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https://mathoverflow.net/questions/29593 | 7 | It is well discussed in many graph theory texts that it is somewhat hard to distinguish non-isomorphic graphs with large order. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. So, it follows logically to look for an algorithm or method that finds all these graphs.
A Google search shows that a paper by [P. O. de Wet](http://www.moreheadstate.edu/files/colleges/science/mcs/mejam/dewet.pdf) gives a simple construction that yields approximately $\sqrt{T\_n}$ non-isomorphic graphs of order n.
( $\{T\_n}$ being the number of labeled graphs of order n.)
So, I have the followings to ponder over:
(1) Are there such algorithms or has there been an improvement on the aforementioned algorithm?
(2) Where can I find a collection of non-isomorphic graphs of a given order?
If you allow me, I would also like to extend my question to connected graphs.
Many thanks.
(I am a beginner in Graph theory, so please give answers in not-very-specialized terms.)
| https://mathoverflow.net/users/5627 | Non-isomorphic graphs of given order. | Acknowledging Timothy’s comment, let me answer the question.
For a diagrammatic list of the non-isomorphic graphs (all in pdfs):
1. <http://keithbriggs.info/images/g4.pdf>
2. <http://keithbriggs.info/images/g5.pdf>
3. <http://keithbriggs.info/images/g6.pdf>
4. <http://keithbriggs.info/images/g7.pdf>
5. <http://keithbriggs.info/images/n8m9.pdf>
6. <http://keithbriggs.info/images/g10-8.pdf>
7. [Small graphs](http://www.research.att.com/~njas/sequences/a000088a.gif)
I [quote](http://keithbriggs.info/cgt.html) “The topologies were computed using the [nauty](http://cs.anu.edu.au/%257Ebdm/nauty) program by Brendan McKay and the layouts created with [graphviz](http://www.research.att.com/sw/tools/graphviz/). I wrote python programs to interface these and produce the pdfs.”
| 3 | https://mathoverflow.net/users/5627 | 29875 | 19,505 |
https://mathoverflow.net/questions/29851 | 2 | The question is about commutator in integral forms. Let $A$ an associative algebra over a field of characteristic zero, $x\in A$ and $k\in \mathbb Z$, we denote $x^{(k)}=\frac{x^{k}}{k!}$.
How to calculate brackets $[x\_\beta^{(n)}, x\_\alpha^{(m)}]$? Can I proceed doing the calculation for $[x\_\beta^{n}, x\_\alpha^{m}]$ and then multiply by $\frac{1}{n!m!}$ and try to rewrite the result in a divided power notation?
ADDED:
Let $g$ be a finite-dimensional simple Lie algebra and $\{\alpha\_i,...,\alpha\_n\}$ the set of positive roots. It is known that there exist an integral form for $U(g)$ and it is generated by $(x\_{\alpha}^-)^{(k)}$ where $k>0$ and $\alpha$ is a positive root.
| https://mathoverflow.net/users/40886 | hyperalgebras (positive characteristic) | As I understand, your question is about relations on the generators of the Garland integral form for the hyper loop algebra of $\mathfrak g$. Maybe the following papers will help you:
H.Garland, The arithmetic theory of loop algebras, J. Algebra 53 (1978), 480--551.
D. Jakelic, A. Moura, Finite-dimensional representations of hyper loop algebras, Pacific J. Math. 233 (2007), 371--402.
| 3 | https://mathoverflow.net/users/7174 | 29899 | 19,517 |
https://mathoverflow.net/questions/29546 | 7 | String topology studies the algebraic structure of the homology of the free loop space $LM = Map(S^1,M)$ of a oriented closed manifold. One aspect of this structure is that the pair $(H\_\ast(LM;\mathbb{Q}),H\_\ast(M;\mathbb{Q}))$ forms a open-closed HCFT with positive boundary (work of Godin). This means that there are operations coming from the homology of moduli space of Riemann surfaces with each connected component having a non-empty outgoing or free boundary. The conjecture (although Blumberg-Cohen-Teleman claim it is a theorem) is that in fact $H\_\ast(M;\mathbb{Q})$ should be seen has $H\_\ast(P(M,M),\mathbb{Q})$, where $P(M,M)$ is the space of paths starting and ending in $M$, and the HCFT-structure can be extended to include open-closed cobordisms which have open boundaries labelled with a closed oriented submanifold $N$ of $M$. This then gives operations not only for $H\_\ast(LM;\mathbb{Q})$ and $H\_\ast(M;\mathbb{Q})$, but also on $H\_\ast(P(N\_1,N\_2);\mathbb{Q})$ for any two closed compact oriented $N\_1$, $N\_2$ in $M$. This is called the full set of branes. For $N\_1 = N\_2$ a single point and $M$ connected, $P(N\_1,N\_2) = \Omega M$, the based loop space. *In general $H\_\ast(P(N\_1,N\_2);\mathbb{Q})$ will therefore not be finite-dimensional.*
On the other hand, Costello has proven a classification theorem of open-closed TCFT. In this we don't work with homology, but chains in the moduli space of Riemann surfaces, and chain complexes instead of (graded) vector spaces. Costello has proven that a open-closed TCFT can be constructed from an open TCFT (cobordism without incoming or outgoing boundary components equal to the circle) and that an open TCFT is equivalent to a Calabi-Yau $A\_\infty$ category. *One of the properties of a Calabi-Yau $A\_\infty$ category is that all hom-spaces are finite-dimensional, forced by a certain non-degenerate pairing.*
One can construct a HCFT from a TCFT by applying homology everywhere. I think this HCFT will in fact be positive (or negative?) boundary, because the TCFT is defined from open-closed cobordisms where each connected component has at least one incoming boundary component. **Is this correct?**
Furthermore, Costello conjectures in his paper that string topology (with the full set of branes) can be constructed as a TCFT, and applying homology then reduces to the HCFT given by Godin. But I can think of two reasons which make this conjecture seems false:
1) The naive choice of $C\_ast(P(N\_1,N\_2);\mathbb{Q})$ as Calabi-Yau $A\_\infty$ category is impossible, because these spaces will certainly be infinite-dimensional.
2) But no choice will work, because the homology of a finite-dimensional cell complex is finite-dimensional and we know some branes must be assigned infinite-dimensional spaces in the HCFT structure.
So my question is: **Is this reasoning enough to make my naive interpretation of Costello's conjecture false? If not, what is the mistake?**
| https://mathoverflow.net/users/798 | Can string topology be a open-closed TCFT with the full set of branes? | No, the conjecture is not false. The mistake is the following: I said "On the other hand, Costello has proven a classification theorem of open-closed TCFT." He proves a classification of open TCFT's and gives a construction of an open-closed TCFT from an open TCFT. String topology can't be obtained by this specific construction.
However, Lurie(-Hopkins) gives a more general construction in section 4.2 of the article on the classification of TQFT's. This can deal with string topology. However, the Calabi-Yau objects in this case only require a non-degenerate cotrace, and string topology fits in now.
| 4 | https://mathoverflow.net/users/798 | 29901 | 19,519 |
https://mathoverflow.net/questions/29913 | 4 | Given a map between two manifolds that induces an isomorphism on integral cohomology in the top dimension, it follows from naturality of the cup product and Poincaré duality and universal coefficient theorem that all maps on cohomology in every dimension are injective with torsion free cokernel.
Is there an example where one of these maps is not surjective?
| https://mathoverflow.net/users/6960 | map of manifolds inducing iso on top cohomology, but not surjective on one other cohomology group | For any orientable manifold M of dimension n, map M to $S^n$ by sending some suitable neighborhood of a point homeomorphically to $\mathbb{R}^n$ and the rest to $\infty$.
| 10 | https://mathoverflow.net/users/4183 | 29916 | 19,528 |
https://mathoverflow.net/questions/29578 | 2 | Given a conflict graph $G = (V, E)$, a man has to transport a set $V$ of items/vertices across the river. Two items are connected by an edge in $E$, if they are conflicting and thus cannot be left alone together without human supervision. The available boat has capacity $b\geq 1$, and thus can carry the man together with any subset of at most $b$ items. A feasible schedule is a finite sequence of triples $(L\_1, B\_1, R\_1),\dots, (L\_s, B\_s, R\_s)$ of subsets of the item set V that satisfies the following conditions (FS1)–(FS3). The odd integer $s$ is called the length of the schedule.
(FS1) For every $k$, the sets $L\_k, B\_k, R\_k$ form a partition of V . The sets $L\_k$ and $R\_k$ form stable sets in $G$. The set $B\_k$ contains at most $b$ elements.
(FS2) The sequence starts with $L\_1 \cup B\_1 = V$ and $R\_1 = \emptyset$, and the sequence ends with $L\_s = \emptyset$ and $B\_s\cup R\_s = V$.
(FS3) For even $k \geq 2$, we have $B\_k\cup R\_k = B\_{k-1} \cup R\_{k-1}$ and $L\_k = L\_{k-1}$. For odd $k \geq3$, we have $L\_k\cup B\_k= L\_{k-1}\cup B\_{k-1}$ and $R\_k = R\_{k-1}$.
Known Result: $VertexCover(G) \geq b \geq VertexCover(G)+1$.
Please help formulate this problem in MSO.
| https://mathoverflow.net/users/7094 | Monadic Second Order (MSO) logic on graphs | Maybe there is a solution. But, for that I assume there is an upper bound in the number of rounds needed, say n, and that the value b is fixed upfront. Then, there is the following EMSO formula,
$\exists L\_{1} \exists B\_{1} \exists R\_{1} ... \exists L\_{n} \exists B\_{n} \exists R\_{n} \phi(L\_{1},B\_{1},R\_{1} ...,L\_{n},B\_{n},R\_{n})$
where $\phi = Seq\_{1} \wedge Seq\_{2} ...\wedge Seq\_{n}$
$Seq\_{1}$=$\forall x (XOR(x\in L\_{1},x \in B\_{1})) \wedge empty(R\_{1}) \wedge$ $lessthanb(B\_1) \wedge IndSet(L\_{1})$
If i is even :
$Seq\_{i} = empty(L\_{i-1}) \vee (equals(L\_{i-1},L\_{i})\wedge
\forall x ((x \in B\_{k-1} \vee x \in R\_{k-1})$
$ \Rightarrow XOR(x \in B\_{k},x \in R\_{k})))
\wedge lessthanb(B\_i) \wedge IndSet(L\_{i}) \wedge IndSet(R\_{i})$
If i is odd (i $\geq$ 3):
$Seq\_{i}$ = $empty(L\_{i-1}) \vee (equals(R\_{i-1},R\_{i})\wedge \forall x ((x \in B\_{k-1} \vee x \in L\_{k-1})$
$ \Rightarrow XOR(x \in L\_{k},x \in B\_{k}))) \wedge lessthanb(B\_i) \wedge IndSet(L\_{i}) \wedge IndSet(R\_{i})$
$Seq\_{n} = empty(L\_{n-1})$
So if we can prove that if there is a solution then there is a solution of maximum n rounds which is dependent on the size of the graph then I think we have a solution.
empty(X) = $\forall x \neg(x\in X)$
lessthanb(X) = $\exists x\_{1} ... \exists x\_{b} (\wedge\_{i \neq j}\neg(x\_{i} = x\_{j})) \forall x (x \in X) \rightarrow (x=x\_{1} \vee... \vee x=x\_{b})$
IndSet(X) = $\forall x,y \in X \neg(R(x,y))$
| 4 | https://mathoverflow.net/users/7176 | 29918 | 19,530 |
https://mathoverflow.net/questions/29921 | 5 | Consider $M$, a positive definite matrix. Let $M^{(1)}$ be the diagonal matrix which agrees with $M$ on the diagonal ($M\_{ii}=M^{(1)}\_{ii}$). We have that $M^{(1)}$ is positive definite because it is diagonalizable and it has non-negative eigenvalues.
What about general bands? Let $M^{(b)}$ be the restriction of $M$ on a band: $M\_{ij}^{(b)}=M\_{ij}$ when $i$ and $j$ differ by less than $b$ in absolute value and $M^{(b)}\_{ij}=0$ for otherwise. Is $M^{(b)}$ positive definite for all $b$?
| https://mathoverflow.net/users/6154 | Are the banded versions of a positive definite matrix positive definite? | No. The matrix
$M = \begin{bmatrix}5 & 4 & 4 \\\\ 4 & 5 & 4 \\\\ 4 & 4 & 5\end{bmatrix} = \begin{bmatrix}2 & 2 & 2\end{bmatrix}\begin{bmatrix}2 \\\\ 2 \\\\ 2\end{bmatrix} + I$
is positive definite, but
$\begin{bmatrix}1 & -\sqrt{2} & 1\end{bmatrix}M^{(2)}\begin{bmatrix}1 \\\\ -\sqrt{2} \\\\ 1\end{bmatrix} = 20 - 16\sqrt{2}<20 - 22.4 < 0$,
so $M^{(2)}$ is not.
| 10 | https://mathoverflow.net/users/5963 | 29925 | 19,532 |
https://mathoverflow.net/questions/29935 | 3 | Let $k$ be a field, $L$, $H$ extension fields of $k$, and $G=L\otimes\_k H$. I wonder why (I want to know the proof but I can't find) the prime ideal of $G$ must be maximal, and its properties:
a) if $L$ is separable over $k$, then $G$ is reduced.
b) if $L$ is algebraic and purely inseparable over $k$, then $G$ has a unique prime ideal.
| https://mathoverflow.net/users/3945 | question about tensor of two fields | For (a): a field extension $L/k$ is separable is $L$ is a separable algebra, that is, if $L\otimes K$ is a semisimple algebra for all field extensions $K/k$. In particular, if $L$ is separable over $k$ and $K$ is an extension of $k$, then $L\otimes K$ will have no nilpotent elements because it is semisimple.
For details, see for example, Pierce's beautiful book Associative algebras.
For (b), for which the argument is more elaborate, see Jacobson's *Lectures in Abstract Algebra*, vol. 3, where the result is part 3 of Theorem 21; he does without algebraicity, by the way.
| 6 | https://mathoverflow.net/users/1409 | 29936 | 19,539 |
https://mathoverflow.net/questions/29825 | 1 | Assume I have a quantile function for an arbitrary probability distribution for random variable **x**.
Would the **x**-value corresponding to the 99th percentile be the same as the **x**-value corresponding to a p-value of 0.01 (one-sided test, right tail)?
**Details for my specific problem:**
I have fitted a gamma distribution to some experimental data and I am trying to calculate p-values (one-sided) for extreme observations in the right tail of the distribution. Since I have learned model parameters for the gamma distribution, I was hoping that I could use `qgamma` in **R** to calculate cutoffs for a given p-value significance threshold. Is this a sane thing to do?
| https://mathoverflow.net/users/7152 | What is the relationship between quantile functions and p-values | First I'll address your initial question without taking into account the details of the specific problem. The answer is "yes" if, and only if, the probability distribution is that of a test statistic, where the null hypothesis will be rejected if the test statistic is too big.
When you get into the details of your specific problem you become unclear. When you say "fitted a gamma distribution to some experimental data", it sounds as if you've got a sample and you're estimating the two parameters. Maybe some of the data points will fall above the 99th percentile. If 1% of them do so, that doesn't sound like a reason to reject any null hypothesis that I can think of. If a substantial proportion of them do so, I'd wonder about your method of fitting. If you have a null hypothesis that says something about the values of the parameters, then there's the question of what you're going to use as a test statistic, and then there's the question of what is the probability distribution of that test statistic if the null hypothesis is true. You haven't told us enough about your specific problem to make any guesses about those things.
| 1 | https://mathoverflow.net/users/6316 | 29943 | 19,542 |
https://mathoverflow.net/questions/29949 | 40 | In short, my question is:
>
> What is the shortest computer program for which it is not known whether or not the program halts?
>
>
>
Of course, this depends on the description language; I also have the following vague question:
>
> To what extent does this depend on the description language?
>
>
>
Here's my motivation, which I am sure is known but I think is a particularly striking possibility for an application to mathematics:
Let $P(n)$ be a statement about the natural numbers such that there exists a Turing machine $T$ which can decide whether $P(n)$ is true or false. (That is, this Turing machine halts on every natural number $n$, printing "True" if $P(n)$ is true and "False" otherwise.) Then the smallest $n$ such that $P(n)$ is false has low [Kolmogorov complexity](http://en.wikipedia.org/wiki/Kolmogorov_complexity), as it will be printed by a program that tests $P(1)$, then $P(2)$, and so on until it reaches $n$ with $P(n)$ false, and prints this $n$. Thus the Kolmogorov complexity of the smallest counterexample to $P$ is bounded above by $|T|+c$ for some (effective) constant $c$.
Let $L$ be the length of the shortest computer program for which the halting problem is not known. Then if $|T|+c < L$, we may prove the statement $\forall n, P(n)$ simply by executing all halting programs of length less than or equal to $|T|+c$, and running $T$ on their output. If $T$ outputs "True" for these finitely many numbers, then $P$ is true.
Of course, the Halting problem places limits on the power of this method.
Essentially, this question boils down to: What is the most succinctly stateable open conjecture?
EDIT: By the way, an amazing implication of the argument I give is that to prove any theorem about the natural numbers, it suffices to prove it for finitely many values (those with low Kolmogorov complexity). However, because of the Halting problem it is impossible to know which values! If anyone knows a reference for this sort of thing I would also appreciate that.
| https://mathoverflow.net/users/6950 | What is the shortest program for which halting is unknown? | There is a 5-state, 2-symbol Turing machine for which it is not known whether it halts. See
<http://en.wikipedia.org/wiki/Busy_beaver>.
| 59 | https://mathoverflow.net/users/2807 | 29955 | 19,550 |
https://mathoverflow.net/questions/29907 | 6 | Consider a connected symplectic manifold $(M, \omega)$ of dimension $m=2n$. A few preliminary reminders (mostly to fix the notation): A vector field $X$ is symplectic if its flow preserves the symplectic form, ie. $L\_X \omega = 0$, where $L\_X$ denotes Lie derivative with respect to $X$. The Cartan formula shows that this is equivalent to the 1-form $i\_X\omega = \omega(X, -)$ being closed. The Hamiltonian vector field associated to a smooth function $f$ is the vector field determined by $\omega(X\_f, -) = df$; any symplectic vector field is locally Hamiltonian. The questions I'm interested in are of local nature, so we don't have to worry about the distinction.
**Question 1:** Which differential forms are invariant under all Hamiltonian flows (meaning $L\_{X\_f}\alpha = 0$ for all smooth functions $f$)?
Clearly, the symplectic form itself generates a truncated polynomial algebra (isomorphic to $\mathbb{R}[x]/(x^{n+1})$) inside $\Omega^\*(M)$ which is invariant under all Hamiltonian flows. But is it possible that there are other than those? I believe I have shown that there are no invariant 1-forms using a horrible calculation in local (Darboux) coordinates, but I'm not sure if this method is suitable for higher degrees. In the even degrees, we know that the answer is not 0, and I can't see how to prove that an invariant $2d$-form is necessarily a constant multiple of $\omega^d$.
**Question 2:** What can one say about more general tensor fields on $M$? I am especially interested in the sections of the symmetric powers of $TM$ (ie. symmetric multi vector fields).
The proof that no $1$-forms are invariant is easily adapted to proving that no vector fields are invariant, but again, I'm not sure if this generalizes.
**Question 3:** Suppose we have a subalgebra $A\subset C^\infty(M)$ with the property that for each $p\in M$, $\{df\_p \mid f\in A\} = T^\*\_pM$ (in other words, the Hamiltonian vector fields associated to the functions in $A$ realize every tangent vector on $M$). Do the answers to Questions 1 and 2 change if we only insist that the forms/tensor fields should be invariant under the Hamiltionian vector fields associated to the elements of $A$?
It is certainly important that we still have a whole algebra of functions available; on $\mathbb{R}^{2n}$, any constant coefficient tensor field is invariant under the Hamiltonian vector fields associated to the coordinate functions $x\_j, y\_j$ (which, up to a sign, are just the corresponding coordinate vector fields $\partial/\partial y\_j, \partial/\partial x\_j$). The proof that no invariant vector fields exists also requires one to consider the Hamiltonians associated to $x\_j^2$ and $y\_j^2$.
| https://mathoverflow.net/users/4747 | Which tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields? | Any symplectic linear transformations in $T\_xM$ is locally realizable as a Hamiltonian vector field, thus for questions 1 and 2, one can profitably use representation theory of the symplectic group.
**FACT** (*Lefschetz decomposition*) Let $W$ be a $2n$-dimensional symplectic vector space, $\bigwedge^\ast W$ its exterior algebra, and $\omega\in\bigwedge^2 W$ the invariant two-form. Exterior multiplication by $\omega$ and the contraction with $\omega$ define a pair of $Sp(W)$-equivariant graded linear transformations $L, \Lambda$ of $\bigwedge^\ast W$ into itself of degrees $2$ and $-2,$ and let $H=\deg-n$ be the graded degree $0$ map acting on $\bigwedge^k$ as multiplication by $k-n.$ Then $L,H,\Lambda$ form the standard basis of the Lie algebra $\mathfrak{sl\_2}$ acting on $\bigwedge^\ast W$ and the actions of $Sp(W)$ and $\mathfrak{sl\_2}$ are the commutants of each other.
See, for example, Roger Howe, *Remarks on classical invariant theory*.
**Corollary** Every homogeneous $Sp(W)$-invariant element of $\bigwedge^\ast W$ is a multiple of $\omega^k$ for some $0\leq k\leq n.$
Since, conversely, every polynomial in $\omega$ is invariant under the Hamiltonian vector fields, this gives a full description of the invariant differential forms.
For question 2, locally every invariant tensor must reduce to an $Sp(W)$-invariant element of the tensor algebra. For the special case of symmetric tensors, the answer is trivial.
**FACT** Under the same assumptions, the $k$th symmetric power $S^k W$ is a simple $Sp(W)$-module (non-trivial for $k>0$).
General case can be handled using similar considerations from classical invariant theory. A more involved question of describing the invariant local tensor *operations* on symplectic manifolds (an analogue of the well-known problem of invariant local operations on smooth manifolds, such as the exterior differential or Schoutens bracket) was considered in an old article by A.A.Kirillov.
| 14 | https://mathoverflow.net/users/5740 | 29976 | 19,563 |
https://mathoverflow.net/questions/29978 | 1 | Do there exist nonconstant real valued functions $f$ and $g$ such that the expression:
$$f(x) -v/g(x)$$
is maximized at $x = v$ for all positive real $v$?
| https://mathoverflow.net/users/7192 | Do there exist nonconstant functions such that... | Take $f(x)=(x+1)e^{-x}$ and $g(x)=e^x$, then $f(x)-v/g(x)=(x+1-v)e^{-x}$ and the derivative with respect to $x$ is $(v-x)e^{-x}$.
| 6 | https://mathoverflow.net/users/5735 | 29985 | 19,569 |
https://mathoverflow.net/questions/24697 | 3 | Let $M$, $N$ be two modules over ring $A$. If $M\oplus M\cong N\oplus N$, can we conclude $M\cong N$? In the case that $M$, $N$ are completely decomposable (e.g. finite-length module by Krull-Schmidt Theorem), it is easy to show this must be true. Does the general case also hold?
| https://mathoverflow.net/users/6103 | Isomorphism between direct sum of modules | There are even counterexamples in the case $A = {\mathbb Z}$: at the end of B. Jónsson’s paper “On direct decompositions of torsion-free abelian groups,” *Math. Scand.* **5** (1957), 230–235, an example is given of torsion-free, finite-rank abelian groups $B \not\cong C$ such that $B \oplus B \cong C \oplus C$.
A further counterexample, which I believe has been pointed out independently by L. S. Levy, R. Wiegand, and R. G. Swan: let $A$ be the coordinate ring of the real 2-sphere and ${}\_AM$ the module for the tangent bundle; then $M \oplus M$ is free of rank $4,$ but $M$ is not free of rank $2$.
In the positive direction, K. R. Goodearl has proved (“Direct sum properties of quasi-injective modules,” *Bull. Amer. Math. Soc.* **82** (1976), no. 1, 108–110, Theorem 3) that if $M$ and $N$ are *quasi-injective* modules over a ring (commutative or not), then $M^n \cong N^n$ implies $M \cong N$ for any positive integer $n$.
Your question is related to an important open problem in noncommutative ring theory, the “separativity” problem for von Neumann regular rings: if $R$ is a von Neumann regular ring (or more generally an exchange ring), and $A$ and $B$ are finitely generated projective left $R$-modules with the property that $A \oplus A \cong A \oplus B \cong B \oplus B$, must we have $A \cong B$? An affirmative answer would resolve several major open problems, as explained in P. Ara, K. R. Goodearl, K. C. O’Meara, and E. Pardo’s paper “Separative cancellation for projective modules over exchange rings,” *Israel J. Math.* **105** (1998), 105–137.
| 9 | https://mathoverflow.net/users/6521 | 29986 | 19,570 |
https://mathoverflow.net/questions/29992 | 3 | Let $X$ be a simplicial set. Let $X\to \Delta^n$ be a right fibration (has the right lifting property with respect to right horn inclusions), and let $$\Delta^{\{n-i\}}\hookrightarrow \Delta^{\{n-i,\dots, n\}}\hookrightarrow \Delta^n$$ (for a fixed $i: 0\leq i\leq n$) be the obvious inclusion maps.
Then why is the induced map:
$$X\times\_{\Delta^n} \Delta^{\{n-i\}} \hookrightarrow X\times\_{\Delta^n}\Delta^{\{n-i,\dots, n\}}$$
a deformation retract? It's not like we can apply Whitehead's theorem, since $\Delta^n$ is not a Kan complex in general.
(This is from the end of the proof of proposition 2.2.3.1 of [HTT](http://arxiv.org/pdf/math/0608040v4). The statement should be true out of the context in the book with the hypotheses I've given, but if not, there's the source.)
| https://mathoverflow.net/users/1353 | Why is the induced map between pullbacks (of inclusions) by a right fibration a deformation retract? | In topology, if a map $Y\to B$ is some sort of fibration then you would think that $B$ being equivalent to a subspace $A$ would imply that $Y$ is equivalent to $Y\_A=Y\times\_BA$. If fibration means Serre fibration and equivalent means weakly, then you might want to use homotopy groups and the Whitehead Theorem. But what if we try to prove directly that a deformation retraction of $B$ to $A$ would yield a deformation retraction of $Y$ to $Y\_A$? Then we will want to assume that $Y\to B$ has the 'relative homotopy lifting property' for the pair $(Y,Y\_A)$, that is, the right lifting property with respect to the inclusion $Y\times 0\cup Y\_A\times I\to Y\times I$.
I suppose that this it how it is in what you are looking at: Any right fibration of simplicial sets has the RLP w.r.t. $K\times \lbrace 1\rbrace \cup L\times \Delta^1\to K\times \Delta^1$ for every $L\subset K$ (but not w.r.t. $K\times \lbrace 0\rbrace \cup L\times \Delta^1\to K\times \Delta^1$), and this is good for lifting the sort of deformation retraction that exists from a simplex to its zeroth vertex.
Something like that.
| 4 | https://mathoverflow.net/users/6666 | 30003 | 19,579 |
https://mathoverflow.net/questions/29508 | 15 | I have a sequence of centered independent random variables $X\_i$ that are all
bounded by one in absolute value. They are not identically distributed, though.
