\prod\_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$
Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.
| 4 | https://mathoverflow.net/users/2384 | 37368 | 24,036 |
https://mathoverflow.net/questions/37370 | 7 | Suppose we have a continuous function $f:R^2\rightarrow R$. I was told of the following remarkable theorem: $f$ can be expressed as the composition of continuous unary functions (that is, functions from $R\rightarrow R$) and addition.
Could anyone give me a reference (or name) for this theorem?
| https://mathoverflow.net/users/8938 | Expressing any f(x,y) using only addition and unary functions? | This is "due in successively more exact forms to Kolmogorov, Arnol'd and a succession of mathematicians ending with Kahane", to quote T.W. Korner on the subject.
I am informed that the proof I met is prepared using:
J.-P. Kahane *Sur le treizieme probleme de Hilbert, le theoreme de superposition de Kolmogorov et les sommes algebriques d'arcs croissants* in the conference proceedings *Harmonic analysis, Iraklion 1978* Springer 1980
G. G. Lorenz, *Approximation of functions* Chelsea Publishing Co. 1986 (First Ed. 1966)
A. G. Vituskin *On the representation of functions by superpositions and related topics* in L'Enseignement Mathématique, 1977, Vol 23, pages 255-320
[This is all from these skeleton notes (no proofs) [here](http://www.dpmms.cam.ac.uk/~twk/CV4.pdf) (Links to pdf; See Chapter 1 and Chapter 11 for references)]
| 5 | https://mathoverflow.net/users/4281 | 37372 | 24,039 |
https://mathoverflow.net/questions/37358 | 10 | This is probably a very naive question from a field that I don't have much background from, but a combination of curiosity and the fact that conceptual questions get very good answers here on MO seemed enough motivation to ask it.
[Solitons](http://en.wikipedia.org/wiki/Soliton) are very interesting objects for a number of special properties they have, such as being "particle like", being [stable](http://terrytao.wordpress.com/2008/02/19/why-are-solitons-stable/) and conserving their geometric features. Some of these properties are related to symmetries related to the notion of solitons, symmetries whose presence is mysterious to me.
My question is, therefore, what is a conceptually satisfying explanation of the non-obvious symmetries of solitons? Where do they come from and/or what causes them? I could not even perform a (fruitful) basic search since I don't know even the key-words necessary for it.
| https://mathoverflow.net/users/2384 | Solitary waves and their symmetries | I have a rather long expository paper in the Bulletin of the American Math. Society called "The Symmetries of Solitons", in which I try to answer just this question. It is aimed at someone without prior familiarity with the field of soliton mathematics, and as well as developing the mathematical tools needed to understand the subject, it also gives an account of its long and interesting history, starting with the first known observation of a soliton by Lord Russell, and describing the remarkable mathematical experiment of Fermi, Pasta, and Ulam, using one of the first electronic computers, an experiment that gave rise to the remarkable modern development of the subject. The article is freely available here (click on the link "Retrieve article in PDF"):
<http://www.ams.org/journals/bull/1997-34-04/S0273-0979-97-00732-5/>
| 17 | https://mathoverflow.net/users/7311 | 37375 | 24,041 |
https://mathoverflow.net/questions/37382 | 4 | Multiplicative intuitionistic linear logic (MILL) has only multiplicative conjunction $\otimes$ and linear implication $\multimap$ as connectives. It has models in symmetric monoidal closed categories.
Compact closed categories are symmetric monoidal closed categories in which every object $A$ has a dual $A^\*$ and $A \multimap B \cong A^\* \otimes B$. Thought of as a resource, $A^\*$ is a debt, owing someone an $A$. Is there a special name for MILL when these conditions hold? ͏ ͏
| https://mathoverflow.net/users/756 | What is the proper name for "compact closed" multiplicative intuitionistic linear logic? | This logic was studied by Masaru Shirahata, ["A Sequent Calculus for Compact Closed Categories"](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.3906). He just calls it "CMLL", but points out it is equivalent in provability to MLL (classical multiplicative linear logic) with tensor and par identified. Note that the direction tensor $\vdash$ par is commonly called "MIX", so this is also MLL + MIX as an isomorphism.
| 3 | https://mathoverflow.net/users/1015 | 37388 | 24,045 |
https://mathoverflow.net/questions/37392 | 23 | When do the notions of totally disconnected space and zero-dimensional space coincide? From what I gather, there are at least three common notions of topological dimension: covering dimension, small inductive dimension, and large inductive dimension. A secondary question, then, would be to what extent and under what assumptions the three different definitions of zero-dimensional coincide. For example, Wikipedia claims that a space has covering dimension zero if and only if it has large inductive dimension zero, and that a Hausdorff locally compact space is totally disconnected if and only if it is zero-dimensional, but I can't track down their source and would like to understand the proofs. I would appreciate any explanation, or a reference, since this is a pretty textbookish question.
| https://mathoverflow.net/users/3544 | totally disconnected and zero-dimensional spaces | Just to agree on notation: A space is zero-dimensional if it is $T\_1$ and has a basis consisting of clopen sets, and totally disconnected if the quasicomponents of all points (intersections of all clopen neighborhoods) are singletons. A space is hereditarily disconnected if no subspace is connected, i.e., if the components of all points
are singletons. (Edit: There seems to be disagreement about the names of these properties.
Often what I call hereditarily disconnected is called totally disconnected and what I call totally disconnected is then called totally separated.)
Note that zero-dimensionality implies Hausdorffness.
Zero-dimensional implies totally disconnected since every point can be separated from every other point by a clopen set.
Totally disconnected implies hereditarily disconnected: given a set $A$ with at least two points, one point is not in the quasi-component of the other and hence the two points can be separated by a clopen set. Hence the set $A$ is not connected.
This shows that the space is hereditarily disconnected.
On the other hand, if $X$ is locally compact and hereditarily disconnected, take $x\in X$
and let $U$ be an open set containing $x$.
By local compactness, find an open neighborhood $V\subseteq U$ of $x$ whose closure $\overline V$ is compact.
In a compact space, components and quasi-components coincide and hence the quasi-component of $x$ in $\overline V$ is $\{x\}$ (you don't need this if you are not interested in hereditary disconnected spaces but just totally disconnected ones). Using compactness again,
there are finitely many clopen subsets $C\_1\dots,C\_n$ of $\overline V$ such that
$x\in C\_1\cap\dots\cap C\_n\subseteq V$. The intersection of the $C\_i$ is closed in $X$ since this intersection is compact.
It is open in $X$ since it is open in $V$ and $V$ is open in $X$.
This shows that the clopen subsets of $X$ form a basis.
Hence $X$ is zero-dimensional.
---
Edit: As suggested by Joseph Van Name, I include a proof that in a compact space the components coincide with the quasi-components.
Let $X$ be a compact space and $x\in X$. The component $C$
of $x$ is the union of all connected subsets of $X$ containing $x$. If $A\subseteq X$ is clopen and $x\in A$, then the component of $x$ is contained in $A$. It follows that the component of $x$ is contained in the quasi-component $Q$ of $x$.
In order to show that the component $C$ and the quasi-component $Q$ coincide,
it is now enough to show that $Q$ is connected.
Observe that $Q$ is closed in $X$ and thus compact.
Now suppose that $Q$ is not connected. Then there are nonempty,
relatively open subsets $A$ and $B$
of $Q$ such that $A\cap B=\emptyset$ and $A\cup B=Q$.
Note that $A$ and $B$ are relatively closed in $Q$ and hence compact.
Hence $A$ and $B$ are closed in $X$.
Two disjoint closed sets in a compact space can be separated by open subsets, i.e.,
there are disjoint open sets $U,V\subseteq X$ such that $A\subseteq U$ and $B\subseteq V$.
We have $$Q=\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}$$
and thus
$$\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}\cap(X\setminus(U\cup V))=\emptyset.$$
By compactness there are finitely many clopen sets $F\_1,\dots,F\_n$ containing $x$ such that
$$F\_1\cap\dots\cap F\_n\cap(X\setminus(U\cup V))=\emptyset.$$
Let $F=F\_1\cap\dots\cap F\_n$.
$F$ is clopen and we have $Q\subseteq F\subseteq U\cup V$.
We have
$$\overline{U\cap F}\subseteq\overline U\cap F=\overline U\cap(U\cup V)\cap F=U\cap F.$$
It follows that $U\cap F$ is clopen in $X$.
We may assume $x\in A$.
Since $B$ is nonempty, there is some $y\in B$.
But now $y\not\in U\cap F$. It follows that $y$ is not in the quasi-component of $x$,
a contradiction.
| 23 | https://mathoverflow.net/users/7743 | 37399 | 24,053 |
https://mathoverflow.net/questions/37383 | 5 | In a recent question of mine I asked whether every infinite group is (isomorphic to) the automorphism group of a graph. The finite case was done by Frucht in 1939.
The first answer to this question pointed out two papers answering my original
question, one by [Sabidussi](http://www.springerlink.com/content/qn32208134273016/)
and one by [de Groot](http://www.springerlink.com/content/w5446363757612w8/).
Reading the 3-page paper by Sabidussi I thought "Wow, these graphs are huge":
Sabidussi realizes a group of size $\kappa$ as the automorphism group of a graph of size $\aleph\_\kappa$.
Indeed, de Groot in his paper notes that every countable group is the automorphism group
of a countable graph, every group of size $\leq 2^{\aleph\_0}$ is the automorphism group of a graph of size $\leq 2^{\aleph\_0}$, and every group of size $\kappa$ is the
the automorphism group of a graph of size $\leq 2^{\kappa}$.
But in general, he doesn't know how large a graph is needed to realize a given group.
Has this issue been resolved? Is there a reason why for a given infinite group $G$ there shouldn't be a graph of size $|G|$ whose automorphism group is isomorphic to $G$?
As I said in my original question, by Frucht's construction (and the constructions of de Groot and Sabidussi) this is related to the question whether there are $\kappa$ many non-isomorphic rigid graphs of size $\kappa$, where a graph is rigid if the identity is the only automorphism.
Is this known? I would guess that there are $2^\kappa$ pairwise non-isomorphic rigid graphs of infinite size $\kappa$, but maybe I am wrong.
| https://mathoverflow.net/users/7743 | Groups as automorphism groups of small graphs and the number of rigid graphs of a given size | It is well-known that every infinite group $G$ can be realized as the automorphism group of a graph of size $|G|$. It is also well-known that for each infinite cardinal $\kappa$, there are $2^{\kappa}$ nonisomorphic rigid graphs of size $\kappa$. For example, both results are easily extracted from Section 4.2 of the following unpublished book:
<http://www.math.rutgers.edu/~sthomas/book.ps>
| 7 | https://mathoverflow.net/users/4706 | 37402 | 24,056 |
https://mathoverflow.net/questions/37384 | 5 | If two Calabi-Yau 3-folds are bi-rational to each other via a Flop , then what is the relation between their mirrors ?
| https://mathoverflow.net/users/5259 | Mirror of Flop? | I assume the question regards the coherent sheaves on these two CY's.
These CY's should be regarded as the "same" complex manifold with two
different choices of complexified symplectic forms ("Kahler form," in
physics terminology).
The mirrors are a "single" symplectic manifold with two different
complex structures on it. There is a curve of complex structures
relating the two.
That's about it. The tricky part is to "parallel transport" the
category of coherent sheaves along this curve, using
a "flat family of categories" defined by stability conditions.
Doing so should provide a preferred isomorphism of the categories.
Examples have been studied, but general statements (like the ones
I have glibly been making) are not proven.
| 7 | https://mathoverflow.net/users/1186 | 37403 | 24,057 |
https://mathoverflow.net/questions/37396 | 4 | The question I want to ask is close to but not exactly what stated in the title:
Fix a language $L$, it is known that a statement $\sigma$ is universal in the language if whenever $M$ satisfies $\sigma $ and $N$ is a substructure of $M$ then $N$ also satisfy $\sigma$. It is also known that a statement $\sigma$ is existential if whenever $M$ satisfies $\sigma $ and $N$ is an extension of $M$ then $N$ also satisfy $\sigma$.
I can not find generalization of these criteria for formulas with more quantifiers. I wonder why this is the case?
| https://mathoverflow.net/users/2701 | Is there a model theoretic realization of the concept of Arithmetical Hierachy? | There is one more well-known equivalence for $\forall \exists$ sentences.
**Theorem** (Chang-Los-Suszko). A theory $T$ is preserved under taking unions of increasing chains of structures if and only if $T$ is equivalent to a set of $\forall \exists$ sentences.
For a proof, see Keisler, "Fundamentals of model theory", *Handbook of Mathematical Logic*, p. 63.
I found a related paper, which is older and doesn't quite answer your question but may be of interest. R. C. Lyndon, "Properties preserved under algebraic constructions", Bull. Amer. Math. Soc. 65 n. 5 (1959), 287-299, [Project Euclid](http://projecteuclid.org/euclid.bams/1183523309)
According to that paper, and MathSciNet, a general solution to your question should be contained in H. J. Keisler, "Theory of models with generalized atomic formulas", J. Symbolic Logic v. 25 (1960) 1-26,
[MathSciNet](http://www.ams.org/mathscinet-getitem?mr=130169), [JStor](http://www.jstor.org/stable/2964333)
| 7 | https://mathoverflow.net/users/5442 | 37406 | 24,060 |
https://mathoverflow.net/questions/35980 | 17 | Suppose that gravity did not follow an inverse-square law, but was instead a central force diminishing
as $1/d^p$ for distance separation $d$ and some power $p$.
Two questions:
1. Presumably the 2-body problem still factors into two independent 1-body problems,
results in planar motion, and can be solved. Have the orbits (the equivalents of elliptical and parabolic orbits for $p=2$) been worked out for other (perhaps specific) values of $p$?
2. In some sense, the 3-body problem for $p=2$ cannot be solved.
Most systems are
choatic; see [this interesting collection](http://faculty.ifmo.ru/butikov/Projects/Collection.html) of Eugene Butikov.
Only a few periodic solutions are known;
see the nice [article](http://www.ams.org/samplings/feature-column/fcarc-orbits1) by Bill Casselman on the discovery of "choreographies."
Is the situation simpler for other values of $p$? Perhaps $p=1$?
References and pointers would be appreciated. Thanks!
**Edit**. Thanks to Agol, Ken, and José. I've now looked at Arnolʹd's *Huygens* and Needham
(but not yet Arnolʹd's *Classical Mechanics*). Indeed, as the commenters say, there is a remarkable
2-body result for $p=1$: the orbits for a linearly attractive force are ellipses, for a linearly repulsive force, hyperbolae. This depends on the Kasner-Arnolʹd theorem stating that for each power law, there is a *dual* power law that maps orbits of one to orbits of the other. Newton proved in *Principia* that elliptical orbits result if and only if the force is inverse-linear or inverse-square. The Kasner-Arnolʹd theorem explains why.
| https://mathoverflow.net/users/6094 | 2- and 3-body problems when gravity is not inverse-square | The answers to question (1) for the 2 body problem are fine, and complete enough.
Regarding (2). The 3 body problem (and N-body) with p =3 is significantly simpler
than with $p \ne 3$. The added simplicity is due to the
occurenc of an additional integral which comes out of the Lagrange Jacobi identity
for the evolution of the total moment of inertia $I$. This identity asserts
that $d^2 I/ dt^2 = 4 H + (4 - 2(p-1)) U$ where $H = K - U$ is the total energy,
with $K$ the kinetic energy and $U$ the NEGATIVE of the total potential energy,
a function which is homogeneous of degree $p-1$. When $p =3$ we get
$d^2 I/ dt^2 = 4H = const.$!.
(The total moment of inertia
is a the squared norm relative to the ``mass inner product' and as such
measures the total size of the system. )
For details on this Lagrange-Jacobi identity and its
use see the first sections of my paper
`Hyperbolic Pants fit a three-body problem', Ergodic Theory and Dynamical Systems, Volume 25, - June 2005, 921-947, which you can also find on my web site
<http://count.ucsc.edu/~rmont/papers/list.html>
or on the arXivs. Also see the references there.
For a study of choreographies with various $p$- potentials see the paper by Fujiwara et al.
`Choreographic Three Bodies on the Lemniscate':J. Phys. A: Math. Gen. 36 (21 March 2003) 2791-2800, available on his web site ( or the ArXivs).
<http://www.clas.kitasato-u.ac.jp/~fujiwara/nBody/nbody.html>
The discoverer of the figure eight, Cris Moore, in his beautiful
2 page paper `Braids and Classical Gravity'
(which Casselman should have a ref. to) found numerically, and argues
convincingly that as one increases $p$ more and more ``braid types'' (and hence choreographies) appear. It is known that
all possible braid types (and so choreography types) occur as soon as $p =2$.
The cases $p \ge 2$ are often called ``strong-force potentials'' and from the
variational perspective are much simpler than $p < 2$ for the reason that
with the strong force potentials all collision paths have infinite action.
This fact regarding action is surprising, since with the strong force condition in
force it seems that almost all bounded solutions end in collision. This "seems" is a
theorem for the 2-body problem, and for the negative energy three body problem
when $p=2$.
| 17 | https://mathoverflow.net/users/2906 | 37413 | 24,064 |
https://mathoverflow.net/questions/37037 | 2 | Let $H$ be a real Hilbert space with
complexification $H\_{\mathbb{C}}$. We denote by $\mathfrak{F}$ the antisymmetric Fock space
over $H\_\mathbb{C}$ ("fermions"). A creation operator is denoted by $c(f)$.
I need a reference for the calculus of
$$
\langle\,\Omega ,\big(c(f\_1)+c(f\_1)^\*\big)...\big(c(f\_{2k})+c(f\_{2k})^\*\big)\Omega\,\rangle\_{\mathfrak{F}}
$$
where $\Omega$ the unit vector, called vacuum.
Thank you.
| https://mathoverflow.net/users/5210 | Reference for Wick product | The answer is provided by the article
* Edward G. Effros and Mihai Popa, *Feynman diagrams and Wick products associated with q-Fock space*, PNAS 100 (15) (2003) 8629-8633, <https://doi.org/10.1073/pnas.1531460100>
However, the authors work in the context of $q$-Fock space. I does not know if there exists an older paper which provides the answer in the less general context of antisymmetric Fock space (i.e. q=1).
| 1 | https://mathoverflow.net/users/5210 | 37414 | 24,065 |
https://mathoverflow.net/questions/37376 | 24 | I'm wondering what the statement is that one has to prove for the Millenium Problem "Quantum Yang-Mills Theory".
According to the official article, it is required to show that for every simple Lie group G there exists a YM quantum field theory for G with a mass gap. Finding a quantum field theory amounts to finding a Hilbert space H, a representation of the restricted Lorentz group by unitary transformations of H and operator valued tempered distributions $\varphi\_1,...,\varphi\_m$ which satisfy density conditions, transformation properties under the Poincare-group, (anti-)commutativity of field operators for test functions of space-like separated supports, an asymptotic completeness property and existence of a unique vacuum state, the field operators acting on this vacuum state span (a dense subspace of) H.
A quantum field theory has a mass gap if the spectrum of the energy operator is contained in $\{0\} \cup [a,\infty)$ for $a>0$.
Now these are straightforward definitions, but what turns a quantum field theory into a YM theory for a group? I know of classical YM theory which gives a Lagrangian and thus equations of motion for the curvature components (fields). Does that mean that $\varphi\_1,...,\varphi\_n$ have to satisfy these equations? In what way? I could not find any reference for this. Is this millenium problem a mathematical statement at all?
| https://mathoverflow.net/users/8794 | Statement of Millenium Problem: Yang-Mills Theory and Mass Gap | The term "Yang-Mills theory" in the mass gap problem refers to a particular QFT. It is believed that this QFT (meaning its Hilbert space of states and its observable operators) should be defined in terms of a measure on the space of connections on $\mathbb{R}^4$; roughly speaking, the moments of this measure are the matrix elements of the operator-valued distributions. (People also use the term "gauge theory" to refer to any QFT, like QCD, which has a Yang-Mills sub-theory.)
The mass gap problem really has two aspects: First, one has to construct an appropriate measure $d\mu$ on some space of connections. Then, one has to work out which functions on the space of connections are integrable with respect to this measure, and show that the corresponding collection of operators includes an energy operator (i.e. a generator of time translations) which has a gap in its spectrum.
You'll have to read the literature to really learn anything about this stuff, but I can make a few points to help you on your way. Be warned that what follows is a caricature. (Hopefully, a helpful one for people trying to learn this stuff.)
About the Measure:
First, the measure isn't really defined on the space of connections. Rather, it should be defined on space $\mathcal{F}$ of continuous linear functionals on the space $\mathcal{S}$ of smooth rapidly-vanishing $\mathfrak{g}$-valued vector fields on $\mathbb{R}^4$, where $\mathfrak{g}$ is the Lie algebra of the gauge group $G$. The space ;$\mathcal{F}$ contains the space of connections, since any connection on $\mathbb{R}^4$ can be written as $d$ plus a $\mathfrak{g}$-valued $1$-form and paired with a vector field via the Killing form, but it also has lots of more "distributional" elements.
We're supposed to get $d\mu$ as the "infinite-volume/continuum limit" of a collection of regularized measures. This means that we are going to write $\mathcal{S}$ as an increasing union of chosen finite-dimensional vector spaces $\mathcal{S}(V,\epsilon)$; these spaces are spanned by chosen vector fields which have support in some finite-volume $V \subset \mathbb{R}^4$ and which are slowly varying on distance scales smaller than $\epsilon$. (You should imagine we're choosing a wavelet basis for $\mathcal{S}$.) Then we're going to construct a measure $d\mu\_\hbar(V,\epsilon)$ on the finite dimensional dual vector space $\mathcal{F}(V,\epsilon) = \mathcal{S}(V,\epsilon)' \subset \mathcal{F}$; these measures will have the form $\frac{1}{\mathcal{Z}(V,\epsilon,\hbar)} e^{-\frac{1}{\hbar}S\_{V,\epsilon,\hbar}(A)}dA$. Here, $dA$ is Lesbesgue-measure on $\mathcal{F}(V,\epsilon)$, and $S\_{V,\epsilon,\hbar}$ is some discretization of the Yang-Mills action, which is adapted to the subspace $\mathcal{F}(V,\epsilon)$. (Usually, people use some version of Wilson's lattice action.)
Existence of the Yang-Mills measure means that one can choose $S\_{V,\epsilon}$ as a function of $V$,$\epsilon$, and $\hbar$ so that the limit $d\mu\_\hbar$ exists as a measure on $\mathcal{F}$ as $vol(V)/\epsilon \to \infty$. We also demand that the $\hbar \to 0$ limit of $d\mu\_\hbar$ is supported on the space of critical points of the classical Yang-Mills equations. (We want to tune the discretized actions to fix the classical limit.)
About the Integrable Functions:
Generally speaking, the functions we'd like to integrate should be expressed in terms of the "coordinate functions" which map the $A$ to $A(f)$, where $f$ is one of the basis elements we used to define the subspaces $\mathcal{S}(V,\epsilon)$. You should imagine that $f$ is a bump vector field, supported near $x \in \mathbb{R}^4$ so that these functions approximate the map sending a $\mathfrak{g}$-valued $1$-form to the value $A\_{i,a}(x)$ of its $(i,a)$-th component.
There are three warnings to keep in mind:
First, we only want to look at functions on $\mathcal{F}$ which are invariant under the group of gauge transformations. So the coordinate functions themselves are not OK. But gauge invariant combinations, like the trace of the curvature at a point, or the holonomy of a connection around a loop are OK.
