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https://mathoverflow.net/questions/26917 | 2 | Let $G$ be a finite group acting on an associative $k$-algebra with unity ($k$ algebraically closed of characteristic zero). The action is called ergodic if $ A^G = \{ a \in A | g \cdot a = a, \forall g \in G \} = k$.
Anyone know an example of a non-semisimple finite dimensional algebra with an ergodic action of a finite group?
| https://mathoverflow.net/users/6517 | A question on group action on algebras | Let $A = k[x,y]/(x^2,xy,y^2)$ and let the generator of the two-element group act by $x \mapsto -x$, $y \mapsto -y$.
More generally, take any finite group acting on $R = k[x\_1, \dots, x\_n]$ with invariant subring $S = k[f\_1, \dots, f\_m]$, and set $A = R/(f\_1, \dots, f\_m)$.
| 8 | https://mathoverflow.net/users/460 | 26921 | 17,605 |
https://mathoverflow.net/questions/25933 | 3 | For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu\_A = \mu\_B$? I want to know because in the bijection between Kleisli triples and monads, given a monad, we define $f^\* := T(f) ; \mu\_B$ if $f : A\to T(B)$ (c.f. [Prop 1.6](http://www.google.com/url?sa=t&source=web&ct=res&cd=1&ved=0CBgQFjAA&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.79.733%26rep%3Drep1%26type%3Dpdf&rct=j&q=notions+of+computation+and+monads&ei=7Br9S9iXK-WT4gah5sihCw&usg=AFQjCNH28ATtI90zSFufGd-GBXGJQu7Drw)), but this needs to be well defined even when $T$ is not an embedding.
Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^\*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu\_B$'s may be different!
| https://mathoverflow.net/users/6082 | Kleisli Monad bijection | In fact, the Kleisli star is a partial map ${}^\* : \mathrm{Mor}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \to \mathrm{Mor}(\mathbf{C})$. So there is no problem!
| 0 | https://mathoverflow.net/users/6082 | 26923 | 17,607 |
https://mathoverflow.net/questions/26927 | 13 | Let $X$ be a (singular) projective variety, in other words something given by a collection of polynomial equations in $\mathbb CP^n$ or $\mathbb RP^n$. How can one prove it is a finite $CW$ complex?
Similar question: Suppose that $X$ affine (i.e. given by polynomial equations in $\mathbb C^n$, or $\mathbb R^n$). How can one prove its one point compactification is a finite $CW$ complex?
These questions are sequel to the discussions here:
[For which classes of topological spaces Euler characteristics is defined?](https://mathoverflow.net/questions/26374/for-which-classes-of-topological-spaces-euler-characteristics-is-defined)
| https://mathoverflow.net/users/943 | How to prove that a projective variety is a finite CW complex? | The Lojasiewicz theorem says that every semi-algebraic subset of $\mathbf{R}^n$ can be triangulated. Moreover, there is a similar statement for pairs of the form (a semi-algebraic set, a closed subset). See e.g. Hironaka, Triangulations of algebraic sets, Arcata proceedings 1974 and references therein (including the original paper by Lojasiewicz).
The case of an arbitrary (not necessarily quasi-projective) complex algebraic variety follows from Nagata's theorem (every variety can be completed) and Chow's lemma (every complete variety can be blown up to a projective one).
| 16 | https://mathoverflow.net/users/2349 | 26929 | 17,611 |
https://mathoverflow.net/questions/26919 | 13 | I'm teaching an undergrad course in real analysis this Fall and we are using the text "Real Mathematical Analysis" by Charles Pugh. On the back it states that real analysis involves no "applications to other fields of science. None. It is pure mathematics." This seems like a false statement. My first thought was of probability theory. And isn't PDE's sometimes considered applied math? I was wondering what others thought about this statement.
| https://mathoverflow.net/users/343 | Real analysis has no applications? | As it happens, I just finished teaching a quarter of undergraduate real analysis. I am inclined to rephrase Pugh's statement into a form that I would agree with. If you view analysis broadly as both the theorems of analysis and methods of calculation (calculus), then obviously it has a ton of applications. However, I much prefer to teach undergraduate real analysis as pure mathematics, more particularly as an introduction to rigorous mathematics and proofs. This is partly as a corrective (or at least a complement) to the mostly applied and algorithmic interpretation of calculus that most American students see first.
Some mathematicians think, and I've often been tempted to think, that it's a bad thing to do analysis twice, first as algorithmic and applied calculus and second as rigorous analysis. It can seem wrong not to have the rigor up-front. Now that I have seen what BC Calculus is like in a high school, I no longer think that it is a bad thing. Obviously I still think that the pure interpretation is important. On the other hand, both interpretations together is also fine by me. I notice that in France, calculus courses and analysis courses are both called "analyse mathématique". I think that they might separate rigorous and non-rigorous calculus a bit less than in the US, and it could be partly because of the name.
In fact, it took me a long time to realize how certain non-rigorous explanations guide good rigorous analysis. For instance, the easy way to derive the Jacobian factor in a multivariate integral is to "draw" an infinitesimal parallelepiped and find its volume. That's not rigorous by itself, but it is related to an important rigorous construction, the exterior algebra of differential forms.
Finally, I agree that Pugh's book is great. As the saying goes, you shouldn't judge it by its cover. :-)
| 25 | https://mathoverflow.net/users/1450 | 26936 | 17,615 |
https://mathoverflow.net/questions/26932 | 3 | I am following the explicit construction of homotopy colimits as described by Dugger in the paper: "Primer on homotopy colimits", which can be found here: <http://www.uoregon.edu/~ddugger/hocolim.pdf>
As described in the appendix of Topological hypercovers and A1-realizations, Mathematische Zeitschrift 246 (2004) in the category of topological spaces no cofibrant-replacement functor is needed when computing the homotopy colimit of a small diagram in $\mathcal{T}op$.
For the index category $\mathcal{I} = \cdot \rightrightarrows \cdot$ and a small diagram $D: \mathcal{I} \rightarrow \mathcal{T}op$ with the image $X \rightrightarrows Y$ where $f, g: X \rightarrow Y$ this yields the space $T := (X \times \nabla^0 \amalg Y \times \nabla^0 \amalg X\_g \times \nabla^1 \amalg X\_f \times \nabla^1) / \sim$ where $\sim$ is given by:
$(x, 1) \sim (x, (0,1)) \in X\_f \times \nabla^n, X\_g\times \nabla^n$, $(f(x), 1) \sim (x, (1,0)) \in X\_f \times \nabla^n$ and $(g(x), 1) \sim (x, (1,0)) \in X\_g \times \nabla^n$ for all $x \in X$.
Notation: $\nabla^n$ is the topogical n-simplex, $X\_f$ and $X\_g$ are just copies of X indexed by a map in the diagram to keep track of all the identifications
1) Are any requirements necessary for $T$ to be homotopy equivalent or weakly homotopy equivalent to $colim\_{\mathcal{I}}{D}$?
2) What are the requirements for a homotopy pushout to be homotopy or weakly homotopy equivalent to the ordinary pushout?
3) What are the requirements for a homotopy colimit of a small diagram from the category obtained from the preorder $(\mathbb{N}, \leqslant)$ to $\mathcal{T}op$ to be be homotopy or weakly homotopy equivalent to the infinite mapping telescope as described in Section 3F (page 312) in the book about algebraic topology by Hatcher?
Since I barely know any model category theory, I would appreciate any elementary answers to this! Thank you very much!!
| https://mathoverflow.net/users/6499 | Simple examples of homotopy colimits | Here's an answer to question 2: A sufficient condition for $\mathrm{colimit}(X \leftarrow A\rightarrow Y)$ to be weakly equivalent to the homotopy colimit, is (a) for one of the maps (say $A\to X$) to have the homotopy extension property. Another sufficient condition is that the diagram $(X\leftarrow A\rightarrow Y)$ is (b) an "excisive triad". Take a look at Chapter 10, section 7 of May's "Consise Course", where (b) is proved to be such a sufficient condition. You can show that (a) is a sufficient condition by showing directly that it is homotopy equivalent to the double mapping cylinder $X\cup A\times \Delta^1\cup Y$, which can be re-analyzed as an "excisive triad".
| 7 | https://mathoverflow.net/users/437 | 26937 | 17,616 |
https://mathoverflow.net/questions/26897 | 21 | I have enough fears that this question might get struck down. Still let me try.
I shall restrict myself to $\frac{\lambda \phi^4}{4!}$ perturbed real scalar quantum field theory and call as "symmetry factor" of a Feynman diagram to be the eventual number by which the power of $\lambda$ is divided in the final integral representation of the diagram.
In that way the symmetry factor of the figure of eight vacuum bubble is 8 and of the "tadpole diagram" it is 2.
One way to get this factor right is to count the number of ways the free arms in the "pre-diagram" can be contracted. But this is more like a cook-book rule than an understanding of how the factor comes.
I believe the most conceptually correct way is to count for every diagram the number of terms in the representation as a functional derivative of the path-integral which give that diagram.
In that picture one has to argue that there were precisely $4!$ terms produced by the functional derivative which produced that figure of eight vacuum bubble. Which I can argue.
But for the tadpole diagram and the product of the vacuum bubble with the free-propagator, I can't find an argument. Like one has to be able to show that in the functional derivative picture there are $4!\times 3! \times (2!)^2$ terms corresponding to the tadpole diagram.
Any help regarding how this counting is done or any general framework which helps compute these symmetry factors correctly?
| https://mathoverflow.net/users/2678 | How to count symmetry factors of Feynman diagrams? | Exactly how you count symmetries depends on your normalizations, and in particular on whether you use divided (i.e. "exponential") power series or ordinary power series. I believe strongly that divided powers are the way to go. To establish notation, I will first review the preliminaries, that I'm sure you already know.
So let me consider Dyson series for integrals of the form:
$$\int\_{x\in X} \exp\left( \sum\_{n\geq 2} \frac{c\_n x^n}{n!} \right) {\rm d}x$$
In actual physics examples, $X$ is an infinite-dimensional space and the measure ${\rm d}x$ does not exist, but no matter. Each coefficient $c\_n$ is a symmetric tensor $X^{\otimes n} \to \mathbb R$ or $\mathbb C$, and we suppose that $c\_2$ is invertible as a map $X \to X^\*$ — in infinite-dimensional settings, this is a very problematic supposition, and leads to questions of renormalization, which I will not address here. (Incidentally, in all actual physical quantum field theories, $c\_2$ is not a priori invertible, because of gauge symmetry; so I am assuming that you have fixed that however suits your fancy.) Finally, Dyson series / Feynman diagrams are by definition *perturbative*, meaning that you need some perturbation parameter. There are various choices for how to do this; ultimately what's important is that ratios $\lvert c\_n\rvert / \lvert c\_2 \rvert^{n/2}$ are infinitesimal for $n \geq 3$. In your example, $c\_2$ is normalized to unity, and $c\_4 = \lambda \ll 1$. Another option is to set all coefficients $\lvert c\_n\rvert \sim \hbar^{-1}$, where $\hbar \ll 1$.
In any case, the Dyson series is correctly calculated by expanding
$$ \int\_{x\in X} \exp\left( \sum\_{n\geq 2} \frac{c\_n x^n}{n!} \right) {\rm d}x = \int\_{x\in X} \exp \left( \frac{c\_2x^2}2\right) \sum\_{m=0}^\infty \left( \sum\_{n\geq 3} \frac{c\_n x^n}{n!} \right)^m {\rm d}x, $$
using "Wick's theorem" for $b\_\ell$ a totally symmetric tensor $X^{\otimes \ell} \to \mathbb R$:
$$ \int\_{x\in X} \exp \left( \frac{c\_2x^2}2\right) \frac{b\_\ell x^\ell}{\ell!} = \begin{cases} 0, & \ell \text{ odd} \\ \text{normalization} \times \frac{b\_\ell (c\_2/2)^{\ell/2}}{(\ell/2)!}, & \ell \text{ even} \end{cases},$$
(where $\text{normalization}$ involves $\det c\_2$, factors of $\sqrt{2\pi}$, and doesn't make really sense in infinite dimensions), and recognizing what all of these exponential generating functions are counting.
When all the dust has settled, what these are counting is (all: disconnected, empty, etc., are allowed) graphs with trivalent-and-higher vertices, mod symmetries. If you throw a $\log$ out in front of the whole integral, then the resulting exponential generating function counts connected graphs, still mod symmetries.
In any case, to actually answer your question for the Dyson series above: the value of each graph is computed by placing a $c\_n$ at each vertex of valence $n$ and a $(c\_2)^{-1}$ on each edge, and contracting these tensors according to the graph. The symmetry factor is precisely the number of automorphisms of the graph, defined as follows:
**Definition:** A *Feynman graph* is a collection $H$ of *half edges* along with two partitions: one (*edges*) into sets of size precisely $2$, and the other (*vertices*) into sets of size at least $3$. (Since each graph will be weighted by the coefficients $c\_n$, if you want graphs for a theory with most $c\_n = 0$, you can restrict to only graphs with the prescribed valences. If $X$ splits as a direct sum $X = X\_1 \oplus X\_2$, you can reasonably define graphs with half-edges colored by $X\_1,X\_2$. If you want to compute an integral with an "observable", you should consider graphs with prescribed "external half-edges".) This is the correct definition, because it gives the correct notion of iso/automorphism. In particular, an *isomorphism of Feynman graphs* is a bijection of half edges that induces bijections on the partitions. An *automorphism* is an isomorphism from a Feynman graph to itself. Then the symmetry factor is precisely the number of automorphism in this sense.
(A different notion of "graph" more commonly used in mathematics is that of an "adjacency matrix", which is a $\mathbb Z\_{\geq 0}$-valued matrix indexed by the vertices of the graph. The natural notion of isomorphism of such things is a bijection of vertices that induces an equality of adjacency matrices, but this is not the right one for counting symmetries of Feynman diagrams, as, for example, it gives "$1$" for each of the figure-eight and tadpole diagrams.)
For small graphs, the automorphism group splits as a direct product. For example, for the figure eight, the automorphism group is $(\mathbb Z/2) \ltimes (\mathbb Z/2)^2$, where the $(\mathbb Z/2)^2$ acts as flips of the two lobes of the figure eight, and the left-hand $\mathbb Z/2$ switches the two lobes. In a "$\phi^3$" theory, the theta graph has automorphism group $(\mathbb Z/2) \ltimes S\_3$, where $S\_3$, the symmetric group on three objects, acts to permute the edges, and the $\mathbb Z/2$ switches the two vertices. So the figure eight has $2\times 2^2 = 8$ symmetries, and the theta has $2\times 3! = 12$ symmetries. For small graphs, you can really just read off these semidirect-product decompositions: self-loops contribute factors of $\mathbb Z/2$, each collection of $k$ parallel edges contributes a factor of $S\_k$, if you permute vertices, that's another factor, etc. For large graphs, the computations become much harder, and I think I can find a graph whose symmetry group is any finite group you want (but don't quote me on that thought). A slow way to do the computation is to consider all possible permutations of half edges, and see if they induce automorphism. This is slow because there are $(\text{lots})!$ many such permutations. The whole point of working with Dyson series is to avoid this. Fortunately, no one ever computes large graphs, because even just contracting all the tensors is so hard for those.
Finally, I cannot help but mention one more fact you probably know. These Dyson series almost never have positive radius of convergence in the perturbation parameters. In particular, even after dividing by symmetry factors, there are still a lot of graphs, and so the coefficients grow as $n!$. A good exercise is to compute the Dyson series for the conditionally-convergent Riemann integral $\int\_{\mathbb R} \exp \frac i \hbar \bigl( \frac{x^2}2 + \lambda \frac{x^3}6 \bigr){\rm d}x$. (The $i$ is there to make the integral conditionally converge; I tend to work with $\hbar$ as my perturbation parameter.) Then the Dyson series grows as $\sum \frac{(6n!)}{(2n)!\,(3n)!} \alpha^n$, where $\alpha$ is linear in $\lambda\hbar$ (something like $\alpha = \lambda\hbar / 6$). The point is that by Stirling's formula $\frac{(6n!)}{(2n)!\,(3n)!} \approx \beta^n n!$ for some positive $\beta$. In any case, this is not at all surprising. Integrals of the form $\int \exp \sum c\_nx^n/n!$ are essentially never analytic at $0,\infty$ in the coefficients — for example, Gauss's formula $\int\_{\mathbb R}\exp(-a^{-1}x^2/2){\rm d}x = \sqrt{2\pi a}$ has a ramified point at $a = 0,\infty$, and so the Riemann integral extends from its domain of convergence (${\rm Re}(a) > 0$) to a double-valued complex function in $a$.
| 19 | https://mathoverflow.net/users/78 | 26938 | 17,617 |
https://mathoverflow.net/questions/26931 | 2 | Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric:
$H^\dagger = H$ and $H^T = -H$.
(T denotes transpose, $\dagger$ denote conjugate transpose. I know that these conditions mean that H is a purely imaginary skew-symmetric matrix.)
It is a textbook result that these two conditions ensure that it's eigenvalues are real, its rank is even and its eigenvalues appear in positive and negative pairs.
If the normalised, orthonormal eigenvectors associated to non-zero eigenvalues are denoted by $u\_i$ and $v\_i$ and the positive eigenvalues are denoted $\lambda\_i$ then we have:
$H u\_i = \lambda\_i u\_i$ and $H v\_i = -\lambda\_i v\_i$
where i=1,2,...,s where 2s is the rank of H. The eigenvectors can be chosen to be complex conjugates of each other: $u\_i = v\_i^\*$.
In terms of these eigenvectors and eigenvalues we can write the eigen-decomposition of H as:
$H = \sum\_{i=1}^s \lambda\_i u\_i u\_i^\dagger - \lambda\_i v\_i v\_i^\dagger$
We can now define a new matrix Q as just the "positive eigenvalue part" of H:
$Q := \sum\_{i=1}^s \lambda\_i u\_i u\_i^\dagger$
This Q is Hermitian, positive semi-definite and satisfies $H = Q - Q^T = Q - Q^\*$. Apparently this Q is also the "closest Hermitian positive semi-definite matrix" to H, as measured in the Frobenius norm (and possibly other norms too).
This all goes through smoothly for finite n x n matrices H.
**My question is**, if H is now a countably infinite dimensional matrix $H\_{xy}$ for x,y=1,2,...,$\infty$ which satisfies the same conditions as before:
$H\_{xy} = H\_{yx}^\*$ and $H\_{xy} = -H\_{yx}$
can we do the same construction and obtain the matrix Q?
It seems like the rigorous way to do this requires treating the infinite matrix as an operator on the $\ell^2$ Hilbert space of infinite square-summable sequences. For general H it seems like this will be an unbounded operator that is probably not defined on the whole Hilbert space (due to issues of convergence when doing the infinite matrix multiplication).
In a best-case scenario we'd like H to define a self-adjoint operator on $\ell^2$. We could then (presumably) apply the spectral theorem and sum the positive eigenvalue part to get a Q operator/infinite-matrix.
What conditions do we have to impose on the infinite H matrix entries to ensure that the Q matrix exists? or to ensure that H defines a self-adjoint operator?
Can we sidestep the use of the eigenvectors and eigenvalues when defining Q and instead seek Q as the "closest positive semi-definite matrix" to H? Does this even make sense in the infinite matrix case?
Can we calculate Q from H if H satisfies the right conditions?
Any thoughts greatly appreciated.
| https://mathoverflow.net/users/4673 | Infinite hermitian matrix | Not an answer really, but a collection of several comments.
1. The "skew-symmetric" condition is not really natural for an operator on a complex Hilbert space, since it isn't preserved by unitary transformations.
2. Do you have a reference for the statement that Q is the closest Hermitian positive semidefinite matrix to H in Frobenius norm? Does this rely in an essential way on H being skew-symmetric?
3. The "positive part" construction would apply to any (possibly unbounded) self-adjoint operator, using an appropriate version of the spectral theorem; self-adjoint operators can be "diagonalized" in a certain general sense. (One version says that, up to a unitary transformation, your Hilbert space is $L^2(X,\mu)$ for some measure space $(X,\mu)$, and your operator is multiplication by some real-valued function $h$ on $X$. So the positive part corresponds to multiplication by the positive part of $h$.)
4. I am not aware of a condition on the entries of an infinite matrix that's equivalent to the corresponding operator being self-adjoint (though I'd be interested to know if there is). Self-adjointness is a fairly delicate property, in general; it requires the domain of the operator to be neither too large nor too small.
5. You could certainly seek the nearest positive semidefinite Hermitian operator to a given one, with respect to some norm. However you are restricting yourself to those operators for which that norm is finite. The Hilbert-Schmidt norm might be natural as it generalizes the Frobenius norm; the operator norm is another choice. The positive semidefinite Hermitian operators are closed under both norms, so looking for a "nearest" one makes sense. Also, Hilbert-Schmidt operators, being compact, are diagonalizable in the more usual sense (there is an orthonormal basis of eigenvectors).
| 1 | https://mathoverflow.net/users/4832 | 26948 | 17,623 |
https://mathoverflow.net/questions/26612 | 5 | Let $(B\_i)$ be a collection of i.i.d. random variables taking values 0 or 1.
Suppose $0 < x^- < x^+ < 1$. Consider two different "success probabilities" for the i.i.d. collection $B\_i$: under measure $P^-$, each $B\_i$ takes value 1 with probability $x^-$, while under measure $P^+$, each $B\_i$ takes value 1 with probability $x^+$.
Consider increasing events which are functions of the collection $(B\_i)$.
I'm looking for a result of the following form: given $u^-\in(0,1)$, there exists
$u^+>u^-$ such that if $A$ is an increasing event with $P^-(A)\geq u^-$,
then $P^+(A)\geq u^+$.
Perhaps I'm missing something simple.... There are easy arguments, for example, when $x^+\geq (x^-)^{1/2}$. (Then one can consider two independent samples of $(B\_i)$ from $P^+$, and take the min of the two samples; this dominates a sample from $P^-$. From this the result follows with $u^+=(u^-)^{1/2}$.)
| https://mathoverflow.net/users/5784 | probabilities of increasing events under different product measures. | I managed to hack out an answer to this question, and get the following result:
For any $k$ and $\epsilon$, if
$
x^+\geq 1-(1-x^-)^{1+1/k}
$
and
$
P\_{x^-}(A)\in (k\epsilon^{1/2}, 1-k\epsilon^{1/2})
$
then
$
P\_{x^+}(A)-P\_{x^-}(A)\geq \epsilon.
$
Given $u^-$, $x^-$ and $x^+$, it's easy to use this to get a $u^+$ of the
form I was asking for, by choosing appropriate $k$ and $\epsilon$.
As it happens, this result is not enough to get a lower bound on the influences that I was asking for under Ryan's post (such a bound would follow for example if one could replace the interval $(k\epsilon^{1/2}, 1-k\epsilon^{1/2})$ by $(k\epsilon, 1-k\epsilon)$ above).
The argument I've got is not deep but too long to post here. It involves considering $k+1$ copies of the collection of Bernoulli random variables, then comparing the probability of the event A occurring based on the max of all $k+1$ of the copies or based on the max of just $k$ of the copies. Looking at an appropriate sequence of conditional probabilities as more information about the Bernoulli collections is revealed, one shows that if there is enough randomness in the sequence for the probability of the event occurring to be significantly away from 0 and from 1, then also there must be a significant chance that the event occurs based on the max of $k+1$ copies but not on $k$ copies. For completeness and just in case anyone happens to care, I've put some messy notes about it at <http://www.stats.ox.ac.uk/~martin/papers/increasing.pdf>
Of course I'd still be very interested to hear of any references where this or something similar has been done. I wouldn't be surprised if a more general approach involving isoperimetry or concentration is possible.....
| 1 | https://mathoverflow.net/users/5784 | 26956 | 17,628 |
https://mathoverflow.net/questions/26959 | 8 | Martin's Axiom implies that $2^{\aleph\_0}$ is a regular cardinal. But can $2^{\aleph\_0}$ be a singular cardinal?
By Konig's Lemma, it can never be $\aleph\_{\omega}$ since cf($2^{\aleph\_0}$)>$\aleph\_0$ but under what conditions can it be $\aleph\_{\omega\_1}$? It is even possible it can be $\aleph\_{\omega\_1}$?
| https://mathoverflow.net/users/3859 | Can the continuum be a singular cardinal? | Yes, but it must have uncountable cofinality. So if it is to be singular, the smallest possibility is $\aleph\_{\omega\_1}$.
The basic fact is that if $\kappa$ is any cardinal such that $\kappa^\omega=\kappa$, then there is a forcing extension $V[G]$ in which $2^\omega=\kappa$. The forcing to achieve this is $\text{Add}(\omega,\kappa)$, which consists of finite partial functions from $\kappa\times\omega$ to $2$.
In particular, if you start with GCH in the ground model, and add $\aleph\_{\omega\_1}$ many Cohen reals, then you will have $2^\omega=\aleph\_{\omega\_1}$ in the forcing extension. The proof uses the following facts: (1) adding any number of Cohen reals is c.c.c. and therefore preserves all cardinals. (2) The forcing clearly adds at least that many reals. (3) A nice name argument shows that every real in the extension has a nice name in the ground model, and there are only $\aleph\_{\omega\_1}$ many such names. So the continuum of the extension is exactly $\aleph\_{\omega\_1}$. The same ideas work for any $\kappa$ for which $\kappa^\omega=\kappa$.
| 13 | https://mathoverflow.net/users/1946 | 26960 | 17,631 |
https://mathoverflow.net/questions/26942 | 55 | Part of what I do is study typical behavior of large combinatorial structures by looking at pseudorandom instances. But many commercially available pseudorandom number generators have known defects, which makes me wonder whether I should just use the digits (or bits) of $\pi$.
A colleague of mine says he "read somewhere" that the digits of $\pi$ don't make a good random number generator. Perhaps he's thinking of the article "A study on the randomness of the digits of $\pi$" by Shu-Ju Tu and Ephraim Fischbach. Does anyone know this article? Some of the press it got (see e.g. <http://news.uns.purdue.edu/html4ever/2005/050426.Fischbach.pi.html> ) made it sound like $\pi$ wasn't such a good source of randomness, but the abstract for the article itself (see <http://adsabs.harvard.edu/abs/2005IJMPC..16..281T> ) suggests the opposite.
Does anyone know of problems with using $\pi$ in this way? Of course if you use the digits of $\pi$ you should be careful not to re-use digits you've already used elsewhere in your experiment.
My feeling is, you should use the digits of $\pi$ for Monte Carlo simulations. If you use a commercial RNG and it leads you to publish false conclusions, you've wasted time and misled colleagues. If you use $\pi$ and it leads you to publish false conclusions, you've still wasted time and misled colleagues, but you've also found a pattern in the digits of $\pi$!
| https://mathoverflow.net/users/3621 | Is pi a good random number generator? | Strictly speaking, there are some known patterns in the digits of $\pi$. There are some known results on how well $\pi$ can be approximated by rationals, which imply (for example) that we know *a priori* that the next $n$ as-yet-uncomputed digits of $\pi$ can't all be zero (for some explicit value of $n$ that I'm too lazy to compute right now). In practice, though, these "patterns" are so weak that they will not affect any Monte Carlo experiments.
The main limitation of using the digits of $\pi$ may be the computational speed. Depending on how many random digits you need, computing fresh digits of $\pi$ might become a computational bottleneck. The further out you go, the harder it becomes to compute more digits of $\pi$.
If you are worried about the quality of random digits that you're getting, then you may want to use *cryptographic* random number generators. For example, finding a pattern in the Blum-Blum-Shub random number generator would probably yield a new algorithm for factoring large integers! Cryptographic random number generators will run more slowly than the "commercial" random number generators you're talking about but you can certainly find some that will generate digits faster than algorithms for computing $\pi$ will.
| 55 | https://mathoverflow.net/users/3106 | 26970 | 17,639 |
https://mathoverflow.net/questions/26983 | 0 | I want to know which isomorphism of vector bundles is the following?