I would like to know if the **central limit theorem** is still true
for such a sequence. Putting $S\_n= X\_1+...+X\_n$, do we have
$$
c\_n = P(\ {S\_n\over\sigma(S\_n)} \in [a,b] ) -
{1\over \sqrt{2\pi}}\int\_a^b exp(-t^2/2) dt \ \rightarrow \ 0\ ?
$$
(let's assume $\sigma(S\_n)$ goes to infinity with n).
I guess it is true but I can't find a reference.
Also, what can be said from the rate of convergence of $c\_n$ ?
Since the $X\_i$ are uniformly bounded, does $c\_n$ goes to zero
exponentially fast ?
| https://mathoverflow.net/users/7082 | Is there a central limit theorem for bounded non identically distributed random variables? | *Theorem* (Billingsley, "probability and measure", example 27.4)
Let X\_i a sequence of independent, uniformly bounded random variables with zero mean, such that $\sigma(S\_n)$ goes to infinity with n. Then $S\_n/\sigma(S\_n)$ converges in law to the normalized Laplace-Gauss distribution.
This follows from the *Lindeberg triangular array theorem*. As pointed out in the other answers, the convergence can be slow. The *Bernstein inequality* may be used to bound the tail. Under the assumption of the previous theorem, for all n, we have
$$P(S\_n>t) \leq exp(-t^2/(\sigma^2(S\_n)+Ct/3))$$
where C is a bound for the |X\_i|'s.
| 10 | https://mathoverflow.net/users/6129 | 30022 | 19,593 |
https://mathoverflow.net/questions/29993 | 30 | What's wrong with defining the rank of a finitely generated module over any (commutative) ring to be just the smallest number of generators? All books I know define rank only locally this way. But why not define it globally?
| https://mathoverflow.net/users/5292 | Rank of a module | Since your profile says you are interested in Algebraic Geometry, here are geometric considerations that might appeal to you.
Consider a projective module $P$ of finite type over a commutative ring $A$. It corresponds to a locally free sheaf $\mathcal F $ over $X=Spec(A)$. The rank of $\mathcal F $ at the prime ideal $\mathfrak p$ is that of the *free* $A\_{\mathfrak p}$-module $\mathcal F\_{\mathfrak p}$.
The rank is then a locally constant function on $X$ and if $X$ is connected (this means that the only idempotents in $A$ are $0$ and $1$) it may be seen as an integer.
If $A$ is a domain, then $X$ is certainly connected and has a generic point $\eta$ whose local ring is the field of fractions $\mathcal O\_\eta=K=Frac(A)$. The rank of $\mathcal F $ or of $P$ is then simply the dimension of the $K$ vector space $P\otimes\_A K$.
Actually, if $A$ is a domain, this formula can be used to define the rank of any $A$-module $M$ (projective or not, finitely generated or not) : $rank(M)=dim\_K ( M\otimes\_A K) $ .
This is the definition given by Matsumura in his book *Commutative Rings*, page 84.
It corresponds to the **maximum** number of elements of $M$ which are linearly independent over $A$.
The **minimum** number of generators of $M$ (which started this discussion) is quite a different, but interesting invariant, which has been studied by Forster, Swan, Eisenbud, Evans,...
Geometrically it corresponds to the minimum numbers of global sections of $\tilde{M}$ which generate this sheaf at each point of $Spec(A)$.
**Elementary example:** Every non-zero ideal of a Dedekind domain is of rank one, can be generated by at most two elements and can be generated by one element iff it is principal.
If the Dedekind domain is not a PID there always exist non free ideals which thus cannot be generated by less than two elements.
**Bibliography**
Ischebeck and Rao have published a [monograph](http://books.google.fr/books?id=P82xkbGioL8C&pg=PA205&lpg=PA205&dq=projective+modules+over+dedekind+rings&source=bl&ots=4ydpUikP4b&sig=XepMnxgKJcTr5xzGEpJEr-Y5_aM&hl=fr&ei=8CwrTMPMFNmXOLaRucgD&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBcQ6AEwADgK#v=onepage&q=projective%20modules%20over%20dedekind%20rings&f=false) *Ideals and reality: projective modules and number of generators of ideals* on exactly this theme
| 43 | https://mathoverflow.net/users/450 | 30024 | 19,594 |
https://mathoverflow.net/questions/30025 | 1 | Let $F=\mathrm{GF}\left(p^k\right)$ be any finite field. Let $G$ be the group of all affine permutations on $F$ (i.e. permutations of form $x\mapsto ax+b$). Then the set of all functions from $F$ to $\bar{F}$ is a linear representation of $G$, where $g(f)(x)=f(gx)$.
What are all sub-representations of this representation? Is it possible to characterize them?
Note: that in this case $\mathrm{gcd}\left(\left|G\right|,F\right)$ not equal to $1$.
| https://mathoverflow.net/users/4246 | Sub-representations of the affine group | As Victor explained consider the functions $X^m$ where $X^m(\alpha)=\alpha^m$. As $m$ runs between $0$ and $p^k-1$, these functions form a basis of your space of functions. This is a nice wavy basis, i.e., its elements span one-dimensional subrepresentations under the multiplicative group.
Now you have to take the additive group into account. All you need to do is to use binomial formula on $(X+\alpha)^m$ and observe which non-zero $X^t$-s, you can get out. This depends on the $p$-th power in $m$.
In particular, as Victor pointed out, polynomials of degree less than $m$ will span a submodule. But there are more, for instance, polynomials of degree $p$ and zero. In general, you will be getting ***spans of $X^t$ with $t\leq m$ and $t$ is divisible by the $p$-th power present in $m$*** as well as the sums of these gadgets.
Hint: $(X+\alpha)^{p^sn}=(X^{p^s}+\alpha^{p^s})^n$
| 3 | https://mathoverflow.net/users/5301 | 30036 | 19,598 |
https://mathoverflow.net/questions/30031 | 25 | I made a passing comment under Max Alekseyev's cute answer to [this question](https://mathoverflow.net/questions/29926/3n-2m-pm-41-is-not-possible-how-to-prove-it) and Pete Clark suggested I raise it explicitly as a different question. I cannot give any motivation for it however---it was just a passing thought. My only motivation is that it looks like fairly elementary number theory but I don't know the answer.
OK so one problem raised in the question linked to above was "prove there are no solutions to $3^n-2^m=41$ in non-negative integers" and Aleksevev's answer was "go mod 60". It was remarked afterwards that going mod 601 or 6553 would also nail it. For example, modulo 6553 (which is prime), 3 has order 39, 2 has order 117, but none of the 39 values of $3^n-41$ modulo 6553 are powers of 2 modulo 6553.
My question (really just a passing remark) is:
Is there an integer $t$ such that the equation $3^n-2^m=t$ has no solutions in non-negative integers $m$, $n$, but for which there are solutions modulo $N$ for all $N\geq1$? (By which of course I mean that for each $N\geq1$ the equation is satisfied mod $N$ for some integers $m,n\geq0$ depending on $N$; I am not suggesting that $m$ and $n$ be taken modulo $N$ or are independent of $N$).
This for me looks like a "Hasse principle" sort of thing---in general checking congruences doesn't give enough information about solvability of the polynomial in integers and there are many examples of such phenomena in mathematics. As exponential Diophantine equations are harder than normal ones I would similarly expect the Hasse Principle to fail here, but others seemed to be more optimistic.
| https://mathoverflow.net/users/1384 | Proving non-existence of solutions to $3^n-2^m=t$ without using congruences | I believe this is closely related to the conjecture by Brenner and Foster [here](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-101/issue-2/Exponential-Diophantine-equations/pjm/1102724775.full). They ask if an exponential Diophantine equation of the form $$\sum \epsilon\_i p\_i^{m\_i}=t$$ where $\epsilon\_i=\pm 1$ can be solved using modular arithmetic. I don't know if your special case is any easier but maybe that's a good place to start looking in case there is anything in literature.
| 6 | https://mathoverflow.net/users/2384 | 30037 | 19,599 |
https://mathoverflow.net/questions/30030 | 26 | I've noticed that the Galois groups associated to Galois field extensions $L$ of a given field $K$ seem remarkably like a sheaf, with the field extensions taking the place of open set, and the Galois group $\mathrm{Gal}(L/K)$ of a field extension $L/K$ is the sections over $L/K$. This makes sense because we have a natural restriction map $\mathrm{Gal}(M/K) \to \mathrm{Gal}(L/K)$ when $M/L/K$. Furthermore, we can define an open covering of an extension $L/K$ as a collection of subextensions whose compositum is $L/K$, and then a collection of elements of the Galois group in each subextension which match up under restriction gives an element of $\mathrm{Gal}(L/K)$. This is a key ingredient in proving that the Galois group of an infinite extension is an inverse limit of finite Galois groups. Is there a way to formalize this?
It also has interesting ramifications for understanding the topology of the Galois groups. Two elements of an infinite Galois group of $L/K$ are close if they agree on larger and larger subextensions of $L/K$, i.e. if they agree on larger "open sets" in this topology. So there's some kind of interesting inverse relation between the topology on the infinite Galois group and the "topology" of the subextensions.
| https://mathoverflow.net/users/1355 | Galois Group as a Sheaf | You might introduce a Grothendieck topology on the Galois subextensions of $L/K$; then the Galois group is indeed a sheaf.
I just want to remark the following: There is a natural homeomorphism between $Gal(L/K)$ and $Spec(L \otimes\_K \overline{K})$. Namely, if $\sigma \in Gal(L/K)$, then the kernel of $L \otimes\_K \overline{K} \to L \otimes\_K \overline{K} \subseteq \overline{K} \otimes\_K \overline{K} \to \overline{K}$ is a prime ideal of $L \otimes\_K \overline{K}$. This is easily checked to be a homeomorphism if $L/K$ is finite, and then also in general. In particular, you can endow the profinite space $Gal(L/K)$ with a sheaf so that we get an affine scheme. If $L'/K$ is a Galois subextension of $L/K$, then the restriction $Gal(L/K) \to Gal(L'/K)$ comes from the inclusion $L \otimes\_K \overline{K} \subseteq L' \otimes\_K \overline{K}$. This provides another reason why $Gal(-/K)$ is a "sheaf", namely because $Spec$ commutes with filtered direct limits.
| 14 | https://mathoverflow.net/users/2841 | 30038 | 19,600 |
https://mathoverflow.net/questions/30042 | 6 | Let $\mathfrak{g} \subset \mathfrak{gl}\_n$ be one of the classical real or complex semisimple Lie algebras. If $g \in \mathfrak{g}$, then $g$ has a Jordan decomposition $g = g\_s + g\_n$ with $g\_s$ semisimple and $g\_n$ nilpotent, and $[g\_s,g\_n]=0$.
The elements $g\_s,g\_n$, which a priori are just in $\mathfrak{gl}\_n$, are both in $\mathfrak{g}$ again. There are various middle-brow general ways to see this (for one, use that $\mathfrak{g}$ is algebraic), but for concrete choices of $\mathfrak{g}$ it's basically elementary, as follows. One knows from the construction of the Jordan decomposition that $g\_s,g\_n$ are both polynomials in $g$ (different polynomials for different $g$, of course), and (EDIT) you can rig the construction so that these polynomials are odd. The Lie algebra $\mathfrak{g}$ is the subspace of $\mathfrak{gl}\_n$ cut out by conditions like $\mathrm{trace}(g)=0$, or $Jg = -g^{t} J$ for some matrix $J$, and so forth. The condition $\mathrm{trace}(g)=0$ is always true for $g\_n$, so it's true for $g\_s$ if true for $g$. The condition $Jg=-g^t J$ is visibly true for odd $p(g)$ if true for $g$, so if true for $g$ then it's true for both $g\_s$ and $g\_n$. Thus $g\_s$ and $g\_n$ visibly satisfy whatever conditions $g$ is required to satisfy, and so are contained in $\mathfrak{g}$.
(This might seem lowbrow but in fact I think this is basically the idea of the proof that Fulton-Harris give for general semisimple Lie algebras.)
Now suppose instead that $G$ is a real or complex linear Lie group with Lie algebra $\mathfrak{g}$. This time the Jordan decomposition is $g = g\_s g\_u$ with $g\_u$ unipotent, and indeed $g\_s$ and $g\_u$ are still in $G$. But if you try to make the same lowbrow argument as in the Lie algebra case, it appears to die horribly (a condition like $g^t = g^{-1}$ certainly need not be preserved by taking a polynomial in $g$). My question is, is there an elementary way to rescue it? (In particular, something other than just the general argument for algebraic groups.) Obviously you're fine for elements $g$ in the image of the exponential map, so the issue is passing to the whole group. A caveat is that I do $\textit{not}$ want to assume that $G$ is connected.
| https://mathoverflow.net/users/379 | Jordan decomposition in a classical group | Give the proof in Humphreys' "Linear Algebraic Groups". It is essentially a context-free version of the argument you give, and hinges only on the fact that if $\rho\_g$ is right-translation by $g$ in $k[\operatorname{GL}\_n]$ and $I$ is the ideal defining $G$ in $\operatorname{GL}\_n$, then $g \in G$ if and only if $\rho\_g(I) \subset I$. It is a simple fact of linear algebra that the semisimple and unipotent parts of $\rho\_g$ stabilize any subspace which $\rho\_g$ itself stabilizes. The intuition, of course, is that $I$ consists of all the "equations" defining $G$ in $\operatorname{GL}\_n$.
This may be the general argument you said you didn't want, in which case I think you should reconsider it as being just the right amount of generality on top of what you have done for Lie algebras.
| 2 | https://mathoverflow.net/users/6545 | 30050 | 19,606 |
https://mathoverflow.net/questions/30035 | 34 | Recall that the scalar curvature of a Riemannian manifold is given by the trace of the Ricci curvature tensor. I will now summarize everything that I know about scalar curvature in three sentences:
* The scalar curvature at a point relates the volume of an infinitesimal ball centered at that point to the volume of the ball with the same radius in Euclidean space.
* There are no topological obstructions to negative scalar curvature.
* On a compact spin manifold of positive scalar curvature, the index of the Dirac operator vanishes (equivalently, the $\hat{A}$ genus vanishes).
The third item is of course part of a larger story - one can use higher index theory to produce more subtle positive scalar curvature obstructions (e.g. on non-compact manifolds) - but all of these variations on the compact case.
I am also aware that the scalar curvature is an important invariant in general relativity, but that is not what I want to ask about here. This is what I would like to know:
1. Are there any interesting theorems about metrics with constant scalar curvature? For example, are there topological obstructions to the existence of constant scalar curvature metrics, or are there interesting geometric consequences of constant scalar curvature?
2. Can anything be said about manifolds with scalar curvature bounds (other than the result I quoted above about spin manifolds with positive scalar curvature), analogous to the plentiful theorems about manifolds with sectional curvature bounds? (Thus additional hypotheses like simple connectedness are allowed)
3. Is anything particular known about positive scalar curvature for non-spin manifolds?
| https://mathoverflow.net/users/4362 | Some questions about scalar curvature | The Kazdan-Warner theorem goes a long way toward answering the first and second questions.
(For notes typed up by Kazdan, see <http://www.math.upenn.edu/~kazdan/japan/japan.pdf>.)
Here's what is says (taken almost verbatim from the notes, page 93): Divide the class of all closed manifolds (edit: of dimension > 2. See comments) into 3 types:
I. Those which admit a metric of nonnegative scalar curvature which is positive somewhere.
II. Those which don't but admit a metric of 0 scalar curvature.
III. All other closed manifolds.
The theorem is that if $M$ is in class I, then any $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric.
If $M$ is in class II, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's identically 0 or negative somewhere.
If M is in class III, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's negative somewhere.
In particular, every closed manifold has a metric of constant negative scalar curvature. Those in class I or II have a metric of 0 scalar curvature, and those in class I have a metric of constant positive scalar curvature.
| 27 | https://mathoverflow.net/users/1708 | 30054 | 19,608 |
https://mathoverflow.net/questions/30051 | 2 | Let $S$ be a locally noetherian scheme, $Y$ a locally noetherien $S$-scheme and $X$ an abelian scheme over $S$. It is known that the map between groups $Hom(Y,X) \to Hom(Pic(X/S),Pic(Y/S)), f \mapsto f^\*$ is quadratic, i.e. we have
$(f+g+h)^\* - (f + g)^\* - (f + h)^\* - (h + h)^\* + f^\* + g^\* + h^\* = 0$.
However, $f \mapsto f^\\*$ is not linear. Is there an easy example for $(f+g)^\* \neq f^\* + g^\*$?
This is related to the order of the functor $X \mapsto Pic(X/S)$. The above inequality would include a nontrivial line bundle on $X \times\_S X$, which is trivial on the both closed subschemes $X \times 0, 0 \times X$. I'm also interested in an easy example for this phenomenon.
| https://mathoverflow.net/users/2841 | Picard functor is not linear | Let $E$ be an elliptic curve; take $f = g = \mathrm{id}\_E$. Then $f+g$ is multiplication by $2$, and has degree $4$; hence if $L$ is an invertible sheaf on $E$, the sheaf $(f+g)^\*L$ has degree $4 \deg L$, while $f^\*L \otimes g^\*L = L^{\otimes 2}$ has degree $2\deg L$.
| 5 | https://mathoverflow.net/users/4790 | 30056 | 19,610 |
https://mathoverflow.net/questions/30058 | 2 | Let $G$ be a finite abelian group. When $\prod\_{g\in G\setminus 1} (1-g)$ vanishes in (say, complex) group algebra of $G$?
It is easy to see that for cyclic group $G$ such product does not vanish, since $G$ may be embedded into a (complex) field.
Oh, it looks like it is obvious: for non-cyclic $G$ there is no exact complex representation of $G$, so for each character $\chi$ there exists $1\ne g\in G$ such that $\chi(1-g)=0$, so our product vanishes.
| https://mathoverflow.net/users/4312 | vanishing of certain product in group algebra | I think the answer is "if and only if the group $G$ is not cyclic". Why?
1) An element of $\mathbb C\left[G\right]$ is zero if and only if it acts as zero on each irreducible representation of $G$ (since $\mathbb C\left[G\right]$ is the direct sum of the endomorphism rings of the irreducible representations).
2) An element of $\mathbb C\left[G\right]$ acts on an irreducible representation of $G$ either as zero or as an automorphism (because each irreducible representation of $G$ is $1$-dimensional, since $G$ is abelian).
Hence, for a product of the form $\prod\_{g\in S}\left(1-g\right)$ to be zero, where $S$ is some ordered list of elements of $G$, it is necessary and sufficient that for each irreducible representation of $G$, there exists some $g\in S$ such that $1-g$ acts as zero on the representation, i. e. that $g$ acts as identity on the representation. Applied to a list $S$ containing all elements of $G\setminus 1$ (maybe several times), this means that the product $\prod\_{g\in S}\left(1-g\right)$ is zero if and only if no irreducible representation of $G$ is faithful. Easy manipulations with roots of unity show this to hold if and only if $G$ is not cyclic.
| 4 | https://mathoverflow.net/users/2530 | 30063 | 19,614 |
https://mathoverflow.net/questions/30064 | 5 | Hi,
I was wondering how much (if anything) $\mathcal{L}\_{PA}$ can express about individual nonstandard elements in a nonstandard model of PA. For instance, presumably it can say that each has $k$-many predecessors, for each $k\in\mathbb{N}$. But:
(a) I can't see that there is any way that the type of one element in a $\mathbb{Z}$-chain differs from the type of any other in that same chain. Is this correct?
(b) Are the types of elements in separate $\mathbb{Z}$-chains also identical? I mean, clearly they won't be the same as those of elements in the initial segment $\mathbb{N}$ - these will have a finite number of predecessors - but in two of the additional chains?
To me, it looks like these questions are straightforwardly true but I could be wrong. Many thanks,
Kate
| https://mathoverflow.net/users/7209 | Are the types of nonstandard natural numbers within a Z-chain identical? | Since the nonstandard numbers believe that every other number is even and every other number is odd, a fact that is expressible in the language you mention, it follows that the types are not the same for every two elements in a $Z$-chain. In fact, more is true: any two nonstandard natural numbers in a common $Z$-chain have distinct types, for if they differ by a finite number $n$, then they will have different residue modulo $n+1$, making their types different.
If one restricts attention only to the order, however, then any two elements in a nonstandard $Z$-chain have the same type, since there are order-automorphisms that shift within this $Z$-chain. And in a countable nonstandard model, the $Z$-chains are ordered like the rationals, and so the order automorphism group acts transitively on these elements. It follows that all the nonstandard elements have the same type in the language containing only the order.
| 13 | https://mathoverflow.net/users/1946 | 30076 | 19,619 |
https://mathoverflow.net/questions/29942 | 4 | You can map whole numbers to combinations when taking them in order. For example, 13 choose 3 would look like:
```
0 --> (0, 1, 2)
1 --> (0, 1, 3)
2 --> (0, 1, 4)
etc...
```
Given a particular combination, such as `(0, 3, 9)`, is there a way to determine which whole number maps to it (26, in this case), short of writing out all the combinations in order until I hit upon the proper one? Furthermore, is there a way of doing this when counting combinations with repetitions?
If anyone is wondering, this isn't homework, but for a personal programming project.
| https://mathoverflow.net/users/1646 | Order of a combination when mapping them to whole numbers | Let $N(n;a\_1,\dots,a\_k)$ where $0\leq a\_1 < a\_2 < \dots < a\_k < n$ be the order number of $(a\_1,\dots,a\_k)$ as a combination from ($n$ choose $k$).
Since there are exactly $\binom{n-1}{k-1}$ combinations with $a\_1 = 0$, we have a recurrence:
if $a\_1 = 0$, then
$$ N(n;a\_1,\dots,a\_k) = N(n-1;a\_2-1,\dots,a\_k-1)$$
if $a\_1 > 0$, then
$$N(n;a\_1,\dots,a\_k) = \binom{n-1}{k-1} + N(n-1;a\_1-1,a\_2-1,\dots,a\_k-1).$$
with initial condition $N(n;)=0$ (i.e., when $k=0$) for any $n$.
For example,
$$N(13;0,1,4) = N(12;0,3) = N(11;2) = \binom{10}{0} + N(10;1)$$
$$= \binom{10}{0} + N(10;1) = \binom{10}{0} + \binom{9}{0} + N(9;0)$$
$$=\binom{10}{0} + \binom{9}{0} + N(8;) = 1 + 1 + 0 = 2$$
as required.
**UPDATE**. In fact, there is a simpler way to enumerate combinations, using [combinatorial number system](https://en.wikipedia.org/wiki/Combinatorial_number_system) of degree $k$. A $k$-combination $0\leq a\_1 < a\_2 < \dots < a\_k < n$ here gets the order number:
$$\binom{a\_1}{1} + \binom{a\_2}{2} + \dots + \binom{a\_k}{k}.$$
The properties of combinatorial number system ensure that this representation is a bijective mapping between $k$-combinations of $n$ and the integers in the interval $[0,\tbinom{n}{k}-1]$. In particular, given an integer $m$ in this interval, its representation in the combinatorial number system of degree $k$:
$$m = \binom{a\_1}{1} + \binom{a\_2}{2} + \dots + \binom{a\_k}{k}$$
uniquely defines numbers $0\leq a\_1 < a\_2 < \dots < a\_k < n$ (the last inequality follows from $m<\tbinom{n}{k}$), i.e., a $k$-combination of $n$.
| 5 | https://mathoverflow.net/users/7076 | 30077 | 19,620 |
https://mathoverflow.net/questions/30072 | 10 | I believe that there is no common theory for finding roots of polynomial sum. In my case I have
$$P\_{n}(x)+AQ\_{n}(x)$$.
I am wondering how roots of this sum depend on $A$?
| https://mathoverflow.net/users/3589 | roots of sum of two polynomials | If they are complex polynomials or can be treated as such, then you could apply [Rouche's theorem](http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem), where the location of the zeros is determined by the dominant polynomial within the sum. (*"Walk the dog on the leash"*)
Possibly related: you could use the [Wronskian](http://en.wikipedia.org/wiki/Wronskian) to determine the values of A that make $P\_n(x)$ and $Q\_n(x)$ linearly independent.
Your question is related to [Mason's theorem](http://mathworld.wolfram.com/MasonsTheorem.html). There are a few papers which explore this specifically
1. *MR1923392 (2003j:30012) Kim, Seon-Hong . Factorization of sums of
polynomials. Acta Appl. Math. 73
(2002), no. 3, 275--284.*
2. *MR2103113 (2005h:30011) Kim,
Seon-Hong . On zeros of certain sums
of polynomials. Bull. Korean Math.
Soc. 41 (2004), no. 4, 641--646*
3. *MR1911767 (2003d:11036) Pintér, Á.
Zeros of the sum of polynomials. J.
Math. Anal. Appl. 270 (2002), no.
1, 303--305.*
| 9 | https://mathoverflow.net/users/5372 | 30086 | 19,628 |
https://mathoverflow.net/questions/30087 | 2 | The motivation behind this is to find the points of intersection between a ray and a level set of a potential function $g$, built in terms of a basic potential function $f$ (the building is explained later). This is a problem in ray tracing where the level set is known as a meta-ball. Normally, the basic potential function $f$ is of the form $\exp(-x)$ or $1/x^n$ where $n \geq 1$ and the the problem is solved by finding the roots of a (generally high order) polynomial by a numerical method. However my question is:
>
> Is there an alternative basic potential function $f$ such that the points of intersection between a ray and a level set of $g$ can always be found explicitly and without requiring numerical methods?
>
>
>
More formally:
Let $f \colon \mathbb{R}\_{\geq 0} \to \mathbb{R}\_{\geq 0} $ be continuous, we say that $f$ is a basic potential function iff:
1) $f$ is non-zero
2) $f$ is strictly decreasing
3) $\lim\_{x \to \infty} f(x) = 0$
Let $f$ be a basic potential function, $c\_1, \ldots, c\_n \in \mathbb{R}^3$ and $a\_1, \ldots a\_n \in \mathbb{R}$, then we build the potential function $g \colon \mathbb{R}^3 \to \mathbb{R}$ in the following way:
$$g(x) = \sum\_{i=1}^n a\_i f(\left| x - c\_i \right|)$$
Let $k \in \mathbb{R}$ then $S = g^{-1}(k)$ is our level set (meta-ball).
We describe a ray $R$ as a pair $(o, d) \in \mathbb{R}^3 \times S^3$, where $o$ is the origin of the ray and $d$ is the direction in which it is 'pointing'. Using this we may see that finding the points in $R \cap S$ is equivalent to solving $$h(t) = g(o + td) = k$$ (with $t \geq 0$)
Traditionally in ray tracing, the choice of basic potential function $f$ would mean that this could be solved by numerical methods, but is there a basic potential function $f$ such that the solutions of $h(t) = k$ can always be found explicitly (or an alternative way of finding the points in $R \cap S$ without requiring numerical methods)?
| https://mathoverflow.net/users/3121 | Root Finding for Raytracing (Ray and Meta-Ball Intersection) | While this isn't precisely high mathematics, I've done a fair bit of metaball modeling in the past. They're not a basic potential functions in the sense you use ($f$ is only non-zero on a bounded interval), but either the classic ease-in ease-out function $f(x) = 1+3x^2-2x^3 (x\leq1)$, $f(x) = 0 (x\geq1)$ (with $r^2$ being used as the value for $x$ here) or Wyvill's similar cubic satisfying $f(0) = 1$, $f(1) = 0$, $f'(1) = 0$, $f(1/4) = 1/2$ (so that when $r=1/2$, $f(r^2)=1/2$) allow for explicit solutions by explicitly solving the (separate) cubics on each piece of a given ray.