Second, when expressing observables in terms of the coordinate functions, you have to be careful, because the naive classical expressions don't always carry over. The expectation value of the function $A \mapsto A\_{i,a}(x)A\_{j,b}(y)$ with respect to $d\mu\_\hbar$ (for $\hbar \neq 0$) is going to be singular as $x \to y$. This is OK, because we were expecting these moments to define the matrix elements of operator-valued distributions. But it means we have to be careful when considering the expectation values of functions like $A \mapsto A(x)^2$. Some modifications may be required to obtain well-defined quantities. (The simplest example is normal-ordering, which you can see in many two-dimensional QFTs.)
Finally, the real problem. Yang-Mills theory should confine. This means, very very roughly, that there are some observables which make sense in the classical theory but which are not well-defined in the quantum theory; quantum mechanical effects prevent the phenomena that these observables describe. In the measure theoretic formulation, you see this by watching the expection values of these suspect observables diverge (or otherwise fail to remain integrable) as you approach the infinite-volume limit.
About the Operators:
In classical Yang-Mills theory, the coordinate observables $A \mapsto A\_{i,a}(x)$ satisfy equations of motion, the Yang-Mills equations. Moreover, in classical field theory, for pure states, the expectation value of a product of observables $\mathcal{O}\_1\mathcal{O}\_2...$ is the product of the individual expectation values. In quantum YM, the situation is more complicated: coordinate observables may not have be well-defined, thanks to confinement, and in any case, observables only satisfy equations of motion in a fairly weak sense: If $\mathcal{O}\_1$ is an expression which would vanish in the classical theory thanks to the equations of motion, then then the expectation value of $\mathcal{O}\_1\mathcal{O}\_2...$ is a distribution supported on the supports of $\mathcal{O}\_2...$.
| 23 | https://mathoverflow.net/users/35508 | 37428 | 24,072 |
https://mathoverflow.net/questions/37417 | 2 | Consider the Banach $^\* $-algebra $\ell^1(\mathbb Z)$ with multiplication given by convolution and involution given by $a^\*(n)=\overline{a(-n)}$.
I would like to find nice necessary and sufficient conditions for an element $b\in\ell^1(\mathbb Z)$ to be positive, that is, to be of the form $a^\* \* a$ for some $a\in\ell^1(\mathbb Z)$.
By now, I have found two necessary conditions. Namely, if $b\in\ell^1(\mathbb Z)$ is positive, then $$b(-n)=\overline{b(n)}$$ and $$\lvert b(n)\rvert\leq b(0)$$ for every $n\in\mathbb Z$.
**Edit:** As t3suji states in his comment below both conditions follow from the more general fact that $a$ is a [positive-definite function](http://en.wikipedia.org/wiki/Positive_definite_function).
>
> Question: Is this condition also sufficient for positivity? If not, what to I have to add?
>
>
>
Good references would also be great.
Motivation: In the end I want to investigate the (failure of) the Gelfand–Naimark theorem for the above non-C\*-algebra.
| https://mathoverflow.net/users/1291 | Characterisation of positive elements in l¹(Z) | Although I am not sure that answering this question will help all that much with your original motivating question/problem, I may as well post a link to Bochner's theorem (see [these remarks on Wikipedia](http://en.wikipedia.org/wiki/Positive-definite_function)).
The passage I have in minds says:
>
> Positive-definiteness arises naturally in the theory of the Fourier transform; it is easy to see directly that to be positive-definite is a necessary condition ... to be the Fourier transform of a function $g$ on the real line with $g(y) \geq 0$.
>
>
> The converse result is Bochner's theorem, stating that a continuous positive-definite function on the real line is the Fourier transform of a (positive) measure.
>
>
>
| 2 | https://mathoverflow.net/users/763 | 37429 | 24,073 |
https://mathoverflow.net/questions/37377 | 6 | Hello,
I do have a collection of trees (mind maps, actually) and want to formally describe this collection of trees.
My first question: how can I describe a tree? Are there any metrics to express how a tree looks like? Of course, it has a height and it might be balanced or not. But are there any other metrics (e.g. to say how balanced a tree is)?
The second question: How can I describe a collection of different trees. Do you know of any metrics that make sense? Of course I could calculate the mean height etc. but are there some more sophisticated metrics?
Best,
Thomas
| https://mathoverflow.net/users/8940 | How to describe a tree? (depth, degree, balance, ... what else?) | Mind maps, as desribed on their wikipedia page, are a way of mapping or *placing a graph structure* onto a collection of data.
Items can be linked together with directed edges and with a label on the edge describing the relationship. Each data item at a vertex can taken on multiple tags (coloring) to describe their type.
If you have a Linux distribution that has the KDE (Kommon Desktop Environment, as opposed to the CDE Common Desktop Environment in Solaris) environment, you can see an implementation of **mind maps** in a note taking and note organizing software package
>
> **BasKet Notepads** <http://basket.kde.org/>
>
>
>
The wikipedia page for BasKet is rather sparse and uninformative at <http://en.wikipedia.org/wiki/BasKet_Note_Pads> and does not really describe the full potential of the note organizing software.
The Mind Map software seems to be made for "rapid collaborating" and "brainstorming", very fuzzy words that seem to match the warm soft fuzzyness of the software. I have played with it, but it is poorly structured and not amazingly useful for organizing my research information.
It is, however, very useful for laying out quick hierarchical diagrams or tree diagrams. It does not easily allow one to export the graph structure in a useful and easy to reuse file format.
As for describing the structure: just look at it as a graph. Do you have any one-way oriented relationships on it? (e.g. links such as PARENT-OF, REFERS-TO, DERIVED-FROM, COMES-AFTER) If so, then you have a **directed graph**, otherwise you have an **undirected graph**.
How many elements are there? That is the **number of vertices**.
How many relations/links are there? That is the **number of edges**.
How many edges are there connecting each vertex? The number of edges on a vertex is the **degree of the vertex**. In a directed graph, you can have **out-degree** for outward-linking edges and **in-degree** for inward linking edges. What is the fewest number of edges? What is the largest number of edges on a vertex?
Draw a histogram of how many vertices have zero edges (free disconnected vertices), how many have one, two, etc. List them in order and you have the **degree sequence** of the graph.
Look at all of the elements; can you reach them all from one to another by following edges? Then you have a **single connected graph**. If you have separate islands that are not linked, count the number of islands as the **number of components** in the graph. Recursively describe each of these islands as listed above.
This is a simple way to. start. Treat the diagram as a graph and describe it in good detail. Could you please provide more details about what you are doing, or perhaps an example structure?
| 2 | https://mathoverflow.net/users/8952 | 37436 | 24,078 |
https://mathoverflow.net/questions/36486 | 20 | Where can I find a clear exposé of the so called "standard reduction to the local artinian (with algebraically closed residue field", a sentence I read everywhere but that is never completely unfold?
---
EDIT: Here, was a badly posed question.
| https://mathoverflow.net/users/8736 | Standard reduction to the artinian local case? | Dear Workitout: The list of comments above is getting unwieldy, so let me post an answer here, now that you have finally identified 1.10.1 in Katz-Mazur as (at least one) source of the question. As I predicted, you'll see that the basic technique to be used adapts to many other settings, and that it is very hard to formulate a "meta-theorem" to cover all cases. The only sure-fire method I know is to read all of sections 8, 9, 11, 12 in EGA IV$\_3$ and sections 17 and 18 in EGA IV$\_4$, and then it becomes really routine. Maybe there's a better method (well I can think of one, but I won't say it here). I will use the terminology and notation around 1.10.1 from Katz-Mazur without explanation below.
Being a "full set of sections" of $Z/S$ is something which is sufficient to check using the constituents of a single open affine cover of $S$, and also a finite generating set of the coordinate ring of $Z$ over each such open. Thus, by working Zariski-locally on $S$ we may assume $S = {\rm{Spec}}(R)$ and that both sides of (1) in KM 1.10.1 have $R$-free coordinate rings (when viewed as finite $R$-schemes), and likewise for their sum (as effective Cartier divisors). Now the assertions (1) and (2) in KM 1.10.1 are identities among finitely many elements of some finite free $R$-modules. For instance, (1) asserts that certain elements in a finite free $R$-module have vanishing image in a finite free $R$-module quotient. This is all now a bunch of identities among finitely many elements of $R$.
OK, finally we come to the part with a real idea. Consider the subring $R\_0$ of $R$ generated over $\mathbf{Z}$ generated by those finitely many elements. It is noetherian. Now unfortunately your initial algebro-geometric setup over $R$ (the smooth separated finite-type curve $C$, the various effective relative Cartier divisors, etc.) probably does not arise via base change from the exact same setup (including flatness properties!) over $R\_0$. But that doesn't matter: what would really be swell is if *some* noetherian subring of $R$ which *contains* $R\_0$ permits such a descent of the situation. Now express $R$ as the direct limit of its finitely generated $R\_0$-subalgebras (all of which are noetherian). Does the entire situation descend to one of those? If it did, we'd be in great shape, since it would then suffice to solve the problem in the case of a noetherian base ring (as that would imply the result over $R$ by suitable base change of the descent from the noetherian subring back up to $R$).
How to implement this strategy of reduction to the noetherian case (after which we'll need to deal with the passage to artin local base with algebraically closed residue field)? OK, so that's where we are all very fortunate that Grothendieck actually wrote out the entire formalism in total detail to handle basically every such situation one could ever want to handle. So it becomes kind of a game in finding the references in EGA (which I admit is hard to do if one doesn't know where to look, but is really easy if one has read the right parts). For your particular situation with some smooth separated curves and some relative effective Cartier divisors, etc., the results you need are: EGA IV$\_3$ 8.3.4, 8.9.1, 8.10.5 (e.g., (v)), 9.2.6.1, 11.2.6(ii), and IV$\_4$ 17.7.9.
I'm not going to say more about how you combine those references here; that is where I again remind you of your pseudonym.
OK, now $R$ is noetherian (even finite type over $\mathbf{Z}$, which is very useful extra stuff to have for other arguments with excellence later in life), and you're trying to prove some finite collection of identities among elements of $R$. To do that it suffices to check in the local rings of $R$, so you can assume $R$ is local. Now a pair of elements of a local noetherian ring are equal if and only if they have equal images in each artinian quotient (Krull intersection theorem). So it suffices to prove the general result over arbitrary artin local rings (really just the artinian quotients of finitely generated $\mathbf{Z}$-algebra, so there's no set-theoretic quantification nonsense going on). Now $R$ is artin local. To check an identity in a ring it suffices to do so in a faithfully flat extension ring. So finally we haul out EGA 0$\_{\rm{III}}$ 10.3.1 to find a faithfully flat artin local extension with an algebraically closed residue field. That's it!
I presume you can now see why anyone who knows how to fill in such arguments never actually writes them out in papers: it is much simpler to say "by standard limit methods from EGA IV$\_3$, sections 8,9, etc." (maybe even to be a bit more specific, as K-M are at the beginning of their 1.8.1), and to trust that the reader will pick up the clue that they should read those parts of EGA if they want to understand what is going in such arguments for themself. It is bad when people don't at least mention the relevance of sections 8, 9, etc., but things could be worse (e.g., Grothendieck could have not written EGA, leaving stuff in a complete mess reference-wise).
| 28 | https://mathoverflow.net/users/3927 | 37450 | 24,087 |
https://mathoverflow.net/questions/37458 | 4 | The celebrated Big Theorem of Picard's is that, in every open set containing an essential singularity of a function $f(z)$, $f(z)$ takes on every value (except for at most one) of $\mathbb{C}$ infinitely often.
Now - is the converse true? Is this a way to characterize the existence of an essential singularity of a function?
For example, if you're given a non-constant function $f(z)$ that is holomorphic on some open set $\Omega$, and you know that there is an accumulation of 0's towards some point $x$ on the boundary of $\Omega$, then do you know that there must be an essential singularity at $x$?
| https://mathoverflow.net/users/5534 | Converse of Picard's Big Theorem? | If the boundary point is not an isolated singularity, then you can't say it is an essential singularity, so the answer to your final question is no. It is quite possible that $f$ cannot be extended to a larger domain that contains a punctured neighborhood of $x$. To be quasi-explicit, take a holomorphic function whose domain of holomorphy is $\Omega$ and multiply by a holomorphic function defined everywhere in the plane except with an essential singularity at $x$.
If $f$ has an isolated singularity at $x$, then yes, this characterizes essential singularities. $f$ goes to $\infty$ at a pole, so in particular is nonzero in a neighborhood of the pole. If the singularity at $x$ is removable, then either $f$ goes to a nonzero value, in which case it is nonzero in a neighborhood of $x$, or $f$ goes to 0. In the latter case, $f$ could be extended to a holomorphic function on a domain containing $x$, so if $x$ were a limit point of the zero set of $f$, then $f$ would be identically zero. And you could easily adjust this to nonzero cases.
You don't need anything near the strength of Picard's theorem. In fact, that is what is so amazing about Picard's theorem: Either $f$ has really nice behavior near an isolated singularity, or it maps everywhere except possibly one point in each neighborhood of the singularity.
| 8 | https://mathoverflow.net/users/1119 | 37459 | 24,094 |
https://mathoverflow.net/questions/37438 | 18 | I recently read the following "open problem" titled "Pennies on a carpet" in "An Introduction To Probability and Random Processes" by Baclawski and Rota (page viii of book, page 10 of following pdf), found here: <http://www.ellerman.org/Davids-Stuff/Maths/Rota-Baclawski-Prob-Theory-79.pdf>
"We have a rectangular carpet and an indefinite supply of perfect pennies. What is the probability that if we drop the pennies on the carpet at random no two of them will overlap? This problem is one of the most important problems in statistical mechanics. If we could answer it we would know, for example, why water boils at 100C, on the basis of purely atomic computations. Nothing is known about this problem.”
I was wondering if this problem goes by a more popular name and whether or not some form of progress has been made on it. In particular, references would be highly appreciated!
| https://mathoverflow.net/users/934 | Pennies on a carpet problem | This is the two-dimensional hard spheres model, sometimes called hard discs in a box.
See Section 4 of Persi Diaconis's recent survey article, [The Markov Chain Monte Carlo Revolution](http://www-stat.stanford.edu/~cgates/PERSI/papers/MCMCRev.pdf). The point here is that even though it very hard to sample a random configuration of nonoverlapping discs by dropping them on the carpet (because the probability of success is far too small for any reasonable number of discs), but it is nevertheless possible to sample a random configuration via Monte Carlo.
| 10 | https://mathoverflow.net/users/4558 | 37465 | 24,099 |
https://mathoverflow.net/questions/13649 | 6 | I've been exposed to various problems involving infinite circuits but never seen an extensive treatment on the subject. The main problem I am referring to is
>
> Given a lattice L, we turn it into a circuit by placing a unit resistance in each edge. We would like to calculate the effective resistance between two points in the lattice (Or an asymptotic value for when the distance between the points gets large).
>
>
>
I know of an approach to solve the above introduced by Venezian, it involves superposition of potentials. An other approach I've heard of, involves lattice Green functions (I would like to read more about this). My first request is for a survey/article that treats these kind of problems (for the lattices $\mathbb{Z}^n$, Honeycomb, triangular etc.) and lists the main approaches/results in the field.
My second question (that is hopefully answered by the request above) is the following:
I noticed similarities in the transition probabilities of a Loop-erased random walk and the above mentioned effective resistances in $\mathbb{Z}^2$. Is there an actual relation between the two? (I apologize if this is obvious.)
| https://mathoverflow.net/users/2384 | Infinite electrical networks and possible connections with LERW | The book by Peres and Lyons, freely available here <http://php.indiana.edu/~rdlyons/prbtree/prbtree.html>, should give you much information at least for the probability part of the question.
| 3 | https://mathoverflow.net/users/4961 | 37471 | 24,102 |
https://mathoverflow.net/questions/37468 | 4 | Suppose $M$ is a parallelizible Riemannian manifold with metric tensor $g\_x(\cdot,\cdot)$ Let $F\_x(\cdot,\cdot)$ denote the flat metric on $M$ that we get from parallelization. Is it true that there exist c, C such that for any $x$ and $v$ in $T\_xM$, $cg\_x(v,v)\leq F\_x(v,v)\leq Cg\_x(v,v)$?
| https://mathoverflow.net/users/nan | Lipschitz equivalence of Riemannian metrics | As Dmitri says any two Riemannian metrics on a **compact** manifold are Lipschitz equivalent. The proof is quite simple.
Consider $g$ and $h$ two metrics on $M$, Let $UM$ be the unit tangent bundle, since $M$ is compact, $UM$ is compact. Then you see that $f:UM\to \mathbb{R}$ defined by $f(x)=\frac{g(x,x)}{h(x,x)}$ is continuous and strictly positive. By compactness, it is bounded above and below by positive constants.
| 8 | https://mathoverflow.net/users/8887 | 37477 | 24,107 |
https://mathoverflow.net/questions/37418 | 2 | Recall that a kernel conditionaly of negative type on a set $X$ is a map $\psi:X\times X\rightarrow\mathbb{R}$ with the following properties:
1) $\psi(x,x)=0$
2) $\psi(y,x)=\psi(x,y)$
3) for any elements $x\_1,...x\_n$ and all real numbers $c\_1,...,c\_n$, with $c\_1+...+c\_n=0$, the following inequality holds:
$$
\sum\_{i=1}^{n}\sum\_{j=1}^{n}
c\_ic\_j\psi(x\_i,x\_j)\leq 0.
$$
Let $G$ be a discrete group.
Recall that a function $G\rightarrow \mathbb{R}$ is conditionally of
negative type if the kernel $\psi$, defined by $\psi(g,h)= \psi(h^{−1}g)$ is conditionaly of negative type.
>
> Does there exist class of discrete groups which admit an explicit description of functions which are conditionaly of negative type?
>
| https://mathoverflow.net/users/5210 | description of functions of conditionally negative type on a group | If you add the condition that Jesse mentioned in the comment above, it is a theorem that such functions are always realized from an affine isometric actions of the group $G$ on a Hilbert space. More precisely, suppose $G$ acts continuously by affine isometries on a Hilbert space $H$. Now, define
$$ \psi (g)= \| g \cdot x-x \|^2 $$
for an arbitrary point $x$ in the affine space $H$. First, it is easy to see that such functions are always negative definite. More difficult is to show that any negative definite function can be obtained in this way.
You can read more about them in Kazhdan's Property (T) by Bekka, de la Harpe and Valette.
| 4 | https://mathoverflow.net/users/3635 | 37480 | 24,110 |
https://mathoverflow.net/questions/37449 | 11 | Is it possible to cover $Z^\infty$ (the infinite direct sum of $Z$'s with the $l\_1$-metric) by a finite set of collections of subsets $U^0,...,U^n$ such that each collection $U^i$ consists of uniformly bounded sets $U\_j^i$ that are 4-disjoint (the distance between any two subsets $U\_j^i$, $U\_k^i$ in each $U^i$ is at least 4)? The motivation is here: <https://arxiv.org/abs/1008.3868> .
| https://mathoverflow.net/users/nan | covers of $Z^\infty$ | The answer is NO even if we replace $4$ by $3$.
Let me sketch a proof. This is based upon the following lemma.
Lemma. Fix $S>0$ and for an integer $k$ conisder in $\mathbb Z^k$ sets $X$ of diameter at most $S$. Denote by $Vol(X)$ the number of points in $X$ and denote by $X1$ the set of points of distance at most $1$ from $X$. Now let $\delta(S,k)$ be the supremum over all $X$ of diameter at most $S$ of the ratio:
$$r(S,k)=sup\_{X\subset \mathbb Z^k}\frac{Vol(X)}{Vol(X1)}.$$
I claim that for a fixed $S$, $lim\_{k\to \infty}r(S,k)=0$.
Let us skip the proof of the lemma and instead deduce the claim. Suppose by contradiction that the answer is positive. Then for every $k$ we will get a solution to the problem in $\mathbb Z^k$ with the fixed number of sets ($U^0,...,U^n$) such that each $U^i\_j$ is of the diameter at most $S$. Now, chose such $k$ that $r(S,k)<\frac{1}{2n}$ and let us deduce the contradiction.
From Lemma it follows that the supremum of asimptotic density of each set $U^i$ in $\mathbb Z^k$ is less than $\frac{1}{n+1}$. Indeed, since the distance between different components of $U^i$ is $4$, every point of $U^i1$ that does not belong to $U^i$ is on distance one from at most one component of $U^i$. And lemma gives us the inequality (that should be understood as assymptotic in $\mathbb Z^k$)
$$Vol(U^i)<\frac{1}{2n} Vol(U^i1)\le \frac{1}{2n}Vol(\mathbb Z^k)$$
Hence $\mathbb Z^k$ can not be covered by $U^0,...,U^n$.
It is clear where this proof breakes if we conisder $2$-disjoint sets. In this case one point of $U^i1$ can be on distance $1$ to many components of $U^i$ and the above inequality will not hold. But for $3$-disjoint sets this works.
As for the proof of the lemma, I think, it should be rather standard.
| 15 | https://mathoverflow.net/users/943 | 37481 | 24,111 |
https://mathoverflow.net/questions/37423 | 12 | A smooth structure on a manifold $M$ can be given in the form of a sheaf of functions $\mathcal{F}$ such that there is an open cover $\mathcal{U}$ of $M$ with every $U\in \mathcal{U}$ isomorphic (along with $\mathcal{F}|\_U$) to an open subset $V$ of $\mathbb{R}^n$ (along with $\mathcal{O}|\_V$, where $\mathcal{O}$ is the sheaf of smooth functions on $\mathbb{R}^n$). I think we might also need to say that this satisfies a smooth-coordinate-change axiom, although maybe that's already tied up in the definition of a sheaf. In any case, here is my question:
>
> Given two smooth manifolds
> $(M,\mathcal{F})$ and
> $(N,\mathcal{G})$, is there an easy
> way to write the sheaf of functions on
> $M\times N$ without reference to
> coordinate neighborhoods?
>
>
>
I'm wondering this because in one of my classes we defined smooth manifolds in this way (and we defined analytic and holomorphic manifolds similarly). It seems like some people are very fond of this alternative definition because it doesn't refer to an atlas, which at first seems like it's an inherent part of the structure of the manifold. So okay fine, everyone loves a canonical definition. However, this is only going to be useful as long as we can tell our whole story in this canonical language. In class, the only way the professor was able to give the sheaf on the product was by breaking down and using coordinates. (Admittedly, he was on the spot and presumably unprepared for the question.)
This also suggests the broader, more open-ended question:
>
> Are there longer-run advantages to the above definition (compared to the usual definition involving an atlas and perhaps a maximal atlas)?
>
>
>
| https://mathoverflow.net/users/303 | Is there an easy way to describe the sheaf of smooth functions on a product manifold? | Smooth manifolds are affine, thus the sheaf of smooth functions is determined by its global sections.
Now C^∞(M×N)=C^∞(M)⊗C^∞(N).
The tensor product here is the projective tensor product
of complete locally convex Hausdorff topological algebras.
| 9 | https://mathoverflow.net/users/402 | 37485 | 24,113 |
https://mathoverflow.net/questions/37483 | 5 | James' theorem states that a Banach space $B$ is reflexive iff every bounded linear functional on $B$ attains its maximum on the closed unit ball in $B$.
Now I wonder if I can drop the constraint that it is a ball and replace it by "convex set". That is, I want to know if every bounded linear functional on a reflexive Banach space $B$ attains its maximum on a closed and bounded convex set in $B$.