[alt text http://2.bp.blogspot.com/\_uGcgLiQvkI8/TAgpIivl6oI/AAAAAAAAAKk/DCg9iK2W3rA/s1600/Capture-52.png](http://2.bp.blogspot.com/_uGcgLiQvkI8/TAgpIivl6oI/AAAAAAAAAKk/DCg9iK2W3rA/s1600/Capture-52.png)
where $G/H$ is the quotient of group $G$ by its subgroup $H$ and $T(G/H)$ is the tangent bundle, $G\times\_H \mathfrak{g}/\mathfrak{h}$ is the bundle associated to the principal bundle $G\to G/H$ via the adjointe representation on $\mathfrak{g}/\mathfrak{h}$
| https://mathoverflow.net/users/2597 | A metric on T(G/H) | There is an obvious map $G\times\mathfrak g\to G\times\_H\mathfrak g/\mathfrak h$, and an isomorphism $TG\to G\times\mathfrak g$. On the other hand, the projection $G\to G/H$ gives a map $TG\to T(G/H)$. Now you can construct a map $T(G/H)\to G\times\_H\mathfrak g/\mathfrak h$ as the composition $$T(G/H)\leftarrow TG\to G\times\mathfrak g\to G\times\_H\mathfrak g/\mathfrak h.$$ Here the backwards arrow means "pick any preimage".
| 5 | https://mathoverflow.net/users/1409 | 26986 | 17,650 |
https://mathoverflow.net/questions/26979 | 30 | Is there a finite group such that, if you pick one element from each conjugacy class, these don't necessarily generate the entire group?
| https://mathoverflow.net/users/799 | Generating a finite group from elements in each conjugacy class | No, this is impossible. This is a standard lemma, but I'm finding it easier to give a proof than a reference: Let $G$ be your finite group. Suppose that $H$ were a proper subgroup, intersecting every conjugacy class of $G$. Then $G = \bigcup\_{g \in G} g H g^{-1}$. If $g\_1$ and $g\_2$ are in the same coset of $G/H$, then $g\_1 H g\_1^{-1} = g\_2 H g\_2^{-1}$, so we can rewrite this union as $\bigcup\_{g \in G/H} g H g^{-1}$. There are $|G|/|H|$ sets in this union, each of which has $|H|$ elements. So the only way they can cover $G$ is if they are disjoint. But they all contain the identity, a contradiction.
**UPDATE:** I found a reference. According to [Serre](http://www.ams.org/journals/bull/2003-40-04/S0273-0979-03-00992-3/home.html), this result goes back to Jordan, in the 1870's.
| 68 | https://mathoverflow.net/users/297 | 26993 | 17,655 |
https://mathoverflow.net/questions/26991 | 4 | JS Milne has a page about common errors in mathematical papers, and one of them is the usage of "verify" to mean "satisfy".
>
> Improper usage: "The set $A$ verifies the condition."
>
>
> Proper usage: "The set $A$ satisfies the condition."
>
>
> Proper usage: "We verify that $A$ satisfies the condition."
>
>
>
Strangely enough, I've only seen this once or twice in a paper or book written in English, but I've seen it in nearly every paper or book written in French that I've read. That is, we have:
>
> L'ensemble $A$ vérifie la condition.
>
>
>
Now, there's another error on Milne's page that he notes, the "associated to" and "associated with" error. If one is attempting to use proper English, "associated with" is the only correct choice. It turns out that this error comes from a mistranslation of the French, "associé à", which means "associated with".
My question then: Is the French usage of "vérifier" to mean "satisfy" acceptable in French, like the usage of "associé à", but not in English, or is it just an error that has propagated to both languages (possibly from a third language where the word for "verify" is the same as the word for "satisfy")?
| https://mathoverflow.net/users/1353 | The origin of the satisfy-verify mixup | Dear Harry, in Serre's collected papers, vol.1, page 183 [or Annals of Math.58(1953) page 270]
you'll find (line -5)
"Soit $\mathcal C$ une classe **vérifiant** (II\_A)..."
and many such examples on the same page, corroborating your testimony on papers and books you read in French. I recoil in horror at the thought that some heretic might not consider this a sufficient proof that the usage of "vérifie" in the sense "satisfies" is more than acceptable in French. Another quote: Bourbaki, in Topologie Générale, Chapitre 1, §6, page 61 (line -15)[Quatrième édition] writes
"Pour qu'un ensemble de parties satisfaisant à (F\_1) **vérifie** aussi..."
| 2 | https://mathoverflow.net/users/450 | 26996 | 17,658 |
https://mathoverflow.net/questions/26985 | 7 | I have a curiosity on the Ergodic decomposition given by the von Neumann's theorem:
$$L^2(X,\Sigma,\mu)=L^2(X,\Sigma\_T,\mu)\oplus\overline{\{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu)\}},$$
that occurs for a measure-preserving map $T$ of a probability space $(X,\Sigma,\mu)$, $\Sigma\_T$ being the sub-σ-algebra of all $T$-invariant measurable sets, and the (orthogonal) projector being the conditional expectation $E(\cdot|\Sigma\_T)$. For all $1\leq p \leq\infty$, the conditional expectation is well-defined as a linear projector of norm 1 on $\textstyle L^p(X,\Sigma,\mu)$, with range the closed subspace $L^p(X,\Sigma\_T,\mu) \subset L^p(X,\Sigma,\mu)$. Therefore it's quite natural to consider the analogue decomposition of the $L^p$ spaces given by the $L^p$ projector , that ''should'' be:
$$L^p(X,\Sigma,\mu)=L^p(X,\Sigma\_T,\mu)\oplus\ {\overline{\{ f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)}\} }^{L^p}.$$
Now if $1\leq p\leq 2$, this splitting actually holds true, and it is easily obtained with a Lp-closure starting from the $L^2$ splitting. For $2\leq p \leq\infty$, a splitting is obtained by restriction to $L^p(X,\Sigma,\mu)$. And here it comes the problem: this way I get the right first factor (the range of the projector) $L^p(X,\Sigma\_T,\mu)=L^2(X,\Sigma\_T,\mu)\cap L^p(X,\Sigma,\mu)$, but I can't see why the kernel of the projector,
$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)$$
should be equal to (and not larger than)
$$\overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}.$$
Maybe it's not a fundamental point (it does not enter in the proof of the main ergodic theorems) but I think that if a complete analogy holds true, it would be nice to state it, and if it doesn't, one would like to know what goes wrong. I checked the main texts of ergodic theory on this point, and found nothing.
**Summarizing**:
is there a (hopefully quick) way to see whether for $2\leq p \leq\infty$ there is an inclusion (hence equality)
$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)\ \subset \ \overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}$$
$$\mathbf{?}$$
| https://mathoverflow.net/users/6101 | Ergodic splitting in L_p | The mean ergodic theorem on L^p spaces is due to F. Riesz (1938) and S. Kakutani (1938). For p in $[1,\infty[$, this is theorem 1.2 ff in the book of Krengel, "Ergodic theorems".
If $p=\infty$, you get the splitting only on the set of functions for which the Birkhoff sums are converging, which, I think is not everything in general. In general Banach spaces, you always have the splitting in restriction to the space of vectors with converging averages (see Krengel, theorem 1.3).
The decomposition is more or less explicit. For all measurable function g, we can write
$g - {1\over n} S\_n(g) = g\_n - g\_n\circ T$
with $S\_n(g)=\Sigma\_0^{n-1}g\circ T^k$ and $g\_n={1\over n}\ \Sigma\_0^{n-1} S\_k(g)$.
If g is in the $L^2$ closure of coboundaries, we know that its conditional expectation w.r.t the invariant $\sigma$-algebra is zero. If moreover g is in $L^p$, $1\leq p < \infty$, we know that ${1\over n} S\_n(g)$ goes to zero in $L^p$ norm, using the $L^p$ ergodic theorem. Also, the functions $g\_n$ are in $L^p$, from their very definition. So g is in the closure of $L^p$ coboundaries.
I am not sure that the result holds for $p=\infty$.
| 2 | https://mathoverflow.net/users/6129 | 27019 | 17,670 |
https://mathoverflow.net/questions/27016 | 12 | In non-relativistic quantum mechanics, what are the necessary conditions on the potential (or on the hamiltonian in general) for the ground state to be non-degenrate?
| https://mathoverflow.net/users/6547 | Non-degeneracy of ground state in quantum mechanics | If a finite number of non-relativistic particles are moving in an infinite potential well, then the combined system has a nondegenerate ground state, regardless of the symmetry of the hamiltonian. I remember this from a long time ago, and I always thought it was impressive. I also remember I was always annoyed that I didn't know how to prove it, or know a reference where I can look it up. If you find one, let me know!
There's probably some sort of fancy entropic argument that you could use to get this result, if that's your thing.
If the potential was bounded above, I can't see immediately why this should create degeneracy on the ground state --- so it's plausible that the theorem holds in this case as well.
Systems containing infinite systems of particles can, and often do, exhibit degeneracy in their ground state.
| 3 | https://mathoverflow.net/users/799 | 27030 | 17,677 |
https://mathoverflow.net/questions/27033 | 3 | Is a subgroup of a finite group uniquely determined, up to conjugation, by the subset of conjugacy classes of the larger group that it intersects?
| https://mathoverflow.net/users/799 | Determining conjugacy class of a subgroup from intersection with conjugacy classes | Let $G$ be the group of affine linear maps over the Galois field $k=GF(16)$
of order $16$. The elements of $G$ are maps from $k$ to itself of the form
$x\mapsto ax+b$ where $a\in k^\*$ and $b\in G$. Those with $a=1$ form
a normal elementary abelian subgroup~$H$. All nontrivial elements of $H$
are conjugate. Then $H$ contains lots of subgroups of order $4$, thirty-five
in all, each consisting of the identity and three elements of this conjugacy class
of involutions. But these are not all conjugate under $G$; it is clear that such a
subgroup has at most fifteen conjugates.
| 5 | https://mathoverflow.net/users/4213 | 27034 | 17,678 |
https://mathoverflow.net/questions/27032 | 14 | Given two convex bodies $A$ and $B$, in $\mathbb R^3$ let's say. We define $A(t)$ and $B(t)$ as $A+xt$ and $B+yt$ where $x,y$ are two arbitrary points. (That is the Minkowski sum, so the two bodies are moving at constant velocity in the $x$ and $y$ directions, and $t$ is the time variable.) Can one show that the function
$$f(t)=\operatorname{Vol}(A(t)\cap B(t))$$ is unimodal? That is nondecreasing up to some point, and then nonincreasing.
| https://mathoverflow.net/users/2384 | Pushing convex bodies together | The sets $\{ (A(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ and $\{ (B(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ are convex, their intersection $K$ is a bounded convex set, and $f(t)$ is the volume of the slice of $K$ at height $t$. By Brunn-Minkowski inequality, this is log-concave, so definitely unimodal.
| 16 | https://mathoverflow.net/users/2368 | 27038 | 17,681 |
https://mathoverflow.net/questions/27042 | 4 | Let $W$ be a standard Brownian motion under given probability space.
For a given constant $a$, $W^a$ is a truncated Brownian motion by stopping time
$T^a = \inf(t>0:W(t) = a)$. That is, $W^a(t) = W(t \wedge T^a)$.
We want to consider the following question: Is the process $W^1$ of [the class DL](http://almostsure.wordpress.com/2009/12/22/class-d-processes/)?
(Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables
$( W(s), 0< s< t)$
is uniformly integrable by definition, since $E [|W^1(t)|] < \infty$.
We provide completely different answer using the following proposition from the Problem 1.5.19 (i) of Book [Karazas and Shereve 98].
[Proposition] A local martingale of class DL is martingale.
(Solution2): No. $W^1$ is strict local martingale, since $E [W^1(T^1)] = 1> E [W(0)]$. By [Proposition], $W^1$ is not of class DL.
In the above, we obtained completely two different solutions. Where is wrong?
| https://mathoverflow.net/users/5656 | Is the truncated Brownian motion of the class DL? | $W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T\_a)$. See also Theorem 3.39 [here](http://books.google.com/books?id=JYzW0uqQxB0C&pg=PA85).
When you write an expression like $\mathbb{E}(W^a(T^a))$ you are implicitly assuming that $W^a(T^a)$ is measurable. This requires $t \ge T\_a$ (and trivializes the expectation).
| 3 | https://mathoverflow.net/users/1847 | 27046 | 17,684 |
https://mathoverflow.net/questions/27039 | 8 | Let $X$ be a stone space, i.e. a compact, totally disconnected hausdorff space. Then $H^1(X,\mathbb{Z}/2)=0$. Here's one way of proving this: $X$ with $\mathbb{Z}/2$ (the constant sheaf) is an affine scheme, now use the vanishing result for quasicoherent sheaves on affine schemes.
What happens if we also allow locally compact, totally disconnected hausdorff spaces? Then $X$ with $\mathbb{Z}/2$ is again a scheme, but it is not affine (unless $X$ is compact). A typical example would be an open subset of a stone space (actually this is generic), or the underlying topological space of a local field.
| https://mathoverflow.net/users/2841 | Sheaf Cohomology on a Stone Space | A search brought up [Sheaf Cohomology of Locally Compact Totally Disconnected Spaces](https://doi.org/10.2307/2035693) by R. Wiegand. [Some topological invariants of Stone spaces](https://doi.org/10.1307/mmj/1029000311) by the same author might also be of interest.
| 3 | https://mathoverflow.net/users/2384 | 27048 | 17,686 |
https://mathoverflow.net/questions/27044 | 2 |
>
> **Possible Duplicates:**
>
> [Free, high quality mathematical writing online?](https://mathoverflow.net/questions/1722/free-high-quality-mathematical-writing-online)
>
> [Most helpful math resources on the web](https://mathoverflow.net/questions/2147/most-helpful-math-resources-on-the-web)
>
>
>
A lot has been said about different kinds of math resources here in MO.
To mention a few:
[Most helpful math resources on the web](https://mathoverflow.net/questions/2147/most-helpful-math-resources-on-the-web)
[Undergraduate Level Math Books](https://mathoverflow.net/questions/761/undergraduate-level-math-books)
[Best online math videos?](https://mathoverflow.net/questions/1714/best-online-math-videos)
So having benefited a lot from free resources on the net, I thought it would be more helpful if I shared my list of important sites and learned from others, too.
Here are my list of sites providing free math resources. Some of them are file hosting and sharing sites and others may need registration.
1. [Math Online](http://mathonline.andreaferretti.it/) Recently launched by Andrea Ferretti
2.[MIT OpenCourseWare](http://ocw.mit.edu/index.htm)
3.[Project Gutenberg](http://www.gutenberg.org/wiki/Main_Page)
4.[2020ok Directory of FREE Online Books and FREE eBooks](http://2020ok.com/)
5.[Freebookcentre.net](http://www.freebookcentre.net/)
6.[Gigapedia](http://gigapedia.com/)
7.[Scribd](http://www.scribd.com/scribid) is a social publishing site, where tens of millions of people share original writings and documents. Scribd's vision is to liberate the written word.
8.[4shared.com - free file sharing and storage](http://www.4shared.com/)
>
> My question here is to find more free
> resources sites possibly specializing
> in a given branch of math. Please
> specify a direct link anf whether
> registration is required.
>
>
>
* PS: One technique I use is to google my desired file with its file name and type as in "Elliptic curves pdf"(which returns a list of pdfs on elliptic curves) or "Random matrices doc" (for word document search) and "Ramanujan history ppt"(which returns a list of presentations or slides).
*You may also suggest such search techniques.*
| https://mathoverflow.net/users/5627 | Websites hosting free math ebooks. | I have found some list of math books here, though not so much advanced.
[onlinecomputerbook](http://www.onlinecomputerbooks.com/free-math-books.php) The name is misleading(There are math books)
| 1 | https://mathoverflow.net/users/5627 | 27053 | 17,690 |
https://mathoverflow.net/questions/27040 | 7 | I have to work with the following variation of Minkowski sum:
>
> Let $\mathbb E$ be a Euclidean space and $K$ be a convex set in $\mathbb E\times \mathbb E$.
> Set
> $$K^+=\{\\,x+y\in\mathbb E\mid(x,y)\in K\\,\}.$$
>
>
>
Note that if $K=K\_x\times K\_y$ for some convex sets $K\_x$ and $K\_y$ in $\mathbb E$ then $K^+$ is the usual Minkowski sum of $K\_x$ and $K\_y$.
**Questions:**
* Did anyone consider this construction?
* Does it have a name?
| https://mathoverflow.net/users/1441 | A variation of Minkowski sum | In additive combinatorics, we call the Minkowski sum the sumset, and write it as ${\mathbb E}+{\mathbb E}$. We call what you're talking about the "sumset along a graph", and write it as ${\mathbb E}+\_K{\mathbb E}$, where $K$ is any graph (you call it a subset of ${\mathbb E}\times {\mathbb E}$ and I call it a graph, but it's the same thing!).
For an example of this terminology in use, check out [this paper](http://arxiv.org/abs/0905.0135) of Alon, Angel, Benjamini, and Lubetzky. Also, a [google scholar search](http://www.google.com/search?q=sumsets+along+a+graph) shows the terminology in action.
| 5 | https://mathoverflow.net/users/935 | 27064 | 17,696 |
https://mathoverflow.net/questions/27004 | 20 | Each of n players simultaneously choose a positive integer, and one of the players who chose [the least number of [the numbers chosen the fewest times of [the numbers chosen at least once]]] is selected at random and that player wins.
For n=3, the symmetric Nash equilibrium is the player chooses m with probability 1/(2^m).
What is the symmetric Nash equilibrium for n=4? Is it known for general n?
| https://mathoverflow.net/users/nan | Lowest Unique Bid | There is some published literature on this problem. See for example the following papers and the references therein.
Baek and Bernhardsson, [Equilibrium solution to the lowest unique positive integer game](http://arxiv.org/pdf/1001.1065v1)
Rapoport et al., [Unique bid auctions: Equilibrium solutions and experimental evidence](http://mpra.ub.uni-muenchen.de/4185/)
Ostling et al., [Strategic thinking and learning in the field and lab: Evidence from Poisson LUPI lottery games](http://en.scientificcommons.org/23204285)
Houba et al., [The Unique-lowest Sealed-bid Auction](http://dare.ubvu.vu.nl//handle/1871/12947)
Apparently, in general, the Nash equilibria are intractable to describe.
| 2 | https://mathoverflow.net/users/3106 | 27069 | 17,701 |
https://mathoverflow.net/questions/27059 | 1 | Does the function $f(x,y) = ((x-1) \mod y)+1$ have an existing name?
`f(1,5) = 1`
`f(2,5) = 2`
`f(3,5) = 3`
`f(4,5) = 4`
`f(5,5) = 5`
`f(6,5) = 1`
`f(7,5) = 2`
| https://mathoverflow.net/users/2644 | What is the name of the function f(x,y) = ((x-1) mod y)+1 ? | In math, as opposed to in computer science, when you apply "mod y" you land in the integers modulo y, denoted Z/yZ, not back in the integers. This means that mod 7, the symbols 1 and 8 *denote the same thing*, i.e. the equivalence class {...,-13,-6,1,8,15,...}.
A more computer-y way to say this is that for mathematicians, "integer mod 7" is a different kind of data class than "integer."
All this is just a long way of saying that this is probably the wrong place to ask your question.
If I were defining this function in a math paper I'd say something like "Let f(x,y) denote the unique number in {1,2,...,y} which is congruent to x modulo y" or "By the division algorithm there exists a unique number in {1,2,...,y} which is congruent to x modulo y, we denote this number f(x,y)."
| 7 | https://mathoverflow.net/users/22 | 27070 | 17,702 |
https://mathoverflow.net/questions/27071 | 2 | Can someone lead me to a method for calculating the number of sequences of length $n$ if the terms of the sequence are chosen (with replacement) from a set with $k$-elements under the condition that all $k$-elements are chosen at least once? It's not my field.
I lead myself to consider the recurrence relation
$$ R(k,n) = k R(k,n-1) + k R(k-1,n-1) $$
where $R(k,k) = k!, R(k,n) = 0$ if $n \lt k$ and $R(1,n) = 1$ for all $n$, but struggled to close it out. Is there a method for this?
$R(n,k)$ is the number I want, so from my point of view I'm asking one question. From your point of view I'm sure I'm asking two. Many Thanks.
| https://mathoverflow.net/users/6556 | Counting sequences - a recurrence relation. | These are related to Stirling numbers. These don't have a "closed form".
Your $R(k,n)=k!S(n,k)$ where $S(n,k)$ is the Stirling number of
the second kind as defined at
<http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind> .
| 7 | https://mathoverflow.net/users/4213 | 27072 | 17,703 |
https://mathoverflow.net/questions/27075 | 106 | What is the oldest open problem in mathematics? By old, I am referring to the date the problem was stated.
Browsing Wikipedia list of open problems, it seems that the *Goldbach conjecture* (1742, every even integer greater than 2 is the sum of two primes) is a good candidate.
The *Kepler conjecture* about sphere packing is from 1611 but I think this is finally solved (anybody confirms?). There may still be some open problem stated at that time on the same subject, that is not solved. Also there are problems about cuboids that Euler may have stated and are not yet solved, but I am not sure about that.
A related question: can we say that we have solved all problems
handed down by the mathematicians from Antiquity?
| https://mathoverflow.net/users/6129 | What is the oldest open problem in mathematics? | Existence or nonexistence of odd perfect numbers.
Update: Goes back at least to Nicomachus of Gerasa around 100 AD, according to [J J O'Connor and E F Robertson](https://web.archive.org/web/20100209232200/http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Perfect_numbers.html). Nichomachus also asked about infinitude of perfect numbers.
(Goes back at least to Descartes 1638 <https://mathworld.wolfram.com/OddPerfectNumber.html> and arguably all the way back to Euclid.)
| 89 | https://mathoverflow.net/users/22 | 27080 | 17,708 |
https://mathoverflow.net/questions/27089 | 2 | Off hand, does anyone know of some useful conditions for checking if a ring (or more generally a semiring) has non-trivial derivations? (By non-trivial, I mean they do not squish everything down to the additive identity.) Part of the motivation for this is that I was thinking about it the other day, and had trouble finding any good example of a semiring with an interesting derivation.
For example, the multiplicative Banach algebra of positive functions is an algebra of the semifield of nonnegative reals. However, the usual definition for derivative breaks down due to the fact that you can have positive functions with negative slope. So, this leads me to wonder if there are any semirings with derivations at all?
As a related question, is there a known classification of all the derivations for an algebra? It feels like this should be a pretty standard thing, but I don't think I've ever encountered it in one of my courses and my initial googling around was not too successful at finding references.
| https://mathoverflow.net/users/4642 | Necessary and sufficient criteria for non-trivial derivations to exist? | There is a notion of a universal derivation for an algebra. I'll assume
everything is commutative for simiplcitity. If $A$ is a $k$-algebra
($k$ a commutative ring) then there is an $A$-module $\Omega\_{A/k}$,
the module of *Kahler differentials* of $A$ over $k$ and a $k$-derivation
$d:A\to\Omega\_{A/k}$ which is universal for derviations of $A$.
That is, if $\delta:A\to M$ is a $k$-derivation from $A$ to an $A$ module,
then $\delta=f\circ d$ for a unique $A$-module homomorphism $f$.
There is an explicit desciption of $\Omega\_{A/k}$ as $I/I^2$
where $I$ is the ideal in the ring $B=A\otimes\_k A$ which is the kernel
of the map $\mu:B\to A$ with $\mu(x\otimes y)=xy$. Then $d$ maps $x\in A$
to $1\otimes x-x\otimes 1$. So to find a derivation from $A$ with values in
your favourite $A$-module $M$ all one has to do is to find an
$A$-homomorphism from $\Omega\_{A/k}$ to $M$. Of course $\Omega\_{A/k}$
may be hard to determine concretely, and even if that is possible, perhaps
it may not be easy to find a homomorphism from that into $M$. Indeed
using this method may be no easier than finding a derivation directly :-)
For details see the commutative algebra texts by Eisenbud or Matsumura.
| 4 | https://mathoverflow.net/users/4213 | 27093 | 17,716 |
https://mathoverflow.net/questions/27076 | 14 | Often undergraduate discrete math classes in the US have a calculus prerequisite.
Here is the description of the discrete math course from my undergrad:
>
> A general introduction to basic
> mathematical terminology and the
> techniques of abstract mathematics in
> the context of discrete mathematics.
> Topics introduced are mathematical
> reasoning, Boolean connectives,
> deduction, mathematical induction,
> sets, functions and relations,
> algorithms, graphs, combinatorial
> reasoning.
>
>
>
What about this course suggests calculus skills would be helpful?
Is passing calculus merely a signal that a student is ready for discrete math?
Why isn't discrete math offered to freshmen — or high school students — who often lack a calculus background?
| https://mathoverflow.net/users/6555 | Why does undergraduate discrete math require calculus? | A significant portion (my observation was about 20-30% at Berkeley, which means it must approach 100% at some schools) of first year students in the US do not understand multiplication. They do understand how to calculate $38 \times 6$, but they don't intuitively understand that if you have $m$ rows of trees and $n$ trees in each row, you have $m\times n$ trees. These students had elementary school teachers who learned mathematics purely by rote, and therefore teach mathematics purely by rote. Because the students are very intelligent and good at pattern matching and at memorizing large numbers of distinct arcane rules (instead of the few unifying concepts they were never taught because their teachers were never taught them either), they have done well at multiple-choice tests.
These students are going to struggle in any calculus course or any discrete math course. However, it is easier to have them all in one place so that one instructor can try to help all of them simultaneously. For historical reasons, this place has been the calculus course.
| 50 | https://mathoverflow.net/users/3077 | 27097 | 17,719 |
https://mathoverflow.net/questions/27090 | 6 | I have a stupid question about the [Metropolis-Hastings sampling algorithm](http://en.wikipedia.org/wiki/Metropolis%25E2%2580%2593Hastings_algorithm).
If I got this right, for every variable $X$ in turn, which currently has value $x\_{old}$, you generate a new sample $x\_{new}$. To do that, you draw $x\_{new}$ from the proposal distribution $Q(x\_{new}\mid x\_{old})$, then you draw a number $\alpha$ uniformly at random from the range between $0$ and $1$. Then, accept $x\_{new}$ if
$\alpha < \min{1,\frac{P(x\_{new})}{P(x\_{old})}\frac{Q(x\_{old}\mid x\_{new})}{Q(x\_{new}\mid x\_{old})}}$
The second ratio does not really make sense to me: Why are we more likely to accept if $Q(x\_{new}\mid x\_{old})$ is low?
| https://mathoverflow.net/users/6561 | Question about this ratio in Metropolis-Hastings MCMC algorithm | From what you're saying, I'm not sure if you want a proof or intuition. As the proof is written up in many places, I'll just guess that you want intuition.
Very informally: the algorithm allows you to, in effect, sample from distribution P using samples from distribution Q. So in a sense we want to take the samples from Q and "remove" statistical properties of these samples that reveal that they come from Q, replacing them with the properties of P. The thing that "gives away" that they came from Q is that they're more likely to come from areas where Q is high. So we want our acceptance probability to be reduced when our samples come from such an area. That's exactly what dividing by $Q(x\_{new}|x\_{old})$ does.
(BTW The $min$ in your expression is redundant.)
| 3 | https://mathoverflow.net/users/1233 | 27101 | 17,721 |
https://mathoverflow.net/questions/27107 | 14 | I heard that the following problem lead to determine the rational points of an elliptic curve:
For which integers $n$ there are integers $x,y,z$ such that $x/y+y/z+z/x=n$. Could anyone show me why this question leads to the theory of elliptic curves?
| https://mathoverflow.net/users/6266 | Transforming a Diophantine equation to an elliptic curve | Nearly 10 years ago, I gave a talk at Wesleyan, and a gentleman named Roy Lisker asked me the same question: Fix an integral solution $(x, \ y, \ z)$ and make the substitution
$$u = 3 \ \frac {n^2 z - 12 \ x}z \qquad v = 108 \ \frac {2 \ x \ y - n \ x \ z + z^2}{z^2}$$
Then $(u, \ v)$ is a rational point on the elliptic curve $E\_n: \ v^2 = u^3 + A \ u + B$ where $A = 27 \ n \ (24 - n^3)$ and $B = 54 \ (216 - 36 \ n^3 + n^6)$. (It actually turns out that $E\_n$ is an elliptic curve whenever $n$ is different from 3, but I’ll discuss this case separately.)
Let me say a little about the structure of this curve for the experts:
This curve has the “obvious” rational point $T=(3 n^2, 108)$ which has order 3, considering the group structure of $E\_n$. It actually turns out that these three multiples correspond to the cases $x = 0$ and $z = 0$, so if such an integral solution $(x, \ y, \ z)$ exists then the rational solution $(u, \ v)$ must correspond to a point on $E\_n$ not of order 3. (Of course, I don’t care about the cyclic permutation $x \to y \to z \to x$.)
In the following table I’m computing the Mordell-Weil group of the rational points on the elliptic curve i.e. the group structure of the set of rational solutions $(u, \ v)$:
$$ \begin{matrix}
n & E\_n(\mathbb Q) \\ \\
1 & Z\_3 \\
2 & Z\_3 \\
3 & \text{Not an elliptic curve} \\
4 & Z\_3 \\
5 & Z\_6 \\
6 & Z\_3 \oplus \mathbb Z \\
7 & Z\_3 \\
8 & Z\_3 \\
9 & Z\_3 \oplus \mathbb Z \\
\end{matrix} $$
Hence when $n =$ 1, 2, 4, 7 or 8 we find no integral solutions $(x, \ y, \ z)$. When $n = 5$, there are only six rational points on $E\_n$, namely the multiples of $(u,v) = (3, 756)$ which all yield just one positive integral point $(x,y,z) = (2,4,1)$.