More broadly, though, I have to ask why you'd rather an explicit solution than numerical methods? Keep in mind that the sqrt operation you're going to get from your floating-point processor (which is likely to be involved in most explicit solutions) is calculated by numerical methods (usually a variant of Newton's algorithm, though there are other good sqrt algorithms) on the processor itself, and likewise something like the log function is probably done with tables, linear approximations and a Newton iteration or two. Even if your own solution is explicit, somewhere behind the works some approximation is likely being done...
| 5 | https://mathoverflow.net/users/7092 | 30095 | 19,632 |
https://mathoverflow.net/questions/30088 | 1 | Does the notion of $\omega$-monoid exist, analogous to the notions of $\omega$-groupoid and $\omega$-category? If so, some references would be appreciated.
This is an attempted rephrasing of question: [Chain/Hierarchy of Monoids](https://mathoverflow.net/questions/24723/chain-hierarchy-of-monoids). My application domain is reasoning about modifiers of modifiers in software product line engineering, thus lacking established mathematical background. I find it easier to adapt existing results, even if the application domain is significantly different, so any help would be appreciated.
| https://mathoverflow.net/users/2620 | $\omega$-monoids | Sure. A monoid is the same as a (pointed) category with a single object.
So an $n$-monoid is the same as a pointed $n$-category with a single object.
These usually go by names like $A\_\infty$-algebras (mostly if they are linear) or similar.
If you want *strict* $\infty$-monoids, then the notion of a strict $\omega$-category with a single object will be all you want. For more general notions, see the links at
<http://ncatlab.org/nlab/show/algebra+in+an+(infinity,1)-category>
| 3 | https://mathoverflow.net/users/381 | 30102 | 19,636 |
https://mathoverflow.net/questions/30026 | 4 | In my recent studies (fourier multipliers on weighted Lp spaces) I have to deal with this kind of transformation: if w is a measurable function on $R^n$, define
$w^\*(x)=\sup\_y \frac{w(x+y)}{w(y)}$.
Has anyone seen this transformation before? Has this interesting properties?
In particular I'm interested in functions such that w=w\* or $w^\*\leq cw$, like the exponential.
Thanks in advance.
UPDATE: I'm interested in $w$ continuous, positive and even
| https://mathoverflow.net/users/4928 | Is anything known about $w^*(x)=\sup_y w(x+y)/w(y)$ for measurable functions w on $R^n$ | Sort of a random thought: if $w = w^\*$, then $w$ must be positive. So take logs on both side. Since log is monotonic, it commutes with sup. Then we have $a = \log w$, and $$a(x) = \sup\_y a(x+y) - a(y)$$
which implies that
$$ a(x) + a(y) \geq a(x+y)$$
so $a$ must be sub-additive. Helge's comment tells us that $a(0) = 0$.
This looks necessary and sufficient. (As for sub-additive functions $a$ such that $a(0) = 0$, $a(x) + a(0) = a(x+0)$ is satisfied, so the sup is attained.)
| 4 | https://mathoverflow.net/users/3948 | 30103 | 19,637 |
https://mathoverflow.net/questions/30112 | 9 | Background
----------
I came up with this while trying to find a sort of high-level exposition of the exterior algebra of a vector space. Let $V$ be a vector space of dimension $n$ over $\mathbb{C}$, and let $k \in \mathbb{N}$. One picture of $\Lambda^k(V)$, the $k^{th}$ exterior power of $V$, is as the space of totally antisymmetric tensors in $V^{\otimes k}$.
This can be constructed as follows. Let
$$ \rho : S\_k \to \mathrm{End}(V^{\otimes k})$$
be the representation given by
$$ \rho\_\pi (v\_1 \otimes \dots \otimes v\_k) = v\_{\pi(1)} \otimes \dots \otimes v\_{\pi(k)},$$
and then let $\sigma$ be the alternating form of this representation, i.e. $\sigma\_\pi = sgn(\pi) \rho\_\pi$. The total antisymmetrizer is the map
$$A\_k = \frac{1}{k!} \sum\_{\pi \in S\_k} \sigma\_\pi.$$
This is the projection onto the space of totally antisymmetric tensors, and so we can calculate the dimension of $\Lambda^k(V)$ simply by taking the trace of the map $A\_k$. It turns out that
$$\mathrm{tr}(\rho\_\pi) = n^{cyc(\pi)},$$
where by $cyc(\pi)$ I mean the number of cycles in the factorization of $\pi$ into disjoint cycles (including cycles of length 1). This can be shown as follows. Take a basis $\{ e\_1, \dots, e\_n \}$ of $V$ and then form the basis for $V^{\otimes k}$ consisting of all vectors $ e\_{i\_1} \otimes \dots \otimes e\_{i\_k}$ such that $1 \le i\_1, \dots, i\_k \le n $. Then
$$ \rho\_\pi(e\_{i\_1} \otimes \dots \otimes e\_{i\_k}) = e\_{i\_{\pi(1)}} \otimes \dots \otimes e\_{i\_{\pi(k)}},$$
so this basis vector contributes 1 to the trace of $\rho\_\pi$ if and only if $i\_j = i\_{\pi(j)}$ for all $1 \le j \le k$, i.e. if and only if all labels are constant over cycles of $\pi$. Since there are $n$ choices for each label, this gives
$$\mathrm{tr}(\rho\_\pi) = n^{cyc (\pi)},$$
and thus
$$ \mathrm{tr}(A\_k) = \frac{1}{k!} \sum\_{\pi \in S\_k} sgn(\pi) n^{cyc (\pi)}.$$
---
>
> Question: does anybody know a simple combinatorial proof that
> $$ \frac{1}{k!} \sum\_{\pi \in S\_k} sgn(\pi) n^{cyc (\pi)} = \binom{n}{k},$$
> where (in case you didn't read the long-winded background that I wrote), $cyc(\pi)$ is the number of cycles in the disjoint cycle factorization of $\pi$.
>
>
>
| https://mathoverflow.net/users/703 | (Elementary?) combinatorial identity expressing binomial coefficients as an alternating sum over permutations. | $n(n-1)...(n-(k-1))$ is the number of injective functions from a set of size $k$ to a set of size $n$. We can count these using inclusion-exclusion: first include all such functions, of which there are $n^k$. Then, for each transposition $(ij)$ in $S\_k$, exclude all the functions such that $f(i) = f(j)$, of which there are $n^{k-1}$. And so forth. This alternating sum cancels out all functions which are invariant under a permutation of the domain, so the only ones left are the injective ones.
There's a much easier way to prove an equivalent identity, which is
$$\frac{1}{k!} \sum\_{\pi \in S\_n} n^{\text{cyc}(\pi)} = {n+k-1 \choose k}.$$
This identity is equivalent because the sign of a permutation is determined by the parity of its number of cycles, and it corresponds to replacing "antisymmetric" by "symmetric" everywhere in your question. But this identity has an obvious proof by Burnside's lemma: the LHS and RHS both count the number of orbits of functions $[k] \to [n]$ under permutation of the domain. (This is a special case of a result I call the [baby Polya theorem](http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/).)
Both identities are in turn a special case of the **exponential formula**, which one can state as a generating function identity for the cycle index polynomials of the symmetric groups. I explain some of this [here](http://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/). The relevant specializations are
$$\frac{1}{(1 - t)^n} = \exp \left( nt + \frac{nt^2}{2} + ... \right)$$
and
$$(1 + t)^n = \exp \left( nt - \frac{nt^2}{2} \pm ... \right).$$
| 16 | https://mathoverflow.net/users/290 | 30114 | 19,641 |
https://mathoverflow.net/questions/30113 | 19 | The classifying space of the nth symmetric group $S\_n$ is well-known to be modeled by the space of subsets of $R^\infty$ of cardinality $n$. Various subgroups of $S\_n$ have related models. For example, $B(S\_i \times S\_j)$ is modeled by subsets of $R^\infty$ of cardinality $i + j$ with $i$ points colored red and $j$ points colored blue. More fun: the wreath product $S\_i \int S\_j \subset S\_{ij}$ has classifying space modeled by $ij$ points partitioned into $i$ sets of cardinality $j$ (but these sets are not "colored").
My question: is there a geometric model, preferably related to these, for classifying spaces of alternating groups? [Note: since any finite group is a subgroup of a symmetric group one wouldn't expect to find geometric models of arbitrary subgroups, but alternating groups seem special enough...]
| https://mathoverflow.net/users/4991 | Geometric model for classifying spaces of alternating groups | $n$ linearly independent points in $R^\infty$ together with an orientation of the $n$-plane which they span.
| 22 | https://mathoverflow.net/users/284 | 30115 | 19,642 |
https://mathoverflow.net/questions/29982 | 18 | Let $n\geq 2$ be a positive integer. For the purposes of this definition, let a *colored graph* be a finite undirected graph in which each edge is colored with one of $n$ colors so that no vertex is incident with two edges of the same color. (Without loss of generality, we suppose that every vertex is incident with *exactly* one edge of each color; add loops wherever necessary.) If $G\_1$ and $G\_2$ are two colored graphs, we define a product $H=G\_1\times G\_2$ as follows.
* The vertices of $H$ are the ordered pairs of vertices $(v\_1,v\_2)$, where $v\_i$ is a vertex of $G\_i$.
* If $(v\_1,w\_1)$ and $(v\_2,w\_2)$ are edges of the same color in $G\_1$ and $G\_2$, respectively, then there is an edge (also of the same color) between $(v\_1,v\_2)$ and $(w\_1,w\_2)$ in $H$.
**Examples.** Consider $n=3$, the simplest interesting case (and the one that interests me most). Let $K'\_3$ be the complete graph on three vertices with an extra loop at each vertex, and let $K\_4$ be the complete graph on four vertices. (There is essentially only one way to to color each of these graphs.)
* $K'\_3\times K'\_3$ has two components: one is a copy of $K'\_3$ and the other has six vertices.
* $K'\_3\times K\_4$ is a connected graph with twelve vertices.
* $K\_4\times K\_4$ has four components, each of which is a copy of $K\_4$.
Does this have a name? Has it been studied? It seems plausible enough to me that this may be a well-studied thing. At the moment, I'm especially interested in necessary/sufficient criteria for the product of connected colored graphs to be connected, or more generally an efficient way to count (or at least estimate) the number of components of the product, but any information that exists is reasonably likely to be useful to me.
My motivation here comes from some problems I'm working on in combinatorial number theory. I have various (colored) graphs that I associate to a positive integer $N$, and in many cases the graph corresponding to $MN$ is the product in this sense of the graphs corresponding to $M$ and $N$ whenever $\gcd(M,N)=1$. The graphs at prime powers are much simpler than the general case.
| https://mathoverflow.net/users/2559 | Has this notion of product of graphs been studied? | If you wanted this for directed graphs, I would say this:
As Andy Putman suggests, look at Stallings Inventiones paper "The topology of finite graphs."
A directed graph colored with $n$ colors admits an obvious map to a colored oriented wedge of $n$ circles, $X$ say.
Given two directed colored graphs, the fiber product of the two maps to $X$ may be defined the same way you are defining your product, but where the condition being that there is an edge going from $(a,b)$ to $(c,d)$ colored $m$ if there is an edge from $a$ to $c$ colored $m$ and an edge from $b$ to $d$ colored $m$ (so the orientation is taken into account).
This graph is smaller than yours, but has the advantage of being the pullback of a simple diagram.
I think that you should be able to take care of the undirected case by thinking of each edge of your graph as two edges, one for each orientation, or some such device. There's a nice way to think about this in Gersten's paper "Intersection of finitely generated subgroups of free groups and resolutions of graphs" in the same issue of Inventiones---he talks about "ghost edges."
| 7 | https://mathoverflow.net/users/1335 | 30118 | 19,645 |
https://mathoverflow.net/questions/30120 | 3 | Hi I'm currently learning Hamiltonian and Lagrangian Mechanics (which I think also encompasses the calculus of variations) and I've also grown interested in functional analysis. I'm wondering if there is any connection between functional analysis and Hamiltonian/ Lagrangian mechanics? Is there a connection between functional analysis and calculus of variations? What is the relationship between functional analysis and quantum mechanics; I hear that that functional analysis is developed in part by the need for better understanding of quantum mechanics?
| https://mathoverflow.net/users/7223 | Functional Analysis and its relation to mechanics | (1) Depends on what you mean by Hamiltonian and Lagrangian mechanics.
If you mean the classical mechanics aspect as in, say, Vladimir Arnold's "Mathematical Methods in ..." book, then the answer is no. Hamiltonian and Lagrangian mechanics in that sense has a lot more to do with ordinary differential equations and symplectic geometry than with functional analysis. In fact, if you consider Lagrangian mechanics in that sense as an "example" of calculus of variations, I'd tell you that you are missing out on the full power of the variational principle.
Now, if you consider instead classical field theory (as in physics, not as in algebraic number theory) derived from an action principle, otherwise known as Lagrangian field theory, then yes, calculus of variations is what it's all about, and functional analysis is King in the Hamiltonian formulation of Lagrangian field theory.
Now, you may also consider quantum mechanics as "Hamiltonian mechanics", either through first quantization or through considering the evolution as an ordinary differential equation in a Hilbert space. Then through this (somewhat stretched) definition, you can argue that there is a connection between Hamiltonian mechanics and functional analysis, just because to understand ODEs on a Hilbert space it is necessary to understand operators on the space.
(2) Mechanics aside, functional analysis is deeply connected to the calculus of variations. In the past forty years or so, most of the development in this direction (that I know of) are within the community of nonlinear elasticity, in which objects of study are regularity properties, and existence of solutions, to stationary points of certain "energy functionals". The methods involved found most applications in elliptic type operators. For evolutionary equations, functional analysis plays less well with the calculus of variations for two reasons: (i) the action is often not bounded from below and (ii) reasonable spaces of functions often have poor integrability, so it is rather difficult to define appropriate function spaces to study. (Which is not to say that they are not done, just less developed.)
(3) See Eric's answer and my comment about Reed and Simon about connection of functional analysis and quantum mechanics.
| 5 | https://mathoverflow.net/users/3948 | 30127 | 19,649 |
https://mathoverflow.net/questions/30130 | 5 | Let $K$ be a field, $\alpha\in\bar{K}$, and $L/K$ a finite extension. How can we determine whether $\alpha\in L$, preferably in as much generality as possible?
Of course, there may be special cases where this is easy, e.g. $K\subset\mathbb{R}$ and $\alpha\in\mathbb{C}\setminus\mathbb{R}$. Another trick, using the field trace, is the described in [exercise 16 of Chapter 2](http://books.google.com/books?id=qMfGgJPCnVMC&lpg=PP1&dq=related%253AISBN0387902791&pg=PA41#v=onepage&q&f=false) of Marcus's *Number Fields*, though as far as I can tell this only works for the special case of radicals (since their traces are always 0).
There's only two general approaches I can think of at the moment: checking whether $\deg(\alpha)\mid [L:K]$, though this may not suffice; or somehow finding the minimal polynomial of $\alpha$ over $L$ (including proving that it is irreducible), which will be of degree 1 iff $\alpha\in L$ (not entirely sure how one would do this).
I'm wondering because I recently had the following messy situation: $K$ is the splitting field of a cubic $g\in\mathbb{Q}(t)[x]$, having $[K:\mathbb{Q}(t)]=3$, and $f\_1$, $f\_2\in K[x]$ are two cubics with splitting fields, $L\_1$ and $L\_2$ respectively, having $[L\_1:K]=[L\_2:K]=3$, and I wanted to know whether $L\_1=L\_2$. It would suffice to show $f\_1$ has a root in $L\_2$ or vice versa (since $L\_1=K($any root of $f\_1)$ and $L\_2=K$(any root of $f\_2)$, which led me to my question. Ultimately (and I still want to double-check my answer), I found that $L\_1=L\_2$, but I depended heavily on the specific properties of my $f\_1$, $f\_2$, and $g$.
| https://mathoverflow.net/users/1916 | Methods of showing an element is / is not in a field | Zev, I think in complete generality there is no good answer to the question. Just think about how to decide if a cubic is reducible. In principle it's easy: find a root! But unless the field has some special structure that makes it feasible to search for roots this may be easier said than done. This can be done on finite fields or ${\mathbf Q}$ (or other quotient rings of known basic UFDs), but on a totally general field what are you going to do?
Here are a few thoughts, but notice in each case they depend on knowing something meaty about the fields, so they're not "general" methods.
1. If you can find a known Galois extension containing your two fields, then check if they have the same fixed subgroup. In your case it may be circular to "use" Galois theory to decide if $L\_1 = L\_2$ since figuring out the Galois group of $L\_1L\_2$ over $K$ requires you already know how much $L\_1$ and $L\_2$ overlap. (A useful example: if I wanted to show ${\mathbf Q}(3^{1/8})$ and ${\mathbf Q}(48^{1/8})$ are not isomorphic then I know that they both lie in the field ${\mathbf Q}(3^{1/8},\zeta\_8)$, and I can compute its Galois group over ${\mathbf Q}$ and then compute the subgroups corresponding to those original two fields. The subgroups turn out to be nonconjugate, hence the two fields are not isomorphic.)
2. Since your fields are coming to us (well, to you) as extensions of ${\mathbf Q}[t]$, exploit possible ring-theoretic properties related to ${\mathbf Q}[t]$ and its integral extensions in your fields, not just plain field theory. In other words, think instead about comparing the integral closures of ${\mathbf Q}[t]$ inside your two fields instead of the fields themselves. I will go out on a limb here and assume your polynomials have coefficients that are integral over ${\mathbf Q}[t]$. Then splitting properties of irreducibles from ${\mathbf Q}[t]$ in the two integral closures might quickly tell them apart. For comparison, if you want to distinguish two finite extensions of ${\mathbf Q}$,
which have the same degree and the same number of real and complex embeddings, the next thing to look at is invariants connected to the integral closure of ${\mathbf Z}$ in the two fields: start comparing the splitting properties of 2, 3, 5, and so on. Usually
when the two fields are not isomorphic you'll quickly find a prime which splits in the two rings of integers in different ways (practically, this means minimal polynomials cutting out the two fields will factor in different ways mod $p$, and separably, for some prime $p$ pretty quickly). This could be described also in terms of $p$-adic embeddings, which has a parallel in your setting of extensions of ${\mathbf Q}(t)$, e.g., maybe one the cubics has a root in ${\mathbf Q}((1/t))$ (assuming $K$ itself can be stuffed in there) and the other does not.
3. Exploit other structures on your specific fields. For example, ${\mathbf Q}(t)$ has a nontrivial derivation on it, which will extend to the larger finite extensions of it.
If there were an intrinsic property of derivations which is not the same on the two fields, then they're not isomorphic.
In your case the two fields turned out to be the same. I don't know how you're going to succeed in proving that without directly showing they're equal (assuming it is circular to use Galois theory in this instance). All this invariant business will tell stuff apart, so I suppose if the discriminants are equal and then the first 30 irreducibles in ${\mathbf Q}[t]$ which you can write down turn out to factor in the same way in the integral closure of ${\mathbf Q}[t]$ in the two fields (yeah, like that'll be easy to do...) then you might suspect the fields are equal and should switch gears to try to show equality instead of nonequality.
| 11 | https://mathoverflow.net/users/3272 | 30134 | 19,654 |
https://mathoverflow.net/questions/30136 | 0 | I was reading a blog post on [a simple derivation of the cross product](http://behindtheguesses.blogspot.com/2009/04/dot-and-cross-products.html). I learned how to determine the area of a [parallelogram enclosed by two vectors $A$ and $B$](http://4.bp.blogspot.com/_D1sP-NndkqU/SfDq13JeaCI/AAAAAAAAHHc/Q_IyEv4AXkY/s1600-h/cross.gif).
First, here is the [proof](http://1.bp.blogspot.com/_D1sP-NndkqU/SfDq4A_2oSI/AAAAAAAAHHk/lWBYwv6Mh4s/s1600-h/decomp.gif) of the solution ($area = A\_x B\_y - B\_x A\_y$)
And here's the [geometric implication](http://4.bp.blogspot.com/_D1sP-NndkqU/SfDrt4Z-ZEI/AAAAAAAAHHs/yYqx4O3SCjQ/s1600-h/3dvec.gif) of the solution.
It bewilders me that the geometric solution is simple but nonintuitive.
One commenter (Tim Poston) on the blog offered an explanation by using the following lemma, which he states can be easily shown using elementary plane geometry, without the use of right angles or base\*height to find areas:
$C(u+w,v) = C(u,v) + C(w,v)$
where $u, v, w$ are arbitrary vectors, and $C$ is the area function of the parallelogram between 2 vectors.
Here's my illustration of an [example implication of this lemma](https://i.imgur.com/RleJK.png) (visually simplified to triangles without loss of generality)
I was unable to prove this lemma, so I couldn't follow the argument. Any thoughts or direction with this lemma or the original problem would be appreciated.
| https://mathoverflow.net/users/7229 | Why does the area function of a parallelogram have a nonintuitive geometric solution? | It is better to see this property if you stay with parallelograms and use the Cavalieri Principle. Here is [page](http://matematica-para-todos.wikispaces.com/Propriedade+de+Determinante+por+Cavalieri) with an animation showing what I mean (clink on the link and accept).
There is also a demonstration of the first formula using triangles and again using only Cavalieri. There was an animation "proof without words" for it in the site "art of problem solving" but I can't find it anymore.
| 0 | https://mathoverflow.net/users/7231 | 30137 | 19,656 |
https://mathoverflow.net/questions/30144 | 2 | This question is related to [Degree of sum of algebraic numbers](https://mathoverflow.net/questions/26832/degree-of-sum-of-algebraic-numbers). Forgive me if
this is a dumb question, but are there two algebraic numbers $a$ and $b$ of degree $3$ and $6$ respectively, such that the sum $a+b$ has degree $12$ ?
Intuitively it would seem that the degree of $a+b$ should divide $3 \times 6=18$, but I was unable to prove this. Hence my question.
| https://mathoverflow.net/users/2389 | algebraic numbers of degree 3 and 6, whose sum has degree 12 | The splitting field $K$ of $x^6-2$ has degree 12 over the rationals. Its Galois group is the dihedral group
$D\_{12}$. This group has subgroups $A$ and $B$, where $A$ has order 2, $B$ has order 4, and $B$ does not contain
$A$; the subgroup generated by $A$ and $B$ together is all of $D\_{12}$. Now let $E$ and $F$ be the fixed fields of $A$ and $B$ respectively; then $E$ has degree 6, $F$ has degree 3, and $F$ is not contained in $E$, so the smallest field containing both $E$ and $F$ is $K$. Now if you pick pretty much any $a$ in $F$ and pretty much any $b$ in $E$, you should have what you want.
| 17 | https://mathoverflow.net/users/3684 | 30145 | 19,661 |
https://mathoverflow.net/questions/30147 | 1 |
>
> **Possible Duplicate:**
>
> [Primes P such that ((P-1)/2)!=1 mod P](https://mathoverflow.net/questions/16141/primes-p-such-that-p-1-21-mod-p)
>
>
>
Motivation comes from comments in [this question](https://mathoverflow.net/questions/30101/composite-pairs-of-the-form-n-1-and-n1), and it is interesting in its own right. These primes are sequence [A055939](http://www.research.att.com/~njas/sequences/A055939) in OEIS.
So, which primes $p$ satisfy $p\\ |\\ (\frac{p-1}{2})! + 1$?
If my calculations (in sage) are correct, the following is true for all primes under 100,000. For $p > 3$:
$$p\\ |\\ (\frac{p-1}{2})! + 1 \iff h(\sqrt{-p})=1 \mod{4}$$
| https://mathoverflow.net/users/2024 | Primes p such that p | ((p-1)/2)! + 1 | Yes, this follows from the analytic class number formula.
See
<http://www.math.niu.edu/~rusin/known-math/97/sign> .
**Added** I have now found a reference. This is a theorem of Mordell:
L. J. Mordell,
The congruence $(p - 1/2)! \equiv \pm 1 (\operatorname{mod} p)$,
*American Mathematical Monthly*, **68** (1961), 145-146.
<http://www.jstor.org/stable/2312481>
| 4 | https://mathoverflow.net/users/4213 | 30148 | 19,663 |
https://mathoverflow.net/questions/30143 | 0 | Let $A$ and $B$ be closed sets (subsets of $\mathbb{R}^m$ and $\mathbb{R}^n$, say), and let $f : A \times B \rightarrow \mathbb{R}$ be a continuous function.
Consider the function $g : A \rightarrow B$ defined by
$g(x) = \underset{y \in B}{\operatorname{arg}\max} f(x,y)$
assuming some tie-breaking strategy for $f(x,y\_1) = f(x,y\_2)$. Clearly, $g$ may have discontinuities (but perhaps only countably many?).
If $A$ and $B$ are intervals of $\mathbb{R}$, $g$ corresponds to looking down the $y$-axis at the 3D graph of $f$, and marking the points on the "skyline".
Does this correspond to some known operation that has a established name?
My motivation is merely that I'm thinking about using this for a fun side project, so I'd like to know if it has a name and any known interesting properties beyond piecewise continuity(?). I've tagged this recreational accordingly; if it's an inappropriate question for MathOverflow, I apologize.
| https://mathoverflow.net/users/nan | Is there a name for the "projection" of a function under argmax? | If you do not use any tie breaking strategy, you simply get the *argmax-correspondence*. That's the name I know. If B is compact, so maximizers actually exist, the argmax-correspondence has nonempty and compact values and is [upper hemicontinuous](http://en.wikipedia.org/wiki/Hemicontinuity#Upper_hemicontinuity). This follows from the Berge maximum theorem. You can find some material [here](http://www.hss.caltech.edu/~kcb/Notes/Correspondences.pdf), the result holds in a much more general context than given in the link. Upper hemicontinuous correspondences do not necessarily allow for continuous selections.
In the special case where the correspondence is single valued (for example if $f(x,\cdot)$ is always strictly concave), you actually get a continuous function, however.
It might be useful to know that the *value function*, the function $v:X\to\mathbb{R}$ given by $v(x)=f(x,y^\* )$ with $y^\*$ being in the argmax of $f(x,y)$, is always continuous.
| 0 | https://mathoverflow.net/users/35357 | 30155 | 19,666 |
https://mathoverflow.net/questions/30186 | 6 | This question is based on the following phrase:
"In a sense, $\textrm{Spec} \ \mathbf{Z}$ looks topologically like a 3-dimensional sphere viewed as the Hopf fibration over $\mathbf{S}^2$."
See page 88 of *Algebraic Geometry II* by Shafarevich.
I find this remark very interesting but I can't seem to parse it.
I always just viewed $\textrm{Spec} \ \mathbf{Z}$ as an arithmetic analogue of $\mathbf{P}^1(\mathbf{C}) = \mathbf{S}^2$. This remark would add "something" to that in a sense.
| https://mathoverflow.net/users/4333 | The ring of integers looks like the 3-dimensional sphere viewed as the Hopf fibration | Various pieces of exposition and references are to be found - [here](http://golem.ph.utexas.edu/category/2007/10/this_weeks_finds_in_mathematic_18.html), [here](http://math.ucr.edu/home/baez/week257.html), [here](http://www.ucl.ac.uk/~ucahmki/baez13.12.pdf), and [here](http://golem.ph.utexas.edu/category/2009/04/afternoon_fishing.html).
| 5 | https://mathoverflow.net/users/447 | 30187 | 19,688 |
https://mathoverflow.net/questions/30179 | 9 | Let $M$ be a compact space and $f,g:M \to M$ whose non-wandering sets satisfy $\Omega(f)=\Omega(g)=M$. Can we have $\Omega(f \circ g)=M$?