By Pietro's answer this is known to be true. Is the maximum unique? In optimization by vector space methods it is know that this is true if the set is the closed ball. This was actually my biggest question since I want to show a optimization problem has a unique solution.
| https://mathoverflow.net/users/5295 | Maximum on unit ball (James' theorem). | A closed and bounded convex set of a reflexive Banach is w\* compact, hence any bounded linear functional does attain its maximum and minimum there. On the other direction, the presence of a closed bounded convex set on which all bounded linear functionals have their maximum, of course, says nothing on the reflexivity of the space (the convex could be a single point).
| 9 | https://mathoverflow.net/users/6101 | 37487 | 24,115 |
https://mathoverflow.net/questions/37497 | 8 | $A$ a commutative Noetherian domain, $M$ a finitely generated $A$-module. How can I show that the kernel of the natural map $M\rightarrow M^{\*\*}$, where $ M^{ \* \*}$ is the double dual (with respect to $A$), is *the* torsion submodule of $M$?
I do know that in this situation torsionlessness coincides with torsion-freeness. According to Auslander this result is ``well-know'' but I can't seem to prove it or find any reference on this.
| https://mathoverflow.net/users/5292 | Torsion submodule | Let $K$ be the fraction field of $A$. Then there is a natural isomorphism
$M^\*\otimes\_A K \cong (M\otimes\_A K)^\*$ (where the dual on the left is the $A$-dual,
and on the right is the $K$-dual).
Thus the double dual map $M \to M^{\* \*}$ becomes an isomorphism after tensoring with $K$
over $A$, and hence its kernel is contained in the kernel of the natural map $M \to
K\otimes\_A M,$ which shows that its kernel is torsion. On the other hand, clearly
the torsion submodule of $M$ is contained in this kernel, since $M^{\* \*}$ is torsion free.
This proves the result.
| 12 | https://mathoverflow.net/users/2874 | 37501 | 24,125 |
https://mathoverflow.net/questions/37502 | 17 | Let $\kappa$ be a cardinal (of uncountable cofinality). A subset $S \subseteq \kappa$ is called stationary if it intersects every club, i.e. closed unbounded subset of $\kappa$. Now my question is basically just: Why do we care about stationary sets? I know some statements, which are independent from ZFC, for example the diamond principle, which involve them, but what was or is the motivation for studying them? Please don't just give an overview of the results, because they can be found in every textbook on cardinals.
Also, what is your intuition for stationary sets? I think clubs are very easy to grasp; they just contain big enough ordinals. Perhaps stationary sets follow the same idea, but in a "second order"; they contain enough big enough ordinals?
| https://mathoverflow.net/users/2841 | What is the idea behind stationary sets? | Some intuition might be given by the following informal analogy with measure theory: if we have a measure space of measure 1, then club sets are analogous to subsets of measure 1, while stationary sets are analogous to sets of positive measure.
In other words, club sets contain "almost all" ordinals, while stationary sets contain "a positive proportion" of them.
| 16 | https://mathoverflow.net/users/51 | 37503 | 24,126 |
https://mathoverflow.net/questions/37510 | 0 | Let $M$ be a smooth manifold. Let's call $M$ quasi-seperated if $M$ has the following property: If $B,C \subseteq M$ are open balls, then $B \cap C \subseteq M$ is a finite(!) union of open balls. By an open ball I mean an open submanifold, which is diffeomorphic to some $D^n$.
Is every manifold quasi-separated? If not, are open balls quasi-separated? Is there a simple characterization of quasi-separated manifolds?
| https://mathoverflow.net/users/2841 | quasi-separated manifolds | Using the Riemann mapping theorem you can easily show that $\mathbb{R}^2$ is not quasi separated.
| 1 | https://mathoverflow.net/users/745 | 37511 | 24,133 |
https://mathoverflow.net/questions/37482 | 2 | I'm planning a short course on few topics and applications of nonlinear functional analysis, and I'd like a reference for a quick and possibly self-contained construction of a structure of a Banach differentiable manifold for the space of continuous mappings $C^0(K,M)$, where $K$ is a compact topological space (even metric if it helps) and $M$ is a (finite dimensional) differentiable manifold.
A construction of a differentiable structure of Banach manifold for this space can be found e.g. in Lang's book *Fundamentals of differential geometry* (1999). The main tools are the exponential map and tubular nbds (having fixed a Riemannian structure on $M$. This is OK but I believe there should be something even more basic.
Does anybody have a reference for alternative constructions (not necessarily elementary) ?
| https://mathoverflow.net/users/6101 | Manifolds of continuous mappings. | Well, there is always my old "Foundations of Global Nonlinear Analysis", which is about just this.
| 6 | https://mathoverflow.net/users/7311 | 37547 | 24,156 |
https://mathoverflow.net/questions/37518 | 26 | Wikipedia describes [Kendall and Smith's 1938 statistical randomness tests](http://en.wikipedia.org/wiki/Statistical_randomness#Tests) like this:
>
> * The **frequency test**, was very basic: checking to make sure that there were roughly the same number of 0s, 1s, 2s, 3s, etc.
> * The **serial test**, did the same thing but for sequences of two digits at a time (00, 01, 02, etc.), comparing their observed frequencies with their hypothetical predictions were they equally distributed.
> * The **poker test**, tested for certain sequences of five numbers at a time (aaaaa, aaaab, aaabb, etc.) based on hands in the game poker.
> * The **gap test**, looked at the distances between zeroes (00 would be a distance of 0, 030 would be a distance of 1, 02250 would be a distance of 3, etc.).
>
>
>
It is not obvious to me that these four particular tests were chosen with any deep understanding of how best to detect nonrandomness. Rather, it seems each one was probably chosen for simplicity.
Well, it's easy to see why early work in the field would be like that. But fast forward to 1995 and George Marsaglia's [Diehard tests](http://en.wikipedia.org/wiki/Diehard_tests) seem, on the surface, just as ad hoc:
>
> * **Birthday spacings:** Choose random points on a large interval. The spacings between the points should be asymptotically Poisson distributed. The name is based on the birthday paradox.
> * **Overlapping permutations:** Analyze sequences of five consecutive random numbers. The 120 possible orderings should occur with statistically equal probability.
>
>
> *(...and so on)*
>
>
>
It is not obvious to me that these tests are really independent (i.e. that none is entirely redundant, rejecting only sequences also rejected by at least one of the other tests), or that there aren't obviously better generalizations of them, much less that they were chosen as a set to try to cover any particular space efficiently.
Ideally, a test suite would be designed to reject as many sequences of low Kolmogorov complexity as possible with minimal computation and false alarms. Is a more theoretical approach to this possible? Why hasn't it happened? —Or is there more to the state of the art than meets the eye?
| https://mathoverflow.net/users/2599 | Why do statistical randomness tests seem so ad hoc? | It's not clear that Marsaglia's tests are really good enough. See [this Stack Overflow discussion](https://stackoverflow.com/questions/584566/pseudo-random-number-generator).
Kolmogorov complexity is not the right criterion for statistical randomness tests, since any pseudorandom sequence has low Kolmogorov complexity. What you really want in a random number generator is for the sequence to be computationally pseudorandom; that is, without knowledge of the seed, no polynomial-time test can distinguish the sequence from a truly random sequence.
In fact, there are a number of random number generators which are believed to be computationally pseudorandom. Nobody uses these, however, partly because they are computationally too expensive, and partly because of inertia and partly because the existing pseudorandom generators we have are good enough most of the time. A while ago, for one of the most common methods of pseudorandom number generation (linear congruential), I ran into cases where they unexpectedly produced the wrong answer.
Marsaglia's tests were developed over a number of years, and I believe each was designed to detect certain flaws that many pseudorandom number generators contained at the time. Once good-enough pseudorandom number generators were available, nobody bothered creating a more stringent series of tests.
| 18 | https://mathoverflow.net/users/2294 | 37557 | 24,163 |
https://mathoverflow.net/questions/37513 | 7 | Let $G$ be a metric group, and let $h$ be the associated Hausdorff dimension function on subsets of $G$. (See for instance Barnea and Shalev, Hausdorff dimension, pro-p groups and Kac-Moody algebras, Trans. AMS 1997.) When do we have $h(AB) = h(A) + h(B) - h(A \cap B)$ for normal subgroups of $G$? If this property fails, is there still a general way to use $h$ (or some similar 'dimension' function) to construct a pseudometric on the lattice of normal subgroups of $G$? What I am looking for here are some informative examples of bad behaviour, particularly for profinite groups.
| https://mathoverflow.net/users/4053 | Hausdorff dimension of products of normal subgroups | In the first conference I ever went to Slava Grigorchuk asked me a similar question and I didn’t have an answer. But when I have got back to Jerusalem I have talked with Elon Lindenstrauss about it and he suggested the following easy counterexample. Take $G=\mathbb{F}\_p[[t]]$. Pick $S$ to be a subset of the integers with density one and with infinite complement $T$. Say $S=\left\{ n\_i \right\}$ and $T=\left\{ m\_i \right\}$. Take $A=\overline{< t^{n\_i}>}$ and take $B=\overline{\left< t^{n\_i} +t^{m\_i} \right>}$. Cleary, $AB=G$, $h(A)=h(B)=1$, but $A \cap B=\emptyset$.
Now, $G$ is not finitely generated, if you would like to have a counterexample which is finitely generated, then you can take $G=SL\_d(F\_p[[t]])$ and construct in a similar way to the above $A$ which is made from upper triangular matrices and $B$ which is made from lower triangular matrices. However, $A$ and $B$ will not be normal any more.
I am not familiar with a counterexample in which $A$ and $B$ are normal and $G$ is finitely generated. I am also not familiar with a counterexample in which $A$ and $B$ are finitely generated. But as you can deduce from my story above this does not mean much.
| 4 | https://mathoverflow.net/users/5034 | 37559 | 24,165 |
https://mathoverflow.net/questions/37551 | 17 | I'm trying to understand the necessity for the assumption in the Hahn-Banach theorem for one of the convex sets to have an interior point. The other way I've seen the theorem stated, one set is closed and the other one compact. My goal is to find a counter example when these hypotheses are not satisfied but the sets are still convex and disjoint. So here is my question:
**Question:** I would like a counter example to the Hahn-Banach separation theorem for convex sets when the two convex sets are *disjoint* but neither has an interior point. It is trivial to find a counter example for the *strict* separation but this is not what I want. I would like an example (in finite or infinite dimensions) such that we *fail* to have any separation of the two convex sets at all.
In other words, we have $K\_1$ and $K\_2$ with $K\_1 \cap K\_2 = \emptyset$ with both $K\_1$ and $K\_2$ convex belonging to some normed linear space $X$. I would like an explicit example where there is **no** linear functional $l \in X^\*$ such that $\sup\_{x \in K\_1} l(x) \leq \inf\_{z \in K\_2} l(z)$.
I'm quite sure that a counter example cannot arise in finite dimensions since I think you can get rid of these hypotheses in $\mathbb{R}^n$. I'm not positive though.
| https://mathoverflow.net/users/8755 | A counter example to Hahn-Banach separation theorem of convex sets. | Here is a simple example of a linear space and 2 disjoint convex sets such that there is no linear functional separating the sets. Note that the notions of convexity and linear functional *do not* require any norm or whatever else. You can introduce them, if you want, but they are completely external to the problem.
The usual trick with taking the difference of the sets shows that it is enough to assume that one set is a point, say, the origin. Now we want to design a convex set $K$ not containing the origin such that the only linear functional $\ell$ that is non-negative on this set is $0$. To this end, take the space $X$ to be the space of all real sequences with finitely many non-zero terms and let $K$ be the set of all such sequences whose last non-zero element is positive. Now, if $x\in X$, choose $y$ to be any sequence whose last non-zero element is $1$ and lies beyond the last non-zero element in $x$. Then, for every $\delta>0$, both $x+\delta y$ and $-x+\delta y$ are in $K$, so $\pm \ell(x)+\delta \ell(y)\ge 0$ with any $\delta>0$ whence $\ell(x)=0$. Thus $\ell$ vanishes identically.
| 38 | https://mathoverflow.net/users/1131 | 37564 | 24,169 |
https://mathoverflow.net/questions/37374 | 19 | In a comment on [Tom Goodwillie's question about relating the Alexander polynomial and the Iwasawa polynomial](https://mathoverflow.net/questions/31250/prime-numbers-as-knots-alexander-polynomial), Minhyong Kim makes the cryptic but tantalizing statement:
> In brief, the current view is that the Iwasawa polynomial=p-adic L-function should be viewed as a path in K-theory space.
As somebody who has been trying, for a long time, to understand the Alexander polynomial, and who believes there is something "more basic" underlying its appearances in mathematics in various guises (my particular interest has to do with why it turns up as the wheels part of the Kontsevich invariant, as per Melvin-Morton-Rozansky), I'd really like to understand this comment.
> **Question**: What is "K-theory space" supposed to be, and why should a path in it be a natural object to be examining? And why should we expect such a thing to give equivalent data to the Iwasawa polynomial/ Alexander polynomial?
| https://mathoverflow.net/users/2051 | What is a path in K-theory space? | I know nothing about Alexander polynomials but let me try to answer the Iwasawa theory part. As is well known, in classical Iwasawa theory one considers cyclotomic $\mathbb{Z}\_p$ extension $F{\infty}$ of $F$. We take the $p$-part of the ideal class group $A\_n$ of the intermediate extension $F\_n$ of $F$ of degree $p^n$. The inverse limit of $A\_n$ with respect to norm maps, say $A$, has an action of $G= Gal(F\_{\infty}/F)$. Since $A$ is pro-$p$, it becomes a $\mathbb{Z}\_p[G]$-module. However, it is not finitely generated over this group ring (and for various other reasons) one considers the completion $\mathbb{Z}\_p[[G]]$ of $\mathbb{Z}\_p[G]$. Since $A$ is compact, it becomes a $\mathbb{Z}\_p[[G]]$-module. As $G \cong \mathbb{Z}\_p$, the ring $\mathbb{Z}\_p[[G]] \cong \mathbb{Z}\_p[[T]]$, the power series ring in variable $T$. There is a nice structure theory for finitely generated modules over $\mathbb{Z}\_p[[T]]$. The module $A$ is a torsion $\mathbb{Z}\_p[[G]]$-module (i.e.$Frac(\mathbb{Z}\_p[[G]]) \otimes A = 0$). For such modules one can define the characteristic ideal using the structure theory. Iwasawa's main conjecture asserts that there is a canonical generator for this ideal called the $p$-adic $L$-function.
In generalised Iwasawa theory (more precisely, to formulate the generalised main conjecture à la Kato), one wants to consider extensions whose Galois groups are not necessarily $\mathbb{Z}\_p$ (but most formulations of the main conjecture still require that the cyclotomic $\mathbb{Z}\_p$-extension of the base field be in the extension). For the completed $p$-adic groups rings of such Galois groups, the structure theory completely breaks down even if the Galois group is abelian.
However, one can still show that $A$ is a torsion Iwasawa module (which again just means that $Frac(\mathbb{Z}\_p[[G]]) \otimes A = 0$. Note that it is always possible to invert all non-zero divisors in a ring even in the non-commutative setting). Hence the class of $A$ in the group $K\_0(\mathbb{Z}\_p[[G]])$ is zero. Strictly speaking, here I must assume that $G$ has no $p$-torsion so that I can take a finite projective resolution of $A$, or I must work with complexes whose cohomologies are closely related to $A$. But I will sinfully ignore this technicality here. Now, since the class of $A$ in $K\_0(\mathbb{Z}\_p[[G]])$ is zero, there is a path from $A$ to the trivial module 0 in the $K$-theory space. In Iwasawa theory this is most commonly written as
There exists an isomorphism $Det\_{\mathbb{Z}\_p[[G]]}(A)$ $\to$ $Det(0)$.
This isomorphism replaces the characteristic ideal used in the classical Iwasawa theory. The $p$-adic $L$-function then is a special isomorphism of this kind. (Well one has to be careful about the uniqueness statement in the noncommutative setting but it is a reasonably canonical isomorphism). Hence the main conjecture now just asserts existence of such a $p$-adic $L$-function.
Thus the $p$-adic $L$-function may be thought of as a canonical path in the $K$-theory space joining the image of Selmer module (or better- a Selmer complex), such as the ideal class group in the above example, and the image of the trivial module.
I hope this answer helps until Minhyong sheds more light on his remarks and relations between $p$-adic $L$-functions and the Alexander polynomials.
[EDIT: Sep. 30th] To answer Daniel's questions below-
1) Take the projective resolution of A to define its class in $K\_0$. In any case the $K$-theory of the category of finitely generated modules is the same as the $K$-theory of finite generated projective modules. 2) I do not know if there are any sensible/canonical multiplicative sets at which to localise $\mathbb{Z}\_p[G]$. The modules considered in Iwasawa theory are usually compact (or co-compact depending on whether you take inverse limit with respect to norms or direct limit with respect to inclusion of fields in a tower) and so the action of the ring $\mathbb{Z}\_p[G]$ extends to an action of the completion $\mathbb{Z}\_p[[G]]$ by which we mean $\varprojlim \mathbb{Z}\_p[G/U]$, where $U$ runs through open normal subgroups of $G$. 3) We do not usually get a loop in the $K$-theory space of $\mathbb{Z}\_p[[G]]$. However, modules which come up from arithmetic are usually torsion as $\mathbb{Z}\_p[[G]]$-modules i.e. every element in the module is annihilated by a non-zero divisor or in other words if $X$ is the module then $Frac(\mathbb{Z}\_p[[G]]) \otimes X=0$. (it is usually very hard to prove that the modules arising are in fact torsion). But once we know that they are torsion we get a loop in the $K$-theory space of $Frac(\mathbb{Z}\_p[[G]])$. If we know that for some multiplicatively closed subset $S$ of $\mathbb{Z}\_p[[T]]$ annihilates $X$ i.e. $(\mathbb{Z}\_p[[G]]\_S) \otimes X=0$, then we already get a loop in the $K$-theory space of $\mathbb{Z}\_p[[G]]$. In noncommutative Iwasawa theory it is often necessary to work with a multiplicative set strictly smaller than all non-zero divisors.
| 19 | https://mathoverflow.net/users/2259 | 37567 | 24,170 |
https://mathoverflow.net/questions/37563 | 9 | LLL and other lattice reduction techniques (such as PSLQ) try to find a short basis vector relative to the 2-norm, i.e. for a given basis that has $ \varepsilon $ as its shortest vector, $ \varepsilon \in \mathbb{Z}^n $, find a short vector s.t. $ b \in \mathbb{Z}^n, \|b\|\_2 < \|c^n \varepsilon\|\_2 $.
Has there been any work done to find short vectors based on other, potentially higher, norms? Is this a meaningful question?
| https://mathoverflow.net/users/4106 | Other norms for lattice reduction techniques (LLL, PSLQ)? | The state of the art (of the possible) is covered in Khot's paper ["Inapproximability Results for Computational Problems on Lattices"](https://doi.org/10.1007/978-3-642-02295-1_14). [Here is a link](https://books.google.com/books?id=9d75m1L_GIgC&pg=PA456#v=onepage&q&f=false) to a brief section on $\ell^p$ norms.
| 6 | https://mathoverflow.net/users/1847 | 37571 | 24,172 |
https://mathoverflow.net/questions/37578 | 12 | Suppose we have $k$ realizations of a random variable uniformly distributed over the unit cube $[0,1]^n$.
What is the probability that their convex hull has all of the $k$ points as extreme points?
If it would be easier, "unit cube" can be replaced by "unit ball".
| https://mathoverflow.net/users/8977 | Convex hull of $k$ random points | Imre Bárány has investigated similar questions, including the asymptotics of $p(k,S)$, the probability that $k$ uniformly chosen points from the convex body $S\subset \mathbb{R}^n$ are in convex position (they are extreme points of their convex hull).
In general one can give the bounds $$c\_1\le k^{2/(n-1)}\sqrt[k]{p(k,S)}\le c\_2$$ for large enough $k$ and constants $c\_1,c\_2$. I don't think closed form formulas are known for all $k$ even for simple convex sets $S$. See [here](https://arxiv.org/abs/math/0304462) and the papers in the references. See [here](https://doi.org/10.1007/BF00334039) for the case when $S$ is the unit ball.
| 13 | https://mathoverflow.net/users/2384 | 37580 | 24,175 |
https://mathoverflow.net/questions/31367 | 9 | I'm working on a set of problems for which I can formulate binary integer programs. When I solve the linear relaxations of these problems, I always get integer solutions. I would like to prove that this is always the case.
I believe that this involves proving that the constraint matrix is totally unimodular. Is there any sufficent conditions for binary matrices to be totally unimodular that might be of use for this?
| https://mathoverflow.net/users/7496 | Proving that a binary matrix is totally unimodular | Here are some common ways of proving a matrix is TU.
1. The incidence matrix of a bipartite graph and network flow LPs are TU; these are standard examples usually taught in every book on TU.
2. The consecutive-ones property: if it is (or can be permuted into) a 0-1 matrix in which for every row, the 1s appear consecutively, then it is TU. (The same holds for columns since the transpose of a TU matrix is also TU.)
3. Every "network matrix," defined as follows, is TU (and they are a fundamental building block of the set of all TU matrices, according to Seymour's theorem). The rows correspond to a tree $T = (V, R)$ each of whose arcs have an orientation (i.e. it is not necessary that exist a root vertex $r$ such that the tree is "rooted into $r$" or "out of $r$").The columns correspond to another set $C$ of arcs on the same vertex set $V$. To compute the entry at row $R$ and column $C = st$, look at the $s$-to-$t$ path $P$ in $T$, then the entry is:
* +1 if arc $R$ appears forward in $P$
* -1 if arc $R$ appears backwards in $P$
* 0 if arc $R$ does not appear in $P$[You can see more in Schrijver's 2003 book.]
5. Ghouila-Houri showed a matrix is TU iff for every subset $R$ of rows, there is an assignment $s : R \to \pm 1$ of signs to rows so that the signed sum $\sum\_{r \in R} s(r)r$ (which is a row vector the same width as the matrix) has all its entries in $\{0, \pm1\}$.
There are other if-and-only-if conditions like Ghouila-Houri too (see Schrijver 1998) but the 4 conditions I gave above have been the most practical for me.
| 13 | https://mathoverflow.net/users/4020 | 37589 | 24,180 |
https://mathoverflow.net/questions/33091 | 4 | Are there results that bound the asymmetry of the duality gap of an integer program? That is to say, if the difference between the LP solution and the IP (primal) solution is $a$, is there a function $f$ so that difference between the LP solution and the IP dual solution is $< |f(a)|$?
| https://mathoverflow.net/users/2588 | Symmetry of the integer gap | Here is an example showing this is impossible, I think. The integrality gap for "independent set" can be up to $n/2$ on a graph with $n$ vertices. But its dual is the naive LP relaxation of edge cover on the same graph; you can show a constant integrality gap upper bound for this by standard methods (and Ojas Parekh proves in his thesis that the best possible bound is 4/3).
This example behaves similarly if you care about the approximability, I have worked on [a paper](http://arxiv.org/PS_cache/arxiv/pdf/0904/0904.0859v5.pdf) which motivated my example above.
| 3 | https://mathoverflow.net/users/4020 | 37591 | 24,182 |
https://mathoverflow.net/questions/37593 | 5 | The decomposition theorem states roughly, that the pushforward of an IC complex,
along a proper map decomposes into a direct sum of shifted IC complexes.
Are there special cases for the decomposition theorem, with "easy" proofs?
Are there heuristics, why the decomposition theorem should hold?
| https://mathoverflow.net/users/2837 | Easy special cases of the decomposition theorem? | Well, it depends on what you mean by "easy". A special case, which I find very instructive, is a theorem of Deligne from the late 1960's.