Something fascinating happens when $n = 6$... The rank is positive (the rank is actually 1) so there are infinitely many rational points $(u, \ v)$. But we must be careful: not all rational points $(u, \ v)$ yield positive integral points $(x, \ y, \ z)$. Clearly, we can scale $z$ large enough to always choose $x$ and $y$ to be integral, but we might not have $x$ and $y$ to both be positive. You’ll note that $x > 0$ if only if $u < 3 \ n^2$, so we only want rational points in a certain region of the graph. Since the rank is 1, this part of the graph is dense with rational points! Let me give some explicit numbers. The torsion part of $E\_n( \mathbb Q)$ is generated by $T = (75, 108)$ and the free part is generated by $(u,v) = (-108, 2052)$. By considering various multiples of this point we get a lot of positive integral -- yet unwieldy! -- points $(x,y,z)$ such that $x/y + y/z + z/x = 6$:
$$\begin{aligned} (x,y,z) & = (12, 9, 2), \\
& = (17415354475, 90655886250, 19286662788) \\
& = (260786531732120217365431085802, 1768882504220886840084123089612, 1111094560658606608142550260961) \\
& = (64559574486549980317349907710368345747664977687333438285188, 70633079277185536037357392627802552360212921466330995726803, 313818303038935967800629401307879557072745299086647462868546) \end{aligned} $$
I’ll just mention in passing that when $n = 9$ the elliptic curve $E\_n$ also has rank 1. The generator $(u,v) = (54, 4266)$ corresponds to the positive integral point $(x,y,z) = (63, 98, 12)$ on $x/y + y/z = z/x = 9$.
What about $n = 3$? The curve $E\_n$ becomes $v^2 = (u – 18) (u + 9)^2$. This gives two possibilities: either $u = -9$ or $u \geq 18$. The first corresponds to $x = z$ while the second corresponds to $(z/x) \geq 4$. By cyclically permuting $x$, $y$, and $z$ we find similarly that either $x = y = z$ or $x/y + y/z + z/x \geq 6$. The latter case cannot happen by assumption so $x = y = z$ is the only possibility i.e. $(x,y,z) = (1,1,1)$ is the only solution to $x/y + y/z + z/x = 3$.
| 36 | https://mathoverflow.net/users/6563 | 27113 | 17,730 |
https://mathoverflow.net/questions/27106 | 4 | Given a set family, what is the best way (empirically) to check whether the set family is equivalent to set of independent sets of some matroid. The input can be either the set family explicitly or bunch of cardinality constraints that must be satisfied. For example, given a universe E, subsets $A\_1,..,A\_k$ of E and positive integers $b\_1,...,b\_k$, a set F is in the family iff $|F\cap A\_i|<= b\_i$ for each $1<= i<= k$. Check whether the family is a matroid.
| https://mathoverflow.net/users/1720 | Checking whether a set family forms a matroid. | One way is to delete an element $x$ from your alleged matroid $M$, recursively check that the smaller structure $M'$ is a matroid, and then check that adding back $x$ gives you a single-element extension of $M'$. An algorithm for testing for single-element extensions is explained in [this paper by Mayhew and Royle](http://arxiv.org/abs/math/0702316), in which they compute all matroids with up to nine elements.
| 4 | https://mathoverflow.net/users/3106 | 27117 | 17,732 |
https://mathoverflow.net/questions/27087 | 3 | I am trying to learn a little Lagrangian Floer theory and I was hoping someone could explain the following calculation. Consider CP^n x CP^n with the form (omega,-omega) and the diagonal Lagrangian L. Now the FH\*(L,L) is isomorphic to the quantum cohomology of CP^n as a ring. How about the higher A-infinity structure on the Floer cochains, CH\*(L,L)? Can we compute this in a reasonable way? I'd be happy to even understand this for CP^1, though I suspect this might be obvious somehow. Is there a way to extract this from the Gromov Witten invariants of CP^n?
| https://mathoverflow.net/users/6986 | The higher structure of the Floer cochains of the diagonal in CP^ x CP^n | Here's an argument that the diagonal Lagrangian correspondence $\Delta$ in $\mathbb{C}P^n \times \mathbb{C}P^n$ is *formal*. That is, its Floer cochains $CF^\ast(\Delta,\Delta)$, as an $A\_\infty$-algebra over the rational Novikov field $\Lambda=\Lambda\_\mathbb{Q}$ (say), are quasi-isomorphic to the underlying cohomology algebra $HF^\ast(\Delta, \Delta)\cong QH^\ast(\mathbb{C}P^n; \Lambda)$ with trivial $A\_\infty$ operations $\mu^d$ except for the product $\mu^2$.
Be critical; I might have slipped up!
Write $A$ for $QH^\ast(\mathbb{C}P^n; \Lambda)=\Lambda[t]/(t^{n+1}=q)$. Here $q$ is the Novikov parameter. I claim that $A$ is *intrinsically formal*, meaning that every $A\_\infty$-structure on $A$, with $\mu^1=0$ and $\mu^2$ the product on $A$, can be modified by a change of variable so that $\mu^d=0$ for $d\neq 2$.
Suppose inductively that we can kill the $d$-fold products $\mu^d$ for $3\leq d\leq m$. Then $\mu^{m+1}$ is a cycle for the Hochschild (cyclic bar) complex $C^{m+1}(A,A)$. The obstruction to killing it by a change of variable (leaving the lower order terms untouched) is its class in $HH^{m+1}(A,A)$. But $A$ is a finite extension field of $\Lambda$ (and, to be safe, we're in char zero). So, as proved in Weibel's homological algebra book, $HH^\ast(A,A)=0$ in positive degrees, and therefore the induction works. Taking a little care over what "change of variable" actually means in terms of powers of $q$, one concludes intrinsic formality.
---
You made a much more geometric suggestion - to invoke GW invariants. If you want to handle $\Delta\_M\subset M\times M$ more generally, I think this is a good idea, though I can't immediately think of a suitable reference. One can show using open-closed TQFT arguments that $HF(\Delta\_M,\Delta\_M)$ is isomorphic to Hamiltonian Floer cohomology $HF(M)$. One could do this at cochain level and thereby show that the $A\_\infty$ product $\mu^d$ of $HF(\Delta\_M,\Delta\_M)$ corresponds to the operation in the closed-string TCFT of Hamiltonian Floer cochains arising from a genus zero surface with $d$ incoming punctures and one outgoing puncture (and varying conformal structure). Via a "PSS" isomorphism with $QH(M)$, these operations should then be computable as genus-zero GW invariants (or at any rate, the cohomology-level Massey products derived from the $A\_\infty$-structure should be GW invariants).
| 6 | https://mathoverflow.net/users/2356 | 27122 | 17,734 |
https://mathoverflow.net/questions/27118 | 3 | I'd be surprised if somebody proved the inverse Galois problem for $\mathbb{C}(x,y)$, but I wanted to make sure.
| https://mathoverflow.net/users/5309 | Is the (regular) inverse Galois problem known for the field C(x,y)? | Surely the inverse Galois problem is known over $\mathbf{C}(x)$: The Galois group of the maximal extension of $\mathbf{C}(x)$ unramified away from $n+1$ given primes of $\mathbf{C}[x]$ is the free profinite group on $n$ generators. Any finite group $G$ is a quotient of such a group, so there exists a finite Galois extension $L/\mathbf{C}(x)$ with Galois group $G$.
Then $L\otimes\_{\mathbf{C}(x)}\mathbf{C}(x,y)$ is a finite Galois extension of $\mathbf{C}(x,y)$ with Galois group $G$.
| 13 | https://mathoverflow.net/users/1114 | 27123 | 17,735 |
https://mathoverflow.net/questions/27120 | 1 | This is the first time I post a question on MO, so I'm shy a liite bit. Can you give a "non-trivial" example of a finite dimensional hereditary algebra which is quotient of an infinite dimensional algebra ?
By "non-trivial" I mean not by killing loops in the path algebra of some quiver, for example
$k[X\_1] \times\ldots \times k[X\_n]/((X\_1)\times\ldots\times (X\_n))$.
| https://mathoverflow.net/users/6565 | Hereditary algebras as quotient algebras | If you use generators and relations, then any algebra is a quotient of an infinite-dimensional algebra, i.e., a quotient of the free associative algebra corresponding to the generators you pick.
| 2 | https://mathoverflow.net/users/321 | 27128 | 17,738 |
https://mathoverflow.net/questions/27099 | 8 | There appear to be a number of rational canonical forms. The best thing about standards is how many there are to choose from. However, the standard I choose seems to have a centralizer that is difficult to describe.
>
> 1) Is there a reference that chooses a specific (hopefully pretty) rational canonical form, and computes its (hopefully pretty) centralizer?
>
>
>
Alternatively,
>
> 2) Is there a pretty description of the centralizer of my choice of canonical form?
>
>
>
Matrices are over commutative (usually finite prime) fields.
Every matrix is similar to a direct sum of matrices whose minimal polynomial is of the form irr ^ pow, for some irreducible polynomial irr and some positive integer pow. While there is some disagreement on how to organize this, a common idea is to have canonical forms associated to pairs [ irr, pow ], (the other being to group coprime irr together: is it Z/2Z × Z/3Z or is it Z/6Z? we choose Z/2Z × Z/3Z).
So given a pair [ irr, pow ], I have seen two main ways to associate the canonical block: either take "the" companion matrix of irr^pow, or take a block diagonal matrix with pow blocks equal to the companion matrix of irr, and then fill in 1s on the sub/sup diagonal you used for the companion matrix. The former more clearly lays out an indecomposable direct decomposition, but it hides a composition series. The latter more subtly lays out the direct sum decomposition, but makes the composition series very clear. Both are quite pretty. In case irr has degree 1, then the latter is a Jordan block, and the former is not.
So I chose the latter. For instance, if irr = x2−x−1 is irreducible and pow = 3, then we get the block B:
`B = $\begin{bmatrix}
.&1&.&.&.&.\\%
1&1&1&.&.&.\\%
.&.&.&1&.&.\\%
.&.&1&1&1&.\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$
If α is a root of x2−x−1, then I expected this guy to be the blow up of b:
`b = $\begin{bmatrix}
\alpha&1&.\\
.&\alpha&1\\
.&.&\alpha\\
\end{bmatrix}$
and so I expected any scalar matrices (blown up) to be in the centralizer of B, since they are in the centralizer of b. In other words, the matrix a:
`a = $\begin{bmatrix}
\alpha&.&.\\
.&\alpha&.\\
.&.&\alpha\\
\end{bmatrix}$
centralizes b, so I expected the matrix A:
`A = $\begin{bmatrix}
.&1&.&.&.&.\\%
1&1&.&.&.&.\\%
.&.&.&1&.&.\\%
.&.&1&1&.&.\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$
to centralize B, but of course AB ≠ BA. For any particular B, one can just solve a bunch of equations, but at least to me the solutions do not seem easy to describe algorithmically. I worry that we might have wanted the slightly uglier canonical form that is actually the blowup of b:
$\tilde B = \begin{bmatrix}
.&1&1&.&.&.\\%
1&1&.&1&.&.\\%
.&.&.&1&1&.\\%
.&.&1&1&.&1\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$
However, I have not found this form listed in any reference, and I'd like to have a reasonable reference to point to to justify my choices (especially if they choose uglier, less sparse matrices).
>
> 1′) What reference uses $\tilde B$ as the rational canonical form of B?
>
>
>
Or alternatively:
>
> 2′) Where are the "scalars" in the centralizer of B?
>
>
>
I'd much prefer to use B, but if I cannot even see the scalar matrices in this form, then it seems like a very poor form indeed.
For those curious the other leading canonical form of B is:
$\hat B = \begin{bmatrix}
.&1&.&.&.&.\\%
.&.&1&.&.&.\\%
.&.&.&1&.&.\\%
.&.&.&.&1&.\\%
.&.&.&.&.&1\\%
1&3&0&-5&0&3\\%
\end{bmatrix}$
It is pretty, but also non-obvious what its composition factors are. This form (along with the Z/6Z style grouping of irreducible factors) is used by GP/Pari. Transposes and alternate groupings of factors allow for a wide variety of canonical forms.
The best answer will address the ease of explicitly writing down generators of the centralizer of B over a finite prime field given just (irr,pow), or will address both writing down the canonical form from (irr,pow) and the centralizer. An inductive answer might prefer to allow non-prime fields, but added generality at the cost of clarity and explicit algorithms is not useful to me.
**Edit:** Both B and $\hat B$ have the nice property that given a generator w for the indecomposable module acted on by the original operator M, a basis realizing the matrix is easy to find. For B this is vi = w⋅Mj⋅irr(M)k where i−1 = k⋅deg(irr) + j and 0 ≤ j < deg(irr), for i = 1, 2, … deg(irr)⋅pow. For $\hat B$ this is vi = w⋅Mi−1, for i = 1, 2, … deg(irr)⋅pow.
In order to use $\tilde B$, I need a similar explicit understanding of the basis defining it.
>
> 1″) How does one express a basis of k[x]/(irr^pow) in terms of the coset of 1 such that the operator corresponding to x has the form $\tilde B$, that is, block Toeplitz with the diagonal the companion matrix of irr, the super-diagonal an identity, and the other diagonals 0?
>
>
>
I am of course still curious about 2′. I think an answer to 1″ would both provide an answer for 2′ and be enough, even without an explicit reference, to finish Robin Chapman's answer to 1.
| https://mathoverflow.net/users/3710 | Centralizers in GL(n,p) | For a start the accepted usage for "rational canonical form" in the literature
is for a diagonal sum $C(f\_1)\oplus C(f\_2)\oplus\cdots\oplus C(f\_k)$ where
$C(f\_i)$ is the companion matrix for a monic polynomial $f$ and
$f\_1\mid f\_2\mid\cdots\mid f\_k$. That said, if I needed to find a centralizer
explcitly it isn't the canonical form I would choose.
As ever we should think of $V=k^n$ as a $k[X]$-module where $X$ acts via $A$.
If the minimum polynomial of $A$ is $u\_1^{a\_1}\cdots u\_k^{a\_k}$ with the
$p\_i$ distinct irreducibles then $V$ splits *uniquely* into a direct sum of submodules
$M\_1\oplus\cdots\oplus M\_k$, where $M\_i$ is annihilated by a power of $u\_i$.
Both $A$ and ts centralizer fix each $M\_i$, so we may
reduce to the case where the minimum polynomial $u^a$ with $u$ irreducible.
Let $u$ have degree $r$ and let $C\in M\_r(k)$ be the companion matrix for $u$
or any conjugate for it. I claim that $A$ is conjugate to a diagonal sum of
``Jordan-style blocks'' like
$$J=\left(\begin{matrix}
C&I&O&O\\\
O&C&I&O\\\
O&O&C&I\\\
O&O&O&C
\end{matrix}\right).$$
This is just a question of checking this has the minimum polynomial $u^a$.
As we are working in $GL(n,p)$ then $C\ne0$ and over a finite field then the diagonal
sum of copies of $C$ is a polynomial in $A$ (indeed $A^{p^rs}$ for suitable $s$).
Thus each matrix in the centralizer of $A$ has a block decomposition into $r$-by-$r$
blocks which commute with $C$ and so are polynomials in $C$. So,
finding the centralizer of $A$ is equivalent to finding the centralizer of the matrix
$A'$ over $k'=k(\alpha)$ where $\alpha$ is a zero of $u$ and $A'$ is obtained by
replacing the above Jordan-style blocks by standard Jordan blocks
$$J'=\left(\begin{matrix}
\alpha&1&0&0\\\
0&\alpha&1&0\\\
0&0&\alpha&1\\\
0&0&0&\alpha
\end{matrix}\right).$$
The centralizer of $A'$ is the same as that of $A'-\alpha I$ which is
a very sparse matrix. At this stage I would just find the centralizer in the matrix
algebra by explicit calculation and extract the nonsingular elements as the centralizer
in the matrix group.
**Added** (6/6/2010) I claimed that $J$ had the minimum polynomial $u^a$.
Let $k'=k(\alpha)$. The matrix $J'$ gives the action of $x$ on a standard
$k'$-basis for the $k'[x]$-module $k'[x]/((x-\alpha)^a)$. Therefore $J$
gives the action of $x$ on a $k$-basis for $k'[x]/((x-\alpha)^a)$ considered as a
$k[x]$-module. To see that $J$ has minimum polynomial $u^a$ it suffices
to show that some element of $k'[x]/((x-\alpha)^a)$ has annihilator $u^a$
over $k[x]$. But the element $1$ has annilhator $(x-\alpha)^a k'[x]$
over $k'[x]$ and so has annihilator
$k[x]\cap (x-\alpha)^a k'[x]=u^a k[x]$ over $k[x]$,
**provided** the extension $k'/k$ is separable (so that $x-\alpha$ is not a repeated
factor of $u$).
So the assertion holds for finite fields, bt not necessarily for general
fields of prime characteristic. It would be interesting to work out the
details for inseparable irreducible polynomials.
| 7 | https://mathoverflow.net/users/4213 | 27136 | 17,741 |
https://mathoverflow.net/questions/27134 | 4 | In mathematics, it is common that theorems/results and problems appearing dull in one generation get revitalized and become the center of research in another one.
Sometimes conjectures that are thought to be untouchable become resolved within years.
I would like to know if there were conjectures that did not appeal to a large number of mathematicians but when they were resolved( maybe partially), the techniques used or the result itself touched upon many areas.
On the other hand, are there also conjectures which were expected to have dramatic impact but upon resolution did not meet to that expectation?
| https://mathoverflow.net/users/5627 | Less-known conjectures of significant influence and the contrary | An example of an important solution to a little-known problem might be
Frank P. Ramsey's "On a problem of formal logic" in Proc. London Math.
Soc. 30 (1930) 264-286. The problem was in logic and not well-known even
to logicians, but Ramsey's solution was taken up by combinatorialists
(notably Erdős and Szekeres) and it grew into the important field now known as
Ramsey theory.
{Added later] An example of the contrary type is [Hilbert's fifth problem](http://en.wikipedia.org/wiki/Hilbert%27s_fifth_problem).
This was a well known and difficult problem, worked on by eminent
mathematicians such as von Neumann and Pontryagin, and it took
more than 50 years to solve. Yet, by the time it was solved it
seemed to be no longer in the mainstream of Lie theory, and books
on Lie theory today make little mention of it.
PS. I agree that this question should be community wiki.
| 12 | https://mathoverflow.net/users/1587 | 27139 | 17,742 |
https://mathoverflow.net/questions/27131 | 1 | A friend of mine and I were trying to answer a question related to his research and he couldn't remember whether or not the special linear group over the complex numbers, SLn(C),was simply connected. (It IS,of course.)
This got me wondering:What are all the simply connected topological subgroups of the general linear group over C? Is there a simple characterization of all of them up to isomorphism? What about thier fundamental groups as topological spaces?Are THEY simply connected if the subgroup is? I would expect them to have fundamental groups as basepoints should be easy to choose via the identity matrix.
So is there such a characterization for the simply connected subgroups of GLn(C)?
| https://mathoverflow.net/users/3546 | The Simply Connected Subgroups of GLn(C)? | There can be no such classification except for small n because that would imply the classification of nilpotent Lie algebras up to isomorphism, which is a well-known wild problem.
By Lie-Engel's theorem, any nilpotent Lie algebra of $n$ by $n$ matrices is a direct sum of a central ideal and a Lie algebra of strictly upper triangular matrices. Every Lie subalgebra $\mathfrak{g}\subseteq\mathfrak{gl}\_n$ consisting of nilpotent matrices is *exponential*: the exponential map $\exp$ is a diffeomorphism. In particular, $G=\exp(\mathfrak{g})$ is a simply-connected Lie subgroup of $GL\_n(\mathbb{C})$ and $G$ and $\mathfrak{g}$ determine each other up to isomorphism.
By Ado's theorem, every $k$-dimensional Lie algebra is linear and, in fact, a subalgebra of $\mathfrak{gl}\_n$ for some $n$ depending on $k$ ($n=k^2+1?$). So classifying simply connected subgroups of $GL\_n$ includes as a subproblem classifying all $k$-dimensional nilpotent Lie algebras up to isomorphism. For $k\geq 7$ there are continuous parameters in this moduli space and for general $k$ the classification is considered impossible (I forgot precise bounds).
Of course, in practice there is Greg's way.
| 14 | https://mathoverflow.net/users/5740 | 27142 | 17,745 |
https://mathoverflow.net/questions/27129 | 13 | Is there a classification of involutions in $\text{GL}\_n(\mathbb{Z})$?
Here's some more details about what I mean. Consider $f \in \text{GL}\_n(\mathbb{Z})$ such that $f^2=1$. Regard $f$ as an automorphism of $\mathbb{Z}^n$. Extend $f$ to an automorphism $g$ of $\mathbb{Q}^n$. Then we can write $\mathbb{Q}^n = E\_1 \oplus E\_{-1}$, where $g$ acts as the identity on $E\_1$ and as $-1$ on $E\_{-1}$. Restricting this decomposition to $\mathbb{Z}^n$, we obtain a finite-index subgroup $A$ of $\mathbb{Z}^n$ and a decomposition $A = F\_1 \oplus F\_{-1}$ such that $f$ acts as the identity on $F\_1$ and as $-1$ on $F\_{-1}$.
However, we definitely cannot assume that $A = \mathbb{Z}^n$. For instance, the matrix whose first row is $(0 1)$ and whose second row is $(1 0)$ (by the way, I can't figure out how to get my matrices to display correctly) is an involution in $\text{GL}\_n(\mathbb{Z})$ that can be diagonalized over $\mathbb{Q}$ but not over $\mathbb{Z}$.
What else can be said here?
| https://mathoverflow.net/users/6567 | Involutions in GL_n(Z) | The problem is equivalent to classifying isomorphism classes of $n$-dimensional integral representations of the cyclic group $C\_2$ of order 2, or $\mathbb{Z}[C\_2]$-modules on $\mathbb{Z}^n$. This group has exactly 3 isomorphism classes of indecomposable free $\mathbb{Z}$-modules:
(1) trivial
(2) sign representation
(3) 2-dimensional with matrix $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}.$
Every $n$-dimensional $\mathbb{Z}[C\_2]$-module is a direct sum of (1), (2), (3) with uniquely determined multiplicities. Thus any involution is conjugate over $\mathbb{Z}$ to a block diagonal matrix with blocks [1], [-1], $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ whose sizes are uniquely determined.
| 16 | https://mathoverflow.net/users/5740 | 27145 | 17,746 |
https://mathoverflow.net/questions/27133 | 4 | I'm studying Farb and Margalit's A primer on mapping class groups and trying to understand Wajnryb's finite presentation of Mod(S). I understand that There exists a finite presentation, but I can't understand how they got explicit Wajnryb's finite presentation. More precisely, I want to know how they knew that those 5 kinds of relation generate all relations.
| https://mathoverflow.net/users/6569 | About the proof of Wajnryb's finite presentation of Mod(S) | It is still an open problem to find a short and simple way to derive a finite presentation for the mapping class group. The book by Farb and Margalit (in the recent preliminary version 4.00) gives a clear sketch of the known derivations, which are rather long and complicated. For the details one must consult the original papers cited in the book. There are two main steps. The first is to construct a certain 2-dimensional cell complex that the mapping class group acts on and prove that this complex is simply-connected. This was originally done in a 1980 paper of Thurston and myself. Basic properties of this complex make it clear that writing down an actual presentation is then just a matter of carrying out some long and difficult calculations, with perhaps some ingenuity to reduce the work. The second step is then to do the work and write down a presentation. Harer carried out this work shortly thereafter, and Wajnryb greatly simplified Harer's presentation in a 1983 paper. In a later 1999 paper Wajnryb gave a more elementary, self-contained exposition of the whole story, but this is a 60-page paper. The exact references are in the Farb-Margalit book. I certainly hope that a more efficient approach is found someday.
| 13 | https://mathoverflow.net/users/23571 | 27148 | 17,749 |
https://mathoverflow.net/questions/26911 | 5 | A subcategory $\mathcal{C}$ of the category $Top$ of topological spaces is a [reflective subcategory](http://en.wikipedia.org/wiki/Reflective_subcategory) if the inclusion functor $i:\mathcal{C}\hookrightarrow Top$ has a left adjoint $R:Top\rightarrow \mathcal{C}$. In other words, for every space $X$, there is a space $RX\in \mathcal{C}$ and a map $r:X\rightarrow RX$ such that for each map $f:X \rightarrow Y$ with $Y\in \mathcal{C}$, there is a unique map $\tilde{f}:RX\rightarrow Y$ such that $\tilde{f}\circ r=f$.
If the reflection map $r$ is always a quotient map, $\mathcal{C}$ is said to be a quotient-reflective subcategory. For example, it seems that the subcategories of $T\_0$ spaces and $T\_2$ spaces are quotient-reflective but the subcategories of completely regular spaces and compact Hausdorff spaces are reflective but not quotient-reflective. I'd like to hear of other examples of quotient-reflective subcategories as well.
There seem to be conditions fully characterizing when a subcategory is reflective. For example, some necessary conditions are mentioned in the answer to this MO [question](https://mathoverflow.net/questions/9504/why-is-top-4-a-reflective-subcategory-of-top-3). Are there also conditions characterizing subcategories which are quotient-reflective?
| https://mathoverflow.net/users/5801 | How do you know when a reflective subcategory of Top is quotient-reflective? | The following theorem (16.8 in Abstract and Concrete Categories, Adamek-Herrlich-Strecker) could be of use.
Let E and M be subclasses of epis and monoes respectively, closed under composition with isomorphisms.
If A is a full subcategory of an (E,M)-factorisable category B, then the following conditions are
equivalent:
(1) A is E-reflective in B.
(2) A is closed under the formation of M-sources in B.
In the case that B has products and is E-co-wellpowered, the above conditions are equivalent to:
(3) A is closed under the formation of products and M-subobjects in B.
In the case of TOP, E would be extremal epis, and M monoes. In topological constructs extremal epis are exactly quotients (final and surjective). TOP is extremal epi-mono factorisable, has products and is extremally co-wellpowered. So a full subcategory is quotient reflective in TOP if and only if it is closed under products and (not necessarily initial) subobjects.
The subcategories of completely regular spaces is for example not closed for "adding more opens", so it is, as you said, not quotient reflective.
| 4 | https://mathoverflow.net/users/2300 | 27160 | 17,758 |
https://mathoverflow.net/questions/27144 | 91 | As you all probably know, Vladimir I. Arnold passed away yesterday. In the obituaries, I found the following statement (AFP)
>
> In 1974 the Soviet Union opposed Arnold's award of the Fields Medal, the most prestigious recognition in work in mathematics that is often compared to the Nobel Prize, making him one of the most preeminent mathematicians to never receive the prize.
>
>
>
Since he made some key results before 1974, it seems that the award would have been deserved. Knowing that the Soviets sometimes forced Nobel laureates not to accept their prizes, I thought at first that the same happened here - but noticing that Kantorovich received his Nobel prize the next year, and that Fields laureates both in 1970 and 1978 were Russians (Novikov and Margulis, respectively), I cannot understand why did the Soviets oppose it in case of Arnold. Can someone shed some light?
EDIT: I googled the resources online in English before asking this question here and found no answers. But after posting it here, I googled it in Russian, and found the [following](https://web.archive.org/web/20100819200230/http://webmath.exponenta.ru/dnu/1/usp/009.htm):
>
> Владимир Игоревич Арнольд был номинирован на медаль Филдса в 1974 году. Далее — изложение рассказа самого Арнольда; надеюсь, что помню его правильно. Всё было на мази, Филдсовский комитет рекомендовал присудить Арнольду медаль. Окончательное решение должен был принять высший орган Международного математического союза — его исполнительный комитет. В 1971 — 1974 годах вице-президентом Исполнительного комитета был один из крупнейших советских (да и мировых) математиков академик Лев Семёнович Понтрягин. Накануне своей поездки на заседание исполкома Понтрягин пригласил Арнольда к себе домой на обед и на беседу о его, Арнольда, работах. Как Понтрягин сообщил Арнольду, он получил задание не допустить присуждение тому филдсовской медали. В случае, если исполком с этим не согласится и всё же присудит Арнольду медаль, Понтрягин был уполномочен пригрозить неприездом советской делегации в Ванкувер на очередной Международный конгресс математиков, а то и выходом СССР из Международного математического союза. Но чтобы суждения Понтрягина о работах Арнольда звучали убедительно, он, Понтрягин, по его словам, должен очень хорошо их знать. Поэтому он и пригласил Арнольда, чтобы тот подробно рассказал ему о своих работах. Что Арнольд и сделал. По словам Арнольда, задаваемые ему Понтрягиным вопросы были весьма содержательны, беседа с ним — интересна, а обед — хорош. Не знаю, пришлось ли Понтрягину оглашать свою угрозу, но только филдсовскую медаль Арнольд тогда не получил — и было выдано две медали вместо намечавшихся трёх. К следующему присуждению медалей родившийся в 1937 году Арнольд исчерпал возрастной лимит. В 1995 году Арнольд уже сам стал вице-президентом, и тогда он узнал, что в 1974 году на членов исполкома большое впечатление произвела глубина знакомства Понтрягина с работами Арнольда.