Or more specifically, if $\Omega(f)=M$, can we have $\Omega(f^n)=M$ for any positive integral number $n$?
Any reference would be helpful. Thanks!
| https://mathoverflow.net/users/3926 | Is the composition of non-wandering maps still non-wandering? | To the first question the answer is negative. There are two homeomorphisms of the circle with irrational rotation number such that their composition is Morse-Smale (in fact, you can multiply two $2\times 2$ matrices with complex eigenvalues to get one hyperbolic matrix, the action on the proyective space does the trick). This implies that $\Omega(f)=\Omega(g)=S^1$ but $\Omega(f\circ g)$ consits of two points.
To the last question, the answer is yes (and it is important that $\Omega(f)=M$), by that I mean $\Omega(f)=\Omega(f^m)=M$ for every $m\geq 1$.
The proof I know goes as follows:
Consider a basis $\{A\_n\}$ of the topology of $M$ and $O\_n=$ {$x\in A\_n \ : \ f^j(x) \in A\_n \ for \ some \ j \geq 1$ } $\cup \overline{A\_n}^c$. The set $O\_n$ is open (because $A\_n$ is open) and dense since every point is non wandering (so, given an open set $U\subset A\_n$ there exists $x\in U$ such that for $j\geq 1$ we have $f^j(x) \in U \subset A\_n$). (Notice that if $\Omega(f)\neq M$ the set $O\_n$ could fail to be dense in $\Omega(f)$).
If $x\in R=\bigcap O\_n$ we get that $x$ is a limit point for $f$ (in fact, it will be recurrent since given a neighborhood $U$ of $x$, there is $x\in A\_n \subset U$, and since $x\notin \overline{A\_n}^c$ and $x\in O\_n$ we get that it has a future iterate in $A\_n$) and thus also for $f^m$. Notice that $x$ may a priori be not recurrent for $f^m$, but it will be a limit point.
Since the [limit set](http://en.wikipedia.org/wiki/Limit_set) is contained in the nonwandering, we obtain the result.
This aplies also to get that if $\Omega(f|\_{\Omega(f)})=\Omega(f)$ then $\Omega(f)=\Omega(f^m)$.
| 10 | https://mathoverflow.net/users/5753 | 30211 | 19,705 |
https://mathoverflow.net/questions/30004 | 2 | On page 43 of the pdf given as reference 2 at <http://en.wikipedia.org/wiki/AKS_primality_test>, the authors mention that this can be done in almost cubic time with Newton's method, although I can't figure out how this would work.
I do know about almost-linear time multiplication.
(this is theoretical enough that I'm guessing it goes here rather than on stackoverflow)
| https://mathoverflow.net/users/nan | Checking if a positive integer is a power other than a first power | I don't know how to do it in cubic time, but I suppose that, to use Newton's Method, you could do the following:
Find floor(log2*n*), and this is the largest "power" that it can be. Then, define:
$f\_k(x) = x^k - n$ where *n* is your number and *k* is the floor value, and iterate Newton's Method until you get a number whose floor is *m* that fits any of the following:
1) *m**k* < *n* && (*m*+1)*k* > *n*
2) *m**k* > *n* && (*m*-1)*k* > *n*
3) *m**k* == *n*
If the third is true, congrats, you've found yourself a root! Otherwise, reduce *k* by 1 and try again.
The problem with this is that I don't know how long it'll take for such a value of *m* to be found. Wikipedia says Newton's Method has a quadratic convergence, and you're making a fixed number of operations each iteration of the Method, so I guess its running-time would be pretty fast.
| 0 | https://mathoverflow.net/users/1982 | 30216 | 19,709 |
https://mathoverflow.net/questions/30217 | 1 | This might be a very silly question, but I just wanted to make sure I have all the right steps.
Suppose we have a univariate continuous random variable $X$, with some pdf and cdf ${{f}\_{X}}(x)$ and ${{F}\_{X}}(x)$, respectively. Now look at the transformation $Y = X + k$, with $k\in \mathbb{R}$. Then, the cdf and pdf of $Y$ are ${{F}\_{Y}}(y)={{F}\_{X}}(y-k)$ and ${{f}\_{Y}}(y)={{f}\_{X}}(y-k)$.
Does this imply that $Y$ has the same distribution type as $X$? In other words, does a translation (shift) either to the left or to the right of the random variable preserve its distribution type (e.g. a translated Normal variate obviously remains Normal, but is this true of any distribution? It would seem obvious that the answer is "yes" (basically I'm taking the shape and moving it without distortions), but I've yet to see a reference on this yet. Any suggestions? Maybe it's so trivial that nobody bothered.
On the other hand, if the answer is "not necessarily", is it just because the domain shifts as well (e.g. shift an Exponential distribution to the right some amount $k$, and now the domain changes from $[0, \infty]$ to $[k, \infty]$, therefore the translated $Y$ is not technically "Exponential", even though the pdf ${{f}\_{Y}}(y)=\lambda e^{-\lambda (y-k)}$ is that of an Exponential r.v.?)
| https://mathoverflow.net/users/7254 | Does the translation of a random variable preserve its distribution type? | Traditionally, the distributions of r.v.'s $X$ and $Y$ are said to be of the same type if there are constants $a>0$ and $b\in\mathbb{R}$ such that the distribution of $aX+b$ coincides with that of $Y$, see, e.g., p.31 in "A modern approach to probability theory" by Bert Fristedt,Lawrence F. Gray (look for it at google books). According to this definition your statement is trivially true. Of course, it is precisely meant to say that dilations and translations should not destroy the type of the distribution.
| 2 | https://mathoverflow.net/users/2968 | 30228 | 19,714 |
https://mathoverflow.net/questions/30191 | 6 | Does anybody know about software that exactly calculates the tree-width of a given graph and outputs a tree-decomposition? I am only interested in tree-decompositions of reasonbly small graphs, but need the exact solution and a tree-decomposition. Any comments would be great. Thanks!
| https://mathoverflow.net/users/6726 | Software for Tree-Decompositions | For general graphs there are no good algorithms known, as the problem of determining the treewidth of a graph is NP-hard. So if your graphs are not from some special class, and instances are small, then a brute force search over all decompositions of small width is a reasonable approach.
As a previous answer suggested, Röhrig's Diplomarbeit ranks highly in a Google search. His rather negative conclusion in 1998 was that when treewidth exceeds $4$, brute force enumeration was essentially the only realistic option; up to $4$ special-purpose algorithms were reasonable. This is not that surprising, as (intuitively speaking) iterating over all choices of bags of up to $k$ elements takes $\Omega(n^k)$ time, so the runtime grows quite fast.
Do your graphs have some special features? The [ISGCI](http://wwwteo.informatik.uni-rostock.de/isgci/) has many special graph classes, for some of which it is possible to find join-trees efficiently. (Join-tree decompositions are another name for tree decompositions, although this term nowadays seems to usually refer to join trees as used in Bayesian networks.)
Taking a really high level perspective, do you really want to compute tree decompositions? If you are decomposing trees because you need to do something with them, then consider an easier-to-compute decomposition. For instance, the modular decomposition can be computed in linear time, and also guarantees fast algorithms for many problems via the modular decomposition tree. There is a [Perl implementation](http://search.cpan.org/~azs/Graph-ModularDecomposition/) of an older algorithm, [Nathann Cohen](http://www-sop.inria.fr/members/Nathann.Cohen/tut/Graphs/) is currently working to incorporate a more recent C implementation into the [Sage](http://www.sagemath.org/) framework, or you could use [Fabien de Montgolfier's C code](http://www.liafa.jussieu.fr/~fm/algos/index.html) directly if you read French (the papers describing the work are in English).
If you really do need tree decompositions, then have a look at the simple approach via induced width, which can be easily implemented (and parallelized) by considering each possible vertex ordering, then checking the induced width it corresponds to. Section 2.3 of Rina Dechter's draft version of her chapter from the Handbook of Constraint Programming is quite useful as a starting point.
| 5 | https://mathoverflow.net/users/7252 | 30230 | 19,715 |
https://mathoverflow.net/questions/3428 | 1 | I am looking for a way to partially "grid" the surface of a sphere to have certain nice properties which will be defined precisely below.
* The areas should be "almost equal".
* It should be possible to calculate in constant time what grid cell any point belongs to (including the boundaries, see below)
* I want to ensure that no point has "too many grid cells that are close".
* I want a constant time lower bound approximation (or exact solution!) for determining minimum distance between any point and a cell
**Definition of what I mean by gridding**
* Each grid cell is an "area" on the sphere. We will assume it is a reasonably well behaved mathematical object - ie. the boundary does not self intersect, the length of the boundary is finite and the grid cells are closed. Does this have a proper mathematical name?
* We wish to cover an area C with grid cells. C consists of "almost all" the sphere, ie. all except for possibly a disc of unspecified size. We are allowed to cover more than just C.
* Each point on the surface of the sphere "belongs" to exactly one grid cell. If it is in the interior of a grid cell, then it "belongs" to that cell. If it is on a boundary, it "belongs" to exactly one of those cells. To clarify, "belonging" is a function from each point on C to the set of grid cells.
**Definition of "too many close grid cells"**
Suppose there are n grid cells. Let r be the radius of a disc with area 1/n of the total area of the sphere. I want only a few grid cells with lower bound approximation <=r from any point. To be precise, I would like this to be asymptotically less than sqrt(n), preferably constant.
**Distance between a cell and a point**
Defined as the minimum distance between a point and any point "belonging" in the cell.
**Almost equal areas**
I want there to be constants `0<e<1<f` so that the area of each grid square is `e/n<a<f/n`
**Observations**
* If I was trying to cover a square on a plane, then normal gridding would trivially solve this. I think that it might be possible if I could wrap a normal grid around a sphere or project it or something. I'm not really sure.
* Latitude and longitude grid lines will fail due to too many squares meeting at the poles.
| https://mathoverflow.net/users/494 | Grid with nice mathematical properties | Unfortunately, the sphere is not a "developable surface". This fact has annoyed map-makers for more than a millennium.
I find your focus on "cells" fascinating.
Most people seem fixated on trying to get points on the globe to correspond with points on a flat image, and don't seem concerned about dividing it up into areas.
The simplest way to convert (lat,long) coordinates to standardized areas is the ["Natural Area Code"](http://en.wikipedia.org/wiki/Natural_Area_Code), but it has the same problems near the poles as latitude and longitude.
Your "too many close grid cells" and "Distance between a cell and a point" criteria reminds me of the "Thomson problem".
I suspect you're trying to rule-out map projections like the Gall–Peters projection that, while they do have nice "equal area" properties, end up having a hundred little squares at the top and bottom of the on the map projected to a hundred long, narrow pie-slices that all touch a pole of the globe.
Perhaps you could pick one of the known [solutions to the "Thomson problem"](http://en.wikipedia.org/wiki/Thomson_problem) to build a nice grid. Most of those solutions look similar to a geodesic sphere -- but there are a few exceptions.
Perhaps the most famous application for "almost equal" patches is the Cosmic Background Explorer (COBE), which has inspired several mappings:
* the [COBE sky cube](http://en.wikipedia.org/wiki/Quadrilateralized_spherical_cube) (quadrilateralized spherical cube) -- an [equal-area projection of the sphere onto a cube](http://www.progonos.com/furuti/MapProj/Normal/ProjPoly/projPoly2.html) -- the 6 large squares divide easily into a nice square grid.
* [An icosahedron-based method for pixelizing the celestial sphere](http://space.mit.edu/home/tegmark/icosahedron.html) -- an equal-area projection of the sphere onto an icosahedron -- the 20 large equilateral triangles divide easily into a nice equilateral triangle grid; or into a nice regular hexagonal grid.
* the HEALPix projection
The COBE "bins" are, as far as I can tell, a synonym for your "cells".
Are those COBE-inspired mappings adequate for you?
| 5 | https://mathoverflow.net/users/7234 | 30232 | 19,717 |
https://mathoverflow.net/questions/30248 | 11 | The title basically says it all.
Is there a group with more than one element that is isomorphic to the group of automorphisms of itself?
I'm mainly interested in the case for finite groups,
although the answer for infinite groups would still be somewhat interesting.
| https://mathoverflow.net/users/nan | Is there a non-trivial group G isomorphic to Aut(G)? | The automorphism group of the symmetric group $S\_n$ is (isomorphic to) $S\_n$ when $n$ is different from $2$ or $6$. In fact, if $G$ is a **complete** group you can ascertain that $G \simeq \mathrm{Aut}(G)$. The reverse implication needn't hold, though.
| 29 | https://mathoverflow.net/users/1593 | 30250 | 19,729 |
https://mathoverflow.net/questions/30224 | 1 | I was not sufficiently clear on my last attempt at asking a similar (but not identical) question. Tom Goodwillie mentioned (in the accepted answer) that the question can be reduced to this one and mentioned a fact of which I was unaware, but after trying to prove this for the past day, I have returned to ask for a sketch of the proof:
Let $X$ be a simplicial set, and let $p:X\to \Delta^n$ be a right fibration (has the right lifting property with respect to all right horn inclusions (equivalently, it has the right lifting property with respect to the maps $$\Delta^1\times A\coprod\_{\{1\}\times A}\{1\}\times B\to \Delta^1\times B$$ for any inclusion $A\hookrightarrow B$). (These generate the same weakly saturated class of maps)).
Let $i\_0:\{0\}\hookrightarrow \Delta^n$ be the inclusion of the 0th vertex of $\Delta^n$. We'd like to show that the pullback of this map by $p$ (i.e. the induced map $f:X\times\_{\Delta^n} \{0\}\hookrightarrow X$) is a deformation retract.
According to Tom, we can use the second characterization of right fibrations to obtain some kind of lifting of the homotopy and retraction for $i\_0$ (which is a deformation retract).
Would somebody mind sketching the proof?
Edit: If this question is answered by 2AM EST (roughly three hours and forty-five minutes from the time of this current edit), I will award the answerer with a 450 point bounty as soon as it becomes possible (one must wait 48 hours from when the question was asked).
| https://mathoverflow.net/users/1353 | Proof Sketch: The pullback of the inclusion of the 0th vertex into the standard n-simplex by a right fibration is a deformation retract (450 point bounty if answered by 2am EST) | Harry, let $Y$ be the fiber of $X\to\Delta^n$ over the 0th vertex. The sense in which $Y$ is going to be a deformation retract of $X$ is going to be the following: There is a map $\Delta^1\times X\to X$ such that (1) on $1\times X$ it's the identity and (2) on $\Delta^1\times Y$ it's the constant homotopy (i.e. projection to $Y$ followed by inclusion) and (3) it takes $0\times X$ to $Y$. This is the same sense in which the 0th vertex is a deformation retract of $\Delta^n$.
So make a square in which the upper left object is $\Delta^1\times Y\cup 1\times X$, the lower left is $\Delta^1\times X$, the left map is inclusion, the right-hand map is the given right fibration $p:X\to \Delta^n$, the upper map is inclusion into $\Delta^1\times X$ followed by projection of $\Delta^1\times X$ to $X$, and the lower map is $Id\times p$ to $\Delta^1\times \Delta^n$ followed by the deformation retraction to $\Delta^n$.
A slanting arrow $\Delta^1\times X\to X$ making the triangles commute will have the desired restriction to $\Delta^1\times Y\cup 1\times X$ because of the upper triangle, and will take $0\times X$ into $Y$ because of the lower triangle.
| 4 | https://mathoverflow.net/users/6666 | 30260 | 19,736 |
https://mathoverflow.net/questions/30290 | 2 | Let $A$ be a $C$-algebra, where $C$ is a commutative ring with $1$, and $M$ be a finitely generated left $A$-module.
Question: Is it true that we can always find a positive integer $n$, a $C$-subalgebra $B$ of $M\_n(A)$ and an ideal $J$ of $B$ such that $B/J$ is isomorphic to $End(M)\ ?$ If not, what other conditions are needed to make the statement true?
By isomorphism I mean a $C$-algebra isomorphism. $M\_n(A)$, as always, is the algebra of $n\times n$ matrices with entries from $A$. By $End(M)$ I mean the algebra of $A$- homomorphisms from $M$ to $M$.
So I'm looking for a homomorphism from some $C$ subalgebra $B$ of $M\_n(A)$ onto $End(M)$. Well, I know that there exists a natural surjection $f$ from $A^n$ to $M$ for some positive integer $n$ because $M$ is finitely generated over $A$.
One way to define a map $g$ from $M\_n(A)$ to $End(M)$ is to define $g(a)(m)=f(ax)$, for all $a \in A$ and $m \in M$, where $x$ is any element of $A^n$ with $f(x) = m$. Ok, this map has obviously the well-definedness issue and that prevents $g$ to be defined on the whole $M\_n(A)$. So, we choose B to be the set of those elements $a \in M\_n(A)$ such that $f(ax)=0$, for all $x$ from the kernel of $f$. Now $g$ is well-defined on $B$ and $B$ is a $C$-subalgebra of $M\_n(A)$. What I'm having trouble with is to show that $g$ is surjective!
PS. I need the above to show that if $S$ is a subalgebra of $R$ and $R$ is finitely generated as $S$-module, then the Gelfand Kirillov dimension of $R$ and $S$ are equal. I didn't know how to prove it directly using the definition. So if anybody knows a direct proof, that would also be great.
Thanks.
| https://mathoverflow.net/users/6941 | Algebra of endomorphisms of f.g. modules as subquotients of matrix algebras | Let $e\_1, \dots, e\_n$ be a basis for $A^{\oplus n}$ and let $m\_1, \dots, m\_n$ be generators for $M$ so that your map $f: A^{\oplus n} \twoheadrightarrow M$ has $f(e\_i) = m\_i$. Now, suppose $\varphi \in \mathrm{End}\_A(M)$.` For each $i$, choose $a\_{i,j} \in A$ such that $$\varphi(m\_i) = \sum\_j a\_{i,j} m\_j.$$
(Obviously, this choice is not necessarily unique.) Now, define $\tilde{\varphi} \in \mathrm{End}\_A(A^{\oplus n})$ by setting
$$\tilde{\varphi}(e\_i) = \sum\_j a\_{i,j} e\_j.$$
You can show that $f \circ \tilde{\varphi} = \varphi \circ f$, which means that $\tilde{\varphi} \in B$ and $g(\tilde{\varphi}) = \varphi$.
| 2 | https://mathoverflow.net/users/6401 | 30295 | 19,753 |
https://mathoverflow.net/questions/30292 | 16 | One can view a random walk as a discrete process whose continuous
analog is diffusion.
For example, discretizing the heat diffusion equation
(in both time and space) leads to random walks.
Is there a natural continuous analog of discrete self-avoiding walks?
I am particularly interested in self-avoiding polygons,
i.e., closed self-avoiding walks.
I've only found reference
(in Madras and Slade,
*The Self-Avoiding Walk*, p.365ff)
to continuous analogs of "weakly self-avoiding walks"
(or "self-repellent walks") which discourage but
do not forbid self-intersection.
I realize this is a vague question,
reflecting my ignorance of the topic.
But perhaps those knowledgeable could point me to the
the right concepts. Thanks!
**Addendum**.
[Schramm–Loewner evolution](https://en.wikipedia.org/wiki/Schramm%E2%80%93Loewner_evolution) is the answer. It is the conjectured scaling limit of the self-avoiding walk and several other related stochastic processes. Conjectured in low dimensions,
proved in high dimensions, as pointed out by Yuri and Yvan. Many thanks for your help!
| https://mathoverflow.net/users/6094 | Random walk is to diffusion as self-avoiding random walk is to ...? | In 2D the scaling limit is believed to be SLE with parameter 8/3. This was conjectured by Lawler, Schramm and Werner and, to the best of my knowledge, still remains open.
| 15 | https://mathoverflow.net/users/2968 | 30297 | 19,755 |
https://mathoverflow.net/questions/30302 | 20 | If an algebraic variety $X$ over a field characteristic p is given by equations $f\_i(x\_1,...,x\_k) = 0$, we can consider the variety $X^{(p)}$ obtained by applying p-th powers to all the coefficients of all $f\_i$'s.
Frobenius morphism, as I understand it, is a morphism $X \to X^{(p)}$,
given on points as raising all coordinates to p-th power.
Can anyone please explain me, what is the geometric Frobenius, as opposed to the arithmetic one?
EDIT: Thanks to Florian and George for the answers. I understand the difference now.
I accepted Florian's answer because he was first and also because I found the last link
<http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf> he provided especially helpful.
| https://mathoverflow.net/users/2260 | Geometric vs Arithmetic Frobenius | Geometric and arithmetic Frobenius live in a Galois group, they are different from the Frobenius morphism. The Galois group of a finite field of cardinality $q$ has a canonical generator $x \mapsto x^q$; this is the arithmetic Frobenius element. Its inverse, i.e., $x \mapsto x^{1/q}$, is the geometric Frobenius element. The Galois group of a non-archimedean local field (i.e., a finite extension of $\mathbb Q\_p$ or $k((x))$ for a finite field $k$) maps surjectively to the Galois group of its residue field (a finite field); an element in the inverse image of an arithmetic/geometric Frobenius is still called arithmetic/geometric Frobenius (but there is no longer a canonical choice).
Finally, I think the reason for the term "geometric" is that for a variety $X/k$ ($k$ a finite field of cardinality $q$), we have a canonical isomorphism $X^{(q)} \cong X$, so the $q$-power Frobenius morphism gives rise to a map $F : X(\bar k) \to X(\bar k)$. The Galois group acts on $X(\bar k)$ as well, and the action of the geometric Frobenius element agrees with $F$.
EDIT: Oops, on $X(\bar k)$ the action of the Frobenius morphism agrees with arithmetic Frobenius, but on the étale cohomology of $X\_{\bar k}$ it agrees with geometric Frobenius. Let me try to find a reference...
Here is one (see p.89). The file name seems to indicate that these are Brian Conrad's, but they are not on his web page as far as I can tell, so I hope he doesn't object to the link!
<http://math.unice.fr/~dehon/CohEtale-09/Elencj_Etale/CONRAD%20Etale%20Cohomology.pdf>
I think I heard that it was Deligne who coined the term "geometric Frobenius element". Deligne's Bourbaki talk in 68/69 doesn't seem to give it a name. (See Jay Pottharst's translation at <http://math.bu.edu/people/potthars/writings/deligne-l-adic.pdf>, in particular Prop. 4.8.) Deligne mentions SGA 5.XV. I don't have time to check further, I guess it has more on the fact I mentioned but not on the terminology.
<http://www.msri.org/publications/books/sga/sga/5/SGA5-page-454.html>
Finally see Katz's "Review of l-adic cohomology" in the Motives volumes.
<http://books.google.at/books?id=v2CuklFFV5IC&pg=PA26&lpg=PA26&dq=%22geometric+frobenius+element%22&source=bl&ots=QUaysRdc3L&sig=4U_nC8QPWQjdg9RUi1-hHXt1Iec&hl=en&ei=WvstTLzbH8P38Aaj1q2fAw&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBwQ6AEwAw#v=onepage&q=%22geometric%20frobenius%20element%22&f=false>
(scroll back one page)
**Update:** I found some expository notes I couldn't find yesterday. Like Brian Conrad's notes they explain why geometric Frobenius has the same action as the Frobenius morphism on étale cohomology. (They use the terminology of arithmetic/geometric Frobenius morphism though.)
<http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf>
| 16 | https://mathoverflow.net/users/1729 | 30309 | 19,758 |
https://mathoverflow.net/questions/30237 | 10 | Somewhere, I don't remember where, I saw a beautiful 3D figure of part a CAT(0) simplicial complex. I am thinking and hoping that this was some finite piece of an affine building of type A2, presumably in characteristic 2. But I'm very frustrated now that I just can't remember exactly what I saw or where I saw it. It was something like part of the neighborhood of radius 1 or radius 2 of a vertex, with enough simplices removed so that the rest fits in $\mathbb{R}^3$. It looked like Mathematica graphics, since (in my recollection) it had good colors to help show the 3-dimensional structure. I'm thinking that I saw it in the AMS Notices, but it could also have been in an AMS calendar or elsewhere. Does anyone remember seeing an image like this, and if so, where?
I'm asking because I'd like to have such an image in a paper that I'm working on, not necessarily the one that I saw but something similar.
---
I got some good answers to my question, both here and in private e-mail to Bill Casselman. But in the end I decided to make my own diagram (with the aid of TikZ and ~~Python~~ SAGE). Here it is.
[alt text http://www.freeimagehosting.net/uploads/e38638d43e.png](http://www.freeimagehosting.net/uploads/e38638d43e.png)
---
For those who are interested in the TikZ and SAGE code, I combined them into one TeX document. I posted both [the TeX source](http://www.math.ucdavis.edu/~greg/a2building.tex) and [its PDF output](http://www.math.ucdavis.edu/~greg/a2building.pdf) (from pdflatex) on my web page.
| https://mathoverflow.net/users/1450 | Looking for figure of part of an A2 affine building | Check out <http://dean.clas.uconn.edu/teitelbaum/colorout.gif>
I made this image long ago for the cover of the AMS Notices.
See <http://www.ams.org/notices/199510/teitelbaum.pdf>
| 6 | https://mathoverflow.net/users/3394 | 30319 | 19,764 |
https://mathoverflow.net/questions/30328 | 11 | Does anyone know where to find an English translation of Riemann's Habilitation Thesis concerning trigonometric series? The German title of the work is "Über die Darstellbarkeit einer Function durch eine trigonometrische Reihe" and the English title is "On the representation of a function as a Trigonometric Series". So far I unfortunately haven't been able to find the English translation, so any suggestions as to where I might find it (if it has been translated) would be greatly appreciated.
| https://mathoverflow.net/users/5431 | English translation of Riemann's Habilitation Thesis | This is XII in Riemann's Werke (1876; 1892) and has been published in English in
Bernhard Riemann, *Collected papers.*
Translated from the 1892 German edition by Roger Baker, Charles Christenson and Henry Orde. Kendrick Press, Heber City, UT, 2004 ISBN: 0-9740427-2-2; 0-9740427-3-0
[MR2121437 (2005m:01028)](http://www.ams.org/mathscinet-getitem?mr=2121437)
| 11 | https://mathoverflow.net/users/5740 | 30336 | 19,774 |
https://mathoverflow.net/questions/30329 | 1 | In answer to the question [Demystifying complex numbers](https://mathoverflow.net/questions/30156/demystifying-complex-numbers), Charles Matthews suggests "finding the points at twice the distance from (-1, 0) that they are from (1, 0)." as a motivation for complex numbers.
Suppose you want to find these points in hyperbolic geometry instead of euclidean geometry.
If this can be done with vectors or complex numbers in R^2, then I reckon it could be done with [gyrovectors or gyro-complex numbers](https://mathoverflow.net/questions/29801/coordinate-algebra-in-hyperbolic-geometry) in the hyperbolic plane, but if you don't use gyro-algebra then how would you find (or describe) the points at twice the distance from (-1, 0) that they are from (1, 0)?