>
> Theorem. $\mathbb{R} f\_\*\mathbb{Q}\cong \bigoplus\_i R^if\_\*\mathbb{Q}[-i]$, when $f:X\to Y$ is a smooth projective morphism of varieties over $\mathbb{C}$. (This holds more generally with $\mathbb{Q}\_\ell$-coefficients.)
>
>
> Corollary. The Leray spectral sequence degenerates.
>
>
>
The result was deduced from the hard Lefschetz theorem.
An outline of a proof (of the corollary) can be found in Griffiths and Harris.
It is tricky but essentially elementary.
A much less elementary, but more conceptual argument, uses
weights. Say $Y$ is smooth and projective, then
$E\_2^{pq}=H^p(Y, R^qf\_\*\mathbb{Q})$ *should* be pure of weight $p+q$ (in the sense of Hodge
theory or $\ell$-adic cohomology). Since
$$d\_2: E\_2^{pq}\to E\_2^{p+2,q-1}$$
maps a structure of one weight to another it must vanish. Similarly for higher differentials.
If $f$ is proper but not smooth, the decomposition theorem shows that $\mathbb{R} f\_\*\mathbb{Q}$ decomposes into sum of translates of intersection cohomology complexes.
This follows from more sophisticated purity arguments (either in the $\ell$-adic setting as in BBD, or the Hodge theoretic setting in Saito's work).
There is also a newer proof due to de Cataldo and Migliorini which seems a bit more geometric.
I have been working through some of this stuff slowly. So I may have more to say in a few months time. Rather than updating this post, it may be more efficient for the people
interested to check
[here](http://www.math.purdue.edu/~dvb/seminar.html) periodically.
| 5 | https://mathoverflow.net/users/4144 | 37604 | 24,189 |
https://mathoverflow.net/questions/37602 | 18 | Let $\mathfrak{g}$ be a simple complex Lie algebra, and let $\mathfrak{h} \subset \mathfrak{g}$ be a fixed Cartan subalgebra. Let $W$ be the Weyl group associated to $\mathfrak{g}$. Let $S(\mathfrak{h}^\*)$ be the ring of polynomial functions on $\mathfrak{h}$. The Weyl group $W$ acts on $\mathfrak{h}$, and this action extends to an action of $W$ on $S(\mathfrak{h}^\*)$. It is a well-known fact that the space of Weyl group invariants $S(\mathfrak{h}^\*)^W$ is generated by $r$ algebraically independent homogeneous generators, where $r$ is the dimension of $\mathfrak{h}$ (equivalently, the rank of $\mathfrak{g}$). The degrees of the generators are uniquely determined, though the actual generators themselves are not.
The degrees of the generators for $S(\mathfrak{h}^\*)^W$ are well-known and can be found, for example, in Humphreys' book "Reflection groups and Coxeter groups" (Section 3.7). When $\mathfrak{g}$ is of classical type (ABCD), it is also not hard to find explicit examples of generators for $S(\mathfrak{h}^\*)^W$ (loc. cit. Section 3.12).
>
> Where, if anywhere, can I find explicit examples of generators for $S(\mathfrak{h}^\*)^W$ when $\mathfrak{g}$ is of exceptional type, specifically, for types $E\_7$ and/or $E\_8$?
>
>
>
I have found explicit examples for types $E\_6$ and $F\_4$ in a paper by Masaru Takeuchi (*On Pontrjagin classes of compact symmetric spaces*, J. Fac. Sci. Univ. Tokyo Sect. I 9 1962 313--328 (1962)). I have probably also come across examples for type $G\_2$, though I don't recall where at this moment. But I have been unable to find anything for types $E\_7$ or $E\_8$.
| https://mathoverflow.net/users/7932 | Polynomial invariants of the exceptional Weyl groups | In Physics these are called the "Casimir operators" and googling this gives the following paper: F. Berdjis and E. Beslmüller [Casimir operators for $F\_4$, $E\_6$, $E\_7$ and $E\_8$](http://jmp.aip.org/resource/1/jmapaq/v22/i9/p1857_s1). For the case of $G\_2$, see the paper in my comment above: [Casimir operators for the exceptional group $G\_2$](http://arxiv.org/abs/hep-th/9306062) by A.M. Bincer and K. Riesselmann.
| 13 | https://mathoverflow.net/users/394 | 37609 | 24,192 |
https://mathoverflow.net/questions/37614 | 1 | Lets define *edge-cycle* in a graph $G$ as a path where the first and the last node are adjacent.
(in contrast with the definition of *cycle* where first and last node are the same).
An *edge-tree* $T$ is a tree with the additional property that doesn't have an edge-cycle.
In a graph we can compute the number of spanning trees by using the [Matrix-Tree](http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem) theorem.
>
> Is there any similar theorem for the computation of the number of edge-trees of a graph?
>
>
>
| https://mathoverflow.net/users/8984 | on counting of special case of trees on a graph | I'll answer a question raised in the comments:
**Problem**: Count the number of induced trees of size $k$.
According to this [paper](http://www.renyi.hu/~p_erdos/1986-08.pdf) by Erdös, Saks and Sos, it is NP-complete to decide given a graph $G$ and an integer $k$, if $G$ contains an induced tree of size $k$. So, it's probably pretty damn hard to count them. Apparently, it remains NP-complete even for bipartite graphs.
Actually, the argument is pretty simple so I'll include it here. Given a graph $H$ and an integer $k$, it is well-known that the problem of deciding if $H$ has an independent set of size $k$ is NP-complete. Suppose that $H$ has $n$ vertices. Let $G$ be the graph obtained from $H$ by first adding a disjoint copy of $P\_n$ (a path on $n$ vertices), and then connecting one end of $P\_n$ to all the vertices in $H$. Clearly, $H$ has an independent set of size $k$ if and only if $G$ contains an induced tree of size $n+k$.
| 3 | https://mathoverflow.net/users/2233 | 37636 | 24,212 |
https://mathoverflow.net/questions/37637 | 6 | First off, I know that there's no general algorithm for determining if there's a solution to a general Diophantine equation, much less a system.
However, I'm wondering if there is an algorithm for solving a Diophantine system of linear and quadratic equations? In fact, I have a system which is "sparse" in some sense (the linear equations are all the sum of three variables equals a number, and all the same number, and the quadratics aren't much worse).
If so, then can it be extended to the case of countably many variables?
| https://mathoverflow.net/users/622 | Algorithms for Diophantine Systems | No. Given any set of diophantine equations $f\_1(z\_1, \ldots, z\_n) = \ldots = f\_m(z\_1, \ldots, z\_n)=0$, we can rewrite in terms of linear equations and quadratics. Create a new variable $w\_{k\_1 \cdots k\_n}$ for each monomial $z\_1^{k\_1} \cdots z\_n^{k\_n}$ which occurs in the $f$'s, or which divides any monomial which occurs in the $f$'s. Turn each $f$ into a linear equation: For example, $x^3 y^2 + 7 x^2 y=5$ becomes $w\_{32} + 7 w\_{21} = 5$. Then create quadratic equations $z\_i w\_{k\_1 \cdots k\_i \cdots k\_n} = w\_{k\_1 \cdots (k\_i +1) \cdots k\_n}$. For example, $x w\_{22} = w\_{32}.$ This shows that the solvability of Diophantine equations is equivalent to that of Diophantine equations of degree $\leq 2$.
I'll also mention a very concrete case. The intersection of two quadrics in $\mathbb{P}^3$ is a genus $1$ curve. To my knowledge, no algorithm is known to test for the existence of rational points even in this case. (But my knowledge is not very large.)
| 14 | https://mathoverflow.net/users/297 | 37640 | 24,213 |
https://mathoverflow.net/questions/37625 | 1 | I suspect that a topic such as this may have been considered before: if so, I hope that someone can point me to a reference on the subject.
I have a graph *G* with an upper bound *d* on its maximum degree. Define an *independent-set cover* for *G* to be a family of independent sets in *G*, whose union is *V(G)*. Is there a non-trivial upper bound for the smallest independent-set cover of *G*? [That is: I would like the number of independent sets in the family to be as small as possible; the size of any particular independent set within that family is unimportant.]
I'm also interested in the case where the degrees of the vertices may differ substantially from the maximum degree.
**Edited to add:** thanks to those who pointed out that the size of the smallest independent-set cover is just the chromatic number (by definition). I'm not sure how I managed to avoid noticing that this is what I was asking about, except that I probably think of colourings too much in terms of vertex-labellings and not often enough in terms of vertex partitions.
| https://mathoverflow.net/users/3723 | Covering of a graph via independent sets | As has already been pointed out, this is the chromatic number $\chi$. (For example, the assertion that all planar graphs have $ \chi \le 4$ is the famous "Four color theorem.")
You say your graph has maximum degree $d$. Then in all cases $\chi \le d+1$, and [Brooks' theorem](http://en.wikipedia.org/wiki/Brooks%2527_theorem) gives that in fact $\chi \le d$ as long as $G$ is not a complete graph or an odd cycle.
| 7 | https://mathoverflow.net/users/4558 | 37648 | 24,219 |
https://mathoverflow.net/questions/37654 | 9 | Dear all,
I am looking for a proof or a reference of the following statement:
Let $f$ be a non-constant polynomial with integer coefficients. Then the sum $\sum \{1/p \mid f \text{ has a root modulo } p\}$ diverges.
I am pretty sure that I saw it somewhere before but I cannot remember and I failed to find it in number theory books. A possible routes that has already been suggested to me is actually showing that the sums of reciprocals for which f even decomposes into linear factors has positive density which should stem from Galois theory. I am however not an expert in Galois theory so that I would prefer a direct proofs or a reference.
Thanks in advance,
Alberto
| https://mathoverflow.net/users/8994 | Sum of reciprocals of primes modulo which a polynomial has a root | This should follow from the Theorem of Frobenius mentioned on p. 7 (PDF numbering) of <http://websites.math.leidenuniv.nl/algebra/chebotarev.pdf>.
| 6 | https://mathoverflow.net/users/6153 | 37656 | 24,223 |
https://mathoverflow.net/questions/37581 | 1 | Let $A \in \mathbb{R}^{m \times n}$ be a random matrix with i.i.d. entries (the distribution is not important), where $m < n$ (i.e. $A$ is a "wide" matrix). I would like a lower bound on
$$
\phi(A) \triangleq \min\_x \frac{\lVert Ax \rVert}{\lVert x \rVert}
$$
that holds with high probability (apologies if the notation $\phi(A)$ conflicts with any established usage).
When $m \geq n$, evidently $\phi(A) = \sigma\_{min}(A)$, the least singular value of $A$ (although I am not certain why this is true). Of course the distribution of the least singular value of a random matrix has been well-studied.
But when $m < n$, it seems that $\phi(A) \neq \sigma\_{min}(A)$ in general. For example, if $m = 1$ and $n > 1$, then $\phi(A) = 0$ (just choose $x$ to be orthogonal to the vector $A$), but $\sigma\_{min}(A)$ is the Euclidean norm of the vector $A$, which usually will not be $0$.
| https://mathoverflow.net/users/8978 | Question about "wide" random matrices | I spoke to someone locally, and we think the issue is which convention is used to define the singular values of a matrix. If one defines the singular values of a matrix $A$ to be the eigenvalues of the matrix
$$
\sqrt{A^TA}
$$
then if $A$ is $m \times n$ with $m < n$ we have $\sigma\_{\min}(A) = 0$ but $\sigma\_{\min}(A^T) \neq 0$ in general. This agrees with the identity $\phi(A) = \sigma\_{\min}(A)$.
However, if one defines the singular values of $A$ to be the diagonal entries of the matrix $\Sigma$ in the singular value decomposition
$$
A = U\Sigma V^T
$$
then $A$ and $A^T$ have exactly the same singular values, and $\phi(A) \neq \sigma\_{\min}(A)$ in general.
| 2 | https://mathoverflow.net/users/8978 | 37659 | 24,224 |
https://mathoverflow.net/questions/37662 | 3 | Let $B \subset C$ be Noetherian integral domains, and $g \in B$. Thus, $\mathrm{Spec} B \to \mathrm{Spec} B\_g$ is an open immersion.
If furthermore $C \subset B\_g$, does it follow that $\mathrm{Spec} C \to \mathrm{Spec} B$ is an open immersion?
| https://mathoverflow.net/users/5094 | If $B \subset C \subset B_g$, is $\mathrm{Spec} C \to \mathrm{Spec} B$ necessarily an open immersion? | No. $B=k[x,y]$, $g=x$, $C=k[x, x^{-1} y]$.
| 7 | https://mathoverflow.net/users/297 | 37663 | 24,227 |
https://mathoverflow.net/questions/37665 | 10 | I do recreational math from time to time, and I was wondering about a couple of graph enumeration issues.
First, is it possible to enumerate all simple graphs with a given degree sequence?
Second, is it possible to enumerate all valid degree sequences for simple graphs with a given number of vertices?
Based on my wikipedia surfing, we can use the Erdos-Gallai theorem to determine if a degree sequence is valid, but this doesn't really lend itself to enumerating valid degree sequences efficiently. Similarly, we can use the Havel-Hakimi algorithm to construct at least one graph for a given valid degree sequence, but this doesn't help to enumerate all graphs for that degree sequence.
My (admittedly uneducated) guess is that it might be possible to work backwards using the Havel-Hakimi condition to construct graphs by building them up in different ways. Any insight would be appreciated :D
| https://mathoverflow.net/users/8998 | Degree Sequences and Graph Enumeration | Regarding the question of enumerating degree sequences. Richard Stanley's paper: [A zonotope associated with graphical degree sequences, in Applied Geometry and Discrete Combinatorics](http://math.mit.edu/~rstan/pubs/pubfiles/83.pdf), DIMACS Series in Discrete Mathematics, vol. 4, 1991, pp. 555-570. Deale with the problem of enumerating graphical degree sequences.
As Chris commented enumerating graphs with presecribed degree sequences can be hard.
A [recent paper of McKay entitled: Subgraphs of dense random graphs with specified degrees](http://arxiv.org/abs/1002.3018) is a good place to start looking of what is known.
(Threshold graphs are precisely those graphs that are unique given their degree sequences.)
| 7 | https://mathoverflow.net/users/1532 | 37667 | 24,229 |
https://mathoverflow.net/questions/37548 | 8 | A *polymatroid* is a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that
1) $d(\varnothing)=0$,
2) $A \subset B$ implies $d(A) \leq d(B)$, and
3) $d(A \cap B) + d(A \cup B) \leq d(A) + d(B)$ for all $A,B \in P(X)$.
A polymatroid is said to be *representable* over $GF(2)$ (the field with two elements), if there exists a collection of subvectorspaces $\lbrace V\_x \mid x \in X \rbrace$ of $GF(2)^{\oplus d(X)}$, such that
$$d(A) = \dim\_{GF(2)} \bigvee\_{x \in A} V\_x, \quad \forall A \in P(X).$$
(It is clear that any function $d$, which is defined this way is indeed a polymatroid.)
A polymatroid is called *matroid* if $d(\lbrace x\rbrace)=1$ for all $x \in X$.
Tutte proved that a matroid is representable over $GF(2)$ if and only if the matroid $U\_{2,4}$ does not appear as a minor. Here, $U\_{2,4}$ is the matroid formed by four points that lie on one line, i.e. the underlying set is $\lbrace x\_1,\dots,x\_4 \rbrace$ and $d(\lbrace x\_i,x\_j\rbrace) = 2$ for $i \neq j$ and $d(\lbrace x\_1,\dots,x\_4\rbrace)=2$. (The necessity of this condition is obvious, since $GF(2)^{\oplus 2}$ has only $3$ non-zero elements.)
>
> **Question:** Is there any useful characterization of the representability of a polymatroid over $GF(2)$ ? What is known about this question ?
>
>
>
| https://mathoverflow.net/users/8176 | Representability of polymatroids over $GF(2)$ | Here's a construction due to Stefan van Zwam.
Let $U\_{2,4}$ be a 4-point line with ground set $[4]$, and rank function $r$. For each $A \subset [4]$, let $\chi(A)$ be 1, if $1 \in A$, and 0 if $1 \notin A$. For each $k \in \mathbb{N}$, let $S\_k$ be the polymatroid with ground set $[4]$ and rank function $r+k\chi$.
**Lemma.** $S\_k$ is not representable over $\mathbb{F}\_2$, for every $k$.
*Proof*. Let $(V\_i : i \in [4])$ be a representation of $S\_k$ over $\mathbb{F}\_2$. Choose a basis $B\_i$ for each $V\_i$. Since {1} has rank $k+1$ and {1,2,3,4} has rank $k+2$, we may assume that $B\_1$ consists of the first $k+1$ standard basis vectors in $\mathbb{F}\_2^{k+2}$. Now, since {1,2}, {1,3}, and {1,4} all have rank $k+2$, it follows that $B\_2, B\_3$, and $B\_4$ all must have last coordinate equal to 1. In particular, $B\_2 + B\_3 +B\_4 \neq 0$. Also, since every two element subset of {2,3,4} has rank 2, it follows that $B\_2, B\_3$ and $B\_4$ are distinct. But now {2,3,4} must have rank 3 in $S\_k$, a contradiction. $\square$
Moreover, it is easy to check that every minor of $S\_k$ is representable over $\mathbb{F}\_2$, where minors of polymatroids are defined in the obvious way. Therefore, the set of binary polymatroids has an infinite set of excluded minors.
| 5 | https://mathoverflow.net/users/2233 | 37681 | 24,237 |
https://mathoverflow.net/questions/37666 | 0 | Hi, I have a paper that I'm reading and they propose an equation,
```
a = exp^{bT},
```
that is fitted to their measurements, and give the value of `b` as well as the coefficient of determination. Is this sufficient information to constrain the value of the standard error, and if so, how might I go about doing that? Could I just add values sampled from a normal distribution with mean of zero; play around with the standard deviation of this distribution until I get something that gives me the target `R^2`, and assume the standard error of that fit? (That's a form of bootstrapping?)
Thanks!
| https://mathoverflow.net/users/5282 | get standard error from correlation coefficient? | Look at the appropriate space in which your data would have a normal distribution, or look at the appropriate space in which your data set becomes linear.
This requires knowing the distribution of your experimental data as a prior value.
The correlation coefficient $R$ works best for linearly related data expressable as $y=mx+b$. While $R$ may have some informative value (positive or negative correlation) for non-linearly related variables, it really doesn't have a good clear meaning in non-linear cases.
Takes the logarithm of both sides and get
$$\log(a)=b \cdot \log(T)$$
and see if you can calculate the correlation and do linear regression to find the best fit linear relation between $\log(a)$ and $\log(T)$. This way, you can see if the correlation coefficient can be informative for you.
But the key thing is in knowing the distribution (or expected distribution) of your data. If it is not normally distributed in the space which you're working in, try to find a transform which moves the data into a space where it **has** a linear normal Gaussian distribution.
| 2 | https://mathoverflow.net/users/8676 | 37686 | 24,240 |
https://mathoverflow.net/questions/37678 | 8 | I was wondering if there are some classical methods to tackle problems in number theory dealing with sums where the primes are not well-"controled". I talk about problems where we want to link a certain sum with information about the primes dividing the elements of the sum: the $abc$ conjecture is an example of such a problem, since we want to link $a$, $b$ and $a+b$ to the prime factors of $abc$, knowing that $a$, $b$ and $a+b$ are coprime. Another "additive" problem is the Goldbach conjecture.
Since the natural way to deal with prime factors is for... factorization, these kinds of problems look way more complicated. Except sieve methods, are there any conclusive methods?
| https://mathoverflow.net/users/8786 | Methods for "additive" problems in number theory | There's the recent [XYZ conjecture](http://arxiv.org/pdf/0911.4147) of Lagarias and Soundararajan. It concerns bounding $\log(\log(A+B))$ in terms of the largest prime $p$ dividing $AB(A+B).$
Also, a great way to understand properties a triple $(A,B,C)$ of nonzero integers satisfying
$$A + B = C$$
is to consider properties of the Frey elliptic curve
$$ E\_{(A,B,C)} : y^2 = x (x-A) (x+B)$$
as well as additional structures (e.g. modular forms, Galois representations) associated with this elliptic curve. This is one of the routes by which the ABC conjecture was originally discovered.
Barry Mazur's notes [here](http://www.math.harvard.edu/~mazur/papers/scanQuest.pdf) are quite informative on these matters.
| 4 | https://mathoverflow.net/users/4872 | 37691 | 24,244 |
https://mathoverflow.net/questions/37584 | 0 | Let X be a complex manifold. Suppose we have holomorphic line bundles $L\_i$ over $U\_i$ where ${U\_i}$ is an open covering of X. Suppose that $L\_i$ and $L\_j$ restrict to the same line bundle over the intersection of $U\_i$ and $U\_j$.
Can we patch these local line bundles into a global holomorphic line bundle L over X? That is, the restriction of L to $U\_i$ is $L\_i$.
| https://mathoverflow.net/users/nan | Can we patch up line bundles? | The cocycle condition for glueing applies to sheaves on any topological space, in particular to line bundles. See for instance Proposition 5.29 of
<http://math.rice.edu/~hassett/teaching/465spring04/CCAGlec5.pdf>.
| 2 | https://mathoverflow.net/users/1149 | 37697 | 24,248 |
https://mathoverflow.net/questions/37638 | 4 | Let $G$ and $H$ be permutation groups on the natural numbers such that the orbits of $G$ and $H$ are all finite. Suppose that for all $\pi \in Sym(\mathbb{N})$, there is some $N$ (depending on $\pi$) such that for all $n \ge N$, the ordered tuple $(\pi(1),\pi(2),\dots,\pi(n))$ has a larger orbit (by a fixed ratio) under $G$ than it has under $H$.
Can $G$ and $H$ be conjugate in $Sym(\mathbb{N})$?
Edit: Answer is 'yes' (see Jim Belk's comment below); indeed $G$ can be conjugate to proper subgroups of itself of finite index, which makes the size of tuple orbit property automatic.
But what if $G$ only has finitely many orbits of size $n$ for each $n \in \mathbb{N}$? This would at least ensure that $G$ cannot be conjugate to one of its own subgroups.
Edit 2: An example would need to have the following property:
There is a tuple $t$, such that for any tuple $u$ for which $G\_u$ is contained in $G\_t$, then the $G$-orbit of $u$ is larger than the $H$-orbit of $u$.
So for instance if we pick a tuple $u$ by saying 'choose a large number $K$, then choose from among the $K$-tuples with no repeats one with smallest possible $G$-orbit', then $G\_u$ would not be contained in $G\_t$ no matter how large $K$ is. I think this rules out examples where the tuple stabilisers of $G$ are totally ordered, for instance if $G$ is cyclic and all orbits have length a power of a fixed prime.
| https://mathoverflow.net/users/4053 | Distinguishing finite-orbit permutation groups by action on tuples | Here's a case where $G$ and $H$ can be conjugate. First some notation: given a sequence $\{k\_n\}$ of positive integers, let $[k\_1,k\_2,\ldots]$ denote the permutation
$$(1,\ldots,k\_1)(k\_1+1,\ldots,k\_1+k\_2)(k\_1+k\_2+1,\ldots,k\_1+k\_2+k\_3)\cdots$$
with cycles of size $k\_1,k\_2,k\_3\ldots$. For example, $[1,1,1,1,\ldots]$ denotes the identity, $[2,2,2,2,\ldots]$ denotes $(1,2)(3,4)(5,6)(7,8)\cdots$, and $[2,3,2,3\ldots]$ denotes $(1,2)(3,4,5)(6,7)(8,9,10)\cdots$.