>
>
>
Translation of the text:
>
> Vladimir Arnold was nominated for the Fields Medal in 1974. The following is Arnold’s own version of the story; I hope I’m remembering it correctly. The matter seemed settled: the Fields Medal Committee recommended to award a medal to Arnold. The final decision was to be taken by the top administrative body of the International Mathematical Union, the Executive Committee. In 1971–1974 the vice-president of the Executive Committee was one of Soviet Union’s (even world’s) greatest mathematicians, Lev Semenovich Pontryagin (also a member of the Soviet Academy of Sciences). On the eve of his visit to the meeting of the EC, Pontryagin invited Arnold to his home for lunch and to talk about Arnold’s work. Pontryagin told Arnold he was instructed not to allow the Fields Medal to be awarded to him. In case the executive committee wouldn’t agree and would still try to award the medal to Arnold, Pontryagin was authorized to threaten that the Soviet delegation would boycott the next International Congress of Mathematicians in Vancouver, or even that the USSR would leave the IMU. However, in order for Pontryagin’s assertions about Arnold’s work to be convincing, Pontryagin said, he would need to know that work very well. That’s why he invited Arnold in order that he describe his work in detail. Arnold did. According to Arnold, the questions Pontryagin asked him were profound, the talk with him interesting, and the meal good. I do not know if Pontryagin had to make his threat known, but Arnold did not receive the medal—and only two medals were awarded instead of the intended three. By the next time the medals were being awarded, Arnold, born in 1937, was over the age limit. In 1995, Arnold himself became vice-president and learned that in 1974 the depth of Pontryagin’s familiarity with Arnold’s work made a great impression on the members of the EC.
>
>
>
| https://mathoverflow.net/users/4925 | Why didn't Vladimir Arnold get the Fields Medal in 1974? | Pontryagin wrote a book "Biography of Lev Semenovich Pontryagin, a mathematician, composed by himself". It is available online at <http://www.ega-math.narod.ru/LSP/book.htm>, in the original Russian. Google does a fairly good job of translation, although it refuses to translate the individual chapters completely because of their length.
In the book, Pontryagin shares a lot about the inner workings of the IMU Executive Board and his own role in holding the Soviet party line there as its vice president. For example, he recounts his version of how France got the IMU presidency in 1974, so that neither the Soviet Union nor the US would dominate.
The only relevant mention of Arnold that I could find in that book is in chapter 5. He states that in 1974 Arnold was not allowed to leave the country to lecture abroad, and that there was a conflict about this with the Executive Board of the IMU, who insisted that he should. From this, you could extrapolate the reasons for blocking Arnold's Fields medal, if the story is true.
| 31 | https://mathoverflow.net/users/1784 | 27171 | 17,764 |
https://mathoverflow.net/questions/27175 | 9 | In his answer to my question [here](https://mathoverflow.net/questions/27129/involutions-in-gl-nz), Victor Protsak quoted the following result:
Let $C\_2$ be a finite cyclic group of order $2$. Then every $\mathbb{Z}[C\_2]$ structure on $\mathbb{Z}^n$ is isomorphic to a direct sum of representations of the following types:
1. The trivial representation
2. The sign representation
3. The representation on $\mathbb{Z}^2$ that permutes the two factors.
This strikes me as a very beautiful result! I know a lot of nice sources for representation theory over various kinds of fields, but a bit of searching does not turn up any books or surveys on representation theory over the integers. Does anyone have anything they recommend?
| https://mathoverflow.net/users/6567 | Representation theory over Z | See Curtis–Reiner's textbook on the Representation Theory of Finite Groups and Associative Algebras ([MR 144979](http://www.ams.org/mathscinet-getitem?mr=144979)), Theorem 74.3, page 507, and especially the introduction starting on page 493.
The result for cyclic groups of prime order, and for order 4 was originally done in:
* Diederichsen, Fritz-Erdmann.
"Über die Ausreduktion ganzzahliger Gruppendarstellungen bei arithmetischer Äquivalenz"
Abh. Math. Sem. Hansischen Univ. 13, (1940). 357–412. [MR2133](http://www.ams.org/mathscinet-getitem?mr=2133).
However, Reiner has written quite a few nice papers on similar subjects. One of his earlier ones is on the same topic:
* Reiner, Irving. "Integral representations of cyclic groups of prime order."
Proc. Amer. Math. Soc. 8 (1957), 142–146.
[MR83493](http://www.ams.org/mathscinet-getitem?mr=83493)
[doi:10.2307/2032829](http://dx.doi.org/10.2307/2032829)
One can also consult texts on things called "crystallographic groups", "space groups", and "p-adic space groups". Plesken has written several nice books using this sort of thing. These give infinite families of nicely related finite groups and of course help crystallographers.
Be careful to distinguish these sorts of representations from ZG-modules. ZG-modules are basically incomprehensible, so instead lots of people focus on ZG-lattices, where the underlying module is projective. This means the idea of using matrices still makes some sense. There is a lot of literature on modules over group rings over nice rings (like Z or Dedekind domains), but a fair amount of it is not applicable to questions about GL(n,Z).
Roughly speaking, even for G=1, ZG modules are too difficult to understand, and adding a G just makes it worse. Another common tack is to look at $\hat {\mathbb{Z}}\_p$ modules, p-adic modules. Again the results are nicest for lattices, but things do not get so bad near so fast there. Reiner's Maximal Orders textbook describes some of the beautiful and well-behaved things you can see there.
| 17 | https://mathoverflow.net/users/3710 | 27178 | 17,769 |
https://mathoverflow.net/questions/27176 | 13 | The thought process that led me to this question is that the identity
$$ \left(\prod\_i \frac1{1-x\_i}\right)\left(\prod\_i {1-x\_i}\right)=1$$
can be understood as expressing exactness of the Koszul complex.
This identity is rewritten by taking $\left(\prod\_i \frac1{1-x\_i}\right)$
as the generating function for the complete symmetric functions $h\_n$
and $\left(\prod\_i {1+x\_i}\right)$ as the generating function for the elementary symmetric functions $e\_n$.
Next we have the Jacobi-Trudi identity which expresses a Schur function as the determinant of a matrix whose entries are complete (or elementary) symmetric functions. Also the Specht module is sometimes constructed as a quotient (or submodule) of the trivial representation of the Young subgroup induced to a representation. This suggests that this is the start of a BGG resolution.
I imagine that if this works then it is well-known. Could I have some references? and where does line of thought lead?
| https://mathoverflow.net/users/3992 | Can the Jacobi-Trudi identity be understood as a BGG resolution? | Look at the short paper MR902299 (89a:17012) 17B10 (20C30)
Zelevinski˘ı, A.V. [Zelevinsky, Andrei] (2-AOS-CY),
Resolutions, dual pairs and character formulas. (Russian)
Funktsional. Anal. i Prilozhen. 21 (1987), no. 2, 74–75, as well as the independent work by Kaan Akin (a former student of David Buchsbaum) including MR1194310 (94e:20059) 20G05
Akin, Kaan (1-OK),
On complexes relating the Jacobi-Trudi identity with the Bernstein-Gel0fand-Gel0fand
resolution. II.
J. Algebra 152 (1992), no. 2, 417–426. A further refinement is given in MR1379204 (97b:20066) 20G05
Maliakas, Mihalis (1-AR),
Resolutions and parabolic Schur algebras.
J. Algebra 180 (1996), no. 3, 679–690.
| 13 | https://mathoverflow.net/users/4231 | 27180 | 17,771 |
https://mathoverflow.net/questions/27163 | 26 | It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.
Every proof I found (e.g. in the classical "Commutative Algebra" by Atiyah and Macdonald) uses Zorn's lemma to prove that $x \notin Nil(A) \Rightarrow x \notin \cap\_{\mathfrak{p}\in Spec(A)} \mathfrak{p}$ (the other way is immediate).
Does anybody know a proof that doesn't involve it?
| https://mathoverflow.net/users/6382 | Nilradicals without Zorn's lemma | Since you asked for a proof, let me complement Chris Phan's answer by outlining a proof that relies only on the [Compactness Theorem](http://en.wikipedia.org/wiki/Compactness_theorem) for propositional logic, which is yet another equivalent to the [Ultrafilter Theorem](http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem) over ZF.
Let A be a commutative ring and let x ∉ Nil(A). To each element a ∈ A associate a propositional variable pa and let T be the theory whose axioms are
* p0, ¬p1, ¬px, ¬px2, ¬px3,...
* pa ∧ pb → pa+b for all a, b ∈ A.
* pa → pab for all a, b ∈ A.
* pab → pa ∨ pb for all a, b ∈ A.
Models of T correspond precisely to prime ideals that do not contain x. Indeed, if P is such an ideal, then setting pa to be true iff a ∈ P satisfies all of the above axioms, and conversely. So it suffices to show that T has a model.
Since xn ≠ 0 for all n, one can verify using ideals over finitely generated subrings of A that the theory T is finitely consistent, i.e. any finite subset of T has a model. (What I just swept under the rug here is a constructive proof of the theorem for quotients of Z[v1,...,vn].) The Compactness Theorem for propositional logic then ensures that T has a model.
| 16 | https://mathoverflow.net/users/2000 | 27187 | 17,777 |
https://mathoverflow.net/questions/23975 | 17 | I'm looking for some examples of actions on Sylow p-groups, and often those actions appear in the case of finite almost simple groups. Given a finite almost simple group, I understand in principle how to calculate its Sylow p-subgroups (here p is usually 2 or 3), but perhaps I am just too slow at doing it. In particular, I am familiar with the papers of Weir and Carter-Fong.
I am not sure how to do the reverse calculation: given a p-group P (possibly described as "the Sylow p-subgroup of the almost simple group X", and X is something explicit like "PGL(3,19)" or "M11"), find all of the almost simple groups that have a Sylow p-subgroup isomorphic to P.
I am pretty sure some people know how to do this, but it's not really clear to me how to go about it. For instance, it would have never occurred to me that PSU(3,8), PSL(3,19), and 3D4(2) have isomorphic Sylow 3-subgroups.
Is there a description of how this is done?
I think there is likely to be a finite answer to the question: Obviously only finitely many (and probably O(p)) alternating groups could work for a given P. We take for granted that only finitely many sporadic groups could work. It seems that, similarly to the alternating case, there are only finitely many ranks (again probably O(p)) of Lie groups that could work, and hopefully for each Lie type (and rank), there are just some congruences on "q" that indicate which ones work and which don't.
However, I've not had much luck doing this calculation in examples, and so I am looking for papers or textbooks where this has been done. I have found some that state the result of doing something like this (post CFSG), and I have found several that do this in quite some detail, but pre-CFSG so they spend hundreds of pages eliminating impossible groups obscuring what should now be an easy calculation. I'm looking for something with the pedagogical style of the pre-CFSG papers, but that doesn't mind using the standard 21st century tools.
Alternatively: does anyone know of a vaguely feasible approach to construct all groups with given Sylow subgroups? Blackburn et al.'s Enumerating book has some upper bounds, but they are pretty outrageous and don't seem adaptable to a feasible algorithm for my problem.
| https://mathoverflow.net/users/3710 | How to find more (finite almost simple) groups with a given Sylow subgroup | If you are only interested in actions on the Sylow p-subgroup, you may not need to find all groups, but just all so-called p-fusion patterns, as formalized in the theory of fusion systems due to Puig, and developed further by Broto-Levi-Oliver and others.
See Broto, Carles; Levi, Ran; Oliver, Bob The homotopy theory of fusion systems. J. Amer. Math. Soc. 16 (2003), no. 4, 779--856, or survey articles eg by Markus Linckelmann or Broto-Levi-Oliver.
Finding the p-fusion patterns on a given p-group is also, implicitly or explicitly, a step in finding all (finite almost simple) groups with a given Sylow p-subgroup. (There is not a one-size-fits-all method for doing this, however...)
For a list of equivalences of different p-fusion patterns in the finite groups of Lie type see:
<http://www.maths.abdn.ac.uk/~bensondj/html/archive/broto-moeller-oliver.html>
which contains some of the examples you mention.
The proof uses homotopy theory, and there does not appear to be a purely group theoretic proof yet (!)
| 10 | https://mathoverflow.net/users/6574 | 27189 | 17,779 |
https://mathoverflow.net/questions/27164 | 33 | I have just completed an introductory course on analysis, and have been looking over my notes for the year. For me, although it was certainly not the most powerful or important theorem which we covered, the most striking application was the [Fourier analytic proof of the isoperimetric inequality](https://mathoverflow.net/a/25076). I understand the proof, but I still have no feeling for why anyone would think to use Fourier analysis to approach this problem. I am looking for a clear reason why someone would look at this and think "a Fourier transform would simplify things here". Even better would be a physical interpretation. Could this somehow be related to "hearing the shape of a drum"? Is there any larger story here?
| https://mathoverflow.net/users/1106 | Why is Fourier analysis so handy for proving the isoperimetric inequality? | Experience with Fourier analysis and representation theory has shown that every time a problem is invariant with respect to a group symmetry, the representation theory of that group is likely to be relevant. If the group is abelian, the representation theory is given by the Fourier transform on that group.
In this case, the relevant symmetry group is that of reparameterising the arclength parameterisation of the perimeter by translation. This operation does not change the area or the perimeter. When combined with the observation (from Stokes theorem) that both the area and perimeter of a body can be easily recovered from the arclength parameterisation, this naturally suggests to use Fourier analysis in the arclength variable.
| 61 | https://mathoverflow.net/users/766 | 27196 | 17,785 |
https://mathoverflow.net/questions/27197 | 29 | The following definitions are standard:
An affine variety $V$ in $A^n$ is a complete intersection (c.i.) if its vanishing ideal can be generated by ($n - \dim V$) polynomials in $k[X\_1,\ldots, X\_n]$. The definition can also be made for projective varieties.
$V$ is locally a complete intersection (l.c.i.) if the local ring of each point on $V$ is a c.i. (that is, quotient of a regular local ring by an ideal generated by a regular sequence).
What are examples (preferably affine) of l.c.i. which are not c.i. ? I have never seen such one.
| https://mathoverflow.net/users/6576 | Local complete intersections which are not complete intersections | (To supplement Alberto's example)
If $V$ is projective, then the gap between being locally c.i and c.i is quite big. In particular, any *smooth* $V$ would be locally c.i., but they are not c.i. typically. For instance, take $V$ to be a few points in $\mathbb P^2$ would give simple examples. In higher dimensions, by Grothendick-Lefschetz, if $V$ is smooth, $\dim V\geq 3$, and $V$ is c.i. then $\text{Pic}(V)=\mathbb Z$, so it is a serious restriction.
The affine case is more subtle. Again one can look at smooth varieties. If $V$ is a smooth affine curve and c.i., then the canonical bundle of $V$ is trivial. So it gives the following strategy: start with a *projective* curve $X$ of genus at least $2$, removing some general points to obtain an (still smooth) affine curve with non-trivial canonical bundle.
For more details on the second paragraph, see this [question](https://mathoverflow.net/questions/9751/is-every-smooth-affine-curve-isomorphic-to-a-smooth-affine-plane-curve), especially Bjorn Poonen's comments. This [paper](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-17/issue-3/Complete-intersections/10.1215/kjm/1250522714.full) contains relevant references, and also an example with trivial canonical bundle.
| 33 | https://mathoverflow.net/users/2083 | 27203 | 17,790 |
https://mathoverflow.net/questions/27181 | 7 | I am currently reading Vickers' text "topology via logic" and Peter Johnstone's "stone spaces", and I understand the material in both of these texts to pertain directly to constructions in elementary topos theory (by which I do not mean 'the theory of elementary topoi). However, these things do not seem to be mentioned explicitly in these texts, at least not to great extent. Where might I avail myself of material which really 'brings home' the notion of topoi as 'generalized spaces' in the context of stone spaces and locales as alluded to in Vickers and Johnstone? I understand that Borceaux's third volume in the 'handbook of categorical algebra' is probably a good place to start...
| https://mathoverflow.net/users/6131 | Stone Spaces, Locales, and Topoi for the (relative) beginner | Many good books have already been mentioned; I like MacLane+Moerdijk as an introduction, and after that both books by Johnstone (in particular, Part C of the Elephant does a good job of connecting locale theory with topos theory). But I also wanted to mention Vickers' paper "Locales and Toposes as Spaces," which I think does a good job of connecting up the topology with the toposes and the logic in a way that isn't directly evident in many other introductions.
| 8 | https://mathoverflow.net/users/49 | 27214 | 17,795 |
https://mathoverflow.net/questions/27198 | 4 | I just want to consider the simplest case:
Let S=[0,0,0,0,1], how to derive the general formula for $Sym^k$ S?
My conjectured formula based on the results of LiE program for finite k values is:
$Sym^k(S)=\overset{\left[k/2\right]}{\underset{i=0}{\oplus}}[i,0,0,0,k-2i]$
But I have no clue how to prove it or even a derivation for the simplest case when k=2.
Any reference or tips would be greatly appreciated.
PS. Sorry for confusion. Now I changed the spinor representation "V" to "S".
| https://mathoverflow.net/users/6577 | How to calculate symmetric tensor products of SO(10) representations? | Based on comments from Eric Rowell and José Figueroa-O'Farrill, I assume that your $V$ is a half-spinor representation of $Spin\_{10}$. The key issue is that your hypothetical decomposition of $S(V)$ is *multiplicity-free*, i.e. any module that appears in it has multiplicity one. There is a beautiful theory of multiplicity-free actions that allows you to do the following.
(1) Test that $V$ is a multiplicity-free $G$-module.
(2) Find the highest weights of the simple summands in $S(V)$.
For (1), the action is multiplicity-free if and only if the Borel subgroup $B$ has an open orbit on $V$. For (2), the highest weights form a free commutative semigroup whose generators are determined from the stabilizer in $B$ of a point in the open orbit.
An excellent survey is Roger Howe's *Schur lectures.*
All multiplicity-free linear actions (i.e. modules) of reductive algebraic groups have been classified. A convenient reference is
Howe and Umeda, *The Capelli identity, the double commutant
theorem and multiplicity-free-actions*, Mathematische Annalen, 290, 565 - 620 (freely available through [GDZ](http://gdz.sub.uni-goettingen.de/), but tricky to link to).
Item (x) of the list (11.0.1) on p.583 is your action $Spin\_{10}\times GL\_1.$ This module arises from the action of a Levi factor of a maximal parabolic of $E\_6$ on its abelian nilradical (remove last node from the Dynkin diagram of $E\_6$ to get $D\_5$). The decomposition is described in 11.10 on p.602: there are two fundamental heighest weights, one in degree 1 corresponding to the Spin module itself and one in degree 2 corresponding to the standard 10-dimensional representation of $SO\_{10}.$ This leads to the decomposition you have stated.
| 9 | https://mathoverflow.net/users/5740 | 27215 | 17,796 |
https://mathoverflow.net/questions/22934 | 36 | In his book *Metric Number Theory*, Glyn Harman mentions the following problem he attributes to Erdős:
Let $f(\alpha)$ be a bounded measurable function with period 1. Is it true that
$$\lim\_{N\rightarrow\infty} \frac{1}{\log N} \sum\_{n=1}^N \frac{1}{n}f(n\alpha) = \int\_0^1 f(x) dx$$
for almost all $\alpha$,
writing "so far as the author is aware, this question remains open."
Harman's book is from 1997. Does anyone know the current status of the problem?
**Motivation, for the curious**
We lose no generality in assuming $f$ has mean $0$. The rough idea is that for almost all $\alpha$, $n\alpha$ will be equidistributed $(\mod 1)$ in a strong enough way to cause a great deal of cancellation in the sum, so in particular we might guess the sum is $o(\log N)$. It is a weaker version of a more classical conjecture of Khintchine that
$$\lim\_{N\rightarrow\infty} \frac{1}{N} \sum\_{n=1}^N f(n\alpha) = \int\_0^1 f(x) dx$$
for almost all $\alpha$, where $f$ is as above. This is known to be false. (Of course, if $f$ is continuous it is true, for all irrational $\alpha$ even.)
| https://mathoverflow.net/users/5621 | A question of Erdős on equidistribution | The statement was shown to be false by J. Bourgain in a paper published in 1988 (Almost Sure Convergence and Bounded Entropy, [doi:10.1007/BF02765022](http://dx.doi.org/10.1007/BF02765022)). Well before either Harman's 1997 book and the 2003 paper Gjergji mentioned in the comment, which both say that it is an open problem!
From page 2 of Bourgain's paper
>
> As further application of our method, a problem due to A. Bellow and a question raised by P. Erdős are settled.
>
>
>
And, further down (using ${\bf T}={\bf R}/{\bf Z}$),
>
> The problem of Erdős mentioned above deals with weaker versions of the Khintchine problem. In particular he raised the question whether given a measurable subset $A$ of $\bf T$, then for almost all $x$ the set $\lbrace j\in{\bf Z }\_+\mid jx\in A\rbrace$ has a logarithmic density, i.e.
> $$
> \frac{1}{\log n}\sum\_{\substack{j\le n\\\\ jx\in A}}\frac1j\to\vert A\vert.
> $$
> We will disprove this fact.
>
>
>
I can't vouch that Bourgain's paper is free of errors, as I have only just found it now and haven't read through it all in detail. However, Bourgain's result is also (very briefly) referred to in this 2004 paper, <http://arxiv.org/abs/math/0409001v1>, so I assume it is considered to be valid.
| 23 | https://mathoverflow.net/users/1004 | 27216 | 17,797 |
https://mathoverflow.net/questions/27229 | 26 | Long time ago there was a [question](https://mathoverflow.net/questions/21003/)
on whether a polynomial bijection $\mathbb Q^2\to\mathbb Q$ exists. Only one attempt
of answering it has been given, highly downvoted by the way. But this answer isn't obviously
unsuccessful, because the following problem (for case $n=2$) remains open.
**Problem.** Let $f$ be a polynomial with rational (or even integer!) coefficients
in $n$ variables $x\_1,\dots,x\_n$. Suppose there exist two distinct points
$\boldsymbol a=(a\_1,\dots,a\_n)$ and $\boldsymbol b=(b\_1,\dots,b\_n)$ from $\mathbb R^n$
such that $f(\boldsymbol a)=f(\boldsymbol b)$. Does this imply the existence
of two points $\boldsymbol a'$ and $\boldsymbol b'$ from $\mathbb Q^n$ satisfying
$f(\boldsymbol a')=f(\boldsymbol b')$?
Even case $n=1$ seems to be non-obvious.
**EDIT.** Just because we have a very nice counter example (immediately
highly rated by the MO community) by Hailong Dao in case $n=1$ and because
for $n>1$ there are always points $\boldsymbol a,\boldsymbol b\in\mathbb R^n$
with the above property, the problem can be "simplified" as follows.
Is it true for a polynomial $f\in\mathbb Q[\boldsymbol x]$ in $n>1$ variables
that there exist two points $\boldsymbol a,\boldsymbol b\in\mathbb Q^n$ such
that $f(\boldsymbol a)=f(\boldsymbol b)$?
The existence of injective polynomials $\mathbb Q^2\to\mathbb Q$ is discussed
in B. Poonen's [preprint](http://arxiv.org/abs/0902.3961)
(and in comments to [this question](https://mathoverflow.net/questions/21003/)).
What can be said for $n>2$?
**FURTHER EDIT.**
The expected answer to the problem is in negative. In other words, there exist injective polynomials
$\mathbb Q^n\to\mathbb Q$ for any $n$.
Thanks to the comments of [Harry Altman](https://mathoverflow.net/users/5583/harry-altman) and
[Will Jagy](https://mathoverflow.net/users/3324/will-jagy), case $n>1$ is now fully
reduced to $n=2$. Namely, any injective polynomial $F(x\_1,x\_2)$ gives rise to the injective
polynomial $F(F(x\_1,x\_2),x\_3)$, and so on; in the other direction, any $F(x\_1,\dots,x\_n)$ in more
than 2 variables can be specialized to $F(x\_1,x\_2,0,\dots,0)$.
In spite of [Bjorn Poonen's verdict](http://arxiv.org/abs/0902.3961)
that case $n=2$ can be resolved by an appeal to
the Bombieri--Lang conjecture for $k$-rational points on surfaces of general type
(or even to the 4-variable version of the $abc$ conjecture), I remain with a hope
that this can be done by simpler means. My vague attempt (for which I search
in the literature) is to start with a homogeneous form $F(x,y)=ax^n+by^n$,
or any other homogeneous form of *odd* degree $n$, which has
the property that only finitely many *integers* are represented by $F(x,y)$ with
$x,y\in\mathbb Z$ relatively prime. In order to avoid this finite set of "unpleasant"
pairs $x,y$, one can replace them by other homogeneous forms $x=AX^m+BY^m$ and
$y=CX^m+DY^m$ (again, for $m$ odd and sufficiently large, say),
so that $x$ and $y$ escape the unpleasant values.
Then the newer homogeneous form $G(X,Y)=F(AX^m+BY^m,CX^m+DY^m)$ will give
the desired polynomial injection. So, can one suggest a homogeneous form $F(x,y)$
with the above property?
| https://mathoverflow.net/users/4953 | Polynomials with rational coefficients | Let $f(x)=x^3-5x/4$. Then for $x\neq y$, $f(x)=f(y)$ iff $x^2+xy+y^2=5/4$ or $(2x+y)^2+3y^2=5$. The last equation clealy have real solutions. But if there are rational solutions, then there are integers $X,Y,N$ such that $(2X+Y)^2+3Y^2=5N^2$. This shows $X,Y,N$ all divisible by $5$, ...
| 32 | https://mathoverflow.net/users/2083 | 27231 | 17,803 |
https://mathoverflow.net/questions/27244 | 17 | Call a tack the one point union of three open intervals. Can you fit an uncountable number of them on the plane? Or is only a countable number?
| https://mathoverflow.net/users/6610 | How many tacks fit in the plane? | First of all, the one-point compactification of three open intervals is not a "tack", it's a three-leaf clover. I think that you mean a one-point union of three closed intervals; of course it doesn't matter if the other three endpoints are there or not. This topological type can be called a "Y" or a "T" or a "simple triod". [R.L. Moore published a solution](http://www.jstor.org/pss/85527) to your question in 1928. The answer is no. It was [generalized in 1944](http://www.ams.org/journals/bull/1944-50-10/S0002-9904-1944-08216-5/home.html) by his student Gail Young: You can only have countably many $(n-1)$-dimensional tacks in $\mathbb{R}^n$ for any $n \ge 2$. For her theorem, the name "tack" makes rather more sense, but she calls it a "$T\_n$-set".
Actually Moore's theorem applies to a more general kind of triod, in which three tips of the "Y" are connected to the center by "irreducible continua", rather than necessarily intervals.
---
I don't know whether I might be spoiling a good question, but here in any case is a solution to the original question (see as both Moore and Young did something more general that takes more discussion). Following domotorp's hint, there is a ~~principle of accumulation onto a countable set of outcomes~~ pigeonhole principle for uncountable sets. If $f:A \to B$ is a function from an uncountable set $A$ to a countable set $B$, then there is an uncountable inverse image $A' = f^{-1}(b)$. If you want to show that $A$ does not exist, then you might as well replace it with $A'$. Unlike the finite pigeonhole principle, which becomes more limited with each such replacement, $A'$ has the same cardinality as $A$, so you haven't lost anything. You are even free to apply the uncountable pigeonhole principle again.
Suppose that you have uncountably many simple triods in the plane. Given a simple triod, we can choose a circle $C$ with rational radius and rational center with the branch point of the triod on the inside and the three tips on the outside. Since there are only countably many such circles, there are uncountably many triods with the same circle $C$. We can trim the segments of each such triod so that they stop when they first touch $C$, to make a pie with three slices (a Mercedes-Benz symbol). Then, given such a triod, we can pick a rational point in each of three slices of the pie. Since there are only countably many such triples of points, there must be uncountably many triods with the same three points $p$, $q$, and $r$. In particular there are two such triods, and a suitable version of the Jordan curve theorem implies that they intersect.
The argument can be simplified to just pick a rational triangle that functions as the circle, and whose corners function as the three separated points. But I think that there is something to learn from the variations together, namely that the infinite pigeonhole principle gives you a lot of control. For instance, with hardly any creativity, you can assume that the triods are all large.
| 45 | https://mathoverflow.net/users/1450 | 27248 | 17,811 |
https://mathoverflow.net/questions/27235 | 1 | Let $k$ be an arbitrary field, $d\in k$, and $d$ is not a square in $k$.