(Defining coordinates in hyperbolic geometry can be done with gyro-algebra, but without it just assume the origin is an arbitrary point, and that (-1,0) is a point of distance 1 from the origin and (1,0) is a point of distance 1 from the origin in the opposite direction.)
| https://mathoverflow.net/users/3537 | Points at twice the distance from (-1, 0) that they are from (1, 0) in hyperbolic geometry | This is a bit basic for MO, but I'll present my solution.
I don't understand what the OP's notation is so I'll
use my favourite model for hyperbolic space, the Poincaré
upper half plane ('cos I like modular forms).
In the upper half plane model the distance satisfies
$$d(a+bi,c+di)=\cosh^{-1}\frac{(a-c)^2+b^2+d^2}{2bd}.$$
As each line in the upper half plane can be transformed
into the imaginary axis, we can take our two points to
lie on this axis. So let $u>v>0$ and seek the $z=x+yi$
with
$$d(z,ui)=2d(z,vi).$$
Using the identity
$$\cosh 2t=2\cosh^2t-1$$
we get
$$\frac{x^2+y^2+u^2}{2yu}=\frac{(x^2+y^2+v^2)^2}{2y^2v^2}-1$$
that is
$$v^2y(x^2+(y+u)^2)=u(x^2+y^2+v^2)^2,$$
a quartic curve.
Is this as easy with gyrovectors (whatever they are)?
| 4 | https://mathoverflow.net/users/4213 | 30337 | 19,775 |
https://mathoverflow.net/questions/30334 | 2 | Let define procedure for converting second order theory to first order:
1. Take any second order theory with equality
2. Invent sort Bool' and new fresh constants F' and T', of sort Bool'
3. Create fresh sort CC
4. Replace each proposition P(x) (where x : XX) except equality to P'(x) == T' where P' : XX -> Bool'
5. Replace each connective to respective function to Bool'
6. For each first order function f : XX -> YY define fresh constant c : CC. We call it the key of function f.
7. For each sort XX and YY, if source theory contained any functions XX -> YY, define new function applyXXYY : CC -> XX -> YY such that if c is key for f then applyXXYY(c,x) = f(x)
8. For each first order function f applied to second order function g replace f by its key:
g(f) replace to g(c) where c is a key for f
9. Each expression with variable containing first order function applied to argument replace application to variable by application to applyXXYY(variable)
(where XX and YY are respective to sort of f).
So, definition g(x,f,y) := x\*f(y)+1-q(f) become g(x,c,y) := x\*apply(c,y)+1-q(c)
Items 6..9 is well known amongst functional programmers as defunctionalization process (roughly).
So, SOL is more expressive than FOL (in strict Felleisen/VanRoy sense) but strictly equal in power.
Is it correct?
Questions:
* Are Second Order Logic really equivalent to First Order Logic?
* Are really any logic of some order can be lowered to FOL?
* Can any higher order Logic be converted to equivalent first order logic?
| https://mathoverflow.net/users/7257 | Lowering order of theory | The answer to your question is that it depends on what semantics you want to use for higher-order logic.
* If you use *full higher-order semantics*, then you cannot reduce your theory to a first-order theory. In these semantics, the higher-order quantifiers range over *all* objects of the appropriate type, and so a model only has to specify the domain of discourse for individuals (elements of the base type).
* If you use *Henkin semantics*, then you can replace your higher-order theory with a completely equivalent multi-sorted first-order theory, using the process you are describing in which you add numerous "Apply" symbols. In Henkin semantics, a model has to provide a separate domain of discourse for each type of quantifier. So it is possible, in a particular Henkin model, that a higher-order quantifier does not range over all objects of its type, only over the ones that were included in that model.
Full higher-order semantics are much more powerful than Henkin semantics. For example, there is a categorical axiomatization of the natural numbers in full second-order semantics, but Henkin semantics are subject to the Lowenheim-Skolem theorem just like first-order semantics. (This is because Henkin semantics really are just first-order semantics.)
One canonical reference for this is Shapiro's *Foundations without foundationalism*.
| 4 | https://mathoverflow.net/users/5442 | 30341 | 19,777 |
https://mathoverflow.net/questions/23724 | 8 | This question is the two-dimensional analogue of [Etale coverings of certain open subschemes in Spec O\_K](https://mathoverflow.net/questions/22883/etale-coverings-of-certain-open-subschemes-in-spec-o-k)
There I asked if one could characterize in a way certain covers of $\textrm{Spec} \ O\_K$. As Cam Mcleman answered, this is basically done by the Galois group of the maximal extension unramified outside $D$. A covering of $U$ is of the form $O\_L[\frac{1}{D}]$, where $L$ is any extension of $K$.
Here I would like to ask the same question, only now for $X=\mathbf{P}^1\_{\mathbf{Z}}$.
Let $D$ be a normal crossings divisor on $\mathbf{P}^1\_{\mathbf{Z}}$ and let $U$ be the complement of its support.
**Q1**. Is there an "equivalence of categories" as Georges Elencwajg mentions in his answer for the analytic case. (See above link.) Basically, is there an arithmetic Grauert-Remmert theorem?
**Q2**. What is known about the etale fundamental group in this case? Is it "finitely generated"? Has anybody studied the maximal pro-p-quotients of these groups?
**Q3**. The analytic analogue would be to consider the same question for $\mathbf{P}^1\_{\mathbf{C}} \times \mathbf{P}^1\_{\mathbf{C}}$.
**Q4** Lars (see above link) mentions a result for $\mathbf{P}\_{\mathbf{Q}}^1$. Is there something similar for $\mathbf{P}^2\_{\mathbf{Q}}$?
| https://mathoverflow.net/users/4333 | Covers of the projective line over Z and arithmetic Grauert-Remmert | Regarding Q3: For any scheme $X$ of finite type over $\mathbb{C}$ the Riemann-Existence Theorem (See SGA1 XII.5) says that the category of finite étale coverings of $X$ is equivalent to the category of finite covering spaces of the associated analytic space $X^{an}$. This implies that the finite quotients of the topological fundametal group of $X^{an}$ are the same as the finite quotients of the étale fundamental group, and one obtains that the étale fundamental group of $X$ is isomorphic to the profinite completion of the topological fundamental group of $X^{an}$.
Q4: The same short exact sequence that I mentioned in your other question is still valid.
As in your other question, I cannot say anything about the situation over $\mathbb{Z}$ :)
| 2 | https://mathoverflow.net/users/259 | 30347 | 19,780 |
https://mathoverflow.net/questions/30243 | 4 | Let $F$ be a function field of "transcendental degree one" over its full constant field $K$. Let $x \in F \backslash K$. We know the divisor of $(x) = (x) - (1/x)$ in $K(x)$. Could you please give me an algorithm to compute the places over two above places in $F$ and the ramification degrees.
If this setting is too abstract, what if we have $F$ is the field of fraction of $K(x)[y]/f(x,y)$ where $K$ is a finite field, could you show me any algorithm to find places over zero place and infinite place of $x$.
As [KConrad](https://mathoverflow.net/users/3272/kconrad) suggested, I'm telling you a little about how I got involved with this problem.
Once upon a time when I was a bit younger (and a bit more stupid but not much less than now) I dared to ask Noam Elkies that how I can represent a curve with an equations of different degree than the one I'm given. For example an elliptic curve of degree 5 (you see, it's not only your time that I waste, so don't take it personal). He wrote me something that time I didn't quite understand at the time but today I went back to the email and fortunately I understood almost all of it:
*start from your sample curve
y^2 + xy + x^3 + 1 = 0 over Z/2Z*
and choose any function of degree 5, say z = x*y. Then eliminate y from the equations by computing the resultant with respect to y of y^2 + xy + x^3 + 1 with the equation
satisfied by x,y,z, which is here z - x*y. This gives z^2 + x*z = x^2 + x^5 with x,z functions of degree 2 and 5 on the curve.*
Sincerely,
--Noam D. Elkies
The only point which wasn't clear for me was "function of degree 5, say z = x\*y". So I assumed it means that the degree of the zero divisor or the pole divisor should be 5. Although I checked it with Magma and it was the case, but I felt the need to compute the divisor for function $z$ in $K(x,y)$ myself. So I tried to compute the divisor of $x$ as the first step. Using the "Extensions = Ramified covers" rule of thumb, and looking at $x$ (the coordinate function) as the covering map to $\mathbb{P}^1$, I said that $(x)$ (the function) correspond to point $x - 0$ in $\mathbb{P}^1$ scheme so I put zero instead of $x$ in my equation and I get my two ramified points $y^2 = 1$. But for the places of over place at infinity downstairs $(1/x)$, I couldn't go that far. I changed the variable $1/\theta = x$ and put zero in $\theta$, I'll get 1=0, unless I replace $y$ with something like $\omega/\theta^2$ as well (which I don't see why) to see my ramification at infinity.
Now my question unfolded is:
1. Do you think what I'm doing makes sense and why it doesn't work for the infinite place.
2. Is there an algebraic/arithmetic way to do what I did instead of the geometric approach of covering space that I used, which I suppose would be more algorithmic friendly.
Sorry I think I gave too much of background.
| https://mathoverflow.net/users/6776 | Computing places over x in F/K(x) | For $x\in F\setminus K$, the degree of $x$ is the degree of the field extension $F/K(x)$. For example, in the $F$ corresponding to your curve, the degree of $x$ is $2$, since the extension $F/K(x)$ is the simple extension corresponding to $y^2 + xy + x^3 + 1 = 0$. Similarly, the degree of $y$ is $3$, since $F/K(y)$ is the simple extension corresponding to $x^3 + yx + y^2 + 1 = 0$.
The degree of $x$ is also the degree of the positive (or negative) part of the divisor of $x$. Thus if $x$ and $y$ are such that the divisor of zeros of each is disjoint from the divisor of poles of the other, then the degree of $xy$ is the sum of the degrees of $x$ and of $y$. These last conditions can be checked as follows: if $y$ is integral over $k[x]$ (i.e. if it satisfies a monic polynomial with coefficients in $k[x]$), then $y$ is finite wherever $x$ is finite. In particular, $y$ never has a pole where $x$ vanishes. Since for your $x$ and $y$, you have $y$ integral over $k[x]$ and $x$ integral over $k[y]$, you know without further calculation that the degree of $xy$ is $5$. Perhaps Elkies had something like this in mind when he wrote you.
Going back to the general case, if you want to compute the actual places where an $x\in F\setminus K$ vanishes (and the vanishing multiplicities), let me recommend
<http://www.cse.chalmers.se/~coquand/place.pdf>, which gives an algorithm. The essential step of the calculation is this: take $K$ to be algebraically closed, and suppose your field $F$ is presented as the fraction field of $K[x,y]/f(x,y)$. Suppose furthermore that you have a solution $(a,b)$ of $f(x,y) = 0$. You must then find the places of $F$ centered at $(a,b)$. If $(a,b)$ is a non-singular point of the affine model, then there is a single place, but generally one needs to resolve the singularity.
For your example, here is how to do the calculations by hand: let's compute the places where $1/x$ vanishes. Let $u = 1/x$ so that $F/K(u)$ is the extension corresponding to $u^3y^2 + u^2y+1 +u^3 = 0$. As you point out, setting $u=0$ gives no solutions. That should not worry us, since $y$ is not integral over $K[u]$, and so we have no reason to expect that $y$ should be finite where $u$ is finite.
Let's try again with $v = 1/y$. We then have $(1+u^3)v^2 + u^2v + u^3 = 0$, which is not monic in $v$, but for which the leading coefficient is invertible when $u$ vanishes. Thus $v$ will be finite when $u$ vanishes. Setting $u=0$, we find $v=0$. We still have a problem here, since
$$
K[u,v]/((1+u^3)v^2 + u^2v + u^3)
$$
is not non-singular at the prime ideal $(u,v)$, as there is no linear term in the defining polynomial, and so it is not clear how many places of $F$ are centered at this prime.
Finally, let's try $w = v/u = x/y$. We then have $(1+u^3)w^2 + uw + u = 0$. As before, $w$ will be finite when $u$ vanishes. We set $u=0$ and find $w=0$, but now we get a place of $F$, since
$$
K[u,w]/((1+u^3)w^2 + uw + u)
$$
is non-singular at the prime ideal $(u,w)$. Furthermore, $w$ is a local uniformizer, since $u = w((1+u^3)w + u)$ implies that $u$ is divisible by $w$. Pulling one more factor of $w$ out on the right, we see $u$ is divisible by $w^2$. Finally, since $u = w^2(\mathrm{unit} + \mathrm{multiple of }w)$, we find that $u$ vanishes to order exactly $2$ at this place. Since $u = 1/x$, we find that $x$ has a pole only at the place of $F$ corresponding to $(1/x,x/y)$ and that the pole order is $2$.
| 2 | https://mathoverflow.net/users/2490 | 30348 | 19,781 |
https://mathoverflow.net/questions/30265 | 10 | Let $G$ be a connected reductive group over a (perfect, why not) field $F$. Let $m$, $pr\_1$, $pr\_2$ denote the multiplication, first, and second projection maps from $G \times G$ to $G$.
Then I'm pretty sure that I can prove the following fact: if $L$ is a line bundle on $G$, then $m^\ast(L)$ is isomorphic to $pr\_1^\ast(L) \cdot pr\_2^\ast(L)$. This (plus Hilbert's 90) implies that $Pic(G)$ classifies the central extensions of $G$ by the multiplicative group $G\_m$, by some stuff in SGA 7, I believe.
The way that I can prove the above fact is by using Kottwitz's isomorphism, which describes $Pic(G)$ in terms of the dual group. I'll probably include this Kottwitzish proof in something I'm writing, but I'm left with the following question:
Is there a proof in the literature that $m^\ast(L)$ is isomorphic to $pr\_1^\ast(L) \cdot pr\_2^\ast(L)$ for line bundles over reductive groups? Someone must have written this up 30 years ago, right? And the implication that $Pic(G)$ classifies central extensions by $G\_m$? Is this published somewhere? It certainly shouldn't require passage to the dual group!
Of course, if I've messed something up, and the above fact is false, I'd appreciate such information too!
| https://mathoverflow.net/users/3545 | Reference for Pic(G) and central extensions. | I can give a reference only for the second part of the question,
namely, about central extensions.
It was answered by Colliot-Thélène in 2008, not 30 years ago!
Colliot-Thélène's paper *Résolutions flasques des groupes linéaires connexes*, J. für die reine und angewandte Mathematik (Crelle) 618 (2008), 77--133,
contains the following corollary (I type it in English):
**Corollary 5.7.** *Let $G$ be a connected linear algebraic group, assumed reductive if char $k > 0$. For any smooth $k$-group of multiplicative type $S$, the natural arrow
Ext$(G,S)\to$ ker$[H^1(G,S)\to H^1(k,S)]$ is an isomorphism*.
Taking $S=\mathbf{G}\_m$, we obtain $H^1(k,\mathbf{G}\_m)=1$ (Hilbert 90) and $H^1(G,\mathbf{G}\_m)=\mathrm{Pic}(G)$
(here $H^1$ means étale cohomology). We obtain an isomorphism Ext$(G,\mathbf{G}\_m)\cong \mathrm{Pic}(G)$.
| 3 | https://mathoverflow.net/users/4149 | 30349 | 19,782 |
https://mathoverflow.net/questions/30020 | 3 | 1-In his article written in German "Über unerreichbare Kardinalzahlen" (On inaccessible cardinals), inside Fund. Math. 1938 (pages 68-89), Alfred Tarski states his axioms A and A' as follows.
Axiom A: "For every set x, there exists a set y satisfying the four following conditions:
* A1: x is a member element of y;
* A2: If z is a member element of y and t is included inside z, then t is a member element of y;
* A3: If z is a member element of y and t is the set having as member elements exactly all sets u included inside z, then t is a member element of y;
* A4: If z is included inside y and if the sets z and y are equipotent, then z is a member element of y."
Axiom A':"For every set x, there exists a set y satisfying the four following conditions:
* A'1: x is included inside y;
* A'2: If z is a member element of y and t is a member element of z, then t is a member element of y;
* A'3: If z is a member element of y, there exists a set w that is a member element of y such that every set t that is included inside z is a member element of w;
* A'4: If z is included inside y and if no set included inside z is equipotent with y, then z is a member element of y."
Tarski asserts, without giving a proof, that axioms A and A' are equivalent.
>
> QUESTION 1: Does anyone know about an explicit proof of the equivalance of A and A' ?
>
>
>
| https://mathoverflow.net/users/30395 | About Tarski's axioms a and A' and around (1) | First, I note that you appear to be missing a *not* in A4,
and you should say that "if $z$ and $y$ are *not*
equipollant", for otherwise we could take $z=y$ and thereby
deduce $y\in y$, contrary to the Foundation axiom.
With this correction, both your axioms are equivalent in
ZFC to the assertion that there is a proper class of
inaccessible cardinals.
For the one direction, if there are such cardinals, then
for any set $x$ we may find an inaccessible cardinal
$\kappa$ such that $x\in V\_\kappa$, and take
$y=V\_\kappa=H\_\kappa$ to fulfill either $A$ or $A'$, which
is easily verified.
Conversely, assume axiom A. Let $x=\alpha$ be an ordinal
and let $y$ arise as in axiom A. Let $\kappa=|y|$. As every
subset of $\alpha$ is in $y$, it follows that
$\alpha\lt\kappa$. If $\beta\lt\kappa$, then there is
subset $z\subset y$ of size $\beta$, and this is an element
of $y$ by A4. We also know $P(z)\subset y$ and $P(z)\in Y$,
so $P(P(z))\subset y$, so $2^\beta\lt\kappa$. Thus,
$\kappa$ is a strong limit. For regularity, suppose that
$\kappa$ singular with cofinality $\gamma\lt\kappa$. Thus,
$y$ is the union of $\gamma$ many subsets, each of size
less than $\kappa$. These subsets are elements of $y$, and
all their subsets are also in $y$. But every subset of $y$
is determined by a similar $\gamma$ sequence of elements of
$y$, and so $y$ will have $2^\kappa$ many elements, a
contradiction. So $\kappa$ is an inaccessible cardinal
above $\alpha$, as desired.
For the other converse direction, assume axiom $A'$.
Consider $x=\alpha+1$, and get $y$ as in $A'$, and again
let $\kappa=|y|$. I claim that $\kappa\subset y$, for
otherwise the least ordinal $\beta$ not in $y$ would be
less than $\kappa$ in size and have all its subsets size
less than $\kappa$, and hence in $y$ by $A4'$. Thus,
$\kappa\subset y$. Now, for any $\gamma\lt\kappa$, every
subset of $\gamma$ is in $y$ and there is an element of $y$
with at least $2^\gamma$ many subsets, all in $y$, so
$2^\gamma\lt\kappa$. So $\kappa$ is a strong limit.
Regularity follows as before, and so $\kappa$ is an
inaccessible cardinal above $\alpha$.
Thus, since the axioms are both equivalent to the assertion that there is a proper class of inaccessible cardinals, they are also equivalent to each other.
Do you have some historical reason to study Tarski's treatment of inaccessibility?
If not, I think you might find the contemporary accounts of large cardinals to be more appealing. You might look at Kanamori's book, *The Higher Infinite*.
If you wanted the equivalence in ZF rather than ZFC, or in
ZF-Foundation, then I wouuld have to think more carefully about it, but I will mention that this issue seems to be remarked on in Solovay's letter mentioned in
your previous question. In particular, without AC there are
competing inequivalent notions of inaccessibility.
| 2 | https://mathoverflow.net/users/1946 | 30356 | 19,785 |
https://mathoverflow.net/questions/30358 | 5 | Related to [A000679](http://www.oeis.org/A000679) (Number of groups of order $2^n$), how many non-Abelian groups of order $2^n$ are there?
| https://mathoverflow.net/users/2391 | Number of non-Abelian groups of order $2^n$ | It is true that there is no known formula for the number of isomorphism classes of groups of order $n$, but there is a very nice asymptotic formula for $p$-groups.
In particular, the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3 + O(n^{8/3})}$. This function grows very rapidly, and there is a folklore conjecture that
"almost all groups are $2$-groups."
<http://en.wikipedia.org/wiki/P-group>
| 13 | https://mathoverflow.net/users/4558 | 30364 | 19,789 |
https://mathoverflow.net/questions/30345 | 16 | On page 283 of Max Karoubi’s book, “K-theory,” he states that for any compact Hausdorff space $X$, the Chern character determines an isomorphism from the group $K^0(X) \otimes Q$ to $H^{even}(X; Q)$, the direct sum of the even-dimensional Cech cohomology groups of X with rational coefficients. In particular, this theorem implies that $K^0(X)$ and $H^{even}(X; Z)$ are isomorphic up to torsion.
>
> Does anyone know of an example (and also a reference, preferably) of a smooth compact manifold $X$ with the property that $K^0(X)$ and $H^{even}(X; Z)$ are not isomorphic?
>
>
>
| https://mathoverflow.net/users/7262 | What is an example of a compact smooth manifold whose K-theory and Cech cohomology are not isomorphic? | For $X=\mathbb RP^4$ the groups $K^0(X)$ and $H^{even}(X)$ respectively are $\mathbb Z\oplus \mathbb Z/4$ and $\mathbb Z\oplus \mathbb Z/2\oplus \mathbb Z/2$. More generally, $K^0(\mathbb RP^{2k})\cong \mathbb Z\oplus \mathbb Z/2^{k}$. These computations of real and complex $K$-groups of real and complex projective spaces can be found in an early paper of J F Adams, I believe the one about vector fields on spheres. Let $k\to\infty$ so that $\mathbb RP^k$ becomes $BG$ for $G$ of order $2$. The Atiyah-Segal Completion Theorem says that for a finite or more generally compact Lie group $G$ the ring $K^0(BG)$ is the completion of the complex representation ring $RG$ with respect to the augmentation ideal (kernel of rank homomorphism $RG\to\mathbb Z$).
Even when the homology groups are torsion-free, so that $K^0$ is abstractly isomorphic to $H^{even}$, there is in some sense a difference between $K^0$ and $H^{even}$; they are then two slightly different lattices in the same rational vector space. For example, if $x$ generates $H^2(\mathbb CP^k,\mathbb Z)$ then a $\mathbb Z$-basis for $H^{even}(\mathbb CP^k)$ is $1,x,\dots,x^k$ while a $\mathbb Z$-basis for (the image under the Chern character of) $K^0(\mathbb CP^k)$ is $1,e^x,\dots,e^{kx}$ (or $1, e^x-1,\dots, (e^x-1)^k$). When $X$ is a sphere $S^{2k}$, they are the same lattice.
| 22 | https://mathoverflow.net/users/6666 | 30365 | 19,790 |
https://mathoverflow.net/questions/30353 | 15 | While digging through old piles of notes and jottings, I came across a question I'd looked at several years ago. While I was able to get partial answers, it seemed even then that the answer should be known and in the literature somewhere, but I never knew where to start looking. So I thought I'd ask here on MO if anyone knows of a reference for this observation.
Here's the notation and background for the question.
Let $(E,\Vert\cdot\Vert)$ be a real, normed vector space (I think the complex case works out to be almost identical). We will see shortly that my question is vacuous unless $E$ is *incomplete*. Let $F$ be the completion of $E$.
Denote by $B(E,\Vert\cdot\Vert)$ the space of all linear maps $T:E\to E$ which ae bounded with respect to $\Vert\cdot\Vert$, i.e. there exists $C$ depending on $T$ such that
$$ \Vert T(x)\Vert \leq C\Vert x\Vert \;\;\hbox{for all $x\in E$.} $$
Clearly each $T\in B(E,\vert\cdot\Vert)$ extends uniquely to a bounded linear operator $F\to F$, and we thus get an injective algebra homomorphism $\imath:B(E,\Vert\cdot\Vert)\to B(F)$.
The question arises: **when does $\imath$ have dense range?**
* It is not hard to show that if $E=c\_{00}$ and $\Vert\cdot\Vert$ is the $\ell\_\infty$ norm then $\imath$ does indeed have dense range.
* On the other hand, if $E=\ell\_1$ and $\Vert\cdot\Vert$ is the $\ell\_\infty$ norm, then by considering "blocks" which have $\ell\_\infty$-norm 1 and large $\ell\_1$-norm, we can construct an isometry on $c\_0$ which is not approximable by operators of the form $\imath(T)$; in particular, $\imath$ does not have dense range in this case.
I can't find the piece of paper where I wrote down the details, but I seem to recall that one obtains the same answer if we replace $\ell\_1$ by $\ell\_p$ for $1\leq p < \infty$ and take $\Vert\cdot\Vert$ to be the $\ell\_r$ norm for any $p<r\leq\infty$.
So here are two explicit questions. I haven't looked at them properly since about 2004/5, so they may well have straightforward solutions.
Q1. Let $\Vert\cdot\Vert$ be any norm on $c\_{00}$, and let $F$ be the completion of $c\_{00}$ in this norm. Does $\imath: B(c\_{00},\Vert\cdot\Vert)\to B(F)$ have dense range?
Q2. Let $(F,\Vert\cdot\Vert)$ be a Banach space with an unconditional basis. Let $E$ be a proper dense subspace of $F$ which is a Banach space under some norm that dominates $\Vert\cdot\Vert$.
Does $\imath: B(E,\Vert\cdot\Vert)\to B(F)$ always have non-dense range?
If the answers to these are known, does anyone know where I might find references to these in the literature?
**Update 5th July 2010:** Q1 has a positive answer, as given by Bill Johnson below (a simmilar approach was also elaborated by Pietro Majer). As pointed out (ibid.) the question can be rephrased/generalized to the following:
>
> given a separable Banach space $F$ and
> a dense linear subspace $E$ of
> countable dimension, can every bounded
> operator on $F$ be approximated by
> operators which take $E$ to $E$?
>
>
>
I'd still be interested to know the answer to Q2, even in the special cases where $F=\ell\_p$ for some $1\leq p < \infty$.
| https://mathoverflow.net/users/763 | Approximating operators on Banach spaces by bounded operators on a proper dense subspace | Q1: Yes. You ask ``If $X$ is a countable dimensional dense subspace of the Banach space $Y$, are the operators on $Y$ which leave $X$ invariant dense in the operators on $Y$?" Use Mackey's argument for producing quasi-complements (just a biorthogonalization procedure, going back and forth between a space and its dual) to construct a fundamental and total biorthogonal sequence $(x\_n,x\_n^\*)$ for $Y$ with the $x\_n$ in $X$; even a Hamel basis for $X$. Now use the principle of small perturbations to perturb an operator on $Y$ to a nearby one that maps each $x\_n$ back into $X$. I am traveling now and so can't provide details or references, but I think that is enough for you, Yemon. The key point is that the biorthogonality makes the perturbation work--if $x\_n$ were only a Hamel basis for $X$ it is hard to keep control.
I have my doubts whether this result appears in print even if oldtimers like me know the result as soon as the question is asked.
EDIT 7/4/10: Once you get the biorthogonal sequence $(x\_n,x\_n^\*)$ with $x\_n$ a Hamel basis for $X$, you finish as follows: WLOG $\|T\|=1$ and normalize the BO sequence s.t. $\|x\_n^\*\|=1$. Define the operator $S$ on $X$, the linear span of $x\_n$, by $Sx\_n=y\_n$, where $y\_n$ is any vector in $X$ s.t. $\|y\_n-Tx\_n\| < (2^{n}\|x\_n\|)^{-1}\epsilon$. On $X$ you have the inequality $\|T-S\|<\epsilon$, so you get an extension of $S$ to $Y$ that satisfies the same estimate on $Y$. In checking the estimate you use the inequality $\|x\| \ge \sup\_n |x\_n^\*(x)|$; i.e., biorthogonality is crucial.