Let
$$g = [1,2,\;\;1,2,4,\;\;1,2,4,8,\;\;\ldots],$$
let
$$h = [1,1,1,\;\;1,1,1,2,2,\;\;1,1,1,2,2,4,4,\;\;\ldots],$$
and let $G$ and $H$ be the cyclic subgroups generated by these elements. Since $g$ and $h$ have the same cycle structure, they are conjuagte in $Sym(\mathbb{N})$, so $G$ and $H$ are conjugate subgroups.
However, for sufficiently large $n$, the orbit of $(\pi(1),\pi(2),\ldots,\pi(n))$ under $G$ will be precisely twice the size of the orbit under $H$.
Of course, in this example $G$ and $H$ both have infinitely many orbits of size $2^k$ for every $k$, so this does not answer the more restrictive version of the question.
| 6 | https://mathoverflow.net/users/6514 | 37701 | 24,251 |
https://mathoverflow.net/questions/33934 | 12 | ### Background
Let $X$ denote a smooth projective curve over $\mathbb{C}$ and let $G$ denote a semi-simple simply connected algebraic group over $\mathbb{C},$ which has associated flag variety $G/B.$
Then we can consider the variety $Maps^d(X, G/B)$ of maps from $X$ to $G/B$ of fixed degree $d$ where $d$ is an $\mathbb{N}$-linear combination of coroots of $G.$ See the top of page 2 of this paper by Alexander Kuznetsov [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019) for the definition of degree. The Plucker embedding of the flag variety into projective space gives an alternative formulation of $Maps^d(X, G/B)$ which can be found in section 1.2 of [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019) or in this survey article of Alexander Braverman [Braverman](http://arxiv.org/abs/math/0603454).
In general, $Maps^d(X, G/B)$ is not compact, but there is a compactification due to Drinfeld, which is referred to as the variety of quasi-maps and denoted $QMaps^d(X, G/B).$ See [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019) or [Braverman](http://arxiv.org/abs/math/0603454).
On the other hand, when $G = SL\_n,$ there is a second compactfication due to Laumon. This is because when $G = SL\_n,$ we have both the Plucker embedding description of the flag variety, but also the description of the flag variety as flags of vector spaces. This latter description gives another formulation of $Maps^d(X, G/B)$ but leads to a compactification known as quasi-flags. Once again, see [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019). When $n>2,$ varieties of quasi-maps and of quasi-flags are different. It turns out that quasi-flags are always smooth, while quasi-maps have singularities.
Broadening our focus somewhat, we could instead consider the representable map of stacks $Bun\_B(X) \to Bun\_G(X),$ and note that the fiber over the trivial $G$-bundle is the union of all the $Maps^d(X, G/B)$ for all possible degrees (note that the degree just tells us which connected component of $Bun\_B$ we live in).
Just as the variety of maps above was not compact, the map $Bun\_B \to Bun\_G$ is not proper. But there exists a relative compactification of $Bun\_B,$ also referred to as the Drinfeld compactification, which I will denote $Bun\_B^D.$ This compactification still maps to $Bun\_G,$ but the map is now proper. The fiber over the trivial bundle of this map coincides with the union of all $QMaps^d(X, G/B).$
As before, when $G = SL\_n,$ there is a second compactification of $Bun\_B$ which I will denote $Bun\_B^L$ whose fiber over the trivial bundle coincides with the union of all the quasi-flags varieties. See this paper by Braverman and Gaitsgory [BG](http://arxiv.org/abs/math/9912097) or this follow-up paper by Braverman, Gaitsgorgy, Finkelberg, and Mirkovic [BGFK](http://arxiv.org/abs/math/0012129) for more details.
### Question
In [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019), Kuznetsov proves that when $X = \mathbb{P}^1$ and $G = SL\_n,$ there is a map from the space of quasi-flags of degree $d$ to the space of quasi-maps of degree $d$ which is a small resolution of singularities.
Later, in [BG](http://arxiv.org/abs/math/9912097), it is asserted that Kuznetsov proved that $Bun\_B^L(X)$ is a small resolution of singularities of $Bun\_B^D(X)$ for any smooth projective curve $X.$
It seems to me that there are two discrepancies here. One has to do with an arbitrary smooth projective curve versus $\mathbb{P}^1.$ The second has to do with moving from the varieties of quasi-maps and quasi-flags to the stacks $Bun\_B^D$ and $Bun\_B^L.$
Does anyone know a reference which explains the bridge between [Kuznetsov](http://arxiv.org/abs/alg-geom/9610019) and the assertions of [BG](http://arxiv.org/abs/math/9912097)? Or perhaps this was just something clear to the experts which never warranted an explanation?
| https://mathoverflow.net/users/916 | Reference Request for Drinfeld and Laumon Compactifications | I guess that formally this is not written anywhere, but it is indeed easy to deduce the general case from Kuznetsov's result. The point is that both Drinfeld and Laumon compactifications consist of G-bundles with some kind of degenerate B-structure, where
the degeneration occurs at finitely many points of the curve. It is easy to see that the
fiber of of the map $Bun\_B^L\to Bun\_B^D$ depends only on the behaviour of everything
in the formal neighbourhood of those points (using this observation you can formally reduce
your question to Kuznetsov's result).
| 10 | https://mathoverflow.net/users/3891 | 37705 | 24,253 |
https://mathoverflow.net/questions/31016 | 6 | Find distinct positive real numbers $x\_1$ , $x\_2$ , ... of least supremum such that, for each positive integer $n$, any two of 0, $x\_1$ , $x\_2$ ,..., $x\_n$ differ by $1/n$ or more.
Note that the hurdle term $1/n$ is optimal in the sense that any replacement for it would need to stay below a constant multiple of it to allow a finite supremum. By a nonconstructive proof, there is a unique solution minimal with respect to the lexicographic ordering of real sequences; so a constructed solution (while eluding me) doesn't seem impossible.
Although I haven't seen this problem anywhere, it looks too simple not to have been posed before. Any pointers would be welcome.
| https://mathoverflow.net/users/7458 | A sequential optimizing task | Belated thanks to Kevin O'Bryant for his pointer to discrepancy theory. This led me eventually to a source where the problem is solved: See Theorem 6.7 in Harald Niederreiter's book *Random Number Generation and Quasi Monte Carlo Methods* (SIAM 1992). The logarithmic sequence described by Tracy Hall is due to Rusza and is indeed optimal.
| 0 | https://mathoverflow.net/users/7458 | 37706 | 24,254 |
https://mathoverflow.net/questions/37708 | 24 | The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$. What can we say in the special case of 2-manifolds? For example, can we always embed a 2-manifold in $R^3$?
| https://mathoverflow.net/users/nan | Nash embedding theorem for 2D manifolds | The Nash-Kuiper embedding theorem states that any orientable 2-manifold is isometrically ${\cal C}^1$-embeddable in $\mathbb{R}^3$.
A theorem of Thompkins [cited below] implies that as soon as one moves to ${\cal C}^2$, even
compact flat $n$-manifolds cannot be isometrically ${\cal C}^2$-immersed in $\mathbb{R}^{2n-1}$.
So the answer to your question for smooth embeddings is: *No*, as others have pointed out.
I believe Gromov reduced the dimension you quote of the space needed for any compact surface to 5,
but I don't have a precise reference for that.
Tompkins, C. "Isometric embedding of flat manifolds in Euclidean space," *Duke Math.J.* **5**(1): 1939, 58-61.
**Edit.** Both Deane Yang and Willie Wong were correct that the Gromov result is in
*Partial Differential Relations*. I believe this is it, on p.298: "We construct here an isometric $\cal{C}^\infty$
($\cal{C}^{\mathrm{an}}$)-imbedding of $(V,g) \rightarrow \mathbb{R}^5$ for all compact surfaces $V$."
$g$ is a Riemannian metric on $V$.
| 22 | https://mathoverflow.net/users/6094 | 37717 | 24,263 |
https://mathoverflow.net/questions/37675 | 6 | In general to prove that a given problem is NP-complete we show that a known NP-complete problem is reducible to it. This process is possible since Cook and Levin used the logical structure of NP to prove that SAT, and as a corollary 3-SAT, are NP-complete. This makes SAT the "first" NP-complete problem and we reduce other canonical NP-complete problems (e.g. CLIQUE, HAM-PATH) from it.
My question is whether there is a way to prove directly from the definition/logical structure of NP that a different problem (i.e. not SAT) is NP-complete. A friend suggested that it would be possible to tailor the proof of the Cook-Levin Theorem to show that, for example, CLIQUE is NP-complete by introducing the reduction from SAT during the proof itself, but this is still pretty much the same thing.
| https://mathoverflow.net/users/8981 | An Alternative to the Cook-Levin Theorem | In his infamously short paper "Average-case complete problems," Leonid Levin uses a tiling problem as the master ("first") NP-complete average-case problem (which means he also automatically uses it as a master NP-complete problem).
UPDATE: Contrary to what I was speculating in my answer previously, in his original paper on the Cook-Levin theorem (I found an English translation linked to from the [Wikipedia Cook-Levin theorem article](http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem)), it's not clear whether Levin uses the tiling problem as a master problem. He lists six NP-complete problems (SAT is number 3, and the tiling problem is number 6), but leaves out the proofs, so it's not completely clear which one is the master problem. Very likely, it was SATISFIABILITY, and so Levin found essentially the same proof as Cook.
| 10 | https://mathoverflow.net/users/2294 | 37718 | 24,264 |
https://mathoverflow.net/questions/37657 | 2 | I was wondering if anything was known about the following:
Let $\mathbb{D}^2=\lbrace x^2+y^2< 1 \rbrace \subset \mathbb{R}^2$ be the open unit disk.
Consider now the Green's functions $G(z; p)$ of this disk. I.e. here $p\in \mathbb{D}^2$ and $G(z;p)$ is smooth and harmonic in $\bar{\mathbb{D}}^2\backslash \lbrace p \rbrace$, vanishes on the boundary and has the property that $H(z;p) =G(z; p)-\log |z-p|$ is smooth.
Now consider the set of functions:
\begin{equation}
S=\lbrace f\in C^\infty(\partial \mathbb{D}^2): f= \sum\_{i=1}^n \lambda\_i \partial\_{\nu} G(z; p\_i), \lambda\_i \in \mathbb{R}, p\_i \in \mathbb{D} \rbrace
\end{equation}
Here $\partial\_{\nu} G(z; p\_i)$ is the normal derivative on $\partial \mathbb{D}^2$.
My question is what can be said about the set $S$? In particular, is there any hope that it is dense in $L^2(\partial \mathbb{D}^2)$?
Playing around with things all I got was a big mess so any references would be appreciated.
| https://mathoverflow.net/users/26801 | The normal derivative of the Green's function | Yes indeed, $S$ is dense in $L^2(\partial \mathbb{D})$.
This is because any $g\in L^2(\partial \mathbb{D})$ has an $L^2$ harmonic extension $h$ to $\mathbb{D}$, and $$h(p)=\frac{1}{2\pi}\int\_{\partial \mathbb{D}} g(z) \partial\_{\nu} G(z; p) dz\;\;\;(\*)$$ by Green's formula. Hence if $g$ is orthogonal to $S$, $h(p)=0$ for all $p$, and $g=0$.
Implicit are the non obvious results that there is a continuous trace from $L^2$ harmonic function in $\mathbb{D}$ to $L^2$ functions on the boundary and that the trace of $h$ given by $(\*)$ is indeed $g$. This should be found in many books treating harmonic analysis or one variable complex analysis, like the references [here](http://en.wikipedia.org/wiki/Poisson_kernel), or Rudin's *Real and complex analysis*.
| 1 | https://mathoverflow.net/users/6451 | 37724 | 24,267 |
https://mathoverflow.net/questions/37726 | 7 | This is a basically an adjusted version of my earlier question about how to define a convolution algebra on a general Riemannian manifold. The motivation for asking such a question of course comes from the observation that if G is a group and X is a manifold and the action of G on X is transitive, then the pullback from each point in X to its orbit is faithful. This then cuts out an ideal in the convolution algebra on G which would (hopefully) correspond to some type of general convolution on X (and that would be pretty handy to have for many obvious reasons).
My intuition is that for 2D surfaces (which is the case I am most interested in right now), the group is going to be something like $PSL(2) / \pi(X)$ with the action obtained by the pushforward of the action of $PSL(2)$ on $RP^2$ by the universal covering of $RP^2 / \pi(X)$. Of course, trying to work all of this out via quotient relations is an enormous pain in the neck, so it would be nice to maybe avoid some headaches and instead try to find maybe some standard references for this sort of thing (if they exist at all).
| https://mathoverflow.net/users/4642 | Which Riemannian manifolds admit a finite dimensional transitive Lie group action? | At least in the compact case, there's a topological obstruction. In a 2005 paper, Mostow proved that a compact manifold that admits a transitive Lie group action must have nonnegative Euler characteristic. Here's the reference:
[MR2174096](http://www.ams.org/mathscinet-getitem?mr=2174096) (2007e:22015)
Mostow, G. D.
A structure theorem for homogeneous spaces.
Geom. Dedicata 114 (2005), 87--102.
Of course, even if X admits a transitive Lie group action, most Riemannian metrics on X will not be homogeneous. (You didn't say whether you wanted actions by isometries, but I assume that's what you're interested in, because otherwise the Riemannian structure on X is irrelevant.) In the 2D case, the only compact, connected, homogeneous Riemann surfaces are the sphere, $RP^2$, the torus (and maybe the Klein bottle?)\*, all with constant-curvature metrics. In general, the group has to be compact, because the isometry group of a compact Riemannian manifold is itself compact.
\*EDIT 3: An earlier paper by Mostow constructed a transitive group action on the Klein bottle, but I doubt that this action preserves a Riemannian metric. It's a complicated construction, so I haven't had a chance to work through it in detail, but here's the reference:
Mostow, G.D., [The Extensibility of Local Lie Groups of Transformations and Groups on Surfaces](http://www.jstor.org/stable/1969437). Ann. Math., Second Series, (52) No. 3 (1950), 606-636.
I don't know what's known in the noncompact case.
| 16 | https://mathoverflow.net/users/6751 | 37729 | 24,268 |
https://mathoverflow.net/questions/37698 | 4 | Cartan's theorem A says that on for a coherent sheaf ${\mathcal{F}}$ on a Stein manifold X, the fibres ${\mathcal{F}}\_x$ over each point x in X are generated by global sections.
I'm wondering if there are compact analogues of these theorem. Here I consider holomorphic line bundles over a compact complex manifold X. Consider the fibre stalk ${\mathcal{O}}\_{X,x}$ of holomorphic germs over some point x. Is this fibre stalk generated by quotients of global holomorphic line bundle sections? That is, given two global sections $s$ and $t$ of the same line bundle $L\to X$. The quotient $s/t$ is a global meromorphic function. Given a holomoprhic function germ $f\_x\in {\mathcal{O}}\_{X,x}$, we can always find such $s$ and $t$ such that $(s/t)\_x=f\_x$?
An analogue for Cartan's theorem B would be nice too. But I can't phrase this precisely.
| https://mathoverflow.net/users/nan | Are there compact analogues of Cartan's theorems A and B? | Colin, I think you have answered your own question in your response to Brian Conrad. The fraction field of $\mathcal O\_{X,x}$ has infinite transcendence degree over $\mathbb C$, while $\mathcal M(X)$, in Elencwajg's notation, has finite transcendence degree. For any global sections $s,t$ of a holomorphic line bundle, the ratio $s/t$ lies in $\mathcal M(X)$. So most $f$ cannot be written as $s/t$.
| 2 | https://mathoverflow.net/users/8726 | 37731 | 24,269 |
https://mathoverflow.net/questions/37709 | 4 | Let $G$ be a connected complex semisimple Lie-group, $T$ a maximal torus and $B$ a Borel subgroup containing it. Let $\phi:G\rightarrow G/B$ denote the projection.
Given a representation ($\theta,V$) of $B$, we can define a $G$-equivariant holomorphic vector bundle over the flag variety $X:=G/B$ by
$$ G\times\_B V :=(G\times V)/\{(g,v)\sim(gb^{-1},\theta(b)v),\forall b\in B\}.$$
Its sheaf of sections $\mathcal{I}(\theta)$ may be described as the holomorphic functions
$$\mathcal{I}(\theta)(U)=\{f:\phi^{-1}(U)\rightarrow V \mid f(gb^{-1}) = \theta(b)f(g)\}. $$
$G$ acts on a section by $(gf)(x)=f(g^{-1}x)$.
An integral weight $\lambda$ of $T$ gives a character $\chi\_\lambda$ of $B$. Let $\theta\otimes\chi\_\lambda$ denote the tensor product of the representations $\theta$ and $\chi\_\lambda$.
Suppose ($\theta,V$) is the restriction of a representation ($\pi,V$) of $G$, then the associated ($G$-equivariant) vector bundle is trivial (i.e. isomorphic to $ X\times V$ with $(g,(x,v))\mapsto (gx,\pi(g)v)$). Is the identity
$$ \mathrm{H}^i(X,\mathcal{I}(\theta\otimes\chi\_\lambda))\simeq \mathrm{H}^i(X,\mathcal{I}(\chi\_\lambda)) \otimes V $$
as $G$ or $\mathfrak{g}$-modules correct?
Background/Motivation
---------------------
I have been reading about the Borel-Weil theorem lately, and
**Edit:**
This question arose from an attempt to fix a mistake in a book.
The identity is indeed correct and I believe I have found the error elsewhere.
Thanks Chuck and Jim!
| https://mathoverflow.net/users/nan | An identity for sheaf cohomology of flag varieties | This may be too naive an answer, but from my experience this kind of question fits comfortably into the foundational material for reductive algebraic groups over an algebraically closed field of arbitrary characteristic. This is for example treated in Part I of J.C. Jantzen's 2003 AMS second edition of *Representations of Algebraic Groups*. There the starting point is the "tensor identity", followed by "generalized tensor identity" for higher derived functors of induction, which translates for the flag variety into the language of vector bundles and sheaf cohomology. This originates basically in classical Frobenius reciprocity for finite groups but becomes quite flexible in situations involving a reductive group, a Borel (or other parabolic) subgroup, and various finite dimensional rational representations. I'm assuming your *V* is finite dimensional. As Chuck Hague points out, no dualization should occur in your formula.
Most of what goes into the classical Borel-Weil theory has a natural formulation in any characteristic, though Bott's theorem can't be imitated so precisely for nondominant line bundles.
P.S. There are quite a few literature sources (papers by H.H. Andersen, Cline-Parshall-Scott, Donkin, etc.), but Chapter II.5 in Jantzen's book gives a fairly comprehensive treatment in algebraic language of the theorems of Borel-Weil and Bott, along with a derivation of Weyl's character formula. To get back to the classical theory over $\mathbb{C}$ does require some translation of the language. The elegant papers of Demazure using algebraic geometry in characteristic 0 are the underlying inspiration for much of this approach to ideas first developed in the setting of complex geometry or compact Lie groups.
| 3 | https://mathoverflow.net/users/4231 | 37733 | 24,271 |
https://mathoverflow.net/questions/37738 | 4 | This question arose while I was trying to work out examples for the second question of this thread: [Reconstruction Conjecture: Group theoretic formulation?](https://mathoverflow.net/questions/34914/reconstruction-conjecture-group-theoretic-formulation)
In the beginning, I considered some computable properties of groups and wondered whether two groups of the same order having equal value for that computable property would necessarily be isomorphic. For instance, take centers of groups and it is not difficult to find many specific examples where two groups have the same order and isomorphic centers but then the two groups are not necessarily isomorphic. Considering lattice of groups, Scott Carnahan has already given a counterexample there.
Are there any two finite groups of the same order that have the same number of subgroups?
| https://mathoverflow.net/users/5627 | Isomorphism and number of subgroups | There are pairs G,H of nonisomorphic p-groups with isomorphic subgroup lattices (and therefore of the same order). The book ``Subgroup Lattices of Groups", by R. Schmidt, is an excellent reference on this subject.
| 12 | https://mathoverflow.net/users/36466 | 37741 | 24,274 |
https://mathoverflow.net/questions/37746 | 1 | Let $f(x\_1,\ldots , x\_n) = \frac{x\_1}{x\_2+x\_3} + \frac{x\_2}{x\_3+x\_4} + \cdots + \frac{x\_n}{x\_1+x\_2}$, defined for $x\_i>0$.
1. Is there $(x\_1, \ldots ,x\_n)\in {\mathbb{R}^\*\_+}^n$ such that $f(x\_1,\ldots , x\_n) < n/2$?
2. Can we find $\inf\_{x\_i>0}f(x\_1,\ldots , x\_n)$?
| https://mathoverflow.net/users/3958 | Inf of a mutivariate function | This is discussed briefly as a generalization of Shapiro's cyclic sum
inequality by J. Michael Steele in his book
[The Cauchy-Schwarz Master Class](http://books.google.co.uk/books?id=7Fm3r9jcbqYC&lpg=PP1&dq=Steele%2520Cauchy&pg=PA104#v=onepage&q=Shapiro%27s&f=false).
He remarks that (1.) holds for $n\ge25$ and refers to this paper:
P. J. Bushell, Shapiro’s “Cyclic Sums",
*Bull. L.M.S.* (1994) **26**, 564–574.
| 6 | https://mathoverflow.net/users/4213 | 37748 | 24,280 |
https://mathoverflow.net/questions/37728 | 4 | An elementary question about Sobolev spaces:
Is there some explicit theorem about embedding relation between spaces $BV(\Omega)$ and $L^p(\Omega)$?
Formulated otherwise: is $BV$ a subset of $L^2$ (i.e. $BV$ possess regularities of $L^2$) ?
| https://mathoverflow.net/users/9014 | Embedding of $BV$ and $L^p$ spaces | **Edit:** The most general imbedding I know of about $L^p$ spaces is that $BV(\Omega) \subset \subset L^{n/n-1}(\Omega)$ where $\Omega \subset \mathbb{R}^n$ and $n > 1$ (replace $n/(n-1)$ with $1$ when $n=1$). This embedding is **compact** for any $p < n/n-1$. Hence for your $n=2$, $n=3$ interest we have $f \in L^{2}(\Omega)$ for $n=2$ and $f \in L^{3/2}(\Omega)$ for $n=3$. On bounded domains you have all *lower* $L^p$ norms as well by an application of Holder's inequality. Sorry for my initial mis-understanding of your question.
| 3 | https://mathoverflow.net/users/8755 | 37752 | 24,283 |
https://mathoverflow.net/questions/37749 | 17 | Is the blowup of an integral normal Noetherian scheme along a coherent sheaf of ideals necessarily normal?
I can show that there is an open cover of the blowup by schemes of the form $\text{Spec } C$, where $B \subset C \subset B\_g$ for some integrally closed domain $B$ and some $g \in B$, but I don't see why this would imply that $C$ is integrally closed. Intuitively, it seems reasonable that a blowup would be at least as "nice" as the original scheme, but that intuition may have more to do with how blowups are generally used than what they are capable of.
| https://mathoverflow.net/users/5094 | Is the blowup of a normal scheme necessarily normal? | For an explicit example, blow up any sufficiently complicated isolated singularity of a surface in affine 3-space, and the result will in general have singularities along curves so is not normal. I think x2+y4+z5 = 0 will do for example: blowing this up gives x2+y4z2+z3 = 0 on one of the coordinate charts, which is singular along the line x=z=0.
(Hypersurfaces in affine space are normal if and only if they are regular in codimension 1.)
| 23 | https://mathoverflow.net/users/51 | 37756 | 24,287 |
https://mathoverflow.net/questions/37757 | 3 | This question is short, and to the point:
Valuation rings are certainly integrally closed, but are they regular?