Consider the binary quadratic form $f(x,y)=x^2-d y^2$
(it is the norm from $k(\sqrt{d})$ to $k$).
I am looking for a reference to a proof of the following fact:
There exist a field extension $K/k$ and a non-zero element $a\in K$
such that $f$ does not represent $a$ over $K$ (that is, there is no
$x,y\in K$ such that $a=x^2-d y^2$).
Edited question (which I really meant): Let $l/k$ be a separable quadratic field extension
(of fields of any characteristic).
Prove that there exists a field extension $K/k$ such that the norm map
$N\colon (l\otimes\_k K)^\times\to K^\times$ is not surjective.
| https://mathoverflow.net/users/4149 | Non-representability by a binary quadratic form | Take $K = k((t))$ (formal Laurent series field) and apply Proposition V.2.3 of Serre's *Local Fields* to the unramified quadratic extension $Kl/K$. (Note that separability of $l/k$ is needed so that $Kl/K$ is unramified.) Then the image of the norm map consists precisely of elements of even $t$-adic valuation: in particular $t$ is not in the image. Having done the local case, it becomes clear that $K = k(t)$ would also work.
[Note that I am leaving the answer as community wiki. This is because I am supposed to be "on vacation" from MO: i.e., concentrating on my own work! I made an exception here because this is something that I knew off the top of my head and because Mikhail is such an eminent mathematician that it is my privilege to help him out, in however small a way. But I better not get back into the reputation game...]
| 3 | https://mathoverflow.net/users/1149 | 27256 | 17,815 |
https://mathoverflow.net/questions/27271 | 31 | There is a lot of good stuff contained in the Problems section of the *American Mathematical Monthly*. One difficulty with extracting that information, however, is that if I see an old *Monthly* problem, it is not always so easy to locate the solution. Sometimes the solution does not appear until many years later, and in some cases the solution has never appeared at all. For those at an academic institution with access to JSTOR, sometimes an electronic search will turn up the solution, but if the search comes up empty, I am left wondering whether the solution was really never published or whether I just did something wrong with my search. Furthermore, even with JSTOR access, there is no easy way to list (for example) all unsolved *Monthly* problems.
Does anybody have, or know of, an index of all the *Monthly* problems and solutions that would address this issue? Failing that, is there any interest in creating such a publicly available resource? If the work is divvied up among a few dozen volunteers it shouldn't take that much work. As I envisage it, the index would (at least initially) not contain the mathematical content of the problems but would just give the bibliographic information for the problem and the solution (if any), and indicate which parts (if any) remain unsolved.
| https://mathoverflow.net/users/3106 | Is there an index for solutions to American Mathematical Monthly problems? | This is somewhat redundant with Igor Pak's P.S., but I found an index for the 1918-1950 range in *The Otto Dunkel memorial problem book*, starting on page 80 of [this pdf](https://drive.google.com/file/d/0B_pEp00B111JNjZjZjE1NGUtN2YwOC00OTUyLWE4NmYtNzEzY2VhMzk5Y2Iz/view?usp=sharing&resourcekey=0-YwwNCk5wkCydkdlbCw9vBQ).
---
I found more: Stanley Rabinowitz and others had the same idea a couple of decades ago, and they started compiling indices, not just for the *Monthly*. Here are Google Books and official links for the two volumes I found:
[1975-1979](http://books.google.com/books?id=KX6D6hefyA0C&lpg=PP1&dq=%2522index%2520to%2520mathematical%2520problems&lr&pg=PP1#v=onepage&q&f=false) (published 1999) [Website](http://www.mathpropress.com/books/1975index/)
[1980-1984](http://books.google.com/books?id=n-F52zK2UAgC&lpg=PR7&ots=cwyXL3xTg4&dq=%2522mathematical%2520monthly%2522%2520problems%2520solutions%2520index&lr&pg=PP1#v=onepage&q&f=false) (published 1992) [Website](http://www.mathpropress.com/books/1980index/), [Errata](http://www.mathpropress.com/errata/1980errata/errata.html)
Googling "Stanley Rabinowitz 1985-1989" led me to [this](http://books.google.com/books?id=RJVii6vc0NAC&lpg=PP11&ots=C4dnFu344h&dq=Stanley%2520Rabinowitz%25201985-1989&pg=PP11#v=onepage&q=Stanley%2520Rabinowitz%25201985-1989&f=false):
>
> The labor involved in classifying and indexing the 5-year indexes has proved to be far greater than anticipated. We hope to have the 1975-1979 Index out later this year and the 1985-1989 Index is well under way. [Stanley Rabinowitz, June 1996]
>
>
>
I don't know if the 1985-1989 index was ever finished, but presumably there's at least a partial draft out there somewhere. The website <http://www.mathpropress.com/> of Rabinowitz's company includes contact information. They say they're looking for authors to publish, among other things, "indexes to mathematical problems or results", but the page apparently hasn't been updated since 2006.
| 22 | https://mathoverflow.net/users/1119 | 27277 | 17,832 |
https://mathoverflow.net/questions/27219 | 17 | Suppose $\mathcal{C}$ is a category with small colimits, and $G \in \mathrm{Ob}(\mathcal{C})$ is a strong generator. (This means that $f: X \to Y$ is an isomorphism iff $f\_{\*}: \mathrm{Hom}(G,X) \to \mathrm{Hom}(G,Y)$ is a bijection.)
>
> How does one prove that every object $X$ is a colimit of copies of $G$ (i.e., a quotient of coproducts of $G$)?
>
>
>
I think $X \simeq \mathop{\mathrm{colim}}\_{G \to X} G$. I can see there is a natural map $e: \mathop{\mathrm{colim}}\_{G \to X} \to X$, and that the induced map $e\_{\*}: Hom(G,\mathop{\mathrm{colim}}\_{G \to X}) \to Hom(G, X)$ is surjective, but I cannot see that it map is injective.
| https://mathoverflow.net/users/2536 | Proof that objects are colimits of generators | I think there are actually *three* possible things that you might be asking, but the answer to all of them is no. Suppose that G is a strong generator in a cocomplete category C. Then you can ask:
1. Is every object X of C the colimit of G over the canonical diagram of shape $(G\downarrow X)$? (If so, then G is called *dense* in C.)
2. Is every object *some* colimit of a diagram all of whose vertices are G? (If so, then G is called *colimit-dense* in C.)
3. Is C the smallest subcategory of itself containing G and closed under colimits? (If so, then G is a *colimit-generator* of C.)
The category of compact Hausdorff spaces is a counterexample to the first two. It is monadic over Set (the monad is the ultrafilter monad, aka Stone-Cech compactification of the discrete topology), and hence cocomplete, and the one-point space is a strong generator. But any colimit of a diagram consisting entirely of 1-point spaces must be in the image of the free functor from Set (the one-point space being its own Stone-Cech compactification), and hence (since that functor is a left adjoint and preserves colimits) must be the free object on some set. However, not every compact Hausdorff space is the Stone-Cech compactification of a discrete set.
As Todd pointed out in the comments, though, CptHaus is the colimit-closure of the one-point space, since every object is a coequalizer of maps between free ones (because the category is monadic over Set); thus it isn't a counterexample to the third question. Counterexamples to the third question are actually much harder to come by, and in fact if you assume additionally that C has finite limits and is "extremally well-copowered", then it is true that any strong generator is a colimit-generator. I think there's a proof of this somewhere in Kelly's book "Basic concepts of enriched category theory," and a brief version can be found [here](https://home.sandiego.edu/~shulman/papers/generators.pdf). However, without these assumptions, one can cook up ugly and contrived counterexamples, such as example 4.3 in the paper "Total categories and solid functors" by Borger and Tholen.
| 19 | https://mathoverflow.net/users/49 | 27287 | 17,838 |
https://mathoverflow.net/questions/27258 | 15 | I recently read the statement "up to conjugacy there are 4 nontrivial finite subgroups of ${\rm SL}\_2(\mathbb{Z})$." They are generated by
$$\left(\begin{array}{cc} -1&0 \\\ 0&-1\end{array}\right),
\left(\begin{array}{cc} -1&-1 \\\ 1&0\end{array}\right),
\left(\begin{array}{cc} 0&-1 \\\ 1&0\end{array}\right),
\left(\begin{array}{cc} 0&-1 \\\ 1&1\end{array}\right) $$
and are isomorphic to $\mathbb{Z}\_2$, $\mathbb{Z}\_3$, $\mathbb{Z}\_4$, and $\mathbb{Z}\_6$, respectively. Does someone know a reference for this statement? (Or, is it easy to see?) My attempt at a Google search turned up this statement, but I wasn't able to find a reference.
| https://mathoverflow.net/users/2669 | Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request) | A finite subgroup will map to a finite subgroup of $PSL\_2(\mathbb Z)$, which is a free product $Z\_2 \* Z\_3$. I believe I have been told that finite subgroups of free products of finite groups are conjugate to subgroups of the factors being free-producted together.
| 23 | https://mathoverflow.net/users/391 | 27290 | 17,841 |
https://mathoverflow.net/questions/27284 | 11 | It is widely believed that mathematicians have a uniform standard of what constitutes a correct proof. However, this standard has, at minimum, changed over time. What are some striking examples where controversies have arisen over what constitutes a correct proof?
Examples of this include:
1. The acceptability of the use of the axiom of choice
2. The acceptability of proofs that rely on assuming that a computer has performed a certain computation correctly
3. The debate over intuitionistic logic versus classical logic
4. Hilbert's re-examination of Euclid's axioms and his discovery of unstated assumptions therein
5. Debates over the use of infinitesimals in calculus, culminating in Weierstrass's epsilons and deltas. There are of course many others.
**Edit:**
The above *re*-formulation of the question was provided by Timothy Chow and copied directly from the meta.mathoverflow [thread](http://mathoverflow.tqft.net/discussion/428/mathematical-controversies/) about this question.
| https://mathoverflow.net/users/6309 | The definition of "proof" throughout the history of mathematics | Paolo Ruffini's work on the impossibility of solving the quintic by radicals did meet a strong passive resistance. Around 1800 he proved the theorem up to a minor gap, that himself or somebody else could have fixed soon, had his book met the attention that deserved. But times were not ready for a such a revolutionary idea as *proving the impossibility*; 20-30 years later this idea had slowly spread and become more natural, and Abel and Galois got more lucky (so to speak).
This is in my opinion a major example of a particular theorem that was met with resistance before being accepted, and in fact it also shows that resistance is not necessarily associated with controversials, but sometimes even with indifference (which may be even worse).
A short and well written account of the story is in J.J.O'Connor and E.F.Robertson's article for the History of Mathematics archive: <http://www-history.mcs.st-and.ac.uk/Biographies/Ruffini.html>
| 12 | https://mathoverflow.net/users/6101 | 27304 | 17,852 |
https://mathoverflow.net/questions/27318 | 5 | I have following two questions
1) Let $S$ be a compact oriented surface with (non-empty) boundary. Also assume that the Euler characteristic of $S$ is negative (Thus, $S$ is not disk or annulus). Then is it possible to put a hyperbolic metric on $S$ such that every boundary component becomes a geodesic?
2) In above case, what is the universal cover? Is it a subset of the unit disk with hyperbolic metric?
| https://mathoverflow.net/users/5538 | Hyperbolic structure on surfaces with boundary | The answers to both questions are positive.
1) If the Euler characteristic of $S$ is negative, then you can find a pants decomposition (where some of the pants boundary components are glued together, while the others are boundary components of $S$). Then you can put hyperbolic metrics on each pair of pants, while prescribing the lengths of the boundary component, so that you can glue them into a hyperbolic metric of $S$ (and you have parameters left for the gluing, just like for a closed surface).
2) You can always, by gluing additional pants, consider $S$ endowed with a hyperbolic metric constructed as above, as a subsurface with geodesic boundary of a closed hyperbolic surface $\bar S$. Then the universal cover of $\bar S$ is the Poincaré disc, and the boundary components of $S$ lift to complete geodesics. The universal cover of $S$ is then one of the connected components of the complement of these complete geodesics.
| 9 | https://mathoverflow.net/users/4961 | 27323 | 17,864 |
https://mathoverflow.net/questions/27316 | 21 | I have no intuition for field theory, so here goes. I know what the algebraic and separable closures of a field are, but I have no feeling of how different (or same!) they could be.
So, what are the differences between them (if any) for a perfect field? A finite field? A number field?
Are there geometric parallels? (say be passing to schemes, or any other analogy)
| https://mathoverflow.net/users/4177 | Separable and algebraic closures? | Geometrically there is a very big difference between separable and algebraic
closures (in the only case where there is a difference at all, i.e., in positive
characteristic $p$). Technically, this comes from the fact that an algebraically
closed field $k$ has no non-trivial derivations $D$; for every $f\in k$ there is
a $g\in k$ such that $g^p=f$ and then $D(f)=D(g^p)=pg^{p-1}D(g)=0$. This means
that an algebraically closed field contains no differential-geometric
information. On the other hand, if $K\subseteq L$ is a separable extension, then
every derivation of $K$ extends uniquely to a derivation of $L$ so when taking a
separable closure of a field a lot of differential-geometric information remains.
Hence I tend to think of a point of a variety for a separably closed field as a
very thick point (particularly if it is a separable closure of a generic point)
while a point over an algebraically closed field is just an ordinary (very thin)
point.
Of course you lose infinitesimal information by just passing to the
*perfection* of a field (which is most conveniently defined as the direct
limit over the system of $p$'th power maps). Sometimes that is however exactly
what you want. That idea first appeared (I think) in Serre's theory of
pro-algebraic groups where he went one step further and took the perfection of
group schemes (for any scheme in positive characteristic the perfection is the limit,
inverse this time, of the system of Frobenius morphisms) or equivalently
restricted their representable functors to perfect schemes. This essentially
killed off all infinitesimal group schemes and made the theory much closer to
the characteristic zero theory (though interesting differences remained mainly
in the fact that there are more smooth unipotent group schemes such as the Witt
vector schemes). Another, interesting example is for Milne's flat cohomology
duality theory which needs to invert Frobenius by passing to perfect schemes in
order to have higher $\mathrm{Ext}$-groups vanish (see SLN 868).
| 48 | https://mathoverflow.net/users/4008 | 27332 | 17,871 |
https://mathoverflow.net/questions/22536 | 8 | For a finite dimensional $k$-algebra $A$, each $A^{\otimes n}, n \geq 0$ is a $k$-algebra ($A^0 = k $). Let $T= \prod\_{n \geq 0} A^{\otimes n}$. This is a $k$-alegbra with unit $(1,1,\dots)$ and multiplication is component-wise. Let $\Delta^{(n)} : A^{\otimes n} \to T \otimes T$ be the deconcatenation map
$$ \Delta^{(n)}(a\_1 \otimes \dots \otimes a\_n) = \sum\_{i=0}^n (a\_1 \otimes \dots \otimes a\_i) \otimes (a\_{i+1} \otimes \dots \otimes a\_n ). $$
I want to extend these $\Delta^{(n)}$ to a comultiplication $\Delta : T \to T \otimes T $. This does not seem to work in a straightforward way because if $t = ( t\_0, t\_1, \dots ) \in T, $ then $\sum\_n \Delta^{(n)}(t\_n)$ may not be a finite sum of pure tensors in $T \otimes T$ (I have not shown this sum can be infinite, but suspect it can be).
>
> Is there a way to make $T$ into a Hopf algebra so that $\Delta(t) = \Delta^{(n)}(t)$ when $t\_i=0$ for $i \ne n$? If not, is there an algebra similar to $T$ where this does work?
>
>
> Is there a standard way to complete the tensor product and instead get a map $\Delta$ from $T$ to the completion? Does this give rise to a genuine Hopf algebra or some generalization of Hopf alegbras?
>
>
>
| https://mathoverflow.net/users/3318 | Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$ | You can indeed complete the tensor product and get a good comultiplication, but it's not strictly speaking a Hopf algebra. An algebraic geometer would call it an affine formal group. If you think of the infinite product $T=\prod\_n A^{\otimes n}$ as a pro-object indexed by $\mathbb{N}$ with $T(n) = \prod\_{i=0}^n A^{\otimes i}$, then you can tensor $T$ with itself and get a pro-object indexed by $\mathbb{N} \times \mathbb{N}$. Take the limit and that's the completed tensor product you want. Alternatively, you can do it with topologized rings and topologically-complete the tensor product.
For the uncompleted tensor product, there is no comultiplication extending your rule. There is, however, a cofree Hopf algebra (as Anton points out, consult [Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint?](https://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-an-adjoint)). That's a sub-Hopf-algebra of the completed Hopf algebra $T$.
| 9 | https://mathoverflow.net/users/4183 | 27335 | 17,873 |
https://mathoverflow.net/questions/27240 | 14 | It seems to me (at least according to books and papers on the subject I read) that the field of automated theorem proving is some sort of art or experimental empirical engineering of combining various approaches, but not a science which tries to explain WHY its methods work in various situations and to find classes of situations for which they work. Or may be I am not aware of some results or the problem is too hard.
Has it been clarified in a rigorous way why, for example, Robinson's resolution works better than Gilmore's saturation?
(intuitive reason of course is that most general unifier already contains all information about infinitely many terms from Herbrand's universe, but it is only intuition, not a concrete reason, as well as the fact that you can not know when to stop searching for a proof of a formula A in the case A and not(A) are both unprovable is not a rigorous explanation of undecidability of first order logic).
I even saw a paper shows that there are cases in which London museum algorithm (brute force search, essentially) performs better than resolution!
Maybe somebody have found some large classes of formulas on which resolution works well?
Maybe some probabilistic analysis of proving methods?
Thanks in advance.
| https://mathoverflow.net/users/6307 | Reasons for success in automated theorem proving | You may be interested in the wonderful little book ``The Efficiency of Theorem Proving Strategies: A Comparative and Asymptotic Analysis'' by David A. Plaisted and Yunshan Zhu. I have the 2nd edition which is paperback and was quite cheap. I'll paste the (accurate) blurb:
``This book is unique in that it gives asymptotic bounds on the sizes of the search spaces generated by many common theorem-proving strategies. Thus it permits one to gain a theoretical understanding of the efficiencies of many different theorem-proving methods. This is a fundamental new tool in the comparative study of theorem proving strategies.''
Now, from a critical perspective: There is no doubt that sophisticated asymptotic analyses such as these are very important (and to me, the ideas underlying them are beautiful and profound). But, from the perspective of the practitioner actually *using* automated theorem provers, these analyses are often too coarse to be of practical use. A related phenomenon occurs with decision procedures for real closed fields. Since Davenport-Heinz, it's been known that general quantifier elimination over real closed fields is inherently doubly-exponential w.r.t. the number of variables in an input Tarski formula. One full RCF quantifier-elimination method having this doubly-exponential complexity is CAD of Collins. But, many (Renegar, Grigor'ev/Vorobjov, Canny, ...) have given singly exponential procedures for the purely existential fragment. Hoon Hong has performed an interesting analysis of this situation. The asymptotic complexities of three decision procedures considered by Hong in ``Comparison of Several Decision Algorithms for the Existential Theory of the Reals'' are as follows:
(Let $n$ be the number of variables, $m$ the number of polynomials, $d$ their total degree, and $L$ the bit-width of the coefficients)
CAD: $L^3(md)^{2^{O(n)}}$
Grigor'ev/Vorobjov: $L(md)^{n^2}$
Renegar: $L(log L)(log log L)(md)^{O(n)}$
Thus, for purely existential formulae, one would expect the G/V and R algorithms to vastly out-perform CAD. But, in practice, this is not so. In the paper cited, Hong presents reasons why, with the main point being that the asymptotic analyses ignore huge lurking constant factors which make the singly-exponential algorithms non-applicable in practice. In the examples he gives ($n=m=d=L=2$), CAD would decide an input sentence in a fraction of a second, whereas the singly-exponential procedures would take more than a million years. The moral seems to be a reminder of the fact that a complexity-theoretic speed-up w.r.t. *sufficiently large* input problems should not be confused with a speed-up w.r.t. *practical* input problems.
In any case, I think the situation with asymptotic analyses in automated theorem proving is similar. Such analyses are important theoretical advances, but often are too coarse to influence the day-to-day practitioner who is using automated theorem proving tools in practice.
(\* One should mention Galen Huntington's beautiful 2008 PhD thesis at Berkeley under Branden Fitelson in which he shows that Canny's singly-exponential procedure can be made to work on the small examples considered by Hong in the above paper. This is significant progress. It still does not compare in practice to the doubly-exponential CAD, though.)
| 12 | https://mathoverflow.net/users/4915 | 27340 | 17,876 |
https://mathoverflow.net/questions/19404 | 9 | I am reviewing and documenting a software application (part of a supply chain system) which implements an approximation of a normal distribution function; the original documentation mentions the same/similar formula quoted [here](http://mathworld.wolfram.com/NormalDistributionFunction.html)
$$\phi(x) = {1\over \sqrt{2\pi}}\int\_{-\infty}^x e^{-{1\over 2} x^2} \ dx$$
This is approximated with what looks like an asymptotic series like [here](http://mathworld.wolfram.com/AsymptoticSeries.html)
however the expression is slightly different:
$$\phi(x) \simeq {1\over x\sqrt{2\pi}} \Bigl( 1 - {1\over x^2} + {3\over x^4} - {15\over x^6} + {105 \over x^8}\Bigr)$$
The original document quotes a *"Cambridge statistical tables"* book which I don't have; also, the software uses a precalculated table for values of $x$ between $[-8.2, 8.2]$ and uses the approximation function for values of $x$ outside that interval.
I need to find some reference that explains why this particular formula was chosen, and how accurate the approximation really is.
(The documentation claims that an error less then the $7$th decimal is "not bad")
Also, the constant $1/\sqrt{2\pi}$ does it have a name? All I was able to find was that "this expression ensures that the total area under the curve $\Phi(x)$ is equal to one"
Does it matter if the constant has only $10$ decimals when the double type in Java (IEEE 754) has a precision of approximately $16$ decimals?
| https://mathoverflow.net/users/4927 | Approximation of a normal distribution function | This is the (divergent) asymptotic development for $
f(x)=\int\_x^{\infty} e^{-{1\over 2}t^2} \ dt
$ given by
$$f(x) \sim
e^{-{x^2\over 2}}\ (\ {1\over x}\ + \
\Sigma\_{k=1}^{\infty}\ {(-1)^k(2k-1)!\over 2^{k-1}(k-1)!}\ {1\over x^{2k+1}}\ )
$$
or
$$f(x) \sim
e^{-{x^2\over 2}}\ (\ {1\over x}\ - {1\over x^3} + {3\over x^5} - {15\over x^7} + {105\over x^9} - {945\over x^{11}}...)$$
It is easily obtained from an integration by parts. The remainder is given by an explicit integral. From its expression, one can check that $f(x)$ is in fact squeezed between two consecutive sums of the series. As a result, we have the bound, for all $x>0$,
$$0 \geq f(x) -
e^{-{x^2\over 2}}\ (\ {1\over x}\ - {1\over x^3} + {3\over x^5} - {15\over x^7} + {105\over x^9}) \geq - e^{-{x^2\over 2}}\ {945\over x^{11}}$$
Divergent series were standard tools at the beginning of the XXe century. I can't provide a reference, but the book of Hardy "*Divergent series*" may be a starting point for a bibliographic search.
The divergent asymptotics was chosen because it is "good" at infinity. When truncated to some power (say 9), it has the correct behavior when $x$ goes to infinity, that is, it goes to zero, just like $f(x)$. So it gives an interesting approximation of $f$ for all sufficiently big value of $x$. This of course is not the case of a development obtained by truncating a converging series in positive powers of $x$.
| 7 | https://mathoverflow.net/users/6129 | 27341 | 17,877 |
https://mathoverflow.net/questions/27297 | 8 | Let's start with a picture: <https://i.stack.imgur.com/mEFct.png>
What you see here are boxes and circles inside the boxes. Each circle is connected to zero or more boxes. One box is the primary box, it's the grey one. Here is another example (the top box is the primary box): <https://i.stack.imgur.com/V0mQS.png> (I apologize for the crappy drawings).
The goal is to select a subset of all circles such that:
1. If a circle is selected, then all the boxes it is connected to must be selected.
2. If a box is selected then *one of* the circles in it must be selected.
3. The primary box is selected.
4. The sum of the numbers in the selected circles is minimal.
In the top picture the optimal selection of circles has been colored blue. In the bottom picture there are two optimal solutions: select the 3 in the top box and the 1 in the right box. Other solution: select the 2 in the top box, the 1 in the left box and the 1 in the right box.
My question is: is this solvable in polynomial time, or if not is it possible to approximate it? If the graph is tree structured then dynamic programming can solve the problem, but diamond structures seem to complicate the problem.
I have asked this problem on stackoverflow.com, but that doesn't seem to get very far and I thought maybe this is a more appropriate place?
| https://mathoverflow.net/users/6210 | Is this problem solvable in polynomial time? | There is a straightforward encoding of 3-SAT into this problem, which means that unless P=NP, no, there cannot be a polynomial-time algorithm that solves it. The encoding can be constructed as follows. Given a formula $\phi=\psi\_1\wedge\dots\wedge\psi\_k$ in conjunctive normal form over propositional variables $v\_1,\dots,v\_n$, you have the following boxes:
1. For each $v\_i$, two boxes $A\_i, B\_i$ representing the literals $v\_i$ and $\neg v\_i$. Both contain a single unit-weight circle.
2. Also for each $v\_i$, a box $C\_i$ representing the law of the excluded middle: This box contains two circles with weight 0, one connected to $A\_i$, the other to $B\_i$.
3. For each clause $\psi\_j=l\_1\vee l\_2 \vee l\_3$, a box $D\_j$ containing three weight-0 circles connected to the corresponding $A\_i$ or $B\_i$.
4. The root box, containing a single weight-0 circle connected to all the $C\_i$ and $D\_j$.
This can be done in time linear in the size of $\phi$.
Then a set of circles satisfying your constraints must contain at least one of $A\_i,B\_i$ for all $i$, and has weight $n$ if and only if it corresponds to a valid valuation of the $v\_i$ which satisfies $\phi$. In particular, satisfiability of $\phi$ can be decided by checking if the minimum weight solution has weight $n$.
| 12 | https://mathoverflow.net/users/6634 | 27342 | 17,878 |
https://mathoverflow.net/questions/27357 | 19 | When you are checking a conjecture or working through a proof, it is nice to have a collection of examples on hand.
There are many convenient examples of commutative rings, both finite and infinite, and there are many convenient examples of infinite non-commutative rings. But I don't have a good collection of finite non-commutative rings to think about. I usually just think of a matrix ring over a finite field.
Do YOU have other examples that you particularly like/find easy to use/find to be a good source of counterexamples?
| https://mathoverflow.net/users/4087 | What are your favorite finite non-commutative rings? | 2 families of examples that are sometimes useful to have in mind:
(1) The group ring of a non-abelian finite group over a finite commutative ring.
and
(2) the incidence algebra of a finite poset over a finite commutative ring (the ring of upper triangular matrices is a basic example of this).
Of course, both of these are special cases of the same more general categorical (or quiver) definition. Before I wrote that I'd never dealt with the more general concept, but that was a lie...
| 15 | https://mathoverflow.net/users/nan | 27363 | 17,889 |
https://mathoverflow.net/questions/27375 | 20 | For a group $G$, is there an interpretation of $\mathbb C[G]$ as functions over some noncommutative space?
If so, what does this space "look like"? What are its properties? How are they related to properties of $G$?
| https://mathoverflow.net/users/2837 | Geometric interpretation of group rings? | The noncommutative space defined by $C[G]$ is (by definition) the dual $\widehat{G}$ of G.
There are as many ways to make sense of this space as there are theories of noncommutative geometry.
(Edit: In particular if G is not finite you have lots of possible meanings for the group ring, depending on what kind of regularity and support conditions you put on G, or equivalently, what class of representations of G you wish to consider, and I will ignore all such issues - which are the main technical part of the subject - below.)
One basic principle is that noncommutative geometry is not about algebras up to isomorphism, it's about algebras up to Morita equivalence -- in other words, it's in fact about categories of modules over algebras (the basic invariant of Morita equivalence). You can think of these (depending on context) as vector bundles or sheaves of some kind on the dual. In this case we're looking at the category of complex representations of G, which are sheaves on the dual $\widehat{G}$. You can think of this as a form of the Fourier transform (modules for functions on G with convolution = modules for functions on the dual with multiplication), though obviously at this level of detail it's a complete tautology. Coarser invariants such as K-theory of group algebras, Hochschild homology etc give invariants, eg K-theory and cohomology, of the dual, noncommutative as it may be. There are many conjectures about this noncommutative topology, most famously the Baum-Connes conjecture relating the K-theory of this "space" to that of classifying spaces associated to G.