To get the biorthogonal sequence, you take any Hamel basis $w\_n$ for $X$ and construct the biorthogonal sequence by recursion so that for all $n$, span $(w\_k)\_{k=1}^n = $ span $(x\_k)\_{k=1}^n$. At step $n$ you choose any $x\_n$ in span $(w\_k)\_{k=1}^n $ intersected with the intersection of the kernels of $x\_k^\*$, $1\le k < n$, and use Hahn-Banach to get $x\_n^\*$.
The Mackey argument I mentioned gives more. If you have any sequence $w\_n$ with dense span in $Y$ and any $w\_n^\*$ total in $Y^\*$, with a back and forth biorthogonalization argument you can build a biorthogonal sequence $(x\_n,x\_n^\*)$ s.t. for all $n$, span $(x\_k)\_{k=1}^{2n}$ contains span $(w\_k)\_{k=1}^n $ and span $(x\_k^\*)\_{k=1}^{2n}$ contains span $(w\_k^\*)\_{k=1}^n $. This is quite useful when dealing with spaces that fail the approximation property; see e.g. volume one of Lindenstrauss-Tzafriri and, for something recent, my papers with Bentuo Zheng, which you can download from my home page.
EDIT 7/11/10: Getting a general positive answer to Q2 would be very difficult. Although not known to exist, it is widely believed that there is a Banach space with unconditional basis upon which every bounded linear operator is the sum of a diagonal operator and a compact operator. On such a space, the operators that map $\ell\_1$ into itself would be dense in the space of all bounded linear operators.
| 11 | https://mathoverflow.net/users/2554 | 30366 | 19,791 |
https://mathoverflow.net/questions/30066 | 6 | M a finitely generated module over a commutative ring A. I can't think of an example of two maximal linearly independent subsets of M having different cardinality. I know that they all have the same cardinality if A is integral domain. Any suggestions are welcome!
| https://mathoverflow.net/users/5292 | Cardinality of maximal linearly independent subset | I found an old paper by Lazarus (Les familles libres maximales d'un module ont-elles le meme cardinal?, Pub. Sem. Math. Rennes 4 (1973), 1-12) which contains the the following result: Let A be a commutative ring with unit and M an A-module. In the following situations, maximal linearly independent subsets of M have the same cardinality:
1. If M is a free A-module of infinite rank.
2. If A is reduced and has only finitely many minimal primes (e.g. integral domain, reduced Noetherian ring)
3. If A is Noetherian and M is a free A-module.
4. If A is Noetherian and M is a submodule of a free A-module of finite rank.
5. If A is Noetherian and M has an infinite linearly independent subset.
6. If A is Noetherian and M is a submodule of a flat A-module.
7. If A is Artin local and the zero ideal $(0)\subset A$ is irreducible.
And the examples given in the paper of modules not satisfying this same cardinality property are highly nontrivial.
| 10 | https://mathoverflow.net/users/5292 | 30369 | 19,794 |
https://mathoverflow.net/questions/30377 | 2 | Let $G$ be a finite Abelian group with endomorphism ring $End(G)$. I am interested in the probability $P(\phi(g\_1) = g\_2)$ for fixed $g\_1,g\_2 \in G$ and a uniformly chosen endomorphism $\phi(\cdot)$ from $End(G)$. Essentially, I want to understand where the set of endomorphisms will take each element $g \in G$. I ran into this question while considering homomorphic compression schemes that compress an $n$-length sequence $g^n$ into a sequence of length $k$ by applying a homomorphism $\phi \colon G^n \rightarrow G^k$. I describe the question in detail below.
Let $\mathbb{Z}\_n$ be the cyclic group of $n$ elements. If $G ={\mathbb{Z}\_{p^r}}$, I understand what is going on and can prove for instance that $\phi(g)$ is uniformly distributed across the smallest subgroup of $\mathbb{Z}\_{p^r}$ that $g$ belongs to as $\phi(\cdot)$ varies over $End(\mathbb{Z}\_{p^r})$. But, I am having trouble understanding what happens in the case of groups of the form $\mathbb{Z}\_{p^r}^k$ such as $\mathbb{Z}\_2^2$ for example. In this case, $\phi(g)$ is uniformly distributed over $\mathbb{Z}\_2^2$ for all non-identity $g$ regardless of which subgroup $g$ belongs to.
>
> Question: Is there a uniform way to write down the probability $P(\phi(g\_1) = g\_2)$ for fixed $g\_1,g\_2 \in G$ and an arbitrary $\phi(\cdot) \in End(G)$ for a finite Abelian group $G$?
>
>
>
I would greatly appreciate any pointers and hope the question isn't too elementary for MO. Please feel free to edit/re-tag the question if needed.
| https://mathoverflow.net/users/2878 | Image of a fixed element under a random endomorphism in an Abelian group | For a group $G=\mathbb{Z}\_{p^r}^k$ this is quite straightforward.
Let $g$ have order $p^r$ in $G$ (if not then we are effectively working
in $\mathbb{Z}\_{p^s}$ where $s < k$). Applying an automorphism of $G$
we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond
to matrices over $\mathbb{Z}\_{p^r}$. The image of $s$ is the first row
(or column if you put the map on the other side) and we see that the image
of $s$ is uniformly distibuted over all of $G$.
Now consider a general finite abelian $p$-group $G$.
Let $g\in G$. We can write $G=\langle h\rangle\times H$
where $g=p^s h\in H$ and $h$ has order $p^m$ for some
$m\ge s$. We can specify an endomorphism of $G$ by mapping
$h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism
from $H$ to $G$. Then the $h'$ are uniformly distributed
amongst the elements of order $\le p^m$ in $G$ and $g'$ is
mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed
over a certain subgroup of $G$.
For general finite abelian $G$ split up $G$ as a product of its Sylow
$p$-subgroups. Then $g\in G$ splits up into its primary components
and each of these behave in the same way, under a random endomorphism,
as in the $p$-group case above.
**Added** I now see that the argument I gave in the prime power case
is valid in the general case too. The key observation is that
a maximal cyclic subgroup of a finite abelian group is a direct summand.
Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup
of order $mn$ containing $\langle g\rangle$. Then the images of
$g$ under random endomorphisms of $G$ are uniformly
distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$
denotes the $mn$-torsion subgroup of $G$.
**Added** (4/7/2010) Thanks to Tom for pointing out my error
above. The argument I had in mind for proving that maximal
cyclic groups are summands doesn't actually work. :-(
As t3suji points out, the images are uniformly distributed
over a subgroup. Identifying this subgroup looks like being
a bit more fiddly than I believed and I lack the patience
to do it now. It seems that reduction to the prime power
case is a good way to proceed.
| 2 | https://mathoverflow.net/users/4213 | 30378 | 19,800 |
https://mathoverflow.net/questions/30288 | 20 | I came across the concept of a hyperring in two recent papers by Connes and Consani ([From monoids to hyperstructures: in search of an absolute arithmetic](http://arxiv.org/abs/1006.4810) and [The hyperring of adèle classes](http://arxiv.org/abs/1001.4260)). It's a weakening of the ring concept, but where the addition is allowed to be multivalued. Indeed the additive part of a hyperring forms a 'canonical hypergroup'.
A canonical hypergroup is a set, $H$, equipped with a commutative binary operation,
$$
+ : H \times H \to P^\*(H)
$$
taking values in non-empty subsets of $H$, and a zero element $0 \in H$, such that
1. $+$ is associative (extended to allow addition of subsets of $H$);
2. $0 + x = {\{x\}} = x + 0, \forall x \in H$;
3. $\forall x \in H, \exists ! y \in H$ such that $0 \in x + y$ (we denote this $y$ as $-x$);
4. $\forall x, y, z \in H, x \in y + z$ implies $z \in x - y$ (where $x - y$ means $x + (-y)$ as usual).
(NB: $x$ may be written for the singleton {$x$}.)
I know that hyperrings occur whenever a ring is quotiented by a subgroup of its multiplicative group, but I'd like to know more about where and how hyperrings and hypergroups have cropped up in different branches of mathematics. How is a canonical hypergroup to be thought of as canonical? Are noncanonical hypergroups important? Is there a category theoretic way to see these hyperstructures as natural?
| https://mathoverflow.net/users/447 | What are hypergroups and hyperrings good for? | While I don't know much about hyperstructures other than hypergroups, I know it is hard to study the history behind them because of the non-consistent terminology attributed to these objects by different authors in different periods. I will say something about hypergroups and hopefully some specialist can come and give better insight.
First some physical intuition for finite commutative hypergroups that I found useful: the simplest way to think of them is to think of a collection of particle types $\{c\_0,c\_1,\cdots,c\_n\}$ where two particles can collide to form a third, however not in a definite manner. Let the structure constants $n\_{ij}^k$ denote the probability that $c\_i+c\_j\rightarrow c\_k$. Now assume $c\_0$ denotes photons and that they get absorbed in every collision. Also assume that for each particle there is a unique antiparticle so that their collision is likely to produce a photon with non-zero probability.
So coming to the actual definitions, call a generalized hypergroup a pair $(\mathcal K, \mathcal A)$ where $\mathcal A$ is a \*-algebra with unit $c\_0$ over $\mathbb C$ and $\mathcal K =\{c\_0,c\_1\dots,c\_n\}$ is a basis of $\mathcal A$ with $\mathcal K ^\*=\mathcal K$ for which the structure constants $n\_{ij}^k$ defined by $$c\_ic\_j=\sum\_k n\_{ij}^k c\_k$$ satisfy the conditions $c\_i^\*=c\_j \iff n\_{ij}^0>0$ and $c\_i^\*\neq c\_j \iff n\_{ij}^0 =0$.
$(\mathcal K,\mathcal A)$ is called Hermitian if $c\_i^\*=c\_i$ for all $i$, commutative if $c\_ic\_j=c\_jc\_i$ for all $i,j$, real if $n\_{ij}^k\in \mathbb R$ for all $i,j,k$, positive if $n\_{ij}^k\geq 0$ for all $i,j,k$ and normalized if $\sum\_k n\_{ij}^k =1$ for all $i,j$. A hypergroup is a generalized hypergroup which is both positive and normalized (if positive is replaced by real you get what's called a signed hypergroup).
Now coming to canonical hypergroups, it is easy to see that associated to any hypergroup you have a new one where the hyperoperation is defined by
$$c\_i\circ c\_j=\{c\_k \quad | \quad n\_{ij}^k\neq 0\}$$ and it is in this sense that they are to be thought of as canonical, and if you accept canonical hypergroups as important then the non-canonical ones are too.
All of the above is written from Wildberger's "Finite commutative hypergroups and applications from group theory to conformal field theory", and let me add here for the ones who can not reach the article a list of mentioned mathematical objects/theories that are very close to the concept of a hypergroup and have been studied under a plethora of different names: Kawada's work on C-algebras, Levitan's work on generalized translation operators, Brauer's work on pseudogroups, Hecke algebras, hypercomplex systems (referring to Berezansky and Kalyushnyi, Vainermann), paragroups (Ocneanu), superselection sectors (Doplicher, Haag and Roberts, Longo), Bose Mesner algebras, Racah Wigner algebras, centralizer algebras, table algebras (Arad and Blau), association schemes and the fusion rules of conformal field theories (Verlinde, Moore and Seiberg). You can look at the article for references.
[Association schemes](http://en.wikipedia.org/wiki/Association_scheme) are for example hypergroups having renormalizations that can be realized by $0,1$-matrices and are very important in algebraic combinatorics and coding theory.
| 16 | https://mathoverflow.net/users/2384 | 30388 | 19,804 |
https://mathoverflow.net/questions/30387 | 3 | An old question that occurred to me again recently: are there any explicit formula known for sequences of irreducible polynomials $g\_{p^n}(X)$ in $Z/pZ[X]$ such that for the finite field with $p^n$ elements
$F\_{p^n} = (Z/pZ)[X] / (g\_{p^n}).$
Such polynomials exist, as anyone who's studied algebra knows, but I've always seen their existence proved nonconsctructively. Ideally, I'm looking for something like
$g\_{p^n}(X) = X^n+c\_{n-1}(p^n)X^{n-1}+...,$
where $c\_{n+1}(p^n)$ is some familiar expression, or something with generating functions, etc.
I'm not an algebraist, so this could have a very obvious answer. One could experiment around with low lying irreducible polynomials to try to come up with a nice pattern, I think, but the fact that I've never seen this presented in an algebra text leads me to suspect that no such formula actually exist and a better answer would come from just asking people in a position to know. In some ways, I realize, it may be like asking for a formula for the nth prime...
| https://mathoverflow.net/users/5621 | Explicit representations of finite fields | There is no known simple explicit formula for an irreducible polynomial
of given degree $n$ over $\mathbb{F}\_p$. However there has been a lot of work
on explicit irreducible polynomials for certain families of $n$, notably
$n$ of the form $rq^s$ where $r$ is fixed and $q$ is a fixed prime. For one
modest contribution to this field see
[this paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WFM-45M8VYC-G&_user=121739&_coverDate=01%2F31%2F1997&_rdoc=1&_fmt=high&_orig=browse&_srch=doc-info%28%23toc%236798%231997%23999969998%23305679%23FLP%23display%23Volume%29&_cdi=6798&_sort=d&_docanchor=&_ct=8&_acct=C000010018&_version=1&_urlVersion=0&_userid=121739&md5=2c4f893bf37032dd8f61bf41c9eda04d) but also see the papers it cites.
| 3 | https://mathoverflow.net/users/4213 | 30389 | 19,805 |
https://mathoverflow.net/questions/30381 | 13 | What is authoritative canonical formal definition of function?
For example,
According to [Wolfram MathWorld](http://mathworld.wolfram.com/Function.html),
$$isafun\_1(f)\;\leftrightarrow\;
\forall a\in f\;(\exists x\exists y \;\langle x,y\rangle = a)
\; \wedge \;
\forall x\forall y\_1\forall y\_2\;((\langle x,y\_1\rangle\in f\wedge\langle x,y\_2\rangle \in f)\rightarrow y\_1=y\_2))$$
According to Bourbaki "Elements de Mathematiques, Theorie des Ensembles",
$$
isafun\_2(f)\;\leftrightarrow\;
\exists d\exists g\exists c\;(\langle d,g,c\rangle=f
\;\wedge\;isafun\_1(g)\;\wedge$$
$$\;\wedge\;
\forall x(x \in d\rightarrow \exists y(\langle x,y\rangle \in g))
\;\wedge\;
\forall x\forall y(\langle x,y\rangle \in g\rightarrow (x \in d\wedge y\in c)))
$$
How to make agree definition of function as triple with extensional equality
$$
\forall f\forall g\;[\;(isafun(f)\wedge isafun(g))
\; \rightarrow \;
[\;(\forall x(\;f(x)=g(x)\;))\leftrightarrow f=g\;]\;]
$$
?
Why such divergences in definitions exist?
Upd: Two additional questions:
1. Why function is not a pair in $isafun\_2$? First component of triple is perfectly derivable from the second.
2. What word `function` exactly means if no underlying theory is specified in context? If I build fully formal knowledge base about mathematics for automated reasoning and want to add notion of contextless function -- how I must describe it?
| https://mathoverflow.net/users/7257 | Definition of Function | The fact is that different subject areas of mathematics use different definitions for this basic concept. The Bourbaki definition is quite common, particularly in many of the areas well-represented here on MO, but other areas use the ordered-pair definition.
For example, if you open any set-theory text, you will find that a function $f$ is a set of ordered pairs having the functional property that any $x$ is paired with at most one $y$, denoted $f(x)$. This definition, which is completely established and much older than the Bourbaki definition, makes a function a special kind of binary relation, which is any set of ordered pairs. The domain of a function is the set of $x$ for which $f(x)$ exists. The range is the set of all such $f(x)$, and so on. The assertion $f:A\to B$ is a statement about the *three* objects, $f$, $A$ and $B$, that $f$ is a function with domain $A$ having its range a subset of $B$. In particular, the same function $f$ can have many different codomains.
Another useful variation of the function concept is the concept of a *partial* function, common in many parts of logic, particularly set theory and computability theory. A partial function on $A$ is simply a function whose domain is included in $A$. In this case, we write $f:A\to B$, but with with *three* dots (my MO tex ability can't seem to do it), to mean that $f$ is a function with $dom(f)\subset A$ and $ran(f)\subset B$. This notion is particularly usefful in computability theory, where one has functions that might not produce an output on all input. But it also arises in set theory, where one often build partial orders consisting of small partial functions from one set to another. The union of a chain of such functions is a function again. It would be silly to insist in the Bourbaki style that there are really invisible functors running through this construction adjusting the domains and co-domains.
One could object that the set-theorists could use the Bourbaki definition, if only they prepared better: in any context where many functions are treated, they should simply delimit an upper bound for the co-domains under consideration and use that co-domain for all the functions. But this proposal bumps into set-theoretic issues. For example, if I consider the class of all functions from an ordinal to the ordinals, then the only common co-domain is the class of all ordinals. But as this is a proper class, it isn't available if I want to consider only set functions. So there are good set-theoretic reasons not to use the Bourbaki definition.
There are numerous other basic concepts that are given different precise meanings in different subjects of mathematics. For example, the concept of *tree*. In graph theory, it is a graph with no loops, whereas in set theory, it is a kind of partial order. In finite combinatorics, it might be a finitte partial order having no diamonds, but in the infinitary theory, one often means a partial order such that the predecessors of every node are well-ordered (making the levels of the tree form a well-ordered hierarchy). The graph-theoretic definition does not allow for the cases of Souslin trees and Kurepa trees, which are central in the other theory.
There are surely numerous other examples where terminology differs.
| 29 | https://mathoverflow.net/users/1946 | 30397 | 19,810 |
https://mathoverflow.net/questions/30307 | 24 | I heard or have read the following nice explanation for the origin of the convention that one uses (almost) always $x,y,z$ for variables. (This question was motivated by question
[Origin of symbol \*l\* for a prime different from a fixed prime?](https://mathoverflow.net/questions/30081/origin-of-symbol-l-for-a-prime-different-from-a-fixed-prime))
It seems this custom is due to the typesetter of Descartes. Descartes used initially other letters
(mainly $a,b,c$) but the typesetter had the same limited number of lead symbols
for each of the 26 letters of the Roman alphabet. The frequent use of variables exhausted his stock and he asked thus Descartes if he could use the last three letters $x,y,z$ of the alphabet (which occur very rarely in French texts).
Does anyone know if this is only a (beautiful) legend or if it contains some truth? (I checked that Descartes uses indeed already $x,y,z$ generically for variables in his printed works.)
| https://mathoverflow.net/users/4556 | Explanation why $x,y,z$ are always variables | You'll find details on this point (and precise references) in Cajori's *History of mathematical notations*, ¶340. He credits Descartes in his *La Géometrie* for the introduction of $x$, $y$ and $z$ (and more generally, usefully and interestingly, for the use of the first letters of the alphabet for known quantities and the last letters for the unknown quantities) He notes that Descartes used the notation considerably earlier: the book was published in 1637, yet in 1629 he was already using $x$ as an unknown (although in the same place $y$ is a known quantity...); also, he used the notation in manuscripts dated earlier than the book by years.
It is very, very interesting to read through the description Cajori makes of the many, many other alternatives to the notation of quantities, and as one proceeds along the almost 1000 pages of the two volume book, one can very much appreciate how precious are the notations we so much take for granted!
| 28 | https://mathoverflow.net/users/1409 | 30414 | 19,821 |
https://mathoverflow.net/questions/28519 | 26 | Hi,
I'm starting to prepare a graduate topics course on Complex and Kahler manifolds for January 2011. I want to use this course as an excuse to teach the students some geometric analysis. In particular, I want to concentrate on the Hodge theorem, the Newlander-Nirenberg theorem, and the Calabi-Yau theorem.
I have many excellent references (and have lectured before) on the Hodge and CY theorems. However, for the Newlander-Nirenberg theorem, I am finding it hard to find a "modern" treatment. I recall going through the original paper in my graduate student days, but I hope that there is a more streamlined version floating around somewhere. (I want to consider the general smooth case, not the easy real-analytic version). Besides the original paper, so far I can only find these references:
J. J. Kohn, "Harmonic Integrals on Strongly Pseudo-Convex Manifolds, I and II" (Annals of Math, 1963)
and
L. Hormander, "An introduction to complex analysis in several variables" (Third Edition, 1990)
Both are easier to follow than the original paper. But my question is: are there any other proofs in the literature, preferably from books rather than papers? The standard texts on complex and Kahler geometry that I have looked at don't have this.
| https://mathoverflow.net/users/6871 | References for "modern" proof of Newlander-Nirenberg Theorem | There is a proof due to Malgrange which can be found in Nirenberg's, Lectures on Linear Partial Differential Equations. I am not sure that one can call the proof modern, but it is the simplest proof that I know.
| 13 | https://mathoverflow.net/users/7300 | 30420 | 19,823 |
https://mathoverflow.net/questions/30392 | 17 | A bit unsure if my use/mention of proprietary software might be inappropriate or even frowned upon here. If this is the case, or if this kind of experimental question is not welcome, please let me know and I'll remove it quickly.
I was experimenting on the distribution of eigenvalues of random matrices via the following Mathematica code
```
dim = 301;
decomplex = {z_Complex -> {Re[z], Im[z]}, A_Real -> {A, 0}}; b =
Eigenvalues[
SparseArray[{i_, i_} -> 1, {dim, dim}] +
Table[RandomReal[{-.9, 1.}], {i, 1, dim}, {j, 1, dim}]] /. decomplex;
ListPlot[b, PlotStyle -> PointSize[0.015], AspectRatio -> Automatic]
```
This simply draws (or should draw) a picture of the complex eigenvalues of a 301 x 301 matrix, with entries uniformly distributed in the real interval [-0.9, 1]. If you try the code you will notice a disk of random eigenvalues, which is expected, plus a single positive eigenvalue at some distance of the disk. Decreasing the lower bound on the entries to -1 makes this phenomenon disappear, while increasing the lower bound makes it more evident.
Question: is this an artifact, a real phenomenon, or maybe even a well known and simple to explain phenomenon?
| https://mathoverflow.net/users/7294 | An experiment on random matrices | The distribution of the bulk of the spectrum is an example of the [circular law](http://mathworld.wolfram.com/GirkosCircularLaw.html). For the model you selected (where each entry is uniformly chosen at random from an interval), the law was first proven [by Bai](http://www.ams.org/mathscinet-getitem?mr=1428519) (at least in the case where the entries are normalised to have mean zero), building upon previous work of Girko; the non-central case (non-zero mean) was recently established [by Chafai](https://mathscinet.ams.org/mathscinet-getitem?mr=2735731).
The non-central case is a rank one perturbation of the central case (by the matrix whose entries are all equal to the mean) and should therefore cause one exceptional eigenvalue.
(In the central case it can be shown that the spectral radius is close to the radius of the disk, so there are basically no exceptional eigenvalues.) If the mean of each entry is $\mu$, then the rank one perturbation has an eigenvalue at $\mu n$, so one expects the exceptional eigenvalue to linger near this number. In your case, $\mu = 0.05$ and $n = 301$, so the exceptional eigenvalue should linger near $15.05$.
There is a [paper of Silverstein](http://www.ams.org/mathscinet-getitem?mr=1284550) which makes the above heuristics precise; see also the earlier [work of Andrew](http://www.ams.org/mathscinet-getitem?mr=1062321).
There is quite a bit of recent literature on the circular law, see for instance [this survey](http://arxiv.org/abs/0810.2994) by Van Vu and myself, or [my lecture notes](http://terrytao.wordpress.com/2010/03/14/254a-notes-8-the-circular-law/) on this topic.
| 21 | https://mathoverflow.net/users/766 | 30433 | 19,831 |
https://mathoverflow.net/questions/30436 | 4 | What kind of 3-manifolds can arise as hypersurfaces $\{ f(x,y,z,w) = 0\} \subset \mathbb{R}^4$? Can they have nontrivial H1 or H2?
| https://mathoverflow.net/users/1358 | What kind of 3-manifolds arise has hypersurfaces in R^4? | A simple construction that bears on the narrow version of John's question: If $M$ is a closed $n$-manifold that embeds in $\mathbb{R}^{n+1}$ (which can only happen if $M$ is orientable), then $M \times S^k$ embeds in $\mathbb{R}^{n+1+k}$. Thicken $M$ in $\mathbb{R}^{n+1}$, then cross with $I^k$ in the new dimensions, and then take the boundary of that. By induction, then, any product of spheres embeds in the next dimension.
On the other hand, Ryan Budney in [arXiv:0810.2346](http://arxiv.org/abs/0810.2346) has both new results and a bibliography of the broad question of which closed 3-manifolds embed in $\mathbb{R}^4$. It is problem 3.20 in Kirby's problem list, it is an interesting open problem, and there have been several partial results.
| 5 | https://mathoverflow.net/users/1450 | 30444 | 19,838 |
https://mathoverflow.net/questions/30425 | 12 |
>
> Let $n>1$. Which smooth manifolds admit a diffeomorphism $f$ of order $n$?
>
>
>
For a closed orientable surface $S\_g$ of genus $g$ and $n=2$ the answer is in the affirmative, since $S\_g$ can be embedded in $\mathbb{R}^3$ in a symmetric way. A similar argument gives a positive answer for $n=3$: $S\_g$ can be embedded in $\mathbb{R}^3$ with rotational symmetry of order 3. The first problem I've encountered in case of $S\_g$ is with $g=2$ and $n=4$: all my candidates for $f$ turned out to have order $2$. And I have no idea as to what happens for $g=2$ and $n=5$.
In the case of a general manifold $M$, the existence of a nontrivial global flow $\phi\_t$ such that $\phi\_0=\phi\_1$ would give a positive answer for every $n$, but it seems to be a much stronger condition.
Is a general answer to this question known? If not, maybe there are some simple examples of manifolds without diffeomorphisms of given order?
| https://mathoverflow.net/users/4698 | Which manifolds admit a diffeomorphism of order $n$? | The Nielsen Realisation Problem asks when a (finite) subgroup of the mapping class group (the group of isotopy classes of diffeomorphisms) of a surface can be realised as a group of diffeomorphisms. Kerckhoff proved in the 80s that **every** finite subgroup of the mapping class group can be realised. (For infinite subgroups, there are various known obstructions, such as the Miller-Morita-Mumford characteristic classes.) Thus, Kerckhoff's theorem implies that a surface admits a diffeomorphism of order n if its mapping class group has an element of order n. Conversely, one can show that any diffeomorphism of a surface must have infinite order if it is isotopic to the identity, every diffeomorphism of order n gives an order n mapping class group element.
If you have a diffeomorphism of finite order on a surface then you can find a complex structure (or a Riemannian metric or a symplectic structure or a conformal structure) for which the diffeomorphism is an automorphism/isometry. This is accomplished by choosing an arbitrary metric and then averaging over all translates of it by powers of the diffeomorphism.
So the point of all this is that in dimension 2 finding an order n diffeomorphism of a genus g surface is the same as finding a complex curve with an order n isometry, or equivalently, a Z/n orbifold point in the moduli space. As Sam and Robin alluded to, there is a bound on the order of n relative to g. Hurwitz's theorem states that the order of the automorphism group of a genus g curve is less than or equal to 84(g−1). There are various other theorems that tell you about what sorts of finite subgroups you can find in mapping class groups.