The motivation is that I'm trying to understand the resolution of singularities of algebraic surfaces as was done originally, and I'm playing around with some of the ideas involved.
| https://mathoverflow.net/users/5309 | Are valuation rings regular? | Regular local rings are Noetherian by definition, but valuation rings are not unless they happen to be discrete valuation rings. So with the usual definitions, most valuation rings are not regular. (It might be possible to come up with a reasonable definition of regularity for non-Noetherian rings, but I have not heard of one.)
| 10 | https://mathoverflow.net/users/51 | 37758 | 24,288 |
https://mathoverflow.net/questions/37682 | 2 | Given a covariance matrix, how can I construct a vector of expressions of randomly distributed variables whose covariance matrix is equal to the given one?
EDIT: All variables are normally distributed.
I have an algorithm that gets the covariances correct, but not the variances on the diagonal:
```
a = [0]*len(r)
for x, row in enumerate(cov_matrix(r)):
for y, item in enumerate(row):
if x > y: continue
v = noise(math.sqrt(abs(item)))
a[x] += v
if item > 0:
a[y] += v
else:
a[y] -= v
```
I feel like this should be simple ...
| https://mathoverflow.net/users/9005 | Computing equivalent vector of random variables from covariance matrix | If $A$ is your target covariance matrix and $LL^T = A$, and $x = (x\_1, \ldots, x\_n)$ is a vector of independent random variables with mean zero and variance 1, then $y = Lx$ has the required covariance. Here $L$ is a matrix and $L^T$ is its transpose. $L$ can just be the Cholesky factor of $A$. ((Check: $\mathrm{cov}(y) = E[yy^T] = E[(Lx)(Lx)^T] = E[Lxx^TL^T] = LE[xx^T]L^T$ (by linearity of expectation) $= L\mathrm{cov}(x)L^T = LIL^T = LL^T = A$. $\mathrm{cov}(y) = E[yy^T]$ because $y$ has mean 0, and likewise for $\mathrm{cov}(x)$.)
That's not too far from a "complete" solution, actually. If you start with a vector $y$ of random variables with mean zero and covariance matrix $A$, then if $A = LL^T$ and $x = L^{-1}y$, then $\mathrm{cov}(x) = I$. That doesn't necessarily imply that the components of $x$ are independent; it means they are uncorrelated. So the most general construction is to begin with a vector $x$ of *uncorrelated* random variables with mean zero and variance 1 and let $y=Lx$. (I only mean that every example can theoretically be obtained that way, not that it's necessarily the best or most computationally efficient way to do it.)
| 3 | https://mathoverflow.net/users/302 | 37764 | 24,294 |
https://mathoverflow.net/questions/37778 | 7 | If $A \subseteq \mathcal B(\mathcal H)$ is an algebra of operators that is closed under adjoint, then its bicommutant $A''$ is a von Neumann algebra, and is the ultraweak closure of $A$; this is one version of von Neumann's bicommutant theorem. Does the theorem hold relative to an arbitrary von Neumann algebra $\mathcal M$? Concretely, what is the truth value of the following statement:
>
> Let $\mathcal M$ be a W\*-algebra, and $A\subseteq \mathcal M$ be a subalgebra closed under adjoint. Then the relative bicommutant $A'' = \{ m \in \mathcal M | \forall x \in \mathcal M. (\forall a \in A. ax = xa) \implies mx=xm \}$ is a W\*-algebra, and is the ultraweak closure of $A$.
>
>
>
A W\*-algebra is C\*-algebra that is isomorphic to a von Neumann algebra. I use the term W\*-algebra to emphasize that the bicommutant is being computed relative to $\mathcal M$ itself rather than relative to a Hilbert space on which $\mathcal M$ is represented.
**Edit:** As Matthew points out below, $A$ should contain the unit of the ambient algebra, i.e., of $\mathcal B(\mathcal H)$ in the bicommutant theorem, and of $\mathcal M$ in the statement in question.
| https://mathoverflow.net/users/2206 | Relative Bicommutant | Let $A\subseteq B(H)$ be a subset, and let $\text{alg}(A)$ be the algebra generated by $A$. Then it's easy to see that $$A' = \text{alg}(A)'.$$ A similarly easy check shows that if $A$ and $B$ are subsets, then $$A' \cap B' = (A\cup B)' = \text{alg}(A\cup B)'.$$
So, for your question, pick some normal representation $M\subseteq B(H)$ (so that $M''=M$), and let $A\subseteq M$ be a subset. Set $$X=\{x\in M:ax=xa \ (a\in A)\} = A'\cap M = A'\cap M'' = (A\cup M')',$$ so your relative commmutatant is `\[\{m\in M:xm=mx \ (x\in X) \} = X' \cap M = X' \cap M'' = (X\cup M')'.\]` So, yes, this is a von Neumann algebra.
In fact, as $A\subseteq M$, clearly $M'\subseteq A'$ and so $A''\cap M = (A'\cup M')' = A''$. So as $X\subseteq A'$, thus $A''\subseteq X'\cap M$.
**Edit:** There is probably an easier example than this... But, let $M=VN(\mathbb F\_2)$, say with canonical generators $a$ and $b$. Let $A$ be a star-algebra generated by $b$ and $a^{-1}ba$: so $A$ is just linear combinations of $b^n$ and $a^{-1}b^na$ for $n\in\mathbb Z$. A bit of combinatorics shows that $X=\mathbb C1$ and so the the relative bicommutant is all of $M$. However, $A$ is not ultraweakly dense in $M$, because we cannot approximate the generator $a$.
**Conclude:** So, if I haven't messed up, this shows that the relative bicommutant is always a W\*-subalgebra of $M$, but that it might be larger than the ultraweak closure of $A$ in $M$.
| 5 | https://mathoverflow.net/users/406 | 37781 | 24,304 |
https://mathoverflow.net/questions/37783 | 0 | How to design or create or generate a bijective ring map?
| https://mathoverflow.net/users/9026 | How to design or create or generate a bijective ring map? | In generality (this is tagged "commutative algebra", so let's talk commutative rings) I wonder if there is more than taking generators of each side and writing the images as polynomials in the generators of the other side. This is a candidate for a bijection of rings, but so far isn't a homomorphism (you need to check the relations hold). In other words express each ring as a quotient of a polynomial ring, set up homomorphisms from the polynomial rings, and then show the kernels are what they should be.
| 0 | https://mathoverflow.net/users/6153 | 37784 | 24,305 |
https://mathoverflow.net/questions/37777 | 14 | In Presburger Arithmetic there is no predicate that can express divisibility, else Presburger Arithmetic would be as expressive as Peano Arithmetic. Divisibility can be defined recursively, for example $D(a,c) \equiv \exists b \: M(a,b,c)$, $M(a,b,c) \equiv M(a-1,b,c-b)$, $M(1,b,c) \equiv (b=c)$. But some predicates which *can* be expressed in Presburger Arithmetic also have recursive definitions, for example $P(x,y,z) \equiv (x+y=z)$ versus $P(x,y,z) \equiv P(x-1,y+1,z)$, $P(0,y,z) \equiv (y=z)$.
How to tell if a predicate, defined recursively without use of multiplication, has an equivalent non-recursive definition which can be expressed in Presburger Arithmetic?
| https://mathoverflow.net/users/2003 | Which recursively-defined predicates can be expressed in Presburger Arithmetic? | Presburger arithmetic admits [elimination of quantifiers](http://en.wikipedia.org/wiki/Quantifier_elimination), if one expands the language to include truncated minus and the unary relations for divisibility-by-2, divisibility-by-3 and so on, which are definable in Presburger arithmetic. (One can equivalently expand the language to include congruence $\equiv\_k$ modulo $k$ for each natural number $k$.) That is to say, every assertion in the language of Presburger arithmetic is equivalent to a quantifier-free assertion in the expanded language.
It follows that the definable subsets of $\mathbb{N}$ in the language of Presburger arithmetic are exactly the eventually periodic sets. These are comparatively trivial sets, of course, and it means that the set of prime numbers and other interesting sets of natural numbers are simply not expressible in the language of Presburger arithmetic. A similar analysis holds in higher dimensions, and for this reason, we usually think of Presburger arithmetic as a weak theory.
The quantifier elimination argument leads directly to the conclusion that Presburger arithmetic is a decidable theory: given any sentence, one finds the quantifier-free equivalent formulation, and such sentences are easily recognized as true or false.
There is an interesting account in [these slides for a talk](http://homepage.divms.uiowa.edu/~tinelli/classes/295/Spring05/notes/quantifier-elim.pdf) by Cesare Tinelli.
| 24 | https://mathoverflow.net/users/1946 | 37786 | 24,306 |
https://mathoverflow.net/questions/37792 | 25 | The homotopy groups $\pi\_{n}(X)$ arise from considering equivalence classes of based maps from the $n$-sphere $S^{n}$ to the space $X$. As is well known, these maps can be composed, giving arise to a group operation. The resulting group contains a great deal of information about the given space. My question is: is there any extra information about a space that can be discovered by considering equivalence classes of based maps from the $n$-tori $T^{n}=S^{1}\times S^{1}\times \cdots \times S^{1}$. In the case of $T^{2}$, it would seem that since any path $S^{1}\to X$ can be "thickened" to create a path $T^{2}\to X$ if $X$ is three-dimensional, the group arising from based paths $T^{2}\to X$ would contain $\pi\_{1}(X)$. Perhaps more generally, can useful information be gained by examining equivalence classes of based maps from some arbitrary space $Y$ to a given space $X$.
| https://mathoverflow.net/users/6856 | A possible generalization of the homotopy groups. | There's always information to be got. But in this case:
* Based homotopy classes of maps $T^2\to X$ don't form a group! To define a natural function $\mu\colon [T,X]\_\*\times [T,X]\_\*\to [T,X]\_\*$, you need a map $c\colon T\to T\vee T$ (where $\vee$ is one point union). And if you want $\mu$ to be unital, associative, etc., you'll want $c$ to be counital, coassociative, etc. For $T=T^n$ with $n\geq2$, there is no $c$ that is counital. (The usual way to see this is to think about the cohomology $H^\*T$ with its cup-product structure.)
* The inclusion $S^1\vee S^1\to T^2$ gives a map
$$r\colon [T^2,X]\_\* \to [S^1\vee S^1,X]\_\*\approx \pi\_1X\times \pi\_1X.$$
The *image* of this map will be pairs $(a,b)$ of elements in $\pi\_1X$ which commute: $ab=ba$. It won't usually be injective; so there might be something interesting to think about the in preimages $r^{-1}(a,b)$.
| 36 | https://mathoverflow.net/users/437 | 37793 | 24,310 |
https://mathoverflow.net/questions/37800 | 12 | [This](https://mathoverflow.net/questions/37792/a-possible-generalization-of-the-homotopy-groups) question got me thinking about what makes the fundamental group (or groupoid) tick. What is so special about the circle? As another possible candidate for generalization, what about taking the one point compactification of the long closed ray R and thinking about homotopy theory with R in place of the interval? Would a theory of "long homotopy" arise? As a follow up, if this doesn't work, are there any other interesting instances of replacing the unit interval with another topological space and getting an interesting homotopy theory out of it? If not is there some characterization of the interval as the unique space which induces a nice homotopy theory?
| https://mathoverflow.net/users/1106 | Long line fundamental groupoid | The compactified long closed ray $\overline R$ will have two endpoints,
but these are distinguishable. One has a neighbourhood
homeomorphic to $[0,1)$ and the other doesn't. This scuppers
"long homotopy" being a symmetric relation.
(Also the transitivity would fail too.)
The standard notion of homotopy relies on the interval $I$
having distinguished points $0$ and $1$, there being a self
map of $I$ swapping $0$ and $1$, and there being a map from
$I\coprod I/\sim$ to $I$ where $\sim$ is the equivalence relation
identifying the $1$ in the first component to the $0$ in the second.
These maps have to satisfy various formal properties. There
is no continuous map of $\overline R$ swapping its "endpoints",
so we can't mimic the classical notion of homotopy.
| 18 | https://mathoverflow.net/users/4213 | 37802 | 24,316 |
https://mathoverflow.net/questions/37808 | 10 | Hi
I hope this question is accurate. Gödel and Cohen could show that the Continuum Hypothesis (CH) is independent from ZFC using models in which CH holds, and fails respectively.
My question now is:
If we take a large enough part of the universe, say $V\_{\omega\_{\omega}}$ , does CH hold in it?
And why can't we go on to argue like this:
If $V\_{\omega\_{\omega}} \models CH$ then there is a bijection between $\omega\_1$ and $2^{\aleph\_0}$, and as $V\_{\omega\_{\omega}}$ is transitive this is 'really' a bijection thus CH holds in ZFC.
If $V\_{\omega\_{\omega}} \models \lnot CH$ then, as any possible bijection between $\omega\_1$ and $2^{\omega}$ is already an element of $V\_{\omega\_{\omega}}$, CH is inside ZFC refuteable?
This are maybe crude argumentations but I can't answer that by myself.
| https://mathoverflow.net/users/8996 | Does the Continuum Hypothesis hold for sets of a certain rank? | The Continuum Hypothesis, viewed as the assertion that every subset of $P(\omega)$ is either countable or bijective with $P(\omega)$, is expressible already in $V\_{\omega+2}$, since that structure has the full $P(\omega)$ and all subsets of it, as well as all functions between such subsets (one should use a flat pairing function for ordered pairs, which doesn't require one to increase rank for pairs). Thus, CH holds if and only if it holds in any (or all) $V\_\alpha$ for $\alpha\geq\omega+2$.
| 11 | https://mathoverflow.net/users/1946 | 37811 | 24,322 |
https://mathoverflow.net/questions/37787 | 6 | While playing around with the fractional calculus, I got stuck trying to show that two different ways of differintegrating the cosine give the same result. DLMF and the Wolfram Functions site don't seem to have this "identity" or something that can obviously be transformed into what I have, so I'm asking here.
The "identity" in question is
$(\alpha-1)\left({}\_1 F\_2 \left(1;\frac{1-\alpha}{2},\frac{2-\alpha}{2};-\frac{x^2}{4}\right)-{}\_1 F\_2 \left(-\frac{\alpha}{2};\frac12,\frac{2-\alpha}{2};-\frac{x^2}{4}\right)\cos(x)\right)\stackrel{?}{=}\alpha x \sin(x)\,{{}\_1 F\_2 \left(\frac{1-\alpha}{2};\frac32,\frac{3-\alpha}{2};-\frac{x^2}{4}\right)}$
Expanding the LHS minus the RHS in a Taylor series shows that the coefficients up to the 50th power are 0; trying out random complex values of $\alpha$ and $x$ seems to verify the identity. I would however like to see a way to confirm the identity analytically. How do I go about it?
| https://mathoverflow.net/users/7934 | Proving a hypergeometric function identity | You can use the great [HolonomicFunctions](http://www.risc.jku.at/research/combinat/software/HolonomicFunctions/index.php) package by Christoph Koutschan to prove this identity in Mathematica. It automatically proves for you that both sides of your identity satisfy the sixth order differential equation
\begin{eqnarray}
0=&&x^2 \left(2 a^2-11 a+18 x^2+14\right) D\_x^6
-2 x \left(2 a^3-19 a^2+18 a x^2+58 a-54 x^2-56\right) D\_x^5 \\\\
&&+\left(2 a^4-25 a^3+28 a^2 x^2+115 a^2-133 a x^2-230
a+90 x^4+154 x^2+168\right) D\_x^4 \\\\
&&-4 x
\left(4 a^3-37 a^2+36 a x^2+115 a-99 x^2-114\right)
D\_x^3 \\\\
&&+4 \left(2 a^4-23 a^3+20 a^2
x^2+96 a^2-71 a x^2-172 a+18 x^4+71 x^2+112\right) D\_x^2 \\\\
&&+8 x \left(4 a^2-34
a+36 x^2+43\right) D\_x
+8 \left(2 a^2-17 a+18 x^2+35\right).
\end{eqnarray}
Together with your check that the first six Taylor coefficients (with respect to x) agree, this proves your identity.
| 7 | https://mathoverflow.net/users/359 | 37836 | 24,337 |
https://mathoverflow.net/questions/37838 | 4 | ### Background
I am reading Tom Blyth's book [*Categories*](http://books.google.co.uk/books?id=mfUZAQAAIAAJ) as I am thinking of using it as a guide for a fourth-year project I'll be supervising this academic year. The books seems the right length and level for the kind of project we require of our final year single honours students in Edinburgh.
In the fourth chapter of the book, a *normal category* is defined as one with zero objects, in which every morphism has a kernel and a cokernel, and in which every monomorphism is a kernel. This last condition can be rephrased as saying that monomorphisms are normal.
My confusion stems from Theorem 4.6 in the book which states that [*a normal category has pullbacks*](http://books.google.co.uk/books?ei=lDKETLrSKMSBswb66rDmCA&ct=result&id=mfUZAQAAIAAJ&dq=%252522A+normal+category+has+pullbacks%252522&q=A+normal+category+has+pullbacks#search_anchor). The proof in the book seems to use that in the diagram
$$\begin{matrix}
& B \cr
& \downarrow \cr
A \rightarrow & C \cr
\end{matrix}
$$
whose limit is the desired pullback, the morphism $A \to C$ is a kernel. This seems to me an unwarranted assumption, since it would seem to imply, in particular, that generic morphisms are monic.
Alas, I have not been able to find an independent proof of the theorem and I am starting to suspect that this may not be true. Googling seems not to be of much help. For one thing one has to wade through a surprising number of hits which have nothing to do with category theory.
Since much of the rest of the chapter seems to depend on this result, I am a little stuck. I have emailed the author, who is an emeritus professor across the firth in St Andrews, but so far no reply. So I thought I would try it here in MO, since I'm sure to get an authoritative answer to my
### Two questions
1. Is the result true? And if so,
2. Can someone point me to a proof?
Thanks in advance!
| https://mathoverflow.net/users/394 | Do normal categories have pullbacks? | If I'm not mistaken, the category “vector spaces of dimension $\leq n$” (for any $n > 0$) is a counterexample? The zero object, kernels, cokernels, and the normality of kernels can all be computed as they normally are for vector spaces; but products (and hence pullbacks) are missing for obvious reasons of dimension. *[See comments for elaboration.]*
The problem is, intuitively, that there's nothing in the definition of “normal” providing a way to build bigger things out of smaller.
On the other hand, I think one can prove “a normal category has pullbacks of monos”; it sounds like that might be what the book is proving here? Maybe that's all that it actually ends up using in the rest of the chapter, and this is just an omission in the statement of the theorem?
Alternatively, if one adds products to the definition of “normal”, then from that together with pullbacks of monos, one can build all pullbacks (the pullback of $f$ and $g$ is the pullback of the appropriate diagonal map (a mono) along $f \times g$).
| 4 | https://mathoverflow.net/users/2273 | 37839 | 24,339 |
https://mathoverflow.net/questions/37853 | 6 | I feel like many of the algorithms that I learned — indeed, that I have taught — in undergraduate linear algebra classes depend sensitively on whether certain numbers are $0$. For example, many a linear algebra homework exercise consists of a matrix and a request that the student calculate a basis for the kernel or image. The standard approach consists of row-reducing the matrix and reading off the answer. The algorithm to row-reduce a matrix has many steps of the form "if $a \neq 0$, do something that involves a division by $a$, and if $a = 0$, do something else". Such a step is unfortunate from many points of view. In particular, it is not even continuous in $a$, so if you only have partial data about the value of $a$ (say, a truncated decimal expansion), then you cannot hope to apply this algorithm.
For the purpose of calculating kernel and image bases, perhaps this is not too surprising. Indeed, the dimensions of the kernel and image of a basis do not depend polynomially on the coefficients — they don't even depend continuously, only semicontinuously — and so there is really no hope in writing an algorithm that computes the coefficients of a basis for either and that is algebraic in the matrix.
On the other hand, even the usual algorithm we teach to compute inverses to invertible matrices again uses row reduction. The final answer is algebraic in the matrix coefficients, and there are algorithms that run algebraically (something to do with minors). So my (slightly ambiguous and open-ended) questions are:
>
> What problems in undergraduate linear algebra can be solved by algorithms that are totally polynomial/algebraic/continuous in the matrix coefficients? In settings where there is not complete information about a matrix — e.g. when the coefficients are given by truncated decimal expansions — what types of algorithms do people actually use to "approximate" the answers to problems that cannot be solved continuously? And how do people who do such problems measure/define how good their "approximations" are?
>
>
>
My motivation for asking this question was the (closed) question [matrix that annihilates matrix](https://mathoverflow.net/questions/37842/matrix-that-annihilates-matrix), in which the asker asks for an algorithm that, when fed an $n\times m$ matrix $A$, computes an $m\times n$ matrix $B$ so that $\ker B = \operatorname{im} A$. This is another example of a problem that is easy to do by row reduction, but it's not clear to me if there are solutions that run "more" continuously than that. (There won't be a completely continuous algorithm. Set $m=n$ and feed in an invertible matrix $A$. Then $B = 0$. But the invertibles are dense among all matrices.)
| https://mathoverflow.net/users/78 | To what extent can algorithms in undergraduate linear algebra be made continuous/polynomial/etc.? | There are perhaps three or four themes lurking in here. NB that "undergraduate linear algebra" is perhaps an artificial construct, and examples set to test whether students understand basic concepts do not really need formal algorithms to do that. What "we teach" may simply be a pedagogic construct. NB the discussion about Cramer's rule already poses the issue of what the point is (formulae or practical answers).
There is a whole area of "numerical linear algebra" for those who need actual answers involving real numbers, that are meaningful and stable.
There is an underlying concept of "bad locus", e.g. the vanishing of the discriminant of the characteristic polynomial of a square matrix, so that most serious problems in linear algebra come out as working well away from the "bad locus". There is probably a way of saying this in algebraic terms, by inverting elements of rings. In other words one can teach things that work well on Zariski-open sets, and the question is then more one of conscience.
But there is another slant, which is that Gaussian elimination and what lies behind it reveal other kinds of serious mathematics. E.g. Schubert cells. Here the combinatorial problems of non-general position are not to be waved away. This material has retained such importance for mathematicians that I doubt it should be de-emphasised in teaching.
| 12 | https://mathoverflow.net/users/6153 | 37858 | 24,349 |
https://mathoverflow.net/questions/37877 | 7 | Given a smooth complete intersection $X=D\_{1} \cap D\_{2} \cap \cdots \cap D\_{k} \subset \mathbb{P}^{n}$ with ${\rm deg}\; D\_i=d\_i$, one can easily show that $\omega\_{X} \simeq \mathcal{O}\_{X}(\sum\_{i=1}^{k} d\_{i} -n-1)$, using induction on the number of hypersurfaces and the usual conormal sequence.
Here is the question.
Suppose $X \subset \mathbb{P}^{n}$ is a smooth projective variety of degree $d$, not necessarily a complete intersection. How to understand $\omega\_{X}$ in terms of the embedding?
Is it even necessarily true that $\omega\_{X}$ is restricted from a line bundle on $\mathbb{P}^{n}$?
Similarly, how to work out the cohomology of $\mathcal{O}\_X$ and $\omega\_X$? Does this only depend on the degree of $X$?
| https://mathoverflow.net/users/9046 | Degree of canonical bundle? | Smooth (or Gorenstein) subvarieties in $\mathbb P^n$ whose canonical bundle is a restriction from $\mathbb P^n$ are known as *subcanonical*, and are very special. A rational twisted cubic in $\mathbb P^3$ is not subcanonical, for obvious reasons of degree.