As to what the dual looks like, this is of course highly dependent on the group.
For completely arbitrary groups I don't know of anything meaningful to say beyond structural things of the Baum-Connes flavor, so you have to pick a class of groups to study.
If G is abelian, the dual is itself a group (the dual group). The formal thing you can say in general is that the dual of G fibers over the dual of the center of G -- this is a form of Schur's lemma, saying irreducible reps live over a particular point of the dual of the center (ie the center acts by evaluation by a character). You might get some more traction by looking at the "Bernstein center" or Hochschild cohomology --- endomorphisms of the identity functor of G-reps. This is a commutative algebra and the dual fibers over its spectrum. In many cases this is a very good approximation to the dual -- ie the "fibers are finite" (this is what happens for say real and p-adic groups).
The orbit method of Kirillov says that for a nilpotent or solvable group, the dual looks like the dual space of the Lie algebra, modulo the coadjoint action. So again that's quite nice.
Very very roughly the Langlands philosophy says that for reductive groups G (in particular over local or finite fields) the dual of G is related to conjugacy classes in a dual group $G^\vee$. This is if you'd like a way to make meaningful the observation that conjugacy classes and irreps are in bijection for a finite group -- you roughly want to say they're in CANONICAL bijection if the two groups are "dual".
Rather than say it this coarsely, it's better to think in terms of the Harish-Chandra / Gelfand philosophy, which (again whittled down to one coarse snippet) says that the dual of a reductive group (over any field) is a union of "series", ie a union of subspaces each of which looks like the dual of a torus modulo a Weyl group. In other words, you look at all conjugacy classes of tori in G, for each torus you construct its dual (which is a group now!), and mod out by the symmetries inherited by the torus from its embedding in G, and this is the dual of G roughly. (This is also very close to saying semisimple conjugacy classes in the dual group of G, which is where the Langlands interpretation comes from).
Anyway this is saying that the dual is a very nice and manageable, even algebraic, object.
Kazhdan formulated this philosophy as saying that the dual of a reductive group is an algebraic object --- the reps of the group over a field F are something like the F-points of one fixed variety (or stack) over the algebraic closure.
Anyway one can go much further, and that's what the Langlands program does.
| 26 | https://mathoverflow.net/users/582 | 27383 | 17,902 |
https://mathoverflow.net/questions/27382 | 10 | Let me begin by noting that I know quite little about category theory. So forgive me if the title is too vague, if the question is trivial, and if the question is written poorly.
Let $\mathcal{C}$ and $\mathcal{D}$ be categories. Say that a functor $T: \mathcal{C} \to \mathcal{D}$ has property X (maybe there is a real name for this property?) if a morphism $f: A \to B$ between objects of $\mathcal{C}$ is an isomorphism whenever $T(f): T(A) \to T(B)$ is an isomorphism. For example, the obvious forgetful functor $CH \to Set$ where $CH$ is the category of compact Hausdorff spaces has property X because a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.
Here is my question. Is there a (nontrivial) functor $T: LCH \to \mathcal{D}$ from the category of locally compact Hausdorff spaces to some category $\mathcal{D}$ with property X? Even better, can we assume that $LCH$ is a subcategory of $\mathcal{D}$ and that $T$ is a forgetful functor?
I don't care to specify what I mean by "nontrivial", except that the "identity" functor from $LCH$ to itself doesn't count. I want it to be genuinely easier to decide whether or not a morphism is an isomorphism in $\mathcal{D}$. If there happen to be lots of ways to do this, perhaps it will help to know that my interest comes from some problems in analysis.
Thanks in advance!
| https://mathoverflow.net/users/4362 | Category Theory / Topology Question | The property $X$, as you call it, is well-known. A functor with this propery is said to "reflect isomorphisms". Another example of such a functor is the geometric realization functor from simplicial sets to compactly generated Hausdorff spaces. There are all sorts of ways of building a category $D$ and a functor $T$ with the properties you want, however, depending on what you want to do, different answers can be more suiting. For example, you can let $D$ be the category $Sh(CH)$ of sheaves on the site of compact Hausdorff spaces and $T$, be "Yoneda": if $Y$ is a $LCH$ space, then $T(Y)$ is the sheaf that assigns each compact Hausdorff space $X$ the set $Hom(X,Y)$. Then, it is a simple exercise to verify that $T$ is fully-faithful and that $T(f)$ is an isomorphism implies that $f$ is (SINCE every locally compact Hausdorff space is compactly generated).
| 6 | https://mathoverflow.net/users/4528 | 27390 | 17,907 |
https://mathoverflow.net/questions/27406 | 9 | Is there a notion of a dimension associated to free resolutions like projective and injective dimensions associated to projective and injective resolutions? My guess is that it coincides with projective dimension but couldn't find any reference talking about this. Any help or tips will be appreciated!
| https://mathoverflow.net/users/5292 | Free resolution dimension? | I'm sort of stealing the idea from t3suji, but here goes:
If the module is projective, i.e. $PD = 0$, then $FD \leq 1$.
If the module is not projective, i.e. $PD > 0$, then $PD = FD$.
The first statement is t3suji's comment. The second statement follows by first observing that if $F$ is the free module on infinitely many generators, there is an exact sequence of length $n+1$,
$0\rightarrow F\rightarrow F\rightarrow\ldots\rightarrow F\rightarrow 0$,
so long as $n\geq 1$. Then if we have a projective resolution
$0\rightarrow P\_n\rightarrow\ldots\rightarrow P\_0\rightarrow M\rightarrow 0$,
we can make the free resolution
$0\rightarrow P\_n\oplus F\rightarrow\ldots\rightarrow P\_0\oplus F\rightarrow M\rightarrow 0$.
| 11 | https://mathoverflow.net/users/5513 | 27415 | 17,923 |
https://mathoverflow.net/questions/27305 | 35 | This is an extract from [Apéry's biography](http://peccatte.karefil.com/PhiMathsTextes/Apery.html)
(which some of the people have already enjoyed in
[this answer](https://mathoverflow.net/questions/25630/major-mathematical-advances-past-age-fifty/25631#25631)).
>
> During a mathematician's dinner in
> Kingston, Canada, in 1979, the
> conversation turned to Fermat's last
> theorem, and Enrico Bombieri proposed
> a problem: to show that the equation
> $$ \binom xn+\binom yn=\binom zn \qquad\text{where}\quad n\ge 3 $$ has
> no nontrivial solution. Apéry left the
> table and came back at breakfast with
> the solution $n = 3$, $x = 10$, $y =
> 16$, $z = 17$. Bombieri replied
> stiffly, "I said nontrivial."
>
>
>
What is the state of art for the equation above? Was it seriously studied?
**Edit.** I owe the following official name of the problem to Gerry,
as well as Alf van der Poorten's (different!) point of view on this story and
some useful links on the problem (see Gerry's comments and response).
The name is *Bombieri's Napkin Problem*. As the [OEIS link](http://www.oeis.org/A010330) suggests,
Bombieri said that
>
> "the equation $\binom xn+\binom yn=\binom zn$
> has no *trivial* solutions for $n\ge 3$"
>
>
>
(the joke being that he said "trivial" rather than "nontrivial"!).
As Gerry indicates in his comments, the special case $n=3$ has a long history
started from the 1915 paper [Bökle, *Z. Math. Naturwiss. Unterricht* **46** (1915), 160];
this is reflected in
[[A. Bremner, *Duke Math. J.* **44** (1977) 757--765]](http://dx.doi.org/10.1215/S0012-7094-77-04433-7).
A related link is [F. Beukers, *Fifth Conference of the Canadian Number Theory Association*, 25--33]
for which I could not find an MR link.
[Leech's paper](http://dx.doi.org/10.1017/S0305004100032850) indicates
the particular solution
$$
\binom{132}{4}+\binom{190}{4}=\binom{200}{4}
$$
and the trivial infinite family
$$
\binom{2n-1}n+\binom{2n-1}n=\binom{2n}n.
$$
| https://mathoverflow.net/users/4953 | A binomial generalization of the FLT: Bombieri's Napkin Problem | Some solutions for $n=3$ can be found at <http://www.oeis.org/A010330> where there is also a reference to J. Leech, Some solutions of Diophantine equations, Proc. Camb. Phil. Soc., 53 (1957), 778-780, MR 19, 837f (but from the review it seems that paper deals with ${x\choose n}+{y\choose n}={z\choose n}+{w\choose n}$).
There are some other solutions at <http://www.numericana.com/fame/apery.htm>
**EDIT** Here are some more references for $n=3$:
Andrzej Krawczyk, A certain property of pyramidal numbers, Prace Nauk. Inst. Mat. Fiz. Politechn. Wrocƚaw. Ser. Studia i Materiaƚy No. 3 Teoria grafow (1970), 43--44, MR 51 #3048.
The author proves that for any natural number $m$ there exist distinct natural numbers $x$ and $y$ such that $P\_x+P\_y=P\_{y+m}$ where $P\_n=n(n+1)(n+2)/6$. (J. S. Joel)
M. Wunderlich, Certain properties of pyramidal and figurate numbers, Math. Comp. 16 (1962) 482--486, MR 26 #6115.
The author gives a lot of solutions of $x^3+y^3+z^3=x+y+z$ (which is equivalent to the equation we want). In his review, S Chowla claims to have proved the existence of infinitely many non-trivial solutions.
W. Sierpiński, Sur un propriété des nombres tétraédraux, Elem. Math. 17 1962 29--30, MR 24 #A3118.
This contains a proof that there are infinitely many solutions with $n=3$.
A. Oppenheim, On the Diophantine equation $x^3+y^3+z^3=x+y+z$, Proc. Amer. Math. Soc. 17 1966 493--496, MR 32 #5590.
Hugh Maxwell Edgar, Some remarks on the Diophantine equation $x^3+y^3+z^3=x+y+z$, Proc. Amer. Math. Soc. 16 1965 148--153, MR 30 #1094.
A. Oppenheim, On the Diophantine equation $x^3+y^3-z^3=px+py-qz$, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 230-241 1968 33--35, MR 39 #126.
| 12 | https://mathoverflow.net/users/3684 | 27425 | 17,930 |
https://mathoverflow.net/questions/27423 | 3 | Let $U : \mathbb R^d \to \mathbb R^d$ be a smooth vector field, and let $F\_t : \mathbb R \times \mathbb R^d \to \mathbb R^d$ be the corresponding smooth flow, defined by the differential equation $$\tfrac{d}{dt} F\_t(x) = U(F\_t(x)).$$ Let $f : \mathbb R^d \to \mathbb R$ be an integrable function, and consider the integral $$\int\_{\mathbb R^d} f( F\_t(x) ) ~dx.$$ To calculate this, I would make the change of variables $y = F\_t(x)$, making the ansatz that there exists some function $\rho(y)$ so that the integral equals $$\int\_{\mathbb R^d} f(y) \rho(y) ~dy.$$ From what I can gather from [the paper I'm reading](http://www.jstor.org/pss/1428307), it's common knowledge that this function $\rho$ has the form $$\rho(y) = \tfrac{1}{J\_t(F\_t^{-1}(y))} = \exp \left(-\int\_0^t (\operatorname{div} U)(F\_{s-t}(y) ) ~ds \right),$$ where of course $J\_t$ stands for the Jacobian of the flow.
Could you please point me to a reference for this formula?
| https://mathoverflow.net/users/238 | A formula for the Jacobian of a flow | This is [the Liouville formula](http://eom.springer.de/l/l059660.htm). It is explained nicely in *Ordinary Differential Equations* by Arnold.
| 4 | https://mathoverflow.net/users/5371 | 27426 | 17,931 |
https://mathoverflow.net/questions/27438 | 1 | In section 1.1 under the subtitle system of classical particles with potential, the authors claim that
"for a system of classical particles with rigid constraints, the configuration space is a Riemannian manifold X with Riemannian structure given by twice the kinetic energy."
I don't quite how the configuration space can be given by a Riemannian manifold, as it is more naturally viewed as a symplectic manifold and there appears to be no natural Riemannian structure on a symplectic manifold. Also the relation between the Riemannian structure and the kinetic energy also eludes me. The best interpretation I can think of is to impose a Riemannian structure on the cotangent bundle via Legendre transform, or the specification of a Lagrangian function. But this is not explcitly given.
| https://mathoverflow.net/users/4923 | Clarification of classical field theory lecture notes by P. Deligne and D. Freed | *Configuration* space is, by definition, the position space of your particles. *Phase* space, on the other hand, is the space of pairs (position, momentum). The latter has a symplectic structure; the former has a Riemannian structure.
Regarding the relationship between kinetic energy and the Riemannian structure: You will recall from your high school physics class that kinetic energy is $\frac{1}{2} mv^2$. Of course the $v^2$ is really the dot product $v \cdot v$, in other words it's $g(v,v)$, where $g$ is the Riemannian metric and $v$ is a tangent vector. The $\frac{1}{2}$ explains the "twice the kinetic energy" part.
| 8 | https://mathoverflow.net/users/83 | 27443 | 17,943 |
https://mathoverflow.net/questions/26608 | 13 | Let $\tau$ be $(-1+\sqrt{5})/2$, let $f(x)$ be $\lfloor (x+1)\tau \rfloor$, let
$s\_n$ be $\tau n (n+1) (n+2) / 6$, and
let $S\_n$ be $$\sum\_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1)
+ (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0).$$
Is $(S\_n-s\_n)/(n \log n)$ bounded for $n > 1$?
(It stays between -0.35 and +0.30 for all $n$ between 2
and $10^6$.)
This is a specific instance of the question
[Dedekind-esque sums](https://mathoverflow.net/questions/24517/dedekind-esque-sums)
that I posted a few weeks ago. It may be an atypical instance in some
ways (since $\tau$ is a pretty atypical real number for Diophantine
approximation problems) but it's the one that interests me most
right now. An affirmative answer to my question would have
implications concerning the "Goldbug machine" described in
* Michael Kleber, *Goldbug Variations*, Mathematical Intelligencer 27 #1 (Winter 2005), pp. 55–63, <https://arxiv.org/abs/math/0501497>.
The plot at the bottom of [http://www.cs.uml.edu/~jpropp/Phi-short.pdf](http://www.cs.uml.edu/%7Ejpropp/Phi-short.pdf)
is a histogram of $(S\_n - s\_n)/n$ for $n$ going from 1 to a million.
As you can see, it doesn't stray very far away from 0.
So perhaps that $\log n$ in the denominator could be replaced by
something smaller, like $\sqrt{\log n}$ or even $\log \log n$ (or
maybe even 1, though I doubt it).
| https://mathoverflow.net/users/3621 | A specific Dedekind-esque sum | I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts".
**Edit**: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's *Works* vol. I p. 145), which depends only on the continued fraction having bounded partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.
| 3 | https://mathoverflow.net/users/6153 | 27448 | 17,947 |
https://mathoverflow.net/questions/27401 | 16 | This is an extremely elementary question but I just can't seem to get things to work out. What I am looking for is a natural definition of the quotient bundle of a subbundle $E'\subset E$ of $\mathbb R$ (say) vector bundles over a fixed base space $B$.
Every source I find on this essentially leaves the construction to the reader. I would like to glue together sets of the form $U\times E\_x/E'\_x$ where $x\in U$ is a locally trivial neighborhood by some sort of transition function derived from those corresponding to $E$ and $E'$, but this doesn't actually make sense in any meaningful way.
While I am tagging this as differential geometry, I would like a construction that works in the topological category (i.e., does not invoke Riemannian metrics) and avoids passing to the category of locally free sheafs.
Sorry if this is a repost (I'm sure it is, but I can't seem to find anything) and thanks in advance.
| https://mathoverflow.net/users/5323 | Defining Quotient Bundles | (I was going to leave this as a comment but decided that it's a bit long for that)
A couple of remarks:
1. You express an aversion to Riemannian metrics because you want to be able to apply this in the *topological* category. That's fine, except for two things: firstly, *Riemannian* metrics would not be explicitly involved in this construction as it is a general construction that applies to all vector bundles, not just tangent bundles. Secondly, having inner products on the fibres of a vector bundle is not something that is special to the smooth category. Using a partition of unity argument (assuming you're working over a sufficiently nice space, or your vector bundle is a pull-back of a universal one - look up "numerable cover" for more on this - but note that *all* the answers to this question tacitly assume this), any finite dimensional vector bundle admits a **continuous** choice of inner product on its fibres. So the standard argument: "choose an orthogonal structure and take the orthogonal complement" works equally well in the continuous category as the smooth one.
2. What is really going on here is a reduction of structure group. The structure group of the big bundle is $Gl(n)$. The inclusion of the subbundle implies that it reduces to the subgroup that preserves $\mathbb{R}^k \subseteq \mathbb{R}^n$. At this point, you should work out what this subgroup consists of - think in terms of matrices if you don't see it immediately. General Nonsense (although for this case, the more junior Lieutenant Nonsense will do) implies that this subgroup is homotopy equivalent to $Gl(k) \times Gl(n-k)$. A reduction to this defines an isomorphism $E \cong E' \oplus E''$, where $E''$ is the required quotient bundle. The two previous answers can be viewed as constructing this reduction. The "standard orthogonality" argument rests on the observation that $Gl(m) \cong O(m)$ for any $m$ so we can reduce everything to the corresponding orthogonal group. Thus we start with $O(n)$, reduce to the subgroup that preserves $\mathbb{R}^k$ and then ... but there is no "and then" because this subgroup is already $O(k) \times O(n-k)$. So either way, you are doing one "reduction of structure group", the difference between the two methods is simply a choice of doing $Gl(n) \to O(n)$ at the start, or doing $Gl(n;k) \to Gl(k) \times Gl(n-k)$ in the middle.
| 11 | https://mathoverflow.net/users/45 | 27450 | 17,948 |
https://mathoverflow.net/questions/27461 | 5 | The title sounds tautological, right? Perhaps I'm missing something completely trivial here ...
Assume $X$ is a compact totally disconnected hausdorff space. It is known that $X$ can be written as directed inverse limit of finite discrete spaces $X\_i$ with surjective transition maps (i.e. $X$ is profinite). How do you prove that every map from $X$ to a finite discrete space factors through some projection $X \to X\_i$?
I know that the fibers of the projections are a basis of the topology of $X$ (not only a subbasis). The corresponding result for profinite groups is true, but I cannot adopt the proof.
Of course you could use Stone duality to reduce the assertions to a completely trivial one (a finite boolean ring in a directed limit of boolean subrings lies in some of these boolean subrings), but I want a direct topological proof.
| https://mathoverflow.net/users/2841 | profinite spaces are the pro-completion of finite sets | Let $f:X \to Z$ be a map to a finite discrete space. Note that each fiber, $f^{-1}(z)$, is both open and closed in $X$. Let $p\_i: X \to X\_i$ be the projection maps.
Fix some $z \in Z$. Since $f^{-1}(z)$ is open, and the fibers of the maps are a basis, there is an open cover of $f^{-1}(z)$ by sets of the form $p\_i^{-1}(x)$, for $x$ in various $X\_i$.
Since $f^{-1}(z)$ is closed in a compact space, it is compact. So we can take a finite subcover of this cover. Thus, there is some single index $i$ for which $f^{-1}(z)$ is covered by sets of the form $p\_i^{-1}(x)$, $x \in X\_i$.
Since $Z$ is finite, there is a single $i$ such that, for every $z \in Z$, the fiber $f^{-1}(z)$ is covered by sets of the form $p\_i^{-1}(x)$, $x \in X\_i$. The map $f$ factors through $X\_i$.
| 7 | https://mathoverflow.net/users/297 | 27466 | 17,958 |
https://mathoverflow.net/questions/26831 | 4 | Hi everybody,
I am dealing with the O' Nan Scott theorem, the classification of finite primitive groups (I am reading "Classes of Finite Groups", by Adolfo Ballester-Bolinches and Luis M. Ezquerro). Here for "primitive group" I mean a finite group $G$ endowed with a core-free maximal subgroup $U$. I say that a primitive group $(G,U)$ is monolithic if it has only one minimal normal subgroup, $N$, which is of the type $S^n$ for some simple group $S$ and positive integer $n$. I know that a monolithic primitive group can have two types of action: diagonal action or product action, and this is recognizable looking at the intersection of $U$ with $N$ (because $U$ turns out to be the normalizer in $G$ of $U \cap N$). This intersection can be either the direct product of some conjugates of a maximal subgroup of $S$ or a direct product of diagonals (a diagonal of $S^m$ is a subgroup containing the elements of the form $(x,x^{\alpha\_2},...,x^{\alpha\_m})$ for some $\alpha\_2,...,\alpha\_m \in Aut(S)$).
Now my question is: what can I say about the primitive groups $G$ for which every maximal core-free subgroup $U$ is of product type (i.e. $(G,U)$ has product action)?
I have an example of such a group: take the semidirect product $(A\_5 \times A\_5) \langle (1,\tau) \varepsilon \rangle$, where $\tau=(12)$ and $\varepsilon$ is the exchange of the two coordinates (I think of it as a subgroup of $Aut(A\_5 \times A\_5) = S\_5 \wr \langle \varepsilon \rangle$): the action is $(x,y)^{(1,\tau)\varepsilon} = (y^{\tau},x)$. Here every subgroup of diagonal type is contained in the maximal subgroup $(A\_5 \times A\_5)\langle(\tau,\tau)\rangle$.
I looked for such groups in some article (by Kovacs, "Primitive permutation groups of simple diagonal type" and "Primitive subgroups of wreath products in product action"), but I didn't find a precise answer.
| https://mathoverflow.net/users/5710 | Monolithic primitive groups without diagonals | You have left out a possbility for a monolithic action. It is possible to have $U\cap N=1$. This can occur when either S is abelian (here G is a subgroup of AGL(d,p) in its usual action on $p^d$ points) or nonabelian.
As for your question, in the case where $S$ is nonabelian, a monolithic primitive group G is a subgroup of $Aut(S) wr S\_n$ which contains $N=S^n$ and transitively permutes the $n$ simple direct factors of $N$. The group $G$ will contain a maximal corefree subgroup of diagonal type if and only if $G/N$ is isomorphic to a subgroup of $(Out(S)\times S\_l) wr S\_k$ where $n=kl$, $l\geq 2$, and the Out(S) occurs as a diagonal subgroup of $Out(S)^l$. When this happens you can find a subgroup $U\_1$ of $N$ which is a direct product of $k$ diagonals and for which $N\_G(U\_1)/U\_1\cong G/N$.
| 5 | https://mathoverflow.net/users/3214 | 27468 | 17,960 |
https://mathoverflow.net/questions/27454 | 4 | Let $G$ be a complex semisimple group, $B$ a Borel subgroup of $G$ and $X=G/B$ the flag variety of $G$. If $G$ is simply connected, then every line bundle $L$ on $X$ can be made $G$-equivariant (see the proof of Theorem 1 [here](http://www.math.harvard.edu/~lurie/papers/bwb.pdf)). Is this also true if $G$ isn't simply connected? By identifying $X$ with the flag variety of the universal covering group $\widetilde{G}$ of $G$, we can at least get a $\widetilde{G}$-action on $L$, but I haven't been able to figure out if this action factors through to yield a $G$-action.
More generally, can an arbitrary vector bundle $V$ on $X$ be made $G$-equivariant? If so, in how many ways?
| https://mathoverflow.net/users/430 | Equivariance of vector bundles over G/B | A vector bundle does not have to be $G$-equivariant. For instance, Horrock-Mumford bundle on $P^4$ is equivariant only with respect to Heisenberg group and not full $SL\_5$. Pull it back to the flag variety of $SL\_5$ to get a bundle of rank 2, not $SL\_5$-equivariant. I am sure someone will tell you easier examples than that.
| 5 | https://mathoverflow.net/users/5301 | 27472 | 17,962 |
https://mathoverflow.net/questions/27361 | 36 | Here is a question in the intersection of mathematics and sociology. There is a standard way to encode a Sudoku puzzle as an integer programming problem. The problem has a 0-1-valued variable $a\_{i,j,k}$ for each triple $1 \le i,j,k \le 9$, expressing that the entry in position $(i,j)$ has value $k$. The Sudoku rules say that four types of 9-sets of the variables sum to 1, to express that each cell is filled with exactly one number, and that each number appears exactly once in each row, column, and $3 \times 3$ box. And in a Sudoku puzzle, some of these variables (traditionally 27 of them) are preset to 1.
It is known that generalized Sudoku, like general integer programming, is NP-hard. However, is that the right model for Sudoku in practice? I noticed that many human Sudokus can all be solved by certain standard tricks, many of which imply a unique *rational* solution to the integer programming problem. You can find rational solutions with linear programming, and if the rational solution is unique, that type of integer programming problem is not NP-hard, it's in P. Traditionally Sudoku puzzles have a unique solution. All that is meant is a unique integer solution, but maybe the Sudoku community has not explored reasons for uniqueness that would not also imply a unique rational solution.
Are there published human Sudoku puzzles with a unique solution, but more than one rational solution? Is there a practical way to find out? I guess one experiment would be to make such a Sudoku (although I don't know how difficult that is), and then see what happens when you give it to people.
| https://mathoverflow.net/users/1450 | Do actual Sudoku puzzles have a unique rational solution? | I wasn't planning on answering here but since someone mentioned my paper in the long comment thread above maybe I should anyway.
When I'm solving problems by hand one of the sets of patterns I frequently use involve uniqueness: something can't happen because it would lead to a puzzle with more than one solution, but a well-posed Sudoku puzzle has only one solution, so it's safe to assume that whatever it is doesn't happen. For instance, it's not possible to have four initially-empty cells in a rectangle of the same two rows, the same two columns, and the same two 3x3 squares, and also for these four cells to all contain one of the same two values, because then the two values they contain could be placed in two different ways and the rest of the puzzle wouldn't notice the change. So if there's only one way to prevent a rectangle like this from forming then the solution has to use that one way.
I have the sense (though I could be wrong) that this sort of inference does not imply a unique rational solution. But of course a puzzle could have a unique rational solution anyway — my computer solver knows these rules too, but uses them less frequently than I do by hand.
ETA: It does indeed seem to be true that these rules don't imply unique rational solutions. With them, my solver easily solves the following puzzle, which has a unique integer solution but does not have a unique fractional solution. Without them, my solver can still solve the same puzzle, but only by using rules that are (in my experience) much more difficult to apply by hand.
```
-----------------------------------
| 3 7 8 | 6 4 5 | 1 2 9 |
| | | |
| 6 9 . | . 7 . | 4 8 5 |
| | | |
| 4 . 5 | 9 . 8 | 3 7 6 |
|-----------------------------------|
| 7 . 9 | 5 3 . | 8 6 4 |
| | | |
| 8 4 . | 7 . . | 5 9 3 |
| | | |
| 5 3 6 | 8 9 4 | 2 1 7 |
|-----------------------------------|
| 1 6 3 | 2 5 7 | 9 4 8 |
| | | |
| 9 8 4 | . . . | 7 5 2 |
| | | |
| 2 5 7 | 4 8 9 | 6 3 1 |
-----------------------------------
```
Or if you want something that looks like the puzzles that get published, here's another one that uses the same mechanism:
```
-----------------------------------
| 1 . . | 8 . 7 | 5 . . |
| | | |
| . 5 . | 6 . . | 9 7 . |
| | | |
| . 6 . | . 4 . | . . . |
|-----------------------------------|
| . . 2 | . . . | . 6 . |
| | | |
| 6 . . | 4 . 3 | . . . |
| | | |
| . 4 . | . . . | 1 . . |
|-----------------------------------|
| . . . | . 1 . | . 5 . |
| | | |
| . 1 3 | . . 6 | . 9 . |
| | | |
| . . 7 | 5 . 4 | . . 1 |
-----------------------------------
```
| 16 | https://mathoverflow.net/users/440 | 27480 | 17,968 |
https://mathoverflow.net/questions/27476 | 11 | Some background: my coauthors and I are working on a problem which deals with the exponential growth rates of certain infinite products of matrices. One of the sub-problems which arises in this project is to prove that a particular function from the unit interval to the reals, which describes the growth rates of a family of related infinite matrix products, does not have any line segments in its graph. Now, the existence of a line segment would imply the existence of a pair of matrices some of whose characteristics coincide in a precise way.
We have found an analytic proof that the graph cannot contain line segments, but it is long and displeasingly ad-hoc, and we would rather like to replace it with a shorter argument! Hence, we have found ourselves wondering whether there is an algebraic obstacle which would prevent these coincidences from occurring.
The question which arises from this research, then, is:
>
> Do there exist matrices $A,B \in SL(2,\mathbb{Z})$ which satisfy the following constraints: the matrices $A$ and $B$ have only positive entries, do not commute, and satisfy
> $$\rho(AB)=\rho(A)\rho(B),$$
> $$\mathbb{Q}\left(\rho(A)\right)=\mathbb{Q}\left(\rho(B)\right) \neq \mathbb{Q},$$
> where $\rho$ denotes spectral radius?