In higher dimensions, it's harder to give a useful answer. If your manifold has a circle action then you are done because $S^1$ contains Z/n for any n. But there are plenty of manifolds around which do not admit circle actions, such as K3 surfaces. The $\widehat{A}$-genus is an obstruction to admitting a circle action. Some nice things are known about the finite groups of automorphisms of K3 surfaces. In fact, I think they are pretty much completely classified into a finite list.
In general, by averaging over translates of a metric, you can still assume that a given finite order diffeomorphism acts by isometries for some metric. Generally, the isometry group of a compact Riemannian manifold will be a finite dimensional compact Lie group (I think this is a theorem of Yau).
| 12 | https://mathoverflow.net/users/4910 | 30445 | 19,839 |
https://mathoverflow.net/questions/30434 | 3 | A friend is looking for a clean proof of the following inequality of Bernstein: If $f: R \to R$ is a bounded function whose Fourier transform has compact support, then
$ \|f'\|\_{\infty} \le C \| f \|\_{\infty} $
where $C$ only depends on the support of the Fourier transform. Any reference would be very much appreciated.
| https://mathoverflow.net/users/3635 | A proof of the Bernstein inequality | The Fourier transform of $f'(x)$ is $i\xi\hat{f}(\xi)$, which has the same support as $\hat{f}(\xi)$. So we can write $i\xi\hat{f}(\xi)$ = $i\xi\hat{f}(\xi)\phi(\xi)$, where $\phi(\xi)$ is a smooth bump function depending on the support of $\hat{f}$, that is equal to one on the support of $\hat{f}$. Taking inverse Fourier transforms, we get $f'(x) = f(x) \star g(x)$, where $g(x)$ is the inverse Fourier transform of $i\xi\phi(\xi)$. From the definition of convolution, one gets $|f'(x)| \leq ||f||\_{\infty}||g||\_1$. Since this holds for any $x$ and $||g||\_1$ depends only on the support of $\hat{f}$, you get the desired inequality with $C = ||g||\_1$.
| 12 | https://mathoverflow.net/users/2944 | 30447 | 19,841 |
https://mathoverflow.net/questions/30357 | 9 | Does anyone know of any lower bounds on the number of nonzero digits that appear in powers of 2 when written to base 3? (Other than the easy "If it's more than 8 it has to have at least 3.") I know there's been some stuff done with powers of 2 written to base 3, but I can't seem to find anything that quite answers this question.
(Are there more general such bounds? Powers of a written to base b? (To avoid triviality, suppose that no power of a is a power of b... or are stricter conditions needed?) From what I can tell, it looks like even specific instances of these are hard, so I suppose I should stick to the specific version.)
| https://mathoverflow.net/users/5583 | Lower bound on # of nonzero digits in ternary expansions of powers of 2? | A nontrivial lower bound can be found in a paper of Cam Stewart (see <http://www.math.uwaterloo.ca/PM_Dept/Homepages/Stewart/Jour_Books/J-reine-ange-Math-1980.pdf>). He proves, more generally, for fixed bases a and b for which $\log a/\log b$ is irrational, that the sum of the number of nonzero digits in the base a and base b digits of an integer n exceeds (essentially) $$\log \log n/ \log \log \log n.$$
| 13 | https://mathoverflow.net/users/7302 | 30449 | 19,843 |
https://mathoverflow.net/questions/30453 | 17 | Possibly the correct answer to this question is simply a pointer towards some recent literature on Tannaka-Krein-type theorems. The best article I know on the subject is the excellent
* André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, *Category Theory, Lecture Notes in Math*, 1991 vol. 1488 pp. 412–492
from which I'm getting most of my facts.
Let me fix, once and for all, a field $\mathbb K$. I will denote by $\text{Vect}$ the usual category of all $\mathbb K$-vector spaces, and by $\text{FinVect}$ its full subcategory of dualizable objects. The notions of (coassociative, counital) **coalgebra** and (right, say, and coassociative and counital) **comodule** are standard; briefly, if $(\mathcal V,\otimes)$ is a monoidal category, a coalgebra in $\mathcal V$ is an object $A$ along with a map $A \to A \otimes A$ satisfying some axioms, and an $A$-comodule is an object $X$ and a map $X \to X\otimes A$ satisfying some axioms. Here's the fast way to say the axioms. Without telling you anything, it's clear what should be a homomorphism of comodules, namely a map $f: X\to Y$ so that the maps $X \overset f \to Y \to Y\otimes A$ and $X \to X\otimes A \overset{f\otimes {\rm id}}\to Y\otimes A$ agree. Then a coalgebra $A$ is **coassociative** iff the comultiplication $A\to A\otimes A$ is a homomorphism of $A$-comodules, and a comodule $X$ is **coassociative** iff $X\to X\otimes A$ is a homomorphism of comodules. (In both cases, the comodule structure on the right is just the comultiplication in the second factor.) I will abuse language so that all coalgebras and comodules are coassociative. Oh, and I haven't said anything about counits, but I want them as well. Given a coalgebra $A$, I will denote its category of right comodules by $\text{Comod}\_A$, and of finite-dimensional right comodules by $\text{FinComod}\_A$.
Let's say that a **Tannakian category** is a $\mathbb K$-linear (i.e. $\text{Vect}$-enriched) abelian category $\mathcal C$ along with a faithful exact $\mathbb K$-linear functor $F: \mathcal C \to \text{FinVect}$. (If this is not the most standard definition, let me know.) The fundamental theorem is the following:
>
> *Theorem (c.f. Op. cit.):* Let $F: \mathcal C \to \text{FinVect}$ be any functor from any category. Then there is a coalgebra $\operatorname{End}^\vee(F) \in \text{Vect}$ given by a certain natural colimit (take the definition of the vector space of natural transformations and turn all arrows around; use the fact that the endomorphism of a finite-dimensional vector space is naturally a coalgebra). The functor $F$ factors through $\text{FinComod}\_{\operatorname{End}^\vee(F)}$ (and its forgetful functor to $\text{FinVect}$), and $\operatorname{End}^\vee(F)$ is universal with respect to this property (if $F$ factors through $\text{forget}: \text{FinComod}\_A \to \text{FinVect}$, then there is a map $\operatorname{End}^\vee(F) \to A$ inducing this). If $(\mathcal C,F)$ is Tannakian, then the map $\mathcal C \to \text{FinComod}\_{\operatorname{End}^\vee(F)}$ is an equivalence of categories.
>
>
>
It follows from the above theorem that there is an equivalence of categories between:
$$ \{\text{Tannakian categories}, \text{strictly commuting triangles}\} \leftrightarrow \{\text{coalgebras}, \text{homomorphisms}\}$$
By a "strictly-commuting triangle" of Tannakian categories $(\mathcal C,F) \overset{f}\to (\mathcal D,G)$, I mean a functor $f: \mathcal C \to D$ so that we have strict equality $F = G\circ f$ as functors $\mathcal C \to \text{Vect}$.
But when talking about functors, etc., it's [evil](http://ncatlab.org/nlab/show/evil) to think about strict equality. Rather, there should be some two-categories floating around. There is a natural two-category whose objects are coalgebras: the one-morphisms are bicomodules, the two-morphisms are homomorphisms of bicomodules, and the one-composition is cotensor product of bicomodules. There are, to my mind, a few different two-categories whose objects are Tannakian categories: perhaps the one-morphisms should be triangles that commute up to specified natural isomorphism ("strong"?), or up to specified natural transformation in one way or in the other ("lax" and "oplax"?).
All together, my question is as follows:
>
> *Question:* What is the correct, non-evil two-category of Tannakian categories so that the last two sentences of the "Tannaka theorem" above expresses an equivalence of two-categories
>
>
> $\{\text{Tannakian categories}\} \leftrightarrow \{\text{coalgebras}, \text{bicomodules}, \text{homomorphisms}\}$?
>
>
> Is this two-category a full sub-two-category of some larger two-category whose objects are pairs
> $(\mathcal C, F:\mathcal C \to \text{Vect})$ with no further conditions? If so, does the full theorem represent some sort of "Tannakaization" of such pairs?
>
>
>
If modifications need to be made (not quite the right two-category of coalgebras, not quite the right definition of Tannakian category for the question to have an answer, etc.) feel free to answer the corresponding question.
| https://mathoverflow.net/users/78 | Does the Tannaka-Krein theorem come from an equivalence of 2-categories? | Your question is dealt with (in a slightly more general setting) in section 11 of Daniel Schäppi's paper, [Tannaka duality for comonoids in cosmoi](https://arxiv.org/abs/0911.0977). Specializing to your setting, he shows that there is a biadjunction (a weak 2-categorical form of adjunction) between the 2-category of $k$-linear categories equipped with a functor to $\operatorname{FinVect}$, where morphisms are triangles commuting up to specified natural isomorphism, and the usual category of coalgebras (thought of as a 2-category with only identity 2-morphisms). I believe this biadjunction should restrict to a biequivalence on the sub-2-category of Tannakian categories (as you have defined them).
This biadjunction is useful because there is a nice tensor product on the 2-category of $k$-linear categories that turns this biadjunction into a *monoidal* biadjunction, which gives us a way of relating things like bialgebras and tensor categories.
| 5 | https://mathoverflow.net/users/396 | 30456 | 19,845 |
https://mathoverflow.net/questions/30457 | 10 | Does every Banach space admit an equivalent strictly convex norm (i.e. such a norm, that a unit sphere does not contain segments)?
| https://mathoverflow.net/users/4312 | Strictly convex equivalent norm |
>
> Every separable Banach space has an equivalent norm which is both strictly convex and smooth. For certain nonseparable spaces, in particular, $\ell\_{\infty}(\Gamma)$ with $\Gamma$ uncountable, there may be no equivalent strictly convex or smooth norm.
>
>
>
[Link.](http://books.google.co.uk/books?id=1A7ppEZPXEIC&pg=PA33&lpg=PA33&dq=equivalent+strictly+convex+norm&source=bl&ots=zx13lvLJXV&sig=bBaKVMIB4be6BpBQlUcrpQaTEUQ&hl=en&ei=aaMvTM2jH9S5jAeY843DBQ&sa=X&oi=book_result&ct=result&resnum=5&ved=0CC8Q6AEwBA#v=onepage&q=equivalent%20strictly%20convex%20norm&f=false)
| 11 | https://mathoverflow.net/users/5371 | 30458 | 19,846 |
https://mathoverflow.net/questions/30455 | 9 | A definition of wedge sum can be found here:
<http://en.wikipedia.org/wiki/Wedge_sum>
My professor has claimed that wedge sums of path connected spaces X and Y are well-defined up to homotopy equivalence, independently of choice of base points x0 and y0. Base point here means the points that are identified under the equivalence relation forming the wedge product out of the disjoint union topology of X and Y.
Recall homotopy equivalence of X and Y means that there is f:X->Y and g:Y->X continuous with gf and fg homotopic to the identity.
With these definitions, please prove my professor's claim, which I have failed to do for a week. (It is left as an exercise in his lecture.)
Thanks.
| https://mathoverflow.net/users/7305 | The Wedge Sum of path connected topological spaces | A counterexample is shown on the cover of the paperback edition of the classic textbook Homology Theory by Hilton and Wylie. This can be viewed on the amazon webpage for the book. The example consists of the wedge of two copies of a cone, the cone on the sequence 1/2, 1/3, 1/4, ... together with its limit point 0. With one choice of basepoints the wedge is not contractible, but with other choices it is.
| 19 | https://mathoverflow.net/users/23571 | 30465 | 19,850 |
https://mathoverflow.net/questions/30474 | 0 | I am looking for reference giving the original definition of prime cycles of coherent sheaves on noetherian scheme. Was it in EGA? I googled, but could not find proper reference.
Thanks in advance
| https://mathoverflow.net/users/1851 | What is Grothendieck associated points(Prime cycles) of coherent sheaves on noetherian scheme? | EGA IV$\_4$, 3.1.1 for any quasi-coherent sheaf on any (pre)scheme: the commutative algebra definition on stalks. So one needs 3.1.2 and 3.1.3 there to get useful alternative formulations of this definition in the locally noetherian case. You can find it by looking for "Point associe" or "Cycle premier associe a un Module" in the index, or "Cycles premiers associes a un Module" in the Table of Contents (only 8 volumes to try...).
| 4 | https://mathoverflow.net/users/6773 | 30475 | 19,854 |
https://mathoverflow.net/questions/30480 | 28 | This question is asked from a point of complete ignorance of physics and the standard model.
Every so often I hear that particles correspond to representations of certain Lie groups. For a person completely ignorant of anything physics, this seems very odd! How did this come about? Is there a "reason" for thinking this would be the case? Or have observations in particle physics just miraculously corresponded to representation theory? Or has representation theory of Lie groups grown out of observations in particle physics?
In short: what is the chronology of the development of representation theory and particle physics (with relation to one another), and how can one make sense of this relation in any other way than a freakish coincidence?
| https://mathoverflow.net/users/5309 | Particle Physics and Representations of Groups | The "chronology" isn't clear to me, and having looked through the literature it seems much more convoluted than it should be. Although it seems like this is basically how things were done since the beginning of quantum mechanics (at least, by the big-names) in some form or another, and was 'partly' formalized in the '30s-'40s with the beginnings of QED, but not really completely *carefully* formalized until the '60s-'70s with the development of the standard model, and not really *mathematically* formalized until the more careful development of things in terms of bundles in the '70s-'80s. (These dates are guesses--someone who was a practicing physicist during those periods is more than welcome to correct my timeline!)
Generally speaking, from a 'physics' point of view, the reason particles are labeled according to representations is not too different than how, in normal quantum mechanics, states are labeled by eigenvalues (the wiki article linked to mentions this, but it's not as clear as it could be).
In normal QM, we can have a Hilbert space ('space of states') $\mathcal{H}$, which contains our 'physical states' (by definition). To a physicist, 'states' are really more vaguely defined as 'the things that we get the stuff that we measure from,' and the Hilbert space exists because we want to talk about measurements. The measurements correspond to eigenvalues of operators (why things are 'obviously' like this is a longer historical story...).
So we have a generic state $| \psi \rangle \in \mathcal{H}$, and an operator that corresponds to an observable $\mathcal{O}$. The measured values are
>
> $\mathcal{O} |\psi\rangle = o\_i | \psi \rangle$.
>
>
>
Because the $o\_i$ are observable quantities, it's useful to label systems in terms of them.
We can have a list of observables, $\mathcal{O}\_j$, (which we usually take to be commuting so we can simultaneously diagonalize), and then we have states $|\psi\rangle$,
>
> $\mathcal{O}\_j | \psi \rangle = {o\_i}\_j | \psi \rangle$.
>
>
>
So, what we say, is that we can uniquely define our normal QM states by a set of eigenvalues $o\_{ij}$.
In other words, the $o\_{ij}$ *define* states, from the physics point of view. Really, this defines a basis where our operators are diagonal. We can--and do!--get states that do not have observables which can be simultaneously diagonalized, this happens in things like neutrino oscillation, and is why they can turn into different types of neutrinos! The emitted neutrinos are emitted in states with eigenvalues which are not diagonal in the operator that's equivalent to the 'particle species' operator. (Note, we could just as well define the 'species' to be what's emitted, and then neutrinos would not oscillate *in this basis*, but would in others!)
This has to do with representations, because when we talk about particles with spin, for example, we're talking about operators which correspond to 'angular momentum.' We have an operator:
>
> $L\_z = i \frac{\partial}{\partial\phi}$
>
>
>
and label eigenvalues by half-integer states which physically correspond to spin. Group theoretically, $L\_z$ comes from the lie algebra of the rotation group, because we're talking about angular momentum (or spin) which has associated rotational symmetries.
Upgrading from here to quantum field theory (and specializing that to the standard model) is technically complicated, but is basically the same as what's going on here. The big difference is, we want to talk there about 'quantum fields' instead of states, and have to worry about crazy things like apparently infinite values and infinite dimensional integrals, that confuse the moral of the story.
But the idea is simply, we want to identify things by observables, which correspond to eigenvalues, which correspond to operators, which correspond to lie algebra elements, which have an associated lie group.
So we define states corresponding to things which transform under physically convenient groups as 'particles.'
If you want a more mathematically careful description, that's still got some physical intuition in it, you can check out Gockler and Schuker's "Differential Geometry, Gauge theory, and Gravity," which does things from the bundle point of view, which is slightly different than I described (because it describes classical field theories) but the moral is similar. At first it might seems surprising that the classical structure here is the same, when it seemed to rely on operators and states in Hilbert spaces, but it only technically relied on it, but morally, what's important is actions under symmetry groups. And that is in the classical theory as well. But it's not as physically clear from the beginning from that point of view.
| 11 | https://mathoverflow.net/users/3329 | 30485 | 19,860 |
https://mathoverflow.net/questions/30497 | 7 | What is the étale fundamental group of projective space over an algebraically closed field?
In char = 0 it is trivial (Lefschetz principle), as well as in dimension 1 (Riemann-Hurwitz).
| https://mathoverflow.net/users/6960 | étale fundamental group of projective space | It is a birational invariant (for smooth proper connected schemes over a field, ultimately due to Zariski-Nagata purity of the branch locus), and its formation is compatible with products (for proper connected schemes over an algebraically closed field), so we can replace projective $n$-space with the $n$-fold product of copies of the projective line to conclude. Likewise, due to limit arguments and invariance of the etale site with respect to finite radiciel surjections (such as a finite purely inseparable extension of a ground field), it suffices to take the ground field to be separably closed rather than algebraically closed. This is all in SGA1.
| 11 | https://mathoverflow.net/users/6773 | 30499 | 19,867 |
https://mathoverflow.net/questions/30494 | 6 | Suppose an operator $T$ is bounded on $L^2$ and also bounded from $L^{1}$ to $L^{1}$-weak. Then by Marcinkewicz interpolation one gets that $T$ is bounded on every $L^{p}$ for p between 1 and 2. Precise versions of the theorem (see the book of L.Grafakos) give an estimate of the norm of $T$ on $L^{p}$, and of course the norm diverges as $p\to 1$. If not, by a simple argument one could obtain that $T$ is bounded also on $L^{1}$, which is clearly false in general (e.g. singular integrals). By the way, even a weaker assumption on the growth of the norm should allow one to conclude that $T$ is bounded on $L^{1}$, as in extrapolation theorems.
Now suppose you (me) are hard-headed and want to use the general machinery of real interpolation, say the K-method as detailed in the book of Bergh and Löfström. Then $T$ is bounded from $L^{1}$ to $L^{1}$-weak which means the Lorentz space $L^{1,\infty}$, with norm $M\_{0}$, and also from $L^{2}$ to $L^{2}$ with norm $M\_{1}$. Then $T$ is bounded on the corresponding real interpolation spaces with norm $M\_{0}^{1-\theta}M\_{1}^{\theta}$. Real interpolates of Lorentz spaces are Lorentz spaces, see Theorem 5.3.1 in B-L. We conclude that $T$ is bounded on every Lorentz space $L^{p,q}$ with p between 1 and 2, and $1\le q\le \infty$, and in particular on $L^{p,p}=L^{p}$ as expected.
Unfortunately, now I have a uniform bound on the norm of $T$ as $p\to 1$, which would allow me to conclude that $T$ is also bounded on $L^{1}$. Where is the mistake? There must be some inaccuracy in one of the steps, but which one exactly?
| https://mathoverflow.net/users/7294 | A puzzling question on real interpolation | (Note: I don't have my copy of B&L handy, so I'm sort of doing this from memory.)
The problem is that $(L\_{p\_0.q\_0},L\_{p\_1,q\_1})\_{\theta,q} = L\_{p\_\theta,q}$ *under equivalent norm*. There's no saying what the constant of equivalency are. Now, if you look at $T: \{X\_1,X\_2\} \to \{Y\_1,Y\_2\}$ with norm $\{M\_1,M\_2\}$, then you have
$$ T: (X\_1,X\_2)\_{\theta,q} \to (Y\_1,Y\_2)\_{\theta,q} $$
with norm $M\_1^{(1-\theta)}M\_2^\theta$. Now put in $X\_1 = L\_{1,1}$, $X\_2 = L\_{2,2}$, $Y\_1 = L\_{1,\infty}$ and $Y\_2 = L\_{2,2}$, while the interpolants $(X\_1,X\_2)\_{\theta,q}$ may be equivalent to $L\_{p\_\theta,q}$, and $(Y\_1,Y\_2)\_{\theta,q}$ to $L\_{p\_\theta,q}$, the constants of equivalency may be different. That's where the degeneracy is hiding.
Observe that if $Y\_1$ were $L\_{1,1}$ (or if $X\_1 = L\_{1,\infty}$) instead, then the two interpolation spaces are actually equal, which is what you expect.
| 8 | https://mathoverflow.net/users/3948 | 30503 | 19,869 |
https://mathoverflow.net/questions/30491 | 3 | Is there any remotely efficient way to determine whether a graph can be disconnected by the removal of fewer than k edges, or even one that has a lower asymptotic complexity than just trying each set of k-1 edges?
If it helps, you can assume the graph is k-regular that k is much smaller than the number of vertices.
| https://mathoverflow.net/users/nan | determining k-edge-connectivity of a graph | Try:
<http://portal.acm.org/citation.cfm?id=122416>
and the references there.
| 2 | https://mathoverflow.net/users/1618 | 30507 | 19,870 |
https://mathoverflow.net/questions/30509 | 1 | In Munkres (Chapter 3, Section 23, p. 148), Munkres shows that if a subspace $Y$ of a space $X$ is not connected, then there are two disjoint open subsets $A,B$ such that the union of $A$ and $B$ contains $Y$. What doesn't make sense is that a separation of $Y$ only requires two open subsets of $Y$ which are disjoint, *as subsets of $Y$*. That is, $A$ and $B$ certainly can't intersect anywhere in $Y$, but who says they have to be disjoint? Or is asking whether a subset $Y$ of $X$ is connected *as a subset of $X$* a different question than saying "okay, here is the topology on $Y$ as a subset of $X$ - now is $Y$ connected under that topology?"
| https://mathoverflow.net/users/1355 | Definition of Connected Subspace | Per your comment, I think you misunderstood what Munkres is trying to say.
>
> If Y is subspace of X, a separation of Y is a pair of disjoint nonempty sets A and B whose union is Y, neither of which contains a limit point of the other. The space Y is connected if there exists no separation of Y.
>
>
>
I read it to mean two definitions are given. Firstly, that he defines what it means to have a separation of a subspace $Y$ inside of the space $X$. Then he defines a space $Y$ as connected if it cannot be separated within itself. ($Y$ is trivially a subspace of itself, so the first definition of a separation can be used.)
Now, in the example I gave where $X = \{a,b,c\}$, $Y = \{a,c\}$, with topology on $X$ generated by $\{a,b\}, \{b,c\}$, the subspace $Y$ is not a connected set in $X$, as it is not a connected space in its subspace topology. But the space $X$ is connected, so the connected component of $\{a\}$ in $X$ is the whole space. (Whereas the connected component in $Y$ is itself.)
This shouldn't be so strange if you consider a more intuitive example: Let $X$ be the open interval $(0,1)$, and $Y$ be the subset $(0,1/4)\cup (3/4,1)$. Then $Y$ is not connected. The connected component containing the point $1/8$ in $X$ is the whole space, whereas the connected component when considered in $Y$ is just the interval $(0,1/4)$.
In other words, the connected component of a point in $X$ is a subspace $Y$ such that $Y$ is connected in the subspace topology and such that $Y$ and $X\setminus Y$ are both open. (And $Y$ of course contains the point in question.)
| 3 | https://mathoverflow.net/users/3948 | 30510 | 19,872 |
https://mathoverflow.net/questions/30487 | 11 | My advisor mentioned to me that he talked to Witten last summer on representation theory, and Witten told him that unitary representations of Kac-Moody algebra are important to working physicists. But he did not explain in details to me.
My question is Why?
Second question: I know there are a lot of people devoting to studying unitary representation of Lie group. But are there papers investigating unitary representations of Kac-Moody algebra?
I am not an expert, so this question might be naive. Thanks
| https://mathoverflow.net/users/1851 | Why do Physicists need unitary representation of Kac-Moody algebra? | As others have mentioned, the reasons lie indeed in two-dimensional conformal field theory and in string theory.
The propagation of string on a compact Lie group $G$ is described by the Wess-Zumino-Witten model, whose dynamical variables are maps $g:\Sigma \to G$ from a riemann surface $\Sigma$ to $G$. The quantisation of that model is difficult in terms of $g$ (although see the [1988 papers](http://www-spires.dur.ac.uk/cgi-bin/spiface/hep/www?rawcmd=FIND+A+GAWEDZKI+AND+A+KUPIAINEN+and+date+1988&FORMAT=www&SEQUENCE=) of Gawedzki and Kupiainen, and also Felder, for a functional integral approach) and one instead chooses to quantise their currents, roughly the (anti)holomorphic components of the pullbacks $g^\*\theta\_L$ and $g^\*\theta\_R$ of the Maurer-Cartan forms on $G$. There is a natural action of two copies of the affine Kac-Moody algebra associated to $G$ on the WZW model which preserves the Poisson structure of the WZW model and gives rise to moment mappings which are, essentially, the currents. In other words, the Poisson bracket of the currents is that of two copies of the affine Kac-Moody algebra of $G$. Hence the quantisation naturally leads one to consider unitary, integrable representations of the affine Kac-Moody algebra. The first "modern" reference for this is a [1986 paper](http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+GEPNER+AND+A+WITTEN&FORMAT=WWW&SEQUENCE=) of Doron Gepner and Edward Witten *String Theory on Group Manifolds*; although there are pioneering papers of Halpern, Bardakci,... in from the late 1960s and early 1970s.
At a more abstract level, we can substitute the group $G$ by any (unitary) two-dimensional (super)conformal field theory with the right central charge. This idea of replacing the geometry by a conformal field theory used to be known as "strings without strings", since one loses the description of strings propagating in some geometry. In this context, it is important to have at one's disposal a number of unitary two-dimensional conformal field theories. The natural ones are those coming from from unitary representations of infinite-dimensional Heisenberg and Clifford algebras (so-called *free fields*) and unitary integrable representations of affine Kac-Moody algebras, but one can also consider constructions (e.g., orbifolds, coset constructions,...) which generate new unitary CFTs from these ones. The first "modern" reference for the coset construction is perhaps the [1986 paper](http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+goddard+and+a+kent+and+a+olive&FORMAT=www&SEQUENCE=) of Peter Goddard, Adrian Kent and David Olive *Unitary representations of the Virasoro and superVirasoro algebras*.
Finally, I should say that although it's the *affine* Kac-Moody algebras which seem to have played the more important rôles thus far, there is also the emergence (in the context of M-theory) of more general Kac-Moody algebras. There's work on this in King's College London (West et al.), Brussels (Henneaux et al.) and Potsdam (Nicolai et al.). I'm not very familiar with this, though.
| 9 | https://mathoverflow.net/users/394 | 30515 | 19,875 |
https://mathoverflow.net/questions/20946 | 35 | Given a polynomial equation $x^n+a\_{n-1}x^{n-1}+\cdots+a\_1x+a\_0=0$, where $n$ is even and all the coefficients $a\_i$ are real, what is the best way to determine whether it has a real root or not?