It is most certainly not true that the cohomologies of $\mathcal O\_X$ and $\omega\_X$ only depend on the degree: for example, consider a twisted cubic as above and a plane cubic embedded in $\mathbb P^3$.
| 13 | https://mathoverflow.net/users/4790 | 37883 | 24,360 |
https://mathoverflow.net/questions/37875 | 4 | A cograph is a graph without induced $P\_4$ subgraphs. I am looking for a reference for a simple exponential bound on the number of distinct unlabeled cographs on $n$ vertices. By [the Mathworld article on cographs](http://mathworld.wolfram.com/Cograph.html) this is the same as the number of series-parallel networks with $n$ unlabeled edges. Judging from the first couple of terms in the list, the bound should be something like $3^n$. Any reference to a simple, exponential bound in closed-form would be much appreciated.
| https://mathoverflow.net/users/5200 | Bound on the number of unlabeled cographs on n vertices | Let your sequence be $a\_n$ for the number of series-parallel networks with $n$ unlabeled edges. The following identity of generating functions holds
$$1+\sum\_{k=1}^{\infty}a\_kx^k=\left[\frac{1}{(1-x)}\prod\_{k\geq 1}\frac{1}{(1-x^k)^{a\_k}}\right]^{1/2}$$ from which the asymptotics
$$a\_n\sim C d^n n^{-3/2}$$ follow, where $C=0.4126...$, $d=3.56083930953894...$
See also the article "Some enumerative results on series parallel networks" by J.W. Moon.
| 6 | https://mathoverflow.net/users/2384 | 37886 | 24,363 |
https://mathoverflow.net/questions/37867 | 0 | I've been working on sorting and factorisation problems on permutations for some time now, and have observed that given a permutation $\pi$ of $n$ elements, the permutation $\pi^\chi=\chi\circ\pi\circ\chi^{-1}$, where $\chi=\chi^{-1}=(n\ n-1\ \cdots\ 2\ 1)$, often has attractive properties (with respect to a particular sorting problem).
Is there a name for this "special" permutation (other than "the conjugate of $\pi$ by $\chi$")?
| https://mathoverflow.net/users/3356 | Name of a particular conjugate permutation | This is the reverse-complement of $\pi$.
In one-line notation, the reverse of a permutation is what you get by writing it backwards and the complement of a permutation is what you get when you replace each entry $i$ by $n -i + 1$. (In other words, one of these operations is multiplication by $\chi$ on the right, the other on the left.) The reverse-complement is what you get by doing both of these operations, or equivalently by giving the permutation matrix a half-turn. (Together with inversion, these operations generate the dihedral group acting on each permutation matrix.)
| 5 | https://mathoverflow.net/users/4658 | 37894 | 24,366 |
https://mathoverflow.net/questions/37888 | -1 | How to define homotopy groups in categories as in Quillen's definition for Higher algebraic K-theory: K\_i(M)=\pi\_{i+1}(BQM, 0), where M is a small category and BQM is the classifying space of QM. thank you.
| https://mathoverflow.net/users/8532 | Definition for fundamental group (higher homotopy groups) for a category? | In this definition BQM can be taken to be a space - the [geometric realization](http://en.wikipedia.org/wiki/Simplicial_set#Geometric_realization) of the [nerve](http://en.wikipedia.org/wiki/Nerve_%2528category_theory%2529) of the category QM. The homotopy groups are then the usual homotopy groups from topology.
There also is a definition of the homotopy group of a simplicial set - you can thus compute the homotopy group of the nerve without passing to the geometric realization first - and the definition you give looks more like that, but the answers are isomorphic.
| 4 | https://mathoverflow.net/users/733 | 37897 | 24,367 |
https://mathoverflow.net/questions/37647 | 25 | Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces. For a based space $X \in \mathcal T$, one has a canonical funtor $S\_X : \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$. The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism.
As is well know, the homotopy groups of $\mathrm{colim} S\_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S\_X = SP^\infty\_h X$ given the stable homotopy of $X$.
Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit?
The motivation for this question comes from thinking about $\infty$-categories. In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict categorical colimits. So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology. In short, is there any $\infty$-categorical analog of the Dold-Thom theorem?
**Update**: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations. I'll just quickly sketch what I have so far:
Let $\mathcal O(\Sigma\_n)$ denote the orbit category. The objects are the homogeneous (discrete) spaces $\Sigma\_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma\_n$, and the morphisms are the $\Sigma\_n$-equivariant maps. There is a canonical functor $$\Sigma\_n \rightarrow \mathcal O(\Sigma\_n)^{op}$$ where we regard $\Sigma\_n$ as a category with one object as usual.
Given a $\Sigma\_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma\_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma\_n/H) = X^H$$ It turns out that the above inclusion is *final* so that it induces an isomorphism of colimits. Hence $\mathrm{colim}\_{\mathcal O(\Sigma\_n)} \tilde X \cong X\_{\Sigma\_n}$, i.e., the coinvariants. It's also not hard to see that the undercategories are copies of $B\Sigma\_n$, hence *not* contractible, so we don't expect an equivalence of homotopy colimits, which is good.
On the other hand, I can now show that when $X$ is *discrete*, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence. My methods here do not generalize to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated. (I think something like this must appear in May's book on equivariant homotopy theory if it's true, but I did not have it available this weekend.)
The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad. (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \mathcal O(\Sigma\_n)$. Take the Grothendieck construction. Some natural diagram on this category might give the right answer.)
| https://mathoverflow.net/users/4466 | The Dold-Thom theorem for infinity categories? | It so happens that Emmanuel Dror Farjoun is visiting the EPFL this week. I figured I'd ask him about this problem at lunch today. What a coincidence! He proved exactly this statement using the exact same techniques. In fact, the construction of $SP^n$ as a homotopy colimit is the subject of Chapter 4 in "Cellular Spaces, Null Spaces, and Homotopy Localization," Lecture Notes in Mathematics, 1622.
It turns out, the idea works more generally so that we can always replace strict colimits with homotopy colimits: define an *orbit* on a category $\mathcal C$ to be a functor $O : \mathcal C \rightarrow \mathcal Set$ such that $\mathrm{colim}\_{\mathcal C} O \cong \*$. There is a category of such functors which we call the orbit category of $\mathcal C$, denoted $\mathcal O(\mathcal C)$. The Yoneda embedding factors through $\mathcal O(\mathcal C)$, and the right Kan extension along this inclusion always results in a "free" diagram.
I still want to play around with the construction a bit to see if there are any wrinkles with $n \rightarrow \infty$, and if I can use this to give easy calculations of homology in the $\infty$-category $\mathcal S$, but I think it's safe to say at this point that answer to my question is yes.
| 15 | https://mathoverflow.net/users/4466 | 37898 | 24,368 |
https://mathoverflow.net/questions/37896 | 4 | Suppose $H$ is a indefinite quaternion algebra over $\mathbb{Q}$. Are there infinitely many quadratic fields that can be embedded into $H$?
| https://mathoverflow.net/users/3945 | finite or infinite many quadratic fields embedding into quaternion algebras? | There are infinitely many. Let $V$ be the subspace of $H$ where the trace is zero. Then norm gives an nondegenerate quadratic form on $V$. For any $v \in V$, the field $\mathbb{Q}(v)$ is isomorphic to $\mathbb{Q}(\sqrt{-N(v)})$. Recall that the fields $\mathbb{Q}(\sqrt{D\_1})$ and $\mathbb{Q}(\sqrt{D\_2})$ are isomorphic if and only if $D\_1/D\_2$ is a square.
So we need to show that $V$ takes infinitely many values in $\mathbb{Q}^\*/(\mathbb{Q}^\*)^2$. For example, if we are dealing with the standard quaternion algebra, we need to show that $p^2+q^2+r^2$ takes infinitely many values in $\mathbb{Q}^\*/(\mathbb{Q}^\*)^2$. This is easy enough that probably any method you think of will work. Here is what I came up with: Take $u$ and $v$ linearly independent members of $V$. Let $a=u^{-1} v$.
First, suppose that $-N(a)$ is a square, say $k^2$. Then, for $s$ and $t \in \mathbb{Q}$, we have $N(su+tv) = N(u) N(s+t a)= N(u) (s-kt)(s+kt)$, and this expression clearly takes infinitely many values in $\mathbb{Q}^\*/(\mathbb{Q}^\*)^2$ as we vary $s$ and $t$.
If $N(a)$ is not a square, let $K=\mathbb{Q}(a)$, this is a subfield of $H$ isomorphic to $\mathbb{Q}[\sqrt{N(a)}]$. For $b \in K$, we have $N(ub) = N\_{K/\mathbb{Q}}(b) N(u)$, and $ub$ is in $V$. Since there are infinitely many primes that split principally in $K$, there are infinitely primes that occur as norms $ N\_{K/\mathbb{Q}}(b)$, and thus we get an infinite subgroup of $\mathbb{Q}^\*/(\mathbb{Q}^\*)^2$.
| 10 | https://mathoverflow.net/users/297 | 37900 | 24,370 |
https://mathoverflow.net/questions/37857 | 4 | Hi, I have a **function defined by an integral** as follows.
$$
z=f(w) = \int\_0^w \frac{(\zeta-a\_1)^{\alpha\_1}(\zeta-a\_2)^{\alpha\_2}...}{(\zeta-b\_1)^{\beta\_1}(\zeta-b\_2)^{\beta\_2}...}\ d\zeta
$$
where $w$ is real, $a\_i$ and $b\_i$ are constants and $\alpha\_i$ are integers. Thus the integrand is a rational function that can be nicely factored into linear terms.
Are there some theorems discussing, relevant properties about, or way of calculating **the inverse**
$$
w=f^{-1}(z)?
$$
All suggestions are welcome.
| https://mathoverflow.net/users/9039 | Inverse of a function defined by an integral | You can always get a (non-linear) ordinary differential equation for $f^{-1}$. It is easy to see that $f$ satisfies a 2nd order linear ODE with polynomial coefficients with no order 0 term [the first order ODE has non-polynomial coefficients, so harder to work with]. From there, it is also mechanical to get an ODE for $f^{-1}$ by interchanging the roles of $f$ and $w$.
But since your original function is (in general, depending on the path of integration) a MeijerG function, very few of these have closed-form inverses. As Piero mentions in the comments, the trigonometric and Elliptic functions are some of the few cases where this 'works'.
It also depends on what you are trying to do 'next' with these functions. If you are looking at numerical evaluation, then there are closed-forms for the Lagrange inverse, some of which translate to closed-forms for the Hermite-Pade approximants, from which efficient approximations can be derived. [J.M.'s method works in general, here it turns out that this can be pushed in closed-form further than usual].
For any given case, there are useful tools in Maple (and Mathematica) to help you carry these computations out. Probably in other CASes as well, but I don't know.
| 7 | https://mathoverflow.net/users/3993 | 37910 | 24,374 |
https://mathoverflow.net/questions/37913 | -2 | There are n red & n blue jugs of different sizes and shapes. All
red jugs hold different amounts of water as the blue ones. For every red jug,
there is a blue jug that holds the same amount of water, and vice versa.
The task is to find a grouping of the jugs into pairs of red and blue jugs that hold the same
amount of water.
Operation allowed: Pick a pair of jugs in which one is red and one is blue, fill the red jug with water and then pour the water into the blue jug. This operation will tell you whether the red or the blue jug can hold more water, or if
they are of the same volume. Assume that such a comparison takes one time unit. Your goal is
to find an algorithm that makes a minimum number of comparisons to determine the
grouping.
You may not directly compare two red jugs or two blue jugs.
>
> 1. Prove a lower bound of Θ(n lg n) for the number of comparisons an algorithm solving
> this problem must make.
> 2. Give a randomized algorithm whose expected number of comparisons is O(n lg n)
>
>
>
| https://mathoverflow.net/users/9054 | Water jug puzzle | The lower bound can be gotten the same way we get a lower bound of $O(n \log n)$ for comparison-based sorting. Each comparison of jugs produces one of three possible outcomes, which is at most $\log\_2 3 = O(1)$ bits of information. But we need to distinguish among $n!$ possible matchings of red and blue jugs, which requires $\log\_2(n!) = O(n \log n)$ bits.
The overall problem can be solved with something very close to quicksort. Pick a red jug arbitrarily. Compare it against every blue jug. This will separate the blue jugs into one that is the same same size as the chosen red jug, a set of smaller blue jugs, and a set of bigger blue jugs. Use the same-size blue jugs to partition the red jugs the same way. Recursively apply this algorithm to the sets of larger and smaller red and blue jugs. The recursion bottoms out when there is only one jug of each color in a set.
| 1 | https://mathoverflow.net/users/7759 | 37915 | 24,377 |
https://mathoverflow.net/questions/37916 | 2 | I'm consideirng the example of
$-\Delta u + V(x) u = 0$ in $\Omega$ with $u = 0$ on $\partial \Omega$. I'm trying to see if it's true that if $-\lambda\_1 < V(x) < -\lambda\_2 < 0$ on $\overline{\Omega}$ that we *do not* have existence of non-trivial solutions to this equation. Here $\lambda\_1$ and $\lambda\_2$ are two distinct eigenvalues of $-\Delta u$, both of which are of course positive.
In the special case that $V$ is constant, this is certainly true since otherwise $V$ would itself be an eigenvalue. I have tried to see if I can use a sort of comparison principle to sohw that $u \equiv 0$ if $-\Delta u + V(x)u = 0$ in the case that $V$ is not constant but I can't seem to establish this.
For simplicity of course assume that $V$ is as smooth and as bounded as you'd like.
| https://mathoverflow.net/users/8755 | Existence of non-trivial solutions to Dirichlet problem with a potential lying between eigenvalues. | I assume $\Omega$ is a bounded open subset of $\mathbb{R}^n$ and, say, $V\in L^\infty(\Omega)$. What you say is correct, but of course you need to assume that the eigenvalues are also *consecutive* (otherwise e.g. $V$ itself could be another eigenvalue in between [edit: as you actually said]). The comparison principle you are seeking comes from the Courant–Fischer–Weyl variational characterization of the eigenvalues of $-\Delta + V,$ by which the $k$-th eigenvalue $\lambda\_k(-\Delta + V\\ )$ in the increasing order is expressed as a certain min-max of the Rayleigh quotient, which is monotone wrto $V:$
$$\frac{\int\_\Omega \left(|\nabla u|^2+V(x)u^2 \right)dx }{\int\_\Omega u^2 dx }$$
(for a precise statement see e.g. Courant-Hilbert, or Reed-Simon, or Gilbarg-Trudinger, &c).
So if we denote $\lambda\_k:=\lambda\_k(-\Delta)$ and assume $-\lambda\_{k+1} < V(x) < -\lambda\_{k}\\ ,$ it follows by the monotonicity
$$\lambda\_k(-\Delta+V)<\lambda\_k(-\Delta - \lambda\_k)=0 =\lambda\_{k+1}(-\Delta - \lambda\_{k+1})< \lambda\_{k+1}(-\Delta +V),$$
so that $0$ is not an eigenvalue of $-\Delta +V.$
| 5 | https://mathoverflow.net/users/6101 | 37928 | 24,384 |
https://mathoverflow.net/questions/37919 | 15 | Does there exist a (noetherian) commutative ring $R$ and an element $a \in R$ such that $a$ is a square in every localization of $R$ but $a$ itself is not a square?
| https://mathoverflow.net/users/4433 | Locally square implies square | OK, I've got it. There is no such local criterion for squareness.
Let $k$ be a field of characteristic not $2$. Take the ring of triples $(f,g,h) \in k[t]^3$, subject to the conditions that $f(1)=g(-1)$, $g(1)=h(-1)$ and $h(1)=f(-1)$. Consider the element $(t^2,t^2,t^2)$. If this were a square, its square root would have to be $(\pm t, \pm t, \pm t)$. But two of those $\pm$'s would be the same sign, and $t$ evaluated at $1$ and at $-1$ are not equal.
Now, to check that $(t^2, t^2, t^2)$ is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, $(-t, t, 1)$ is a square root of $(t^2, t^2, t^2)$.
For the suspicious, an algebraic proof. Set $u\_1=(0, (1+t)/2, (1-t)/2)$ and let $u\_2$ and $u\_3$ be the cyclic permutations thereof. We have $u\_1+u\_2+u\_3=1$ so, in any local ring, one of the $u\_i$ must be a unit. WLOG, suppose that $u\_1$ is a unit. Notice that $u\_1 (1, -t, t)^2 = u\_1 (t^2, t^2, t^2)$. So, in a local ring where $u\_1$ is a unit, $(1,t,-t)$ is a square root of $(t^2, t^2, t^2)$.
| 20 | https://mathoverflow.net/users/297 | 37931 | 24,387 |
https://mathoverflow.net/questions/37926 | 19 | When I think about $\mathcal{D}$-Modules, I find it very often useful to envison them as vectorbundles endowed with a rule that decides whether a given section is flat. Or alternatively a notion of parallel transport.
Now my question is, what are good ways to think about modules over sheavers of twisted differential operators?
| https://mathoverflow.net/users/2837 | What is a twisted D-Module intuitively? | One way to think of twisted $D$-modules that I like is to view them as monodromic $D$-modules (see Beilinson, Bernstein *A Proof of Jantzen Conjectures* section 2.5, available as number 49 on [Bernstein's web page](http://www.math.tau.ac.il/%7Ebernstei/Publication_list/Publication_list.html) ). Let $T$ be a torus, and let $\pi: \tilde{X} \to X$ be a $T$-torsor. The sheaf of algebras $\tilde{D} = (\pi\_\* D\_{\tilde{X}})^T$ has center $U(\mathfrak{t}) = S(\mathfrak{t})$, and its category of modules is the category of weakly $T$-equivariant $D$-modules on $\tilde{X}$. For any $\chi \in \mathfrak{t}^\vee$, there is a maximal ideal $m\_\chi \subset S(\mathfrak{t})$, and the algebra of $\chi$-twisted differential operators is $\tilde{D}/m\_\chi \tilde{D}$.
If you want to twist by a fractional power $c$ of a line bundle $L$, then you can let $T$ be the usual one dimensional split torus, $\tilde{X}$ be the total space of $L$ with the zero section removed, and $\chi = c$. For ordinary differential operators, set $\chi = 0$. Intuitively, I think of a ($\mathcal{O}$-coherent) twisted $D$-module as a vector bundle on the total space of the torsor such that flat sections obey a fixed monodromy when parallel transported in the torus direction.
| 17 | https://mathoverflow.net/users/121 | 37939 | 24,391 |
https://mathoverflow.net/questions/37940 | 5 | Question 1: Given a smooth Riemannian surface $M$ in $R^3$ (i.e., a smooth Riemannian 2-manifold embedded in $R^3$) and a diffeomorphism $f: M\rightarrow M$ of class $C^{k\geq 2}$, does $f$ admit a smooth extension $\tilde{f}$ to all of $R^3$? If not always, then are there sufficient conditions?
Question 2: If the answer to Q1 is affirmative, then given two diffeomorphisms $f,g: M\rightarrow M$ of class $C^{k\geq 2}$ which are close in the $C^2$-topology, can we find extensions $\tilde{f}, \tilde{g}$ which are also close in the $C^2$-topology?
Edit 1: I should add that $M$ carries the induced metric (from $R^3$).
Edit 2: We can ask a more general question. Say $M$ is a smooth Riemannian $m$-manifold. Embed $M$ in $R^N$ isometrically. Say $f: M\rightarrow M$ is a diffeomorphism of class $C^k$. Can we extend $f$ smoothly to $R^N$?
| https://mathoverflow.net/users/nan | Extending diffeomorphisms of Riemannian surfaces to the ambient space | Q1: Definately not always. More like "almost never". If the automorphism extends to $\mathbb R^3$, then the bundle $S^1 \ltimes\_f M$ would embed in $S^4$. $S^1 \ltimes\_f M$ is the bundle over $S^1$ with fiber $M$ and monodromy $f$. The most-commonly used obstructions to embedding in this case are things like the Alexander polynomial, and Milnor signatures.
I don't see where the metric on $M$ plays a role for this.
If you want to see automorphisms that extend (and do not extend) for your Q1, take a look at my [arXiv paper](https://arxiv.org/abs/0810.2346). You'll also find some references to several Jonathan Hillman papers that explore such obstructions.
In the case that your surface is unknotted -- bounding handlebodies on both sides (thinking of $M \subset S^3$) then the automorphisms of $M$ that extend in this case are well-known. They're called the mapping class group of a Heegaard splitting of $S^3$. It's an infinite group. Generators are known for it (if I recall, they're the automorphisms induced by handle slides) but off the top of my head I'm not sure how much is known about the structure of the group. Do a little Googling on "mapping class group of a Heegaard splitting of S^3" and you should start finding relevant material.
To respond to your 2nd edit, if the co-dimension is high enough all automorphisms extend. This is a theorem of Hassler Whitney's. The basic idea is this, let $f : M \to M$ be an automorphism. Let $i : M \to \mathbb R^k$ be any embedding. So you have two embeddings $i \circ f$ and $i$ of $M$ in $\mathbb R^k$. Any two maps $M \to \mathbb R^k$ are isotopic provided the co-dimension is large enough $k \geq 2m+3$ suffices, for example. So isotope your standard inclusion to the one pre-composed with $f$. The Isotopy Extension Theorem gives you the result.
For example, if $\Sigma$ is a Heegaard splitting / the surface is unknotted, $\Sigma \subset \mathbb R^3$ (or $\subset \mathbb S^3$) and you have an automorphism $f : \Sigma \to \Sigma$ a neccessary and sufficient condition for $f$ to extend to $\mathbb R^3$ (or a side-preserving automorphism of the pair $(S^3,\Sigma)$ in the $S^3$ case) is that if $C \subset \Sigma$ is a curve that bounds a disc on either the inside or outside of $\Sigma$ respectively, then $f(C)$ bounds a disc on the inside or outside of $\Sigma$ respectively (here I'm using inside/outside re the Jordan-Brouwer separation theorem). Since the fundamental group of the complement is just a free product of infinite cyclic groups this is something that can be checked rather easily provided you know the map $f$ well enough.
| 6 | https://mathoverflow.net/users/1465 | 37941 | 24,392 |
https://mathoverflow.net/questions/37933 | 52 | What are some applications of the [Implicit Function Theorem](https://en.wikipedia.org/wiki/Implicit_function_theorem) in calculus? The only applications I can think of are:
1. the result that the solution space of a non-degenerate system of equations naturally has the structure of a smooth manifold;
2. the Inverse Function Theorem.
I am looking for applications that would be interesting to an advanced US math undergraduate
or 1-st year graduate student who is seeing the Implicit Function Theorem for the first or second time. The context is I will be explaining this result as part of
a review of manifold theory.
| https://mathoverflow.net/users/5337 | What is the Implicit Function Theorem good for? | The infinite-dimensional implicit function theorem is used, among other things, to demonstrate the existence of solutions of nonlinear partial differential equations and parameterize the space of solutions. For equations of standard type (elliptic, parabolic, hyperbolic), the standard version on Banach spaces usually suffices, but you have to be clever about which Banach space to use. There is a generalization of the implicit function theorem, due to Nash who used it to demonstrate the existence of isometric embeddings of Riemannian manifolds in Euclidean space, that works for even more general types of PDE's. Moser stated and proved a simpler version of the theorem. There is a beautiful survey article by Richard Hamilton (who originally used the Nash-Moser implicit function theorem to prove the local-in-time existence of solutions to the Ricci flow) on the Nash-Moser implicit function theorem.
| 43 | https://mathoverflow.net/users/613 | 37945 | 24,393 |
https://mathoverflow.net/questions/37880 | 17 | Let $p$ be an odd prime and let $k\in[2,p-3]$ be an even integer such that $p$ divides (the numerator of) the Bernoulli number $B\_k$ (the coefficient of $T^k/k!$ in the $T$-expansion of $T/(e^T-1)$). This happens for example for $p=691$ and $k=12$.