>
>
>
The non-existence of such a pair of matrices would imply the non-existence of the line segments mentioned above.
One of my coauthors suspects that the spectral radius constraints on $A$ and $B$ cannot be met unless $A=C^k$, $B=C^\ell$ for some $C \in SL(2,\mathbb{Z})$ and $k,\ell \in \mathbb{N}$, which would contradict the hypothesis that $A$ and $B$ do not commute. Any solutions which end up being used in our paper will of course be gratefully acknowledged!
| https://mathoverflow.net/users/1840 | Existence of a pair of matrices in SL(2,Z) satisfying certain constraints on the spectral radius | This is true: if one has two matrices $A, B \in SL\_2 C$, and one knows the three traces $tr(A),tr(B),tr(AB)$,
then this uniquely determines the matrices $A,B$ up to conjugacy *if $A$ and $B$ generate a non-elementary discrete group* (except in
a few degenerate cases which won't occur for positive matrices). See for example [Maclachlan-Reid](http://books.google.com/books?id=yrmT56mpw3kC&lpg=PP1&dq=maclachlan%20reid&pg=PA128#v=onepage&q&f=false). Since $tr(X) = \rho(X)+\rho(X)^{-1}$ for a positive matrix $X$, this implies that the three matrices have the same traces as two commuting matrices with the same traces,
and therefore $A,B$ must be conjugate to a pair of commuting matrices.
In fact, as your coauthor suspects, then they must be powers of the same matrix.
Addendum: I'll give some details of the argument, and I forgot a condition ($A$ and $B$ might generate a solvable group otherwise, although this is not possible for discrete groups).
Assume that $A, B \in SL\_2 Z$. We want to show that $tr(A),tr(B),tr(AB)$
uniquely determine $A$ and $B$ up to conjugacy in $SL\_2 C$.
Since $A$ is a positive matrix in $SL\_2 Z$, we may conjugate it in $SL\_2 R$ to be of the form
$$
A=\begin{pmatrix} \lambda & 0 \\\ 0 & \lambda^{-1} \end{pmatrix}
$$
where $\lambda >1$ (because $tr(A)>2$). We conjugate $B$ by the same matrix, and relabel so that
$$
B=\begin{pmatrix} a & b \\\ c & d \end{pmatrix}.
$$
If $A$ and $B$ commute, then we check that $b=c=0$, so in particular $ad=1$. Conversely, if $ad=1$, then we conclude that $bc=0$, and therefore $b=c=0$ since $A,B$ cannot generate a noncommutative solvable group since $SL\_2 Z$ is discrete (see below).
From now on, let's assume that $A$ and $B$ don't commute.
The centralizer of $A$ in $SL\_2 C$ consist of matrices of the
form
$$
\begin{pmatrix} z & 0 \\\ 0 & z^{-1} \end{pmatrix}.
$$
We may conjugate $B$ by the centralizer of $A$ to be of the form
$$
B=\begin{pmatrix} a & 1 \\\ ad-1 & d \end{pmatrix}.
$$
Here I'm making use of the fact that $A,B$ are conjugate into $SL\_2 Z$ to conclude that they
don't generate a solvable group. This follows from the Margulis lemma (or an easy exercise).
If $B$ had off-diagonal entries which were
zero, then $\langle A, B\rangle$ would be solvable. The point is that then we may
conjugate by elements of the centralizer of $A$ in $SL\_2 C$ to arrange the upper
right entry to be 1.
Thus, we have
$$tr(A)=\lambda+\lambda^{-1}, tr(B)=a+d, tr(AB)=\lambda a +\lambda^{-1} d.$$
Now we compute
$$a= \frac{\lambda tr(AB)-tr(B)}{\lambda^2-1} , d= \frac{\lambda^{-1} tr(AB)-tr(B)}{\lambda^{-2}-1}.$$
Thus, $tr(A), tr(B), tr(AB)$ canonically determine $A,B$ up to conjugacy (of course $\lambda$ is uniquely determined by $tr(A)$ given $\lambda >1$).
Now, we need to check that $\rho(AB) \neq \rho(A)\rho(B)$, where $\rho(A)=\lambda, \rho(B)=\mu$. If not, $tr(AB)=\lambda\mu +(\lambda\mu)^{-1}$ (again since $AB$ is positive).
Plugging into the above equations, we conclude that $a=\mu, d=\mu^{-1}$, a contradiction.
I'm sure this argument can be simplified, but I wanted to carry through the sketch of the argument I gave the first go around.
Also, one may see that any subgroup of commuting matrices in $SL\_2 Z$ is either powers of a parabolic element, a hyperbolic element, or a finite-order element.
| 10 | https://mathoverflow.net/users/1345 | 27488 | 17,974 |
https://mathoverflow.net/questions/27490 | 6 | Motivation
----------
The common functors from topological spaces to other categories have geometric interpretations. For example, the fundamental group is how loops behave in the space, and higher homotopy groups are how higher dimensional spheres behave (up to homotopy in both cases, of course). Even better, for nice enough spaces the (integral) homology groups count $n$-dimensional holes.
---
A groupoid is a category where all morphisms are invertible. Given a space $X$, the fundamental groupoid of $X$, $\Pi\_1(X)$, is the category whose objects are the points of $X$ and the morphisms are homotopy classes of maps rel end points. It's clear that $\Pi\_1(X)$ is a groupoid and the group object at $x \in X$ is simply the fundamental group $\pi\_1(X,x)$. My question is:
>
> Is there a geometric interpretation $\Pi\_1(X)$ analogous to the geometric interpretation of homotopy groups and homology groups explained above?
>
>
>
| https://mathoverflow.net/users/343 | Geometric interpretation of the fundamental groupoid | I'm not sure how to answer this, because it already seems pretty geometric to me. So let me
answer a slightly different question: what is the fundamental groupoid good for?
Since one knows that the fundamental group and groupoid are equivalent as categories for
path connected spaces, it's tempting to view the groupoid as giving nothing new.
But in fact, there are situations when it seems more natural. For example, if a group
$G$ acts continuously on a space $X$, then unless one knows something more, we only get an outer action of $G$ on $\pi\_1(X,x)$ i.e. it's only well defined up to inner automorphisms.
However, $G$ will act on the fundamental groupoid $\Pi\_1(X)$ on the nose, and in fact,
the previous statement becomes easier to see from this point of view.
| 17 | https://mathoverflow.net/users/4144 | 27492 | 17,975 |
https://mathoverflow.net/questions/27471 | 3 | Let $L/K$ be a finite Galois extension of function fields, with Galois groups $G$. I want to look at the ramification of primes in the extension, i.e. to get $e\_p$ and $f\_p$ for a prime $p$ in the base field $K$ (since the extension is Galois, the ramification index and inertia degree are independent of the choice of the prime lying above $p$). From Serre's 'Local Fields', it is clear that if we fix a prime $q$ in $L$ which lies over $p$, then we can look at the decomposition group associated to $q$, say $G\_q$, and its inertia group, say $(G\_q)\_0$ (please forgive me for the notation :P), and an immediate result is that $e\_p = \left|{(G\_q)\_0} \right|$ and $f\_p = \left| {G\_q/(G\_q)\_0} \right|$.
And here is my problem. Is there any nice way to compute the decomposition groups, inertia groups or just the cardinalities? If not, can we do something in some special cases? For example, when $G$ is cyclic?
| https://mathoverflow.net/users/3849 | on the computation of decomposition groups | Henri Cohen's A Course in Computational Algebraic Number Theory contains quite a bit of information. Chapters 4.8, 6.2 and 6.3 combined result in algorithms that compute decomposition groups. Note that if you want to relate different primes you will have to first compute the galois group (6.3) and fix a presentation.
| 2 | https://mathoverflow.net/users/2024 | 27499 | 17,980 |
https://mathoverflow.net/questions/27494 | 15 | The two definitions alluded to in the title can be found here: <http://en.wikipedia.org/wiki/Separable_sigma_algebra> (one is that the $\sigma$-algebra is countably generated, the other is pretty much the standard usage of the word separable wrt the semi-metric given by the measure). Why are they equivalent?
| https://mathoverflow.net/users/5498 | Separable sigma-algebra: equivalence of two definitions | The two notions are not equivalent. Indeed, they are not
equivalent even when one considers completing the measure
by adding all null sets with respect to any countably
generated $\sigma$-algebra. Nevertheless, the forward
implication holds.
First, let me explain the forward implication. Suppose that
$S$ is a $\sigma$-algebra generated by a countable
subfamily $S\_0$ and $\mu$ is a finite measure defined on
$S$. The semi-metric on $S$ is defined by
$d(A,B)=\mu(A\triangle B)$. Let $S\_1$ be the collection of
finite Boolean combinations of sets in $S\_0$. This is a
countable family, and I claim it is dense in the
semi-metric. To see this, let $S\_2$ be the closure of $S\_1$
in the semi-metric, that is, the sets $A\in S$ that are
approximable by sets in $S\_1$, in the sense that for any
$r\gt 0$ there is $B\in S\_1$ such that $d(A,B)\lt r$. Note
that $S\_2$ contains $S\_1$ and is closed under complement
since the measure was finite. I claim it is also closed
under countable unions: if each $A\_n$ is approximable by
$B\_n$ to within $r/2^n$, then $\cup\_n A\_n$ is approximated
by $\cup\_n B\_n$ to within $r$, and so one may find an
approximating finite union. So $S\_2$ is actually a
$\sigma$-algebra, and since it contains $S\_0$, it follows
that $S\_2=S$. That is, every set in $S$ is approximable by
sets in $S\_1$, and so $S\_1$ is a countable dense set in the
semi-metric, as desired.
Let's turn now to the reverse implication, which is not
generally true. The easiest counterexample for this in the
strict sense of the question is to let $X$ be a set of size
continuum and $S=P(X)$, the full power set of $X$. This is
a $\sigma$-algebra, but it is easily seen not to be
countably generated on cardinality grounds. Fix any $p\in
X$ and let $\mu$ be the measure placing mass $1$ at $p$ and
0 mass outside {p}. In this case, the family {emptyset, X}
is dense in the semi-metric, since every subset is
essentially empty or all of $X$, depending on whether it
contains $p$. So the semi-metric is separable, but the
$\sigma$-algebra is not countably generated.
Note that in this counterexample, the $\sigma$ algebra is
obtained from the counting measure on {p} by adding a large
cardinality set of measure $0$ and taking the completion.
Similar counterexamples can be obtained by adding such
large cardinality set of measure $0$ to any space and
taking the completion.
At first, I thought incorrectly that one could address the
issue by considering the completion of the measure, and
showing that the $\sigma$-algebra would be contained within
the completion of a countably generated $\sigma$-algebra.
But I now realize that this is incorrect, and I can provide
a counterexample even to this form of the equivalence.
To see this, consider the filter $F$ of all sets $A\subset
\omega\_1$ that contain a closed unbounded set of countable
ordinals. This is known as the *club filter*, and it is
closed under countable intersection. The corresponding
ideal $NS$ consists of the *non-stationary sets*, those
that omit a club, and these are closed under countable
union. It follows that the collection $S=F\cup NS$, which
are the sets measured by a club set, forms a
$\sigma$-algebra. The natural measure $\mu$ on $S$ gives
every set in $F$ measure $1$ and every set in $NS$ measure
$0$. This is a countably additive 2-valued measure on $S$.
Note that every set in $S$ has measure $0$ or $1$; in
particular, there are no disjoint positive measure sets. It
follows that the family {emptyset,$\omega\_1$} is dense in
the semi-metric, since every set in $S$ either contains or
omits a club set, and hence either agrees with emptyset or
with the whole set on a club. Thus, the semi-metric is
separable. But for any countable subfamily $S\_0\subset S$,
we may intersect the clubs used to decide the members of
$S\_0$, and find a single club set $C\subset\omega\_1$ that
decides every member of $S\_0$, in the sense that every
member of $S\_0$ either contains or omits $C$. This feature
is preserved under complements, countable unions and
intersections, and therefore $C$ decides every member of
the $\sigma$-algebra generated by $S\_0$. The completion of
the measure on the $\sigma$-algebra generated by $S\_0$ is
therefore contained within the principal filter generated
by $C$ together with its dual ideal. This is not all of
$S$, since there are club sets properly contained within
$C$, such as the set of limit points of $C$. Thus, this is
a measure space that has a separable semi-metric, but the
$\sigma$-algebra is not contained in the completion of any
countably generated $\sigma$-algebra.
| 12 | https://mathoverflow.net/users/1946 | 27502 | 17,981 |
https://mathoverflow.net/questions/27367 | 14 | The homology algebra $H\_\*( K(\mathbb{Z},2n); \mathbb{Z})$ contains a
divided polynomial algebra on a generator $x$ of dimension $2n$.
I suppose I could read through the Cartan seminar for a proof, but I'm hoping
someone knows of a nice simple argument for this fact.
| https://mathoverflow.net/users/3634 | Good reference for homology of $K(\mathbb{Z}, 2n)$? | We need to do three things: (1) show that the homology contains a polynomial algebra, (2) show that powers of the generator are sufficiently divisible, and (3) show that torsion doesn't interfere.
Let me repeat Hatcher's path through (1), since we need the particular generator to show that it is divisible. According to Dold-Thom (Hatcher 4.K; cf Dold-Kan1), a model for $K(Z,m)$ is the free commutative monoid on pointed2 $S^m$. According to James (Hatcher 4.J), the homology of the free associative monoid $JS^m$ on $S^m$ is a polynomial algebra on one generator in degree $m$. Thus there is a map of monoids $JS^m\to K(Z,m)$ so its map on homology is compatible with the ring structure. It is easy to see (eg, by the Serre spectral sequence) that the rational homology of $K(Z,m)$ is the signed-commutative algebra free on a generator in degree $m$, so this map is a rational isomorphism if $m=2n$.
Now for step (2), where we divide the James class by $k!$. If we restrict to words of length $k$ in these monoids, the affect of imposing commutativity is to quotient by the $k$th symmetric group. This piece of the James construction is a pseudo-manifold representing the class in degree $nk$ which is the $k$th power of the generator. The action of the symmetric group is generically free, so this cycle has multiplicity $k!$ when it lands in $K(Z,2n)$. Thus it is divisible. In other words, choose a fundamental domain for the action of the symmetric group on the $k$th filtration of the James construction. This yields is an $kn$-chain which when symmetrized is the whole of the $k$th filtration, and thus a cycle. Projecting it to the quotient is an alternative way of symmetrizing, so it is a cycle for $K(Z,2n)$.
I will leave step (3) as an exercise. What this shows so far is that the integral homology hits the divided power algebra in the rational homology, but the canonical choice of the divided element should make it plausible that these divisions are the right choices, ie, $x^{(k)}x^{(l)}=\binom {k+l}k x^{(k+l)}$.
1If one accepts the fragment of Dold-Kan of the agreement of the two things one can do to a simplicial abelian group (1) homology of (either) associated chain complex and (2) the homotopy groups of the geometric realization; then a version of Dold-Thom follows trivially. Indeed, the very definition of simplicial homology is the free abelian group (ie, simplicial group) on the simplicial set.
2 by "the free (commutative) monoid on a pointed set," I mean that we impose on the free object the relation that the basepoint is the identity for the monoid.
| 11 | https://mathoverflow.net/users/4639 | 27506 | 17,985 |
https://mathoverflow.net/questions/27508 | 17 | Hi,
I have no idea where to look for, so I'm hoping you can give me some pointers.
I'm interested by numbers of form $p-1$ when $p$ is a prime number. Do they have a name, so that I can google them?
More precisely, I'm interested in their factors. Ok, obviously 2 is a factor, but what about the others? Are there a lot of small factors? Do we know the rate of the growth of its larger factor, is it linear, logarithmic?
Thanks.
| https://mathoverflow.net/users/6673 | Factors of p-1 when p is prime. | Two extremes of this problem are Fermat primes and Sophie Germain primes. If either class had infinitely many members, that would contribute to an answer of your question. There is literature about the distribution of prime factors (cf. Riesel and Knuth), but I do not know the literature regarding the restriction to numbers of the form prime - 1 .
If I had to start anywhere for questions like this, I would choose one or all three of the following:
Richard Guy's Unsolved problems in Number Theory,
Hans Riesel's book on computer methods for primality proving and factorization,
Prime Numbers: a Computational perspective, by Crandall and Pomerance.
Forgive my memory if the titles or authors are missing or incorrectly spelled.
Gerhard "Ask Me About System Design" Paseman, 2010.06.08
| 8 | https://mathoverflow.net/users/3528 | 27510 | 17,987 |
https://mathoverflow.net/questions/27481 | 4 | A balanced smooth rational curve in a calabi-Yau X is a smooth rational curve whose normal bundle is $O(-1)\oplus O(-1)$.
We usually like these curves because of their rigidity.
But, Is there any theorem that guaranty the existence of at least one such curve. For example for Quintic? Or for any other example?
| https://mathoverflow.net/users/5259 | balanced curves in Calabi-Yau 3-folds | Perhaps you already know this: but we don't even know how to show that there are finitely many rational curves of a given degree $d$ on the general quintic threefold. This was originally conjectured by Clemens. However, for low degrees (up to $d = 11$ or something close to that), the conjecture is verified in the "strong form": any smooth rational curve of low degree (again, at most $11$) on the general quintic has normal bundle $O(-1) \oplus O(-1)$. As far as I know, that's the current state of affairs. It can often be difficult to produce rational curves with the "expected" normal bundle in any given situation!
| 5 | https://mathoverflow.net/users/397 | 27513 | 17,989 |
https://mathoverflow.net/questions/27516 | 4 | I am reading the paper:
The Hausdorff dimension of horseshoes of Diffeomorphism of surfaces, Bulletin Brazilian Mathematical Society, Ricardo Mañe,(1990). Roughly speaking the author state that, the unstable and stable foliation of horseshoe can be extended to a $C^1$ foliation on a neighborhood of the horseshoe. I can not locate a reference for this statement and I was not able to prove it. More precisely:
Let $M$ a two-dimensional compact manifold and $f:M\to M$ a diffeomorphism $C^r$, $r\geq 2$.
Suppose that $\Lambda\subset M$ is a horseshoe for $f$.
Why the stable and unstable foliations defined by $W^u(x)$ and $W^s(x)$, when $x$ varies
$\Lambda$, extend to $C^1$ foliation of a neighborhood of $\Lambda$ ?
Could you help me with the argument or a reference for this statement ?
| https://mathoverflow.net/users/2386 | Local product structure horseshoes | In page 166 of [Palis-Takens](http://books.google.com/books?id=pwydPA23KVUC&printsec=frontcover&dq=palis+takens+hyperbolicity&hl=es&cd=1#v=onepage&q&f=false) book that is also stated. In the discusion before, the outline is given (see Appendix 1).
| 2 | https://mathoverflow.net/users/5753 | 27518 | 17,993 |
https://mathoverflow.net/questions/27511 | 5 | Let $M$ be a compact Riemannian manifold with metric $g$ and let $f \in Diff(M)$.
Under what circumstances is there a natural metric $g\_f$ s.t. the associated smooth measure $\nu\_f$ is preserved by $f$, and how can such a $g\_f$ be obtained?
| https://mathoverflow.net/users/1847 | When is there a natural Riemannian metric whose measure preserves a self-diffeomorphism? | Let $\Omega$ be the standard volume on your Riemannian manifold,
and $\phi$ a smooth function on M. A quick computation shows that
$e^\phi \Omega$ is invariant by f if and only if the following cohomological equation is satisfied:
$$ \phi(f^{-1}(x))-\phi(x)=log\ Jf(x)$$
where Jf is the jacobian of f. This implies for example that $Jf^n(x)=1$ for all $x\in Fix(f^n)$.
This later condition is in fact sufficient for C2 transitive Anosov diffeomorphisms (see e.g. Katok-Hasselblatt th 19.2.7). For these diffeos, this is also equivalent to saying that the SRB measure for f and the SRB measure for the inverse of f are equal (this is interesting because transitive Anosov diffeos always admit SRB measures, but of course not always smooth invariant measures).
As pointed out by Deane, any volume form can be realised as the volume associated to a Riemannian metric. Embed your manifold M in R^n, extend your volume form $f dvol\_{eucl}$ to a neighborhood of the manifold, then take the restriction of the metric $f^{2/m}g\_{eucl}$ to M . In fact, it is even possible to find a smooth conjuguate of f that preserves any given riemannian volume: this is the Moser trick.
Let M be a compact riemannian manifold, $\Omega\_0$ and $\Omega\_1$ two volume forms with the same volume. Then there is a diffeo g such that $g^\*\Omega\_0=\Omega\_1$.
| 6 | https://mathoverflow.net/users/6129 | 27520 | 17,994 |
https://mathoverflow.net/questions/21511 | 10 | ### Background
Zev Chonoles recently asked the question "[which number fields are monogenic?"](https://mathoverflow.net/questions/21267/which-number-fields-are-monogenic-and-related-questions). The answers say that for a specific number field the question is hard. So, I thought, how about looking at all of them.
### Question
There are two natural ways, that I can think of, to count number fields:
(1) By discriminant: what is the asymptotic behavior of $$ \frac{\{K:K \text{ is monogenic, } |\Delta\_K| < x\}}{\{K :|\Delta\_K|<x\}}$$
(2) By minimal polynomial of an element: what is the asymptotic behavior of $$\frac{\{K: K \text{ is monogenic, } \exists \alpha \in K \text{ with } ||m(\alpha)||\_{\infty} < x\}}{\{K: \exists \alpha \in K \text{ with } ||m(\alpha)||\_{\infty} < x\} }$$
where $m(\alpha)$ is the minimal polynomial of $\alpha$ over $\mathbb{Z}$.
| https://mathoverflow.net/users/2024 | Density of monogenic number fields? | A recent article of Bhargava and Shankar, "Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves" (<http://arxiv.org/abs/1006.1002>), addresses, among many other related questions, the density of monogenic cubic orders, counted by the height (slightly modified) of their invariants $I$, $J$.
This height is almost the same as the height of the discriminant of the polynomial to which they are associated with. So this answers the first case above for cubic fields.
Theorem 4.1 (Bhargava, Shankar):
>
> Let $N\_3^{(0)}(X,\delta)$ (resp. $N\_3^{(1)}(X,\delta)$) denote
> the number of cubic submonogenized rings $(C,x)$ of index $n$ with
> positive (resp. negative) discriminant such that $H(C,x)\lt X$ and
> $n\lt X^\delta$, where $\delta \leq 1/4$. Then we have
> $$N\_3^{(0)}(X,\delta)=\displaystyle\frac{4}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6});$$
> $$N\_3^{(1)}(X,\delta)=\displaystyle\frac{16}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6}).$$
>
>
>
If I understand correctly, this means that the density for case (1) above (restricted to cubic fields) is in fact 0.
The article goes on to study quartic rings, but the result, Theorem4.8, is a bit more difficult to understand. But if I do, it says (a bit more than) that **the density of monogenic quartic fields counted by discriminant is, once more, 0**. Note that in the theorem the count is by the invariants of resolvent, which is like the discriminant of the resolvent, which is a constant factor away from the discriminant of the quartic polynomial to which it is associated with.
So, an educated guess is that the answer to (1) above is 0 for all degrees, and then in general (as stated above without degree). On the other hand, the discriminant of an $n$ degree polynomial is a multivariate polynomial, and I suspect that sieve theory can prove that for every constant degree there is a positive density (counted as in (2)) of polynomials that have discriminant with bounded squareful part. This in turn would prove that there is a positive density of polynomials giving orders of bounded index in their respective maximal orders. Taking this educated guess a little further, I suspect the answer to (2) is greater than 0, contrary to (1).
| 5 | https://mathoverflow.net/users/2024 | 27523 | 17,995 |
https://mathoverflow.net/questions/27522 | 6 | Hi,
We have a real symmetric matrix M with diagonal elements 0's, the eigenvalues and eigenvectors of M are computed.
Now we wish to change its diagonal elements arbitrarily to minimize the sum of absolute eigenvalues. Does there exist a way to find such modifications?
If we add a constraint : keep Tr(M)=0, would that become easier?
Is there some topics about these?
Thank you for your help
Zhi Ming
| https://mathoverflow.net/users/6679 | minimize the sum of absolute eigenvalues | The sum of the absolute value of the eigenvalues is the same (since the matrix is real and symmetric) as the sum of the singular values. This sum is called the nuclear norm of the matrix. So what you are saying is that you have an affine space of matrices (a "matrix pencil") over which you would like to minimize the nuclear norm. This is the case whether you add the trace constraint or not.
This is a convex optimization problem. A google search for "nuclear norm" will show how this can be converted to a semidefinite program and solved that way. You'll also get results about more specialized interior point methods which are optimized for just this problem.
| 9 | https://mathoverflow.net/users/5963 | 27526 | 17,998 |
https://mathoverflow.net/questions/27521 | 4 | I must be missing something trivial here.
Let $G$ be, say, a reductive Lie group (or more generally any locally compact Hausdorff unimodular topological group). A *unitary Hilbert space representation* of $G$ is a group homomorphism from $G$ to the group of unitary endomorphisms $U(H)$ of a Hilbert space $H$, with the property that for all $v\in H$ the map $G\to H$, defined by sending $g$ to $gv$, is continuous. A unitary Hilbert space representation is *irreducible* if $H\not=0$ and $0,H$ are the only closed $G$-invariant subspaces of $H$.
I am ploughing through the book by Jacquet and Langlands and have made it to the all-important last chapter. On page 497 they make an assertion which seems to me to be the following:
if $V$ and $W$ are irreducible Hilbert space reps of $G$, and $i:V\to W$ is a continuous $G$-equivariant map, then either $i=0$ or $V$ and $W$ are isomorphic.
But I cannot rule out the possibility that $V$ and $W$ are not isomorphic and yet $i$ is an injection with dense image.
What am I missing? Of course I might have misunderstood the assumptions being made---they are in an explicit situation with $G$ equal to $GL(2,A)$ for $A$ some adele ring---but I cannot imagine that this can make much difference. Another assumption I think one can make is that any $f\in C\_c(G)$ acts on $V$ and $W$ via Hilbert-Schmidt operators. But my gut feeling is that there's a one-line observation that kills this.
I am supposed to be lecturing on this in about 14 hours so I had intended offering a big bounty! But I see from the FAQ that I have to wait 2 days before I can do so. *grr*
| https://mathoverflow.net/users/1384 | Injection between non-isomorphic irreducible Hilbert space reps? | No conditions are needed on the group $G$, or on the continuity of the representation, you do need the assumption that $i$ is continuous however. Since $i:V \to W$ is continuous (equivalent to being bounded) it has a continuous adjoint $i^\* : W \to V$ which is also $G$ equivariant, hence $i^\* i:V \to V$ is $G$ equivariant and since $G$ acts on $V$ irreducibly must be a scalar multiple of the identity (if not the spectrum would contain more than one point and you could take a spectral projection of $i^\* i$ and obtain a closed non-trivial $G$-invariant subspace). This shows that $i$ is a scalar multiple of an isometry and hence the image must be closed. If $G$ also acts irreducible on $W$ then $ii^\*$ is also a scalar multiple of the identity and hence if $i \not= 0$ it must be a non-zero scalar multiple of a unitary between $V$ and $W$.
| 12 | https://mathoverflow.net/users/6460 | 27528 | 18,000 |
https://mathoverflow.net/questions/27519 | 5 | Hello **MathOverflow**, my first question,
*I apologize for the LaTeX, it works in preview but not in Safari once posted.*
There are many methods out there that generate a list of unique unit fractions that sum to some rational number. One of the simplest is called the "Splitting Algorithm" which uses the identity $\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a(a+1)}$. Any fraction $\frac{a}{b}$ can be represented as $\sum^a \frac{1}{b}$. The method looks at all the fractions that are duplicates, keeps one of them, and applies the identity again.
For example
$\frac{2}{3}=\frac{1}{3}+\frac{1}{3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}$
Each of the denominators can be looked at as a composition of functions $f(a)=a+1$ and $g(a)=a(a+1)$. The method works in spite of ~~$f,g,f\circ g,g\circ f,\cdots$ are never equal.~~ fractions like $\frac{5}{2}$. When the method is applied to fractions like this duplicates will appear although it has been proved they can be removed by subsequent applications of the method so that no remaining compositions are equal.
---
The Question
------------
Can you think of any other areas where compositions of functions are used like this? I know that on its own that is too general; used in the list of sums, products, useful identities, number theory? The best case scenario would be that there are other problems like this and they can be searched for under a common name.
---
**Edit:** *method produces unique unit fractions; clarified 'equal', when I wrote that I was thinking of cases like $\frac{b-1}{b}$ that may not lead to duplicates*
| https://mathoverflow.net/users/1150 | other examples of composition of functions | Computing a continued fraction representation for a real number x can be seen as a repeated application of two functions. Starting with some real number x in [0,1[, apply 1/x, then x+1 the correct amount of time to come back in [0,1[, then 1/x again and so on.
The theory of fuchsian groups makes use of these codings to connect geometric properties of hyperbolic spaces to arithmetic properties of real numbers. Continued fractions representation for example is related to the geodesics on the modular surface $SL\_2(R)/SL\_2(Z)$.
| 6 | https://mathoverflow.net/users/6129 | 27533 | 18,003 |
https://mathoverflow.net/questions/27536 | 10 | Recently I noticed an intriguing talk by Kollár at the [MAGIC conference](http://www.nd.edu/%7Emagic/MAGIC%2710/). The abstract says:
>
> Title: Cohomology groups of structure sheaves
>
>
> Abstract: I will discuss the behavior of cohomology groups of the structure sheaf and of the dualizing sheaf under deformations and birational maps.
>
>
>
This sounds very interesting to me and can be potentially helpful for my research. However, I could not find any further information about the talk, except for some [pictures](http://www.flickr.com/photos/coffmanadam/4566336704/in/set-72157623963787382/), from which I shamefully failed to reconstruct the materials!
Would any expert who happened to be there please point me to some references for which the talk is based on, or better yet, explain briefly the motivations and key results? Many thanks.
| https://mathoverflow.net/users/2083 | A recent talk by Kollar on cohomology of structure sheaves | I would guess, he was lecturing about this article: <http://arxiv.org/abs/0902.0648>
| 7 | https://mathoverflow.net/users/3822 | 27555 | 18,016 |
https://mathoverflow.net/questions/27572 | 24 | A problem is said to be complete for a complexity class $\mathcal{C}$ if a) it is in $\mathcal{C}$ and b) every problem in $\mathcal{C}$ is log-space reducible to it. There are natural examples of NP-complete problems (SAT), P-complete problems (circuit-value), NL-complete problems (reachability), and so on.
Papadimitrou states that "semantic" complexity classes like (I think) BPP are less likely to admit complete problems. Then again, we do not actually know whether P = BPP or not, and if so, there would be BPP-complete problems. (There exist IP-complete problems, since IP=PSPACE, and determining whether a quantified boolean formula is satisfiable is PSPACE-complete.)
>
> **Question:** Are there any natural complexity classes that can be shown *not* to have complete problems?
>
>
>
I think this question has to be modified slightly, because I would imagine $Time(n)$ has no complete problems because log-space reductions can introduce a polynomial factor.
So, in the word "natural," I include the assumption that the complexity class should be invariant under polynomial transformations (I don't know how to make this precise, but hopefully it is clear).
(Also, time or space bounds should be at least $\log n$, of course.)
**Edit:** A commenter has pointed me to an interesting result of Sipser that $\mathrm{BPP}^M$ for $M$ a suitable oracle does not have complete problems. Is the same true for a (less fancy) class of the form $\bigcup\_f TIME(f)$, where the union is over a class of recursive functions $f$ that are all polynomially related to each other? (Same for $\bigcup\_f NTIME(f)$, and ditto for space.
| https://mathoverflow.net/users/344 | Are there complexity classes with provably no complete problems? | The [zoo of complexity classes](http://qwiki.stanford.edu/wiki/Complexity_Zoo) extends naturally into the realm of [computability theory](http://en.wikipedia.org/wiki/Computability_theory) and beyond, to [descriptive set theory](http://en.wikipedia.org/wiki/Descriptive_set_theory), and in these higher realms there are numerous natural classes which have no members that are complete, even with respect to far more generous notions of reduction than the one you mention.
* For example, consider the class Dec of all decidable sets of natural numbers. There can be no decidable set $U$ such that every other decidable set $A$ reduces to it in time uniformly bounded by any computable function $f$ (even iterated exponentials, or the Ackerman function, etc.). In particular, it has no member that is complete in your sense. If there were such a member, then we would be able to consruct a computable enumeration $A\_0$, $A\_1$, $\ldots$ containing all decidable sets, which is impossible since then we could diagonalize against it: the set of $n$ such that $n\notin A\_n$ would be computable, but it can't be on the list.
* The class of all arithmetically definable sets is obtained by closing the decidable sets (or much less) under projection from $\mathbb{N}^{n+1}\to \mathbb{N}^n$ and under Boolean combinations. The members of this class are exactly the sets that are defined by a first order formula over the structure $\langle\mathbb{N},+,\cdot,\lt\rangle$, and the hierarchy is stratified by the complexity of these definitions. This class has no member that is complete with respect to any computable reduction, and even with respect to any arithmetically definable reduction of bounded complexity, for in this case the hierarchy would collapse to some level $\Sigma\_n$, which is known not to occur.
* A similar argument works for the hyperarithmetic hiearchy, which can have no universal member hyperarithmetic reductions of any fixed complexity.
* And similarly for the projective hiearchy on sets of reals.
The general phenomenon is that there are numerous hierarchies growing from computability theory into descriptive set theory which are all known to exhibit strictly proper growth in such a way that prevents them from having universal members.
| 13 | https://mathoverflow.net/users/1946 | 27575 | 18,025 |
https://mathoverflow.net/questions/27579 | 15 | Let $p \ne 2$ be a prime and $a$ the smallest positive integer that is a primitive root modulo $p$. Is $a$ necessarily a primitive root modulo $p^2$ (and hence modulo all powers of $p$)? I checked this for all $p < 3 \times 10^5$ and it seems to work, but I can't see any sound theoretical reason why it should be the case. What is there to stop the Teichmuller lifts of the elements of $\mathbb{F}\_p^\times$ being really small?
| https://mathoverflow.net/users/2481 | Is the smallest primitive root modulo p a primitive root modulo p^2? | It is not true in general. See <http://primes.utm.edu/curios/page.php/40487.html> for the example, 5 mod 40487^2.
| 26 | https://mathoverflow.net/users/3710 | 27581 | 18,029 |
https://mathoverflow.net/questions/27592 | 25 | This question may sound ridiculous at first sight, but let me please show you all how I arrived at the aforementioned 'identity'.
Let us begin with (one of the many) equalities established by Euler:
$$ f(x) = \frac{\sin(x)}{x} = \prod\_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) $$
as $(a^2-b^2)=(a+b)(a-b)$, we can also write: (EDIT: We can not write this...)
$$ f(x) = \prod\_{n=1}^{\infty} \Big(1+\frac{x}{n\pi}\Big) \cdot \prod\_{n=1}^{\infty} \Big(1-\frac{x}{n\pi}\Big) $$
We now we arrange the terms with $ (n = 1 \land n=-2)$, $ (n = -1 \land n=2$), $( n=3 \land -4)$ , $ (n=-3 \land n=4)$ , ..., $ (n = 2n \land n=-2n-1) $ and $(n=-2n \land n=2n+1)$ together.
After doing so, we multiply the terms accordingly to the arrangement. If we write out the products, we get:
$$ f(x)=\big((1-x/2\pi + x/\pi -x^2/2\pi^2)(1+x/2\pi-x/\pi - x^2/2\pi^2)\big) \cdots $$
$$
\cdots \big((1-\frac{x}{(2n)\pi} + \frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(n-1))^2\pi^2})(1+\frac{x}{2n\pi} -\frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(2n-1))^2\pi^2)})\big) $$
Now we equate the $x^2$-term of this infinite product, using
[Newton's identities](http://en.wikipedia.org/wiki/Newton%27s_identities) (notice that the '$x$'-terms are eliminated) to the $x^2$-term of the Taylor-expansion series of $\frac{\sin(x)}{x}$. So,
$$ -\frac{2}{\pi^2}\Big(\frac{1}{1\cdot2} + \frac{1}{3\cdot4} + \frac{1}{5\cdot6} + \cdots + \frac{1}{2n(2n-1)}\Big) = -\frac{1}{6} $$
Multiplying both sides by $-\pi^2$ and dividing by 2 yields
$$\sum\_{n=1}^{\infty} \frac{1}{2n(2n-1)} = \pi^2/12 $$
That (infinite) sum 'also' equates $\ln(2)$, however (According to the last section of [this](http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf) paper).
So we find $$ \frac{\pi^2}{12} = \ln(2) . $$
Of course we all know that this is not true (you can verify it by checking the first couple of digits). I'd like to know how much of this method, which I used to arrive at this absurd conclusion, is true, where it goes wrong and how it can be improved to make it work in this and perhaps other cases (series).
Thanks in advance,
Max Muller
(note I: 'ln' means 'natural logarithm)
(note II: with 'to make it work' means: 'to find the exact value of)
| https://mathoverflow.net/users/93724 | Why is $ \frac{\pi^2}{12}=\ln(2)$ not true? | You cannot split
$$\left(1-\left(\frac{x}{n}\right)^2\right)\tag{1}$$
into
$$\left(1 -\frac{x}{n}\right) \left(1 + \frac{x}{n}\right)\tag{2}$$
since the products no longer converge.
| 44 | https://mathoverflow.net/users/3983 | 27596 | 18,040 |
https://mathoverflow.net/questions/27578 | 7 | In the general case, quiver cycles are of the form of orbit closures of $GL\cdot V\_{\vec{r}}$, where $GL= \prod\_{i=0}^n GL\_{r\_i}$ is the possible changes of basis on all of the vector spaces on each of the vertices and $V\_{\vec{r}}$ is any representation of the quiver with fixed dimension vector $\vec{r}$. In the equioriented An case, these are well understood by Zelevinsky and Lakshmibai-Magyar by showing them isomorphic to open sets in Schubert varieties. Bobinski and Zwara claim to reduce the non-equioriented case to the equioriented case, but I don't see how they are doing that.
In the introduction to ``Normality of Orbit Closures for Dynkin Quivers" (manuscripta math. 2001), Bobinski and Zwara say that they will generalize the result that equioriented An quivers have the same singularities as Schubert varieties to non-equioriented An quivers. They claim that they will do this by reducing the non-equioriented case to the equioriented case. So far, so good. But then, they say that this result follows from the proposition that they will prove, which I don't see has to do with the theorem at all.
The proposition is about a Dynkin quiver, Q, of type Ap+q+1 with p arrows in one direction and q arrows in the other and Q' an equioriented Dynkin quiver of type Ap+2q+1, their respective path algebras B=kQ and A=kQ', and respective Auslander-Reiten quivers &GammaB and &GammaA over the category of finite dimensional left modules over A and B. The proposition says ``Let A=kQ' and B=kQ be the path algebras of quivers Q' and Q, respectively, where Q and Q' are Dynkin quivers of type A. Assume there exists a full embedding of translation quivers $F: \Gamma\_B \to \Gamma\_A$. Then there exists a hom-controlled exact functor $\mathcal{F}: \text{mod }B \to \text{mod }A$."
Can anyone tell me how (or if) their results translate into a result that tells me a recipe for constructing a Kazhdan-Lustzig variety from my non-equioriented quiver? (By K-L variety, I mean a Schubert variety intersect an opposite Bruhat cell.) Alternately, is there a way to see which particular sub-variety of the representation variety of equioriented Ap+2q+1 I get out of this theorem and how that is (maybe a GIT quotient away from) a Kazhdan-Lustzig variety?
Thanks,
Anna
| https://mathoverflow.net/users/6353 | Why do non-equioriented A<sub>n</sub> quivers have singularities identical to the singularities of Schubert varieties? | The relevance of hom-controlled functors comes from Zwara's paper "Smooth morphisms of module schemes" (Theorem 1.2). The definition there is that two schemes with basepoints $(X,x)$ and $(Y,y)$ have identical singularities if there is a smooth morphism $f \colon X \to Y$ such that $f(x) = f(y)$.
Let $F$ be a hom-controlled functor. He shows that when we're dealing with module varieties, and $X$ is an orbit closure $\overline{O}\_M$ with basepoint $x$ some closed point of $O\_N$ (so $x$ represents the isomorphism class of a module $N$), then $(\overline{O}\\_M, x)$ has identical singularities as $(\overline{O}\\_{FM}, y)$ where $y$ is a closed point of $O\_{FN}$.
So if one starts with non-equioriented ${\rm A}\_n$ and picks an orbit closure $\overline{O}\_M$ together with a closed point in it, then one knows that there is a smooth morphism to some orbit closure in a bigger ${\rm A}\_m$. The orbit closure is just the image of $M$ under the hom-controlled functor constructed in Bobinski and Zwara's paper (though this construction is long and I don't remember the details). Then one can use Lakshmibai–Magyar to get a smooth morphism from this orbit closure to some Schubert variety.
So it's enough to understand how to construct $F$, which I remember being explicit but requiring quite a few steps, if we just want the varieties together with the singularities, but constructing the smooth morphism itself would take a lot more digging to construct explicitly.
| 2 | https://mathoverflow.net/users/321 | 27604 | 18,046 |
https://mathoverflow.net/questions/27606 | 3 | Suppose $f: \mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is has continuous partial derivatives and
$$4f(x,y)=f(x+\delta,y+\delta)+f(x-\delta,y+\delta)+f(x-\delta,y-\delta) + f(x+\delta,y-\delta)$$
for all $(x,y)$ in $\mathbb{R}\times\mathbb{R}$ and all $\delta$ in $\mathbb{R}$.
I don't believe that $f$ is necessarly harmonic but I cannot construct a counter-example.
Is $f$ harmonic?
| https://mathoverflow.net/users/6700 | Harmonic Functions | Yes, the Taylor series works. Actually $C^2$ suffices for the remainder term, although my sophomore calculus book gives the proof using $C^3.$ I get
$$ 4 f(x\_0, y\_0) = 4 f(x\_0, y\_0) + \left( 2 f\_{xx}(x\_0, y\_0) + 2 f\_{yy}(x\_0, y\_0) \right) \delta^2 \; + \; o( \delta^2 ) $$
and
$$ \left( 2 f\_{xx}(x\_0, y\_0) + 2 f\_{yy}(x\_0, y\_0) \right) \delta^2 \; = \; o( \delta^2 ) $$
and
$$ 2 \left( f\_{xx}(x\_0, y\_0) + f\_{yy}(x\_0, y\_0) \right) \; = \; 0 $$
LATER EDIT: unless I am vastly mistaken this argument still works if we put in the caveat $ | \delta | < \Delta = \Delta(x\_0, y\_0), $ that is we only require your equation for small $\delta$ and even say that the allowable size of $\delta$ depends on the position of the center point that I am calling $(x\_0, y\_0).$ But with this change we can build an easy discontinuous example of your relation, take
$$ f(x\_0, y\_0) = 1, \; \; if \; \; y\_0 > 0, $$
$$ f(x\_0, y\_0) = 0, \; \; if \; \; y\_0 = 0, $$
$$ f(x\_0, y\_0) = -1, \; \; if \; \; y\_0 < 0. $$
Then your relation holds for $ | \delta | < | y\_0 | $ when $y\_0 \neq 0$ and holds for all $\delta$ when $ y\_0 = 0.$
| 4 | https://mathoverflow.net/users/3324 | 27608 | 18,049 |
https://mathoverflow.net/questions/27424 | 21 | This question is motivated by this recent [question](https://mathoverflow.net/questions/27406/free-resolution-dimension). Suppose $R$ is commutative, Noetherian ring and $M$ a finitely generated $R$-module. Let $FD(M)$ and $PD(M)$ be the shortest length of free and projective resolutions (with all modules f.g.) of $M$ respectively.
We will be interested in two conditions on $R$:
(1) For all f.g. $M$, $PD(M)<\infty$ if and only if $FD(M)<\infty$.
(2) For all f.g. $M$, $PD(M)=FD(M)$.
It is not hard to see that one is equivalent to (3): *all f.g. projective modules are stably free*. Also, (2) is equivalent to (4): *all f.g projectives are free*. It is natural to ask whether (1) and (2) are equivalent. I don't think they are, but can't find a counter-example. Thus:
>
> Can we find a commutative Noetherian ring which satisfies (3) but not (4)?
>
>
>
Some thoughts: (3) is equivalent to the Grothendick group of projective $K\_0(R)=\mathbb Z$. Well-known class of rings satisfying (3): local rings or polynomial rings over fields.
Of course, there are well known rings which fails (4): coordinate rings of $n$-spheres $R\_n=\mathbb R[x\_0,\cdots,x\_n]/(x\_0^2+\cdots+x\_n^2-1)$ for $n$ even. Unfortunately, for those rings we have $K\_0(R\_n)=\mathbb Z^2$. (except for $n=8r+6$ which do provide examples, see below)
UPDATE: it seems to me that Tyler's answer suggests $R\_5$ may work, but one needs to check some details.
UPDATE 2: I think one can put together a more or less complete solution, based on Tyler and Torsten's answers, the comments and some digging online.
**Claim**: $R\_n$ satisfies (3) but not (4) if $n=5,6$ or $n=8r+k$ with $r>0$, $k\in \{3,5,6,7\}$
**Proof**: Let $S^n$ be the $n$-sphere and $C(S^n)$ be the ring of continuous functions.
To show that $R\_n$ satisfies (3), it suffices to show $K\_0(R\_n)=\mathbb Z$ (Lemma 2.1, Chapter 2 of Weibel's [K-book](http://www.math.rutgers.edu/%7Eweibel/Kbook.html)) or the reduced group $\tilde K\_0(R\_n)=0$. But one has isomorphism (known to Swan, see for example 5.7 of [this paper](http://www.math.ku.edu/%7Emandal/talks/bottPP1.pdf)):
$$\tilde K\_0(R\_n) \cong \tilde KO\_0(S^n) $$
the [reduced $K$-groups of real vector bundles](https://math.ucr.edu/home/baez/octonions/node10.html), and it is well-known that $\tilde KO\_0(S^n) =0$ if $n\equiv 3,5,6,7 (\text{mod} 8)$.
It remains to show our choices of $R\_n$ fail (4). Let $T\_n$ be the kernel of the surjection $R\_n^{\oplus n+1} \to R$ defined by the column of all the $x\_i$s. $R$ is projective, so the sequence splits, thus $T\_n$ is stably free.
On the other hand, $R\_n$ embeds in $C(S^n)$ and tensoring with $C(S^n)$ turns $T\_n$ into the tangent bundle of $S^n$. If $T\_n$ is free, said bundle has to be trivial. But $S\_n$ [has trivial tangent bundle if and only if $n=1,3,7$](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-64/issue-3.P1/On-the-parallelizability-of-the-spheres/bams/1183522319.full).
While it is not surprising that topological machinery is helpful, I would still be interested in seeing a pure algebraic example, so if you have one, please post.
| https://mathoverflow.net/users/2083 | A ring such that all projectives are stably free but not all projectives are free? | The more canonical example probably is the standard universal example for such a question. So, let $R\_n=k[x\_i,y\_i]/\sum x\_iy\_i=1$ where $k$ is any field and there are $2n$ variables. By localization one easily checks that $K\_0(R\_n)=\mathbb{Z}$ for any $n$. But the projective module given by the presentation,
$$0\to R\_n\stackrel{(x\_i,\ldots,x\_n)}{\to} R\_n^n\to P\to 0$$
is clearly stably free but not free if $n\geq 3$. An algebraic proof (given by myself and Madhav Nori) can be found in the article of Swan (Annals of Math studues, vol 113, pp 432-522).
| 21 | https://mathoverflow.net/users/9502 | 27611 | 18,051 |
https://mathoverflow.net/questions/27586 | 4 | I need to count the number of monomials of degree $n$ in $k$ variables, $x\_1,\ldots ,x\_k$, that contain at least one variable with a power of 1. The monomials need not include all the variables. Their powers just need to some to $n$ and they must be divisible by $x\_i$, but not $x\_i^2$, for some $i$.
| https://mathoverflow.net/users/6694 | Number of A Subset of Monomials | Another formula (almost without alternating signs) can be obtained as a variation of the comment of David Speyer. Namely, for each $S\subset\{1,\ldots,k\}$ we can consider the set of all monomials that depend precisely on all $x\_k$ with $k\in S$ and \emph{do not satisfy the property we are studying}. Such a monomial is divisible by $\prod\_{k\in S}x\_k^2$, so the number of such monomials is $\binom{|S|+n-2|S|-1}{|S|-1}$. The number of choices for $S$ is $\binom{k}{|S|}$, so altogether the number of ``unwanted'' monomials
is $$\sum\_{s=1}^{k}\binom{k}{s}\binom{n-s-1}{s-1},$$
and the number of monomials you want to compute is
$$\binom{k+n-1}{k-1}-\sum\_{s=1}^{k}\binom{k}{s}\binom{n-s-1}{s-1}.$$
| 5 | https://mathoverflow.net/users/1306 | 27614 | 18,053 |
https://mathoverflow.net/questions/27618 | 4 | Hello.
Looking at a set S of N points in the plane, I call a subset B of S "legal" if the set of points contained in the convex hull of B is exactly B itself. In other words, a subset B of S in legal if there exists some polygon P such that the set of points in and on P is exactly B.
Given N, can you bound from above the number of legal subsets of a set of N points in the plane?
Refinement: Given N, can you bound from above the number of legal subsets of size K of a set of N points in the plane?
Thank you very much.
| https://mathoverflow.net/users/nan | Upper bound for the number of subsets of N points which exhaust their convex hull | Why can't you just put the points in a circle? Doesn't that make all subsets legal?
| 12 | https://mathoverflow.net/users/1459 | 27620 | 18,057 |
https://mathoverflow.net/questions/27625 | 3 | P. Erdős and Leon Alaoglu proved in [**1**] that for every $\epsilon > 0$ the inequality $\phi(\sigma(n)) < \epsilon \cdot n$ holds for every $n \in \mathbb{N}$, except for a set of density $0$.
C. L. mentioned in [**2**] that as a consequence of the previous result one can ascertain that $\displaystyle \lim\_{n \to \infty} \frac{\phi(\sigma(n)) }{n} = 0.$
Does anybody know how it is that C. L. proceeded in order to arrive at such a conclusion?
Clearly enough, the fact that an inequality of the type $a\_{n} < \epsilon \cdot n$ holds for every $\epsilon > 0$ and a subset of $\mathbb{N}$ of density $1$ does not imply, in general, that the sequence $\displaystyle \frac{a\_{n}}{n}$ goes to $0$ as $n \to \infty$.
Hope you guys can shed some light on this inquiry of mine. Let me thank you in advance for your continued support.
**References**
[**1**] L. Alaoglu, P. Erdős: *A conjecture in elementary number theory*, Bull. Amer. Math. Soc. 50 (1944), 881-882.
[**2**] *Mathematical Reflections*, Solutions Dept, Issue #3, 2009, page 23.
| https://mathoverflow.net/users/1593 | A limit involving the totient function | Everyone knows (but no one can prove) that there are infinitely many primes $p$ such that $q=2p-1$ is also prime. $\sigma(q)=q+1=2p$, $\phi(\sigma(q))=\phi(2p)=p-1$, $\phi(\sigma(q))/q=(p-1)/(2p-1)\to1/2$ as
$q\to\infty$.
**Edit**: I don't know why it didn't occur to me to look at Guy, Unsolved Problems In Number Theory. Under B42, he writes, "Makowski and Schinzel prove that $\limsup\phi(\sigma(n))/n=\infty$. The reference is A. Makowski, A. Schinzel, On the functions $\phi(n)$ and $\sigma(n)$, Colloq Math 13 (1964-65) 95-99, MR 30 #3870. I haven't found the paper on the web, but it's in Volume 2 of Schinzel's Selecta, 890-894. I don't have the energy to write out the proof in full, but here's the idea. Given $M$, choose $t$ such that
$$\prod\_{i=1}^t{p\_i\over p\_i-1}>M$$. Then given $p$ (and it's not clear to me whether $p$ is meant to be a prime), and letting $$n=\sigma\left(\prod\_{i=1}^tp\_i^{p-1}\right),$$ we get $$\sigma(n)=\prod\_{i=1}^tN(p\_i,p),$$ where $N(a,p)=(a^p-1)/(a-1)$. Now you prove $\limsup\_{p\to\infty}\phi(\sigma(n))/n\gt M$, using along the way a lemma which says that $$\lim\_{p\to\infty}{\phi(N(a,p))\over N(a,p)}=1.$$
| 7 | https://mathoverflow.net/users/3684 | 27628 | 18,060 |
https://mathoverflow.net/questions/3697 | 14 | Is there a 'classification' of the endofunctors F: Δ --> Δ where Δ denotes the simplex category with objects [n] and the weakly monotone maps [m] --> [n] as morphisms (Actually, I don't know if 'simplex category' is the right name)?
For instance there is a shift functor S: Δ --> Δ defined by S([n])=[n+1] on objects and S(d): [m+1] --> [n+1] being d on [m] and mapping m+1 to n+1 for a morphism d: [m]-->[n]. Hence for a simplicial set X one gets a path-object XoS.
| https://mathoverflow.net/users/467 | What are the endofunctors on the simplex category? | To carry Charles' train of thought further:
By an 'interval' let us mean a finite ordered set with at least two elements; let $Int$ be the category of intervals, where a morphism is a monotone map preserving both endpoints.
The simplicial set $\Delta^1$ can be viewed as a simplicial interval. That is, this functor $\Delta^{op}\rightarrow Set$ factors through the forgetful functor from $Int$ to $Set$. In fact, the resulting functor $\Delta^{op}\rightarrow Int$ is an equivalence of categories.
This extra structure (ordering and endpoints) on $\Delta^1$ is inherited by Charles' $K\_1=F^\*\Delta^1$; it, too, is a simplicial interval.
There aren't that many things that a simplicial interval can be. Its realization must be a compact polytope with a linear order relation that is closed. That makes it at most one-dimensional, and makes each component of it either a point or a closed interval. Simplicially each of these components can be either a $0$-simplex, or a $1$-simplex with its vertices ordered one way, or a $1$-simplex with its vertices ordered the other way, or two or more $1$-simplices each ordered one way or the other and stuck together end to end.
The three simplest things that a simplicial interval can be are: two points, a forward $\Delta^1$, and a backward $\Delta^1$. These arise as $F^\*\Delta^1$ for three examples of functors $F:\Delta\rightarrow \Delta$, the only examples that satisfy $F([0])=[0]$, namely the constant functor $[0]$, the identity, and "op".
It's clear that any functor with $F([0])=[n]$ has the form $F\_0\coprod\dots\coprod F\_n$ where $F\_i[0]=[0]$ for each $i$. This means that the corresponding simplicial interval can be made by sticking together those which correspond to the $F\_i$. For example, the 'shift' functor mentioned in the question is $id\coprod [0]$; Reid mentioned $id\coprod id$ and $op\coprod id$. These correspond respectively to: a $1$-simplex with an extra point on the right, two $1$-simplices end to end, and two $1$-simplices end to end one of which is backward. As another example, the constant functor $[n]$ corresponds to $n+1$ copies of (two points) stuck together end to end, or $n+2$ points.
In short, every functor $F:\Delta\rightarrow \Delta$ is a concatenation of one or more copies of $[0]$, id, and op. I can more or less see how to prove this directly (without toposes or ordered compact polyhedra).
| 17 | https://mathoverflow.net/users/6666 | 27638 | 18,064 |
https://mathoverflow.net/questions/27634 | 9 | This question is motived by this recent [question](https://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free/).
$K\_{0}(R)=\mathbb{Z}$ is often used as a euphemism for saying that every finitely generated projective module is stably free; however, there are some subtleties involved.
The statement that every finitely generated projective module is stably free is equivalent to saying that $K\_{0}(R)$ is generated by the isomorphism class of $R$. To show that the two are not the same, consider $R = M\_{2}(\mathbb{C})$, $2\times 2$ matrices over the complex numbers. Devissage tells us that $K\_{0}(R)=\mathbb{Z}$, with the unique simple module as a generator. However, every stably free projective has to have even length as an $R$ module, as length($R$)=2, and thus the unique simple module (projective as $R$ is semisimple) is not stably free.
Are there any commutative examples of this phenomena? More precisely, is there a commutative ring with $K\_{0}(R)=\mathbb{Z}$, but with f.g. projectives which aren't stably free? A commutative noetherian one?
| https://mathoverflow.net/users/3613 | $K_{0}(R) =\mathbb{Z}$ but some f.g. projective not stably free? | If $R$ is a commutative ring with $K\_{0}(R)=\mathbb{Z}$, then $\mathop{\rm Spec} R$ is connected, because otherwise $R$ would split as a product, and $K\_{0}(R)$ would contain a copy of $\mathbb{Z} \oplus \mathbb{Z}$. Hence there is a well defined surjective rank homomorphism $K\_{0}(R) \to \mathbb{Z}$, which must then be an isomorphism. Since $R$ has rank 1, it follows that the class of $R$ generates. This implies that every projective module is stably free.
| 11 | https://mathoverflow.net/users/4790 | 27645 | 18,068 |
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