Ribet (Inventiones, 1976) then provides an everywhere-unramified degree-$p$ cyclic extension $E$ of the cyclotomic field $K={\bf Q}(\zeta)$ (where $\zeta^p=1$, $\zeta\neq1$) which is galoisian over $\bf Q$ and such that the resulting conjugation action of $\Delta={\rm Gal}(K|{\bf Q})$ on the ${\bf F}\_p$-line $H={\rm Gal}(E|K)$ is given by the character $\chi^{1-k}$, where $\chi:\Delta\to{\bf F}\_p^\times$ is the ``mod-$p$ cyclotomic character''.
Kummer theory then tells us that there are units $u\in{\bf Z}[\zeta]^\times$ such that $E=K(\root p\of u)$. Which units ?
More precisely, there is an ${\bf F}\_p$-line $D\subset K^\times/K^{\times p}$ such that $E=K(\root p\of D)$. Which line ?
| https://mathoverflow.net/users/2821 | Kummer generator for the Ribet extension | Let me first add that Herbrand wasn't the first to publish his result; it was obtained (but with a less clear exposition) by Pollaczek (*Über die irregulären Kreiskörper der $\ell$-ten und $\ell^2$-ten Einheitswurzeln*, Math. Z. 21 (1924), 1--38).
Next the claim that the class field is generated by a unit is true if $p$ does not divide the class number of the real subfield, that is, if Vandiver's conjecture holds for the prime $p$.
**Proof.** (Takagi)
Let $K = {\mathbb Q}(\zeta\_p)$, and assume that the class number of
its maximal real subfield $K^+$ is not divisible by $p$. Then any
unramified cyclic extension $L/K$ of degree $p$ can be written in
the form $L = K(\sqrt[p]{u})$ for some unit $u$ in $O\_K^\times$.
In fact, we have $L = K(\sqrt[p]{\alpha})$ for some element
$\alpha \in O\_K$. By a result of Madden and Velez, $L/K^+$ is normal
(this can easily be seen directly). If it were abelian, the subextension
$F/K^+$ of degree $p$ inside $L/K^+$ would be an unramified cyclic
extension of $K^+$, which contradicts our assumption that its class
number $h^+$ is not divisible by $p$.
Thus $L/K^+$ is dihedral. Kummer theory demands that
$\alpha /\alpha' = \beta^p$ for some $\beta \in K^+$, where
$\alpha'$ denotes the complex conjugate of $\alpha$.
Since $L/K$ is unramified, we must have $(\alpha) = {\mathfrak A}^p$.
Thus $(\alpha \alpha') = {\mathfrak a}^p$, and since $p$ does not
divide $h^+$, we must have ${\mathfrak a} = (\gamma)$, hence
$\alpha \alpha' = u\gamma^p$ for some real unit $u$.
Putting everything together we get $\alpha^2 = u(\beta\gamma)^p$,
which implies $L = K(\sqrt[p]{u})$.
If $p$ divides the plus class number $h^+$, I cannot exclude the possibility that the Kummer generator is an element that is a $p$-th ideal power, and I cannot see how this should follow from Kummer theory, with or without Herbrand-Ribet.
If $p$ satisfies the Vandiver conjecture, the unit in question can be given explicitly, and was given explicitly already by Kummer for $p = 37$ and by Herbrand for general irregular primes satisfying Vandiver: let $g$ denote a primitive root modulo $p$, and let $\sigma\_a: \zeta \to \zeta^a$. Then
$$ u = \eta\_\nu = \prod\_{a=1}^{p-1} \bigg(\zeta^\frac{1-g}{2}\
\frac{1-\zeta^g}{1-\zeta}\bigg)^{a^\nu \sigma\_a^{-1}}, $$
where $\nu$ is determined by $p \mid B\_{p-\nu}$.
Here is a [survey](http://www.rzuser.uni-heidelberg.de/~hb3/publ/pcft.pdf) on class field towers based on my (unpublished) thesis on the explicit construction of Hilbert class fields that I have not really updated for quite some time. Section 2.6 contains the answer to your question for primes satisfying Vandiver.
| 12 | https://mathoverflow.net/users/3503 | 37951 | 24,397 |
https://mathoverflow.net/questions/37938 | 3 | Grothendieck (if it was him) said somewhere :
>
> This XXX, at least, is an idea that will not be used in physics.
>
>
>
Q1 : Is XXX an n-groupoid? a stack? Can someone supply the precise quote, either in French or in English?
Q2: Predecessors of this quote in the same vein would also be of interest.
Thank you.
| https://mathoverflow.net/users/3005 | A mathematical idea "abstract enough to be useless for physics" | Dear Jérôme, I doubt that Grothendieck ever said that.
However, in an analogous vein, Jean Leray, a brilliant French mathematician, was taken prisoner by the Germans in 1940 and sent to Oflag XVIIA ("Offizierslager", officers' prison camp) in Edelsbach (Austria), where he remained for five years till the end of WW2.
He managed to hide from his captors that he was an expert in fluid dynamics and mechanics, lest they would force him to contribute to their war effort (submarines, planes).
Instead, he organized a course, attended by his fellow prisoners, on the foundations of Algebraic Topology, a harmless subject for applications in his eyes. It is in these courses that he introduced sheaves, cohomology of sheaves and spectral sequences.
His strategy worked out fine since these discoveries didn't play any role in the construction of weapons by the German enemy, who never cared about Leray's courses and findings. On the other hand, these theoretical tools have had a non entirely negligible role in pure mathematics since.
| 14 | https://mathoverflow.net/users/450 | 37954 | 24,399 |
https://mathoverflow.net/questions/37959 | 2 | Definitions:
Recall the [definition](https://mathoverflow.net/questions/19363/the-join-of-simplicial-sets-finale) of the join of two simplicial sets. We may regard the functor $-\star Y$ as a functor $i\_{Y,-\star Y}:Set\_\Delta\to (Y\downarrow Set\_\Delta)$ by replacing the resulting simplicial set $X\star Y$ with the canonical map $i\_{Y,X\star Y}: Y\cong \emptyset\star Y \to X\star Y$. It is not hard to show that $i\_{Y,-\star Y}$ commutes with colimits in $(Y\downarrow Set\_\Delta)$, and therefore that it admits a right adjoint. We call the adjoint functor the *overcategory* or *over-simplicial-set* functor. We can give an explicit description of this simplicial set: Given an arrow $f:Y\to S$, define $(S\downarrow f)\_n:=Hom\_{(Y\downarrow Set\_\Delta)}(i\_{Y,\Delta^n\star Y},f)$.
As with all adjunctions, we have a unit and counit natural transformation $\eta\_X:X\to (X\star Y\downarrow i\_{Y,X\star Y})$, and $\epsilon\_f: i\_{Y,(X\downarrow f)\star Y}\to f$ respectively.
Then the question: Intuitively, the unit map should map $X$ to the simplicial subset of $(X\star Y \downarrow i\_{Y,X\star Y})$ spanned by the "original" vertices of $X$, and I have seen this applied before (specifically in the case where $X$ is a simplex). However, I can't seem to figure out why this should be true formally. Is it true, and if so, how can we prove it?
Edit: As Tim mentions, the augmentation is given by the empty presheaf, that is, $\Delta^{-1}:=\emptyset$, which gives $X(-1):=Hom(\Delta^{-1}, X)=Hom(\emptyset,X)=\{\ast\}$
| https://mathoverflow.net/users/1353 | Computing the image of the unit map of the join/overcategory adjunction for simplicial sets | It may help to look at Phil Ehlers thesis ``Algebraic Homotopy in simplicially Enriched groupoids'' Bangor 1993. He used some ideas from Duskin and van Osdol and write down the details of the right adjoint.(It can be found on the n-Lab at
<http://ncatlab.org/nlab/files/Ehlers-thesis.pdf> )
I do not know if that will answer your question, since it is quite a long time since I read it, but he did describe the adjunction explicitly (see his chapter 3). The restriction placed in the answer to your earlier MO question (namely that $X(-1) = \*$).
The right adjoint is given by $[Y,Z]\_{n-1} = Set\_{\Delta\_a}(Y,Dec^nZ)$. The derivation is just jiggling about using the end calculus.
| 2 | https://mathoverflow.net/users/3502 | 37969 | 24,405 |
https://mathoverflow.net/questions/37944 | 12 | Is there ever a practical difference between the notions induction and strong induction?
Edit: More to the point, does anything change if we take strong induction rather than induction in the Peano axioms?
| https://mathoverflow.net/users/8871 | Induction vs. Strong Induction | The terms "weak induction" and "strong induction" are not commonly used in the study of logic. The terms are commonly used only in books aimed at teaching students how to write proofs.
Here are their prototypical symbolic forms:
* weak induction: $(\Phi(0) \land (\forall n) [ \Phi(n) \to \Phi(n+1)]) \to (\forall n) \Phi(n)$
* strong induction: $(\Psi(0) \land (\forall n) [ (\forall m \leq n) \Psi(m) \to (\forall m \leq n+1) \Psi(m)]) \to (\forall n) \Psi(n)$
The thing to notice is that "strong" induction is almost exactly weak induction with $\Phi(n)$ taken to be $(\forall m \leq n)\Psi(n)$. In particular, strong induction is not actually stronger, it's just a special case of weak induction modulo some trivialities like replacing $\Psi(0)$ with $(\forall m \leq 0 )\Psi(m)$. Of course you can write variations of the symbolic forms, but the same point applies to all of them: "strong" induction is essentially just weak induction whose induction hypothesis has a bounded universal quantifier.
So the question is not why we still have "weak" induction - it's why we still have "strong" induction when this is not actually any stronger.
My opinion is that the reason this distinction remains is that it serves a pedagogical purpose. The first proofs by induction that we teach are usually things like $\forall n \left [ \sum\_{i=0}^n i = n(n+1)/2 \right ]$. The proofs of these naturally suggest "weak" induction, which students learn as a pattern to mimic.
Later, we teach more difficult proofs where that pattern no longer works. To give a name to the difference, we call the new pattern "strong induction" so that we can distinguish between the methods when presenting a proof in lecture. Then we can tell a student "try using strong induction", which is more helpful than just "try using induction".
In terms of logical strength in formal arithmetic, as you can see above, the two forms are equivalent over some weak base theory as long as you are looking at induction for a class of formulas that is closed under bounded universal number quantification. In particular, all the syntactic classes of the analytical and arithmetical hierarchies have that property, so weak induction for $\Sigma^0\_k$ formulas is the same as strong induction for $\Sigma^0\_k$ formulas, weak induction for $\Pi^1\_k$ formulas is the same as strong induction for $\Pi^1\_k$ formulas, and so on. This equivalence will hold under any reasonable formalization of "strong" induction - I chose mine above to make the issue particularly obvious.
---
**Addendum** I was asked in a comment why
$$
(1)\colon (\forall t)[(\forall m < t)\Phi(m) \to \Phi(t)]
$$
implies
$$
(2)\colon (\forall n)[(\forall m \leq n)\Phi(m) \to (\forall m \leq n+1) \Phi(m)].
$$
I'm going to give a relatively formal proof to show how it goes. The proof is *not* by induction, instead it just uses universal generalization to prove the universally quantified statements.
For the proof, I will assume (1) and prove (2). Working towards that goal, I fix a value of $n$ and assume:
$$
(3)\colon (\forall m \leq n)\Phi(m).
$$
I first want to prove $(\forall m < n+1)\Phi(m)$, which is an abbreviation for $(\forall m)[m < n+1 \to \Phi(m)]$. Pick an $m$. If $m < n+1$, then $m \leq n$, so I know $\Phi(m)$ by assumption (3). So, by universal generalization, I obtain $(\forall m < n+1)\Phi(m)$.
Next, note that a substitution instance of (1) gives $(\forall m < n+1)\Phi(m) \to \Phi(n+1)$. I have proved $(\forall m < n+1)\Phi(m)$ so I can assert $\Phi(n+1)$.
So now I have assumed $(\forall m \leq n)\Phi(m)$ and I have also proved $\Phi(n+1)$. Another proof by cases establishes $(\forall m \leq n+1)\Phi(m)$.
By examining the proof, you can see which axioms I need in my weak base theory. I need at least the following two axioms:
* $(m \leq t) \leftrightarrow (m < t) \lor (m=t)$
* $(m < t+1) \to (m \leq t)$
I believe those are the only two axioms I used in the proof.
| 21 | https://mathoverflow.net/users/5442 | 37971 | 24,406 |
https://mathoverflow.net/questions/37963 | 47 | I am looking for a copy of the following
W. Thurston, Groups, tilings, and finite state automata, AMS Colloquium Lecture Notes.
I see that a lot of papers in the tiling literature refer to it but I doubt it was ever published. May be some notes are in circulation ?
Does anyone have access to it? I would be extremely grateful if you can send me a copy or tell me where can I find it.
| https://mathoverflow.net/users/6766 | Lecture notes by Thurston on tiling | Unfortunately, the original to this is hard to locate. It was distributed by the AMS at the time of the colloqium lectures, but they apparently didn't keep the files they used. At one time it was distributed as a Geometry Center preprint, but the Geometry Center is now defunct. I've lost track of the source files through multiple moves, computer crashes, etc.
What I have is a scanned version of the Geometry Center version that is legible but not beautiful, which I can forward by email
Bill Thurston
ADDENDUM:
Renaud Dreyer told me of an online scanned version I wasn't aware of, which appears to be better quality than the one I have:
<http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf>
| 138 | https://mathoverflow.net/users/9062 | 37973 | 24,407 |
https://mathoverflow.net/questions/37972 | 6 | Consider rational functions $F(x)=P(x)/Q(x)$ with $P(x),Q(x) \in \mathbb{Z}[x]$. I'd like to know when I can expect $F(k) \in \mathbb{Z}$ for infinitely many positive integers $k$. Of course this doesn't always happen ($P(x)=1, Q(x)=x, F(x)=1/x$). I am particulary interested in answering this for the rational function $F(x)=\frac{x^{2}+3}{x-1}$.
| https://mathoverflow.net/users/6254 | When does a rational function have infinitely many integer values for integer inputs? | If $F=P/Q$ is integral infinitely often then $F$ is a polynomial.
Write $$P(x)=f(x)Q(x)+R(x)$$ for some polynomial $R$ of degree strictly less than the degree of $Q$. If you have infinitely many integral $x$ so that $P/Q$ is integral then you get infinitely many $x$ so that $NR/Q$ is integral, where $N$ is the product of all denominators of the coefficients in $f$. However $R/Q\to 0$ as $x\to \pm \infty$ so $R\equiv 0$ and so $Q(x)$ is a divisor of $P(x)$.
Now, as pointed out by Mark Sapir below, not all polynomials with rational coefficients take on integer values infinitely often (at integers), but you can check this in all practical cases by seeing if $dF$ has a root $\pmod{d}$, where $d$ is the common denominator of the coefficients in $F$.
| 11 | https://mathoverflow.net/users/2384 | 37975 | 24,409 |
https://mathoverflow.net/questions/37984 | 4 | Let $\rho$ be irreducible representation of group $G$.
How one can characterize all subgroups $H< G$ such that $\rho$ can be embedded into permutation representation $F^X$, where $X=G/H$.
| https://mathoverflow.net/users/4246 | Embedding into Permutation Representation | There is the following adjointness (a form of Frobenius reciprocity):
$Hom\_G(\rho,F^X) = Hom\_H(\rho,trivial).$
Thus $\rho$ embeds in $F^X$ if and only if $\rho$ admits a non-trivial $H$-fixed quotient.
(If $H$ is finite and $F$ has characteristic zero, or at least prime to the order
of $H$, so that $\rho$ is semi-simple
as an $H$-representation, then this is equivalent to requiring that $\rho$ have a
non-trivial $H$-fixed subrepresentation.)
(Note also that a non-zero $G$-equivariant map out of $\rho$ is automatically injective,
because $\rho$ is irreducible.)
| 11 | https://mathoverflow.net/users/2874 | 37987 | 24,415 |
https://mathoverflow.net/questions/37989 | 8 | I have proved a certain result for all 2-connected graphs apart from those that fit the following criteria:
1. They are "minimally 2-connected", that is, deleting any vertex will produce a graph which is no longer 2-connected, and
2. They have circumference less than $\frac{n+2}{2}$, where $n$ is the number of vertices.
I have not been able to come up with an example of such a graph. Can anyone help?
Of course the best possible outcome would be that they do not exist!
| https://mathoverflow.net/users/4078 | Does this graph exist? | There are lots of examples. For $t>5$, let $P\_{1},..., P\_{t}$ be internally disjoint paths with length $3$ such that each path has the vertices $x$ and $y$ as endpoints.
| 9 | https://mathoverflow.net/users/9069 | 37995 | 24,417 |
https://mathoverflow.net/questions/37970 | 3 | I'm currently interested in the following result:
Let $f: X \to Y$ be a fpqc morphism of schemes. Then there is an equivalence of categories between quasi-coherent sheaves on $Y$ and "descent data" on $X$. Namely, the second category consists of quasi-coherent sheaves $\mathcal{F}$ on $X$ with an isomorphism $p\_{1}^\*(\mathcal{F}) \simeq p\_2^\*(\mathcal{F})$, where $p\_1, p\_2: X \times\_Y X \to X$ are the two projections. Also, a diagram involving an iterated fibered product is required to commute as well (the cocycle condition).
In Demazure-Gabriel's *Introduction to Algebraic Geometry and Algebraic Groups,* it is proved (under the name ffqc (sic) descent theorem) that the sequence
$$ X \times\_Y X \to^{p\_1, p\_2} X \to Y$$ is a coequalizer in the category of locally ringed spaces under the above hypotheses. If I am not mistaken, this is the same as (or very closely equivalent to) the theorem that says that representable functors are sheaves in the fpqc topology. On the other hand, D-G give a fairly explicit description of the quotient space.
**Question:** For a coequalizer diagram of ~~(locally) ringed spaces~~ schemes,
$$A \rightrightarrows^{f,g} B \to C,$$
is there a descent diagram for quasi-coherent sheaves on $A,B,C$? In particular, does the D-G form of the descent theorem directly, by itself, imply the more general one for quasi-coherent sheaves?
My guess is the answer is no. First, I've heard that the coequalizer condition above is actually very weak. So, suppose instead we have that $A \rightrightarrows^{f,g} B \to C,$ is a coequalizer *and* any base-change of it is a coequalizer. Does that imply that there is a descent diagram for quasi-coherent sheaves?
My guess is that the answer to the modified question is still no, for the meta-reason that Vistoli in *FGA Explained* spends much more time on proving descent for quasi-coherent sheaves than proving that the fpqc topology is subcanonical. On the other hand, I'd like to see a counterexample.
(N.B. I initially asked this question on [Math.SE](https://math.stackexchange.com/questions/4045/do-cokernels-in-ringspc-automatically-lead-to-descent). I was advised to re-ask it here.)
| https://mathoverflow.net/users/344 | Do coequalizers in RingSpc automatically lead to descent? | Initial question has a negative answer even for affine schemes. Let $B$ = Spec($R$) equipped with an action by a finite group $G$, and define $R' = \prod\_{g \in G} R$ and $A$ = Spec($R'$). Let $A \rightrightarrows B$ be the natural maps. Then $C$ := Spec($R^G$) is easily checked to be the coequalizer in the category of schemes (even in the category of locally ringed spaces), yet the descent diagram (if true) would say that for any $G$-equivariant $R$-module $M$ the natural map $R \otimes\_{R^G} M^G \rightarrow M$ is an isomorphism (since $M^G$ corresponds to the equalizer sheaf for $M$ under $A \rightrightarrows B$). Or even if you don't want to be so abstract, merely positing that $M = R \otimes\_{R^G} N$ for *some* $R^G$-module $N$ forces $N = M^G$ when $R$ is faithfully flat over $R^G$ and $N$ is a finite free $R^G$-module.
For example, let $R$ be the ring of integers of a quadratic field $K$, $G$ the order-2 Galois group, so $R^G = \mathbf{Z}$ is a PID. Take any finite torsion-free (= locally free) $R$-module $M$. Then automatically if $M$ were to have the form $R \otimes\_{R^G} N$ we see that $N$ must be torsion-free and hence free over $\mathbf{Z} = R^G$. In particular, if $M$ is an invertible $R$-module then $N$ must have rank 1 and hence it would be forced that $M$ is free of rank 1 over $R$. So to make a counterexample, we just need a nontrivial line bundle $M$ over $R$ to admit a $G$-action over the one on $R$. Take $M$ to be a non-principal ramified prime ideal (which always exists when the discriminant is not prime, or 4 times a prime, or whatever the condition to deal with 2...you know what I mean).
[It is no coincidence that such an example is related to the non-etale locus for the finite flat map $\mathbf{Z} \rightarrow O\_K$, since away from that the map is exactly the "Galois" instance of finite etale descent, for which everything does work exactly as in the familiar case of Galois descent for vector spaces over fields.]
| 5 | https://mathoverflow.net/users/3927 | 37996 | 24,418 |
https://mathoverflow.net/questions/37980 | 4 | Is every real vector bundle over the circle necessarily trivial? If yes - could you please point to a reference. If no - what are sufficient conditions?
I am particularly concerned with the case of a smooth map $\gamma:S^1\rightarrow Q$ and the vector bundle $\gamma^\* TQ$.
| https://mathoverflow.net/users/3509 | Is every real vector bundle over the circle necessarily trivial? | In the spirit of Theo's comment, I'll say something about sufficient conditions.
A real vector bundle over the circle is trivial if and only if it is orientable. I discussed this a bit [here.](https://mathoverflow.net/questions/21649/how-to-prove-that-w-1ew-1dete/21660#21660)
The main point is that up to isomorphism, every real vector bundle over the circle is either trivial, or the Whitney sum of a trivial bundle with the Mobius bundle. The latter is not orientable.
The other answers at the question I linked to above may also be helpful to you.
| 14 | https://mathoverflow.net/users/4042 | 38000 | 24,421 |
https://mathoverflow.net/questions/37992 | 5 | It is a standard fact that smashing with a fixed spectrum $Z$ preserves cofiber sequences. So if I have a cofiber sequence $$X \xrightarrow{f} Y \rightarrow C\_f$$ then there is also a cofiber sequence $$Z \wedge X \rightarrow Z \wedge Y \rightarrow Z \wedge C\_f$$
If more generally I have a map $Z \xrightarrow{g} W$, is there any formula for the cofiber of the map $$Z \wedge X \xrightarrow{g \wedge f} W \wedge Y$$ in terms of $C\_f$ and $C\_g$? (The above discussion corresponding to $g = \mathrm{id}\_Z$).
| https://mathoverflow.net/users/4466 | The homotopy cofiber of the smash product of two maps of spectra | By factoring $g\wedge f$ as $X\wedge f$ followed by $g\wedge Y$ you see that it's in the middle of a cofiber sequence $Z\wedge C\_f\to \ ?\to C\_g\wedge Y$. Similarly it's in the middle of a cofiber sequence $C\_g\wedge X\to\ ?\to W\wedge C\_f$. It's also in the middle of a cofiber sequence $(Z\wedge C\_f)\vee (C\_g\wedge X)\to \ ?\to C\_g\wedge C\_f$. We could probably come up with more. But no formula in the sense that I think you mean. It may be instructive to think of the special case where $X$ and $W$ are both contractible.
| 8 | https://mathoverflow.net/users/6666 | 38002 | 24,423 |
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