I know [Sturm's theorem](http://en.wikipedia.org/wiki/Sturm%27s_theorem), but I am wondering if it's possible to determine this by the sign of some form of resolvent (*e.g.*, [discriminant](http://en.wikipedia.org/wiki/Discriminant) works for $n=2$, but not 4 or beyond)?
Please point me to some reference if this has already been studied. Thanks for your time!
---
I apologize for not searching hard enough - I just realized there is a similar question discussed [here](https://mathoverflow.net/questions/9431/when-does-a-real-polynomial-have-a-pair-of-complex-conjugate-roots) before, and has already got lots of useful comments. My hope is to find a function of these real coefficients and whether the polynomial has real root or not is determined by its sign.
| https://mathoverflow.net/users/3350 | Criteria to determine whether a real-coefficient polynomial has real root? | There is indeed an easy way to check if a univariate poly with real coefficients has a real root, without computing the roots.
Note that the answer for odd degree polynomials is always yes. For an even degree polynomial $p(x)$ do the following:
1. Compute the Hermite form of the polynomial. This is a symmetric matrix defined e.g. [here](http://stellar.mit.edu/S/course/6/sp10/6.256/courseMaterial/topics/topic2/lectureNotes/lecture-05/lecture-05.pdf): (on page 4, near the bottom denoted by $H\_1(p)$). The entries of this matrix can be filled using [Newton identities](http://en.wikipedia.org/wiki/Newton%27s_identities).
2. The number of real roots of $p(x)$ is equal to the signature of the Hermite matrix $H\_1(p)$, i.e., the number of positive eigenvalues minus the number of negative eigenvalues.
3. Since the Hermite form is symmetric, its characteristic poly has
only real roots. Therefore, we can apply [Descartes' rule of signs](http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs) to its char.
polynomial, which would give us exactly the number of positive and negative
eigenvalues of the Hermite form.
This process gives you the exact number of real roots of $p(x)$ without computing the roots, and in particular you can see if the polynomial has a real root.
Hope this helps.
-Amirali Ahmadi
| 49 | https://mathoverflow.net/users/50854 | 30533 | 19,888 |
https://mathoverflow.net/questions/30538 | 1 | I want to generate a code on n bits for k different inputs that I want to classify. The main requirement of this code is the error-correcting criteria: that the minimum pairwise distance between any two encodings of different inputs is maximized. I don't need it to be exact - approximate will do, and ease of use and speed of computational implementation is a priority too.
In general, n will be in the hundreds, k in the dozens.
Also, is there a reasonably tight bound on the minimum hamming distance between k different n-bit binary encodings?
| https://mathoverflow.net/users/942 | Algorithm for generating a size k error-correcting code on n bits | My answer that I gave on StackOverflow:
The problem of finding the exact best error-correcting code for given parameters is very hard, even approximately best codes are hard. On top of that, some codes don't have any decent decoding algorithms, while for others the decoding problem is quite tricky.
However, you're asking about a particular range of parameters where n ≫ k, where if I understand correctly you want a k-dimensional code of length n. (So that k bits are encoded in n bits.) In this range, first, a random code is likely to have very good minimum distance. The only problem is that decoding is anywhere from impractical to intractible, and actually calculating the minimum distance is not that easy either.
Second, if you want an explicit code for the case n ≫ k, then you can do reasonably well with a [BCH code](http://en.wikipedia.org/wiki/BCH_code) with q=2. As the Wikipedia page explains, there is a good decoding algorithm for BCH codes.
Concerning upper bounds for the minimum Hamming distance, in the range n ≫ k you should start with the [Hamming bound](http://en.wikipedia.org/wiki/Hamming_bound), also known as the volume bound or the sphere packing bound. The idea of the bound is simple and beautiful: If the minimum distance is t, then the code can correct errors up to distance floor((t-1)/2). If you can correct errors out to some radius, it means that the Hamming balls of that radius don't overlap. On the other hand, the total number of possible words is 2n, so if you divide that by the number of points in one Hamming ball (which in the binary case is a sum of binomial coefficients), you get an upper bound on the number of error-free code words. It is possible to beat this bound, but for large minimum distance it's not easy. In this regime it's a very good bound.
| 4 | https://mathoverflow.net/users/1450 | 30541 | 19,890 |
https://mathoverflow.net/questions/30544 | 2 | Lusztig's theory of character sheaves gives a geometric
way to obtain character tables of finite groups of Lie type (coming
from reductive groups). I am interested to know if there is a similar
theory for other kinds of algebraic groups. Thanks so much.
| https://mathoverflow.net/users/6277 | Is there a version of character sheaves for non reductive algebraic groups? | Boyarchenko and Drinfeld have developed a theory of character sheaves for unipotent groups. See [*A motivated introduction to character sheaves and the orbit method for unipotent groups in positive characteristic*](https://arxiv.org/abs/math/0609769) for an introduction (as well as subsequent papers by Boyarchenko and Boyarchenko-Drinfeld).
| 5 | https://mathoverflow.net/users/396 | 30547 | 19,892 |
https://mathoverflow.net/questions/30548 | 3 | I haven't seen this mentioned in Jech or Kunen, sorry if these are basic questions but I am not understanding these points.
If $\kappa$ is regular uncountable then in ZFC every stationary subset of $\kappa$ is the disjoint union of $\kappa$ stationary sets. So this holds for every successor $\gamma^+$. we have strictly less than $2^{\kappa}$ of them but is there a way to count how many stationary and c.u.b sets we can have?
Is there a way of knowing which subsets of say $\omega\_1$ are stationary? Do we know if a stationary set can contain a c.u.b set or the other way around or maybe it does not matter?
What about for a singular cardinal, for instance for $\aleph\_{\omega}$, can we count the stationary subsets and the c.u.b sets? Would the GCH help in this case?
| https://mathoverflow.net/users/3859 | Counting stationary and c.u.b sets | For every infinite cardinal $\kappa$, the number of club subsets of $\kappa$ is $2^\kappa$, fully as large as it could possibly be. Since every club set is stationary, this means also there are fully $2^\kappa$ many stationary sets also.
To see that there are this many club sets, observe that there are $\kappa$ many successor ordinals below $\kappa$. For any set $A\subset\kappa$, let $A+1 = \{\alpha+1| \alpha\in A\}$' be the set of successors to elements of $A$. Let $C\_A$ be the set consisting of $A+1$ together with all limit ordinals below $\kappa$. This is a club subset of $\kappa$, and it is easy to see that if $A\neq B$, then $C\_A\neq C\_B$. Since there are $2^\kappa$ many subsets $A$, it follows that there are $2^\kappa$ many club sets. This argument does not need GCH.
For stationary sets, you could have used Solovay's theorem directly. Since every stationary subset of $\kappa$ is a union of $\kappa$ many disjoint stationary sets, we can take any subunion corresponding to any subset of the index set, to get $2^\kappa$ many distinct stationary sets.
About your latter considerations. Every club set is stationary; the intersection of two clubs is club and the intersection of a stationary and a club is stationary, and these are usually counted among the elementary facts about club and stationary sets. We usually do not consider the concept of stationary on singular cardinals of cofinality $\omega$, because in this case, the clubs do not form a filter. When $\kappa$ has uncountable cofinality, then the club filter makes sense, and the stationary concept is robust. The right way to think about it is: club means having measure $1$ with respect to the club filter, and stationary means not having measure $0$. Thus, stationary sets have outer measure $1$ with respect to club filter, although if they are co-stationary, then they will have inner measure $0$.
One interesting issue is that whether or not a set $A$ is stationary can be affected by forcing. For example, for every stationary set $A\subset\omega\_1$, there is an $\omega\_1$-preserving forcing extension where it contains a club. Thus, if the complement of $A$ was stationary in the ground model, it becomes nonstationary in the extension.
| 5 | https://mathoverflow.net/users/1946 | 30550 | 19,893 |
https://mathoverflow.net/questions/30542 | 5 | As a complex affine variety or projective variety, is it possible it is a manifold with boundary?
| https://mathoverflow.net/users/2391 | Is it possible a variety be a manifold with boundary | No, this is not possible (unless you allow the boundary to be empty). If $X$ is a complex algebraic variety (affine or projective, this doesn't matter), there are two possibilities. If $X$ is smooth (as an algebraic variety), then $X$ is a smooth manifold with empty boundary. Otherwise let $Y$ be the singular locus of $X$. Then $X\setminus Y$ is a smooth variety and hence a smooth manifold with empty boundary. On the other hand, the \emph{real} codimension of $Y$ in $X$ is at least two (because the complex codimension is at least one). However, if $M$ is a (real) manifold with nonempty boundary, it is not possible to find a subset $Y$ of $M$ that has real codimension two such that $M\setminus Y$ has empty boundary. (The boundary $\partial M$ of $M$ is either empty or has real codimension one in $M$.) This implies that $X$ is not a manifold with boundary.
| 7 | https://mathoverflow.net/users/4384 | 30553 | 19,895 |
https://mathoverflow.net/questions/30549 | 4 | Is Deligne-Mumford space could also be defined in the complex geometry context? I check wiki, it says we can similarly define Riemann surface with nodes and stability condition, I am wondering if there is any reference providing more details about this aspect. Thanks!
| https://mathoverflow.net/users/2391 | Deligne-Mumford space defined in complex geometry category | I'm not sure where to point you for full details of this, but quite a few details are in some old research announcements of Bers. See his papers
MR0361051 (50 #13497)
Bers, Lipman
Spaces of degenerating Riemann surfaces. Discontinuous groups and Riemann surfaces (Proc. Conf., Univ. Maryland, College Park, Md., 1973), pp. 43--55. Ann. of Math. Studies, No. 79, Princeton Univ. Press, Princeton, N.J., 1974.
and
MR0361165 (50 #13611)
Bers, Lipman
On spaces of Riemann surfaces with nodes.
Bull. Amer. Math. Soc. 80 (1974), 1219--1222.
and
MR0374496 (51 #10696)
Bers, Lipman
Deformations and moduli of Riemann surfaces with nodes and signatures.
Collection of articles dedicated to Werner Fenchel on his 70th birthday.
Math. Scand. 36 (1975), 12--16.
---
EDIT : Sorry to resurrect this ancient thread, but I heard a lovely talk from Sarah Koch a few weeks ago in which she described a recent paper that she wrote with John Hubbard in which they give a complex-analytic construction of the DM compactification of the moduli space of curves and prove that (as a complex analytic space) it is isomorphic to the analyticification of the usual one. In particular, this gives all the missing details in Bers's papers above (along with much more). See their paper "An analytic construction of the Deligne-Mumford compactification of the moduli space of curves" available from Sarah's webpage [here](http://www.math.harvard.edu/~kochs/papers.html).
| 5 | https://mathoverflow.net/users/317 | 30554 | 19,896 |
https://mathoverflow.net/questions/30555 | 7 | Hello all, I would appreciate comments on the following question:
A main theorem of symmetric functions might be formulated: Let k be a field of char. 0. Then $k[x\_1,...,x\_n]^{S\_n} = k[s\_1,...,s\_n]$, i.e. symmetric polynomials can be written as polynomials in the elementary symmetric polynomials. Moreover, $s\_1,...,s\_n$ satisfy no polynomial relations.
I want to see how to prove it using Galois theory. I thus consider the field $M=k(x\_1,...,x\_n)$ and its subfields $K=k(s\_1,...,s\_n)$ and $L$, the subfield of symmetric functions. Thus $K \subset L \subset M$. I then consider the polynomial $G(t)=(t-x\_1)...(t-x\_n)$. It has coefficients in $K$. $M$ is the splitting field of $G$ over $K$. Hence $[M:K] \leq n!$. From this we already see that $s\_1,...,s\_n$ satisfy no polynomial relations. On the other hand, $M$ has $n!$ different automorphism over $L$, which are permuting the $x\_i$. Hence from Galois theory we can conclude that $L=K$.
My question is: How can I deduce the claim $k[x\_1,...,x\_n]^{S\_n} = k[s\_1,...,s\_n]$ from the corresponding one for rational functions: $L=K$.
Thanks.
| https://mathoverflow.net/users/2095 | Symmetric polynomials theorem | I hope the following works. Let $A=k[x\_1,\ldots,x\_n]^{S\_n}$, and let $B=k[s\_1,\ldots,s\_n]$. The polynomial algebra $k[x\_1,\ldots,x\_n]$ is an integral extension of $B$, and hence, a fortiori, $A$ is integral over $B$. I think your argument proves that $A$ and $B$ have the same fraction field. However, since $B$ is (isomorphic to) a polynomial algebra, it must be integrally closed, whence $A=B$.
| 6 | https://mathoverflow.net/users/4384 | 30558 | 19,898 |
https://mathoverflow.net/questions/30557 | 7 | Suppose $U'\cup U''=X$ is an open cover $U$ of a topological space $X$ and $F$ is a sheaf on $X$ with values in abelian groups. There is a special instance of the Grothendieck spectral sequence relating Cech to sheaf cohomology:
$$E\_2^{p,q}=\tilde{H}^p(U,H^q(-,F))\Rightarrow H^{p+q}(X,F)$$
I would like to see, how this implies the Mayer-Vietoris sequence for this easy cover $U$. Drawing the $E\_2$-page, I get so far that only the first two columns $p=0,1$ are non-zero. Therefore this page equals the $E\_\infty$-page.
| https://mathoverflow.net/users/7316 | Grothendieck spectral sequence and Mayer-Vietoris sequence | Recall that the Cech-to-derived functor spectral sequence is constructed as follows. We start with a sheaf $F$ and an open cover $\mathfrak{U}$. Then we can write the Cech resolution of the sheaf; take an injective (or Godement or...) resolution thereof to get a double complex. Let $C^{\ast,\ast}$ be the resulting complex of global sections and take the filtration $F^i=\bigoplus C^{\geq i,\ast}$. See e.g. Godement, Th\'eorie des faisceaux, 5.2. The rows of the $E\_1$ sheet are precisely the Cech cochain complexes constructed from the open cover $\mathfrak{U}$ and the presheaves $U\mapsto H^i(U,F)$ (see Godement, ibid, just before theorem 5.2.4).
If $\mathfrak{U}$ has just two elements, $U$ and $U''$, then the $E\_1$ term has two columns, the 0-th and the 1-st ones. Applying e.g. theorem 4.6.1 from Godement, ibid, one gets the long exact sequence
$$\cdots\to E\_1^{1,i-1}\to H^i(X,F)\to E\_1^{0,i}\to E^{1,i}\_1\to\cdots$$
where the last arrow is the $d\_1$ differential, $E\_1^{1,j}=H^j(U'\cap U'',F)$ and $E\_1^{0,j}=H^j(U',F)\oplus H^j(U'',F)$.
| 10 | https://mathoverflow.net/users/2349 | 30564 | 19,900 |
https://mathoverflow.net/questions/28829 | 4 | I sometimes read $\int\_{E(\mathbf{R})} \frac{dx}{2y + a\_1x + a\_3}$ and sometimes $\int\_{E(\mathbf{R})} |\frac{dx}{2y + a\_1x + a\_3}|$. Furthermore, one has to choose an orientation on $E(\mathbf{R})$.
So what's the correct definition for the constant appearing in the BSD conjecture?
| https://mathoverflow.net/users/nan | How is the period of an elliptic curve defined exactly? | The comments above give already the answer, but for the sake of completeness let us be a bit more precise.
Let $E/\mathbb{Q}$ be an elliptic curve. Let $\Omega^{+}$ be the smallest positive element in the period lattice $\Lambda$. Then the conjecture of Birch and Swinnerton-Dyer predicts that
$$
\frac{L^{\*}(E,1)}{[E(\mathbb{R}):E(\mathbb{R})^{o}] \cdot \Omega^{+}} = \frac{\prod\_{p} c\_p \cdot | Sha| \cdot Reg}{| E(\mathbb{Q})\_{tors}|^2}
$$
The denominator on the left, where the index is the number of connected components of $E(\mathbb{R})$, can also be written as the *absolute value* of $\int\_{E(\mathbb{R})}\omega\_E$ where $\omega\_E$ is a invariant differential of a global minimal Weierstrass model.
A better way of formulating the conjecture especially if $E$ is no longer defined over $\mathbb{Q}$ but over an arbitrary global field was given Tate. (See for instance conjecture 2.1 in [Dokchitser's paper](http://arxiv.org/abs/math/0610290) for a formulation). Since there are no global minimal models anymore one has to make either a conjecture that is invariant of the choice of a model or work with the Néron model.
| 5 | https://mathoverflow.net/users/5015 | 30568 | 19,903 |
https://mathoverflow.net/questions/30572 | 9 | Sorry if this is an easy one, I'm a little rusty on my group theory. My first guess was that it's simply the inverse limit of the Aut($\mathbb{Z}/p^i\mathbb{Z})$, with the map when $i\leq j$ given by taking $\sigma\in$ Aut$(\mathbb{Z}/p^j\mathbb{Z})$ to the map $\tilde{\sigma}:\mathbb{Z}/p^i\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ defined by solving $\phi\circ\sigma=\tilde{\sigma}\circ\phi$ where $\phi:\mathbb{Z}/p^j\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ is the reduction map, but that seems too optimistic - I couldn't think of any reason $\tilde{\sigma}$ would be well-defined, much less be an automorphism of $\mathbb{Z}/p^i\mathbb{Z}$.
Also, barring a full answer to my question, I would be interested in whether Aut$(\mathbb{Z}\_p)$ is a $p$-group. If not, what can we say about the elements $\sigma\in$ Aut$(\mathbb{Z}\_p)$ with order a power of $p$?
| https://mathoverflow.net/users/1916 | What is the automorphism group of the additive group of the p-adic integers? | First the $p$-adic integers are finitely generated (actually cyclic) pro-$p$ group therefore from a result of Serre all automorphisms are continuous. Now as it cyclic it is enough to see what happens to $1$. It has to go to another generator, i.e. any element of the form $a\_0+a\_1p+a\_2p^2+\cdots$, where $0 \leq a\_i < p$ for all $i$ and $a\_0 \ne 0$. Hence, every element in $\mathbb{Z}\_p$ is just multiplied by $a\_0+a\_1p+a\_2p^2+\cdots$. Thus, the automorphism group is the multiplicative group of $\mathbb{Z}\_p$.
It is of course not a pro-$p$ group, but it contains a subgroup of index $p-1$ which is pro-$p$ and is actually again a cyclic pro-$p$ group, i.e. isomorphic to $\mathbb{Z}\_p$ and thus have no elements of finite order.
| 11 | https://mathoverflow.net/users/5034 | 30574 | 19,908 |
https://mathoverflow.net/questions/30559 | -1 | Let (δ;U) is a proximity space.
I will call a set A connected iff for every partition {X,Y} of the set A holds X δ Y.
Is the following true? (I need a proof or a counter-example.)
**Conjecture** If S is a collection of connected sets and ∩S≠∅ then ∪S is connected.
Note that instead of proximity we may consider the more general case of δ being limited only by the axioms (a subset of axioms of proximity space):
¬(∅δB), ¬(Aδ∅), X∪YδB⇔XδB∨YδB, AδX∪Y⇔AδX∨AδY.
| https://mathoverflow.net/users/4086 | Union of proximally connected sets | Let $\{X,Y\}$ be a partition of the union of the family $S$, and let p be a point in the intersection of $S$ (which you've assumed is nonempty). Without loss of generality, p is in $X$. But, since a partition can't contain the empty set, $Y$ contains some point q from some set $A$ in the family $S$. Then $\{A\cap X, A\cap Y\}$ is a partition of $A$. By assumption, $A$ is connected, so $(A\cap X)\delta(A\cap Y)$. Therefore $X\delta Y$ as required. (All that was used about $\delta$ is that it's monotone in both arguments, which follows from your assumptions about unions.)
(Could this be homework? Do proximity spaces show up in classes?)
| 3 | https://mathoverflow.net/users/6794 | 30576 | 19,909 |
https://mathoverflow.net/questions/30581 | 11 | Fix a prime number $p$. Suppose that I have a valuation $v\_p: \mathbb{Q} \to \mathbb{Q}$ on the rationals $\mathbb{Q}$. That is, $v\_p( p^n(\frac{a}{b})) = p^{-n}$ where each of $a,b$ is not divisible by $p$.
How can I extend $v\_p$ to $v$ on the reals $\mathbb{R}$ such that $v|\_\mathbb{Q} = v\_p$? I am looking for an explicit description of $v$, if that is possible. I know for a fact that one can extend valuation on any field extension.
Thank you,
| https://mathoverflow.net/users/7328 | Extension of valuation | As you point out, it follows on general grounds that there is an extension of $v\_p$ to a valuation on ${\mathbb R}$ (in fact, there are uncountably many such extensions), but it is impossible to give an "explicit" description. Indeed, not only will any such extension by discontinuous with respect to the usual Euclidean topology on the reals, but it will not be a measurable function.
| 15 | https://mathoverflow.net/users/5147 | 30582 | 19,914 |
https://mathoverflow.net/questions/30599 | 3 | Let A be an artin ring which is also a finitely generated algebra over Z.
Show that $|A|<\infty$.
If A would have been a field then I know how to prove it. I know that A is a product of local rings, so I could restrict the question to Local artin rings that are finitely generated algebra over Z. But how does this help?
Thanks,
Yatir
| https://mathoverflow.net/users/7332 | Finitely-generated algebra over Z | Take $A$ local (you already reduced to it), with $m$ the max. ideal. I claim that $A/m$ is a finite field. Suppose first that it has char. 0. Then we get injections $\mathbb Z \to \mathbb Q \to A/m$. By Zariski's lemma, $\mathbb Q \to A/m$ is finite, since it is of finite type.
Now (unfortunately I don't have it on me), Atiyah-Macdonald have a beautiful lemma which says that if $A \subset B \subset C$ are (comm.) rings, $A$ noetherian, $A \subset C$ of finite type, $B \subset C$ finite, then $A \subset B$ is of finite type.
In our case, $\mathbb Z \to \mathbb Q$ is of finite type, contradiction. Thus $\mathbb Z/p \to A/m$ is of finite type, hence finite for some prime number $p$. So $A/m$ is a finite field. Also $m^n = 0$ for some $n$ since $A$ is artin local. Finally, $m^i/m^{i+1}$ is a f.d. $A/m$-vector space (since $A$ is noetherian), so it is finite as well. And $|A| = \sum |m^i/m^{i+1}|$.
| 6 | https://mathoverflow.net/users/1729 | 30603 | 19,924 |
https://mathoverflow.net/questions/30598 | 1 | I recently learned about Dirichlet problems and was wondering if there were similar solutions in the case where only few temperature points are known instead of a continuous temperature boundary.
For instance, say the temperature is known at the three points of an equilateral triangle, and is assumed to be at a steady state. Is it possible to derive a differential (and of closed form) equation that describes the temperature within the triangle?
Additionally, what are these type of problems called?
| https://mathoverflow.net/users/7168 | Point boundary problems | I think you may have gotten things backwards.
The point of differential equations is to describe macroscopic (global) phenomena via microscopic (local) physical laws, as differential operators are strictly local objects. Solving a differential equation one often finds families of solutions, which can live in various different function spaces of different regularity. The study of well-posedness of a differential equation often becomes the study of *under what conditions can we obtain the existence of a unique solution*. It is often found that the "degrees of freedoms" in the families of solutions can be restricted by prescribing boundary data of sufficient regularity. This is the case of elliptic operators and leads to the Dirichlet problem.
Now, the amount of data to be prescribed at the "boundary" is not the same for every problem. For elliptic type problems one only need to give a Dirichlet or Neumann type condition, but in general giving both conditions may lead to non-existence of a solution (over-determined problem). But for a different type of boundary and a different type of equation (say hyperbolic/wave equation in the initial value problem formulation), it is necessary to prescribe both the "Dirichlet" and the "Neumann" conditions for the question to be well-posed.
What you are asking is sort of an opposite problem: you are asking that given the value of a function on some subset of, say, the plane, what differential equations are well-posed for this data. This problem has too many solutions. Just to give a few
* As I mentioned in the comments, the equation $\partial^2f = 0$ (vanishing Hessian)
* Or, let $v,w$ be arbitrary unit vectors not parallel to the sides of the triangle, then you can set $v(f) = w(w(w(f))) = 0$. The solution is constant in the $v$ direction, and along integral lines of $w$ must be quadratic, which is determined by three constants.
* Or, let $v$ be the unit vector going from point 1 to point 2, and let $w$ be the unit vector that connects point 3 to the line formed by point 1 and point 2 perpendicularly. Let $a$ be a number that is not a multiple of the distance between points 1 and 2, and $b$ be a number that is not a multiple of the distance between point 3 and the line, then take $v(v(f)) = - a^2 f$ and $w(w(f)) = - b^2 f$. The general solution is a trigonometric function depending on one translation coordinate and one scaling parameter.
But if you ask that also the function represents the steady state of a solution to the classical heat equation (in other words a solution to the Laplace equation), then the answer is no: you can extend the temperature profile on the boundary of your triangle arbitrarily from the three fixed data points you gave. For every extension (say differentiable) there is a corresponding solution to the Laplace equation. In other words, there are many, many steady-states whose temperature are as given at those three points on the triangle. So the map from your data to admissible steady-state temperature distributions is necessarily non-unique. So unless you prescribe additional conditions to pick out which of the many possible solutions you want, it is in general impossible (by definition) to write down a well-posed differential equation doing what you want it to do.
*Edit: I just saw your answer to Qiaochu's comment*
Yes, the devil is in the details on how you insert the constraint though. The Dirichlet problem is well posed. The three-point version isn't. By counting dimensions your constraint needs to be strong enough to mod-out a infinite dimensional set. For example, my first example of an equation $\partial^2 f = 0$, is one possibility. A solution to that equation most certainly solves the Laplace equation. It is equivalent to extending the data to be linear along the boundary of the triangle, and solving the Laplace equation. It also happens to be the one that also minimizes the $H^2$ norm among all solutions to the Laplace equation. Is it a meaningful one? I dunno, what do you think? But it certainly is optimal when considering one metric.
In any case, any conditions you can impose that leads to a unique solution most certainly will be equivalent to one that leads to a unique set of compatible boundaries. Then you can impose differential conditions (ask that $f$ solves a second order ODE along each segment of the boundary) or algebraic conditions. Unless you have a physical justification, optimality really is in the eye of the beholder.
| 3 | https://mathoverflow.net/users/3948 | 30605 | 19,925 |
https://mathoverflow.net/questions/30529 | 3 | What's the name for a digraph such that for each pair of vertices $u,v$, there is either a path from $u$ to $v$ or a path from $v$ to $u$? I'd call it just connected, since this is an intermediate property between weak and strong connectivity, and is in fact equivalent to the existence of a path containing all vertices. However, I'm not an expert of the subject, and I was unable to find any reference about this, so far.
| https://mathoverflow.net/users/7320 | Digraph intermediate connectivity | Just `connected' is fine. For example, [Wikipedia](http://en.wikipedia.org/wiki/Connectivity_%28graph_theory%29) and [Tutte](http://books.google.com/books?id=uTGhooU37h4C&lpg=PP1&dq=graph%2520theory&pg=PA132#v=onepage&q&f=false) agree. However, since "the number of systems of terminology presently used in graph theory is equal, to a close approximation, to the number of graph theorists," (R.P. Stanley, 1986) you might want to include the definition anyway.
| 5 | https://mathoverflow.net/users/840 | 30615 | 19,931 |
https://mathoverflow.net/questions/30597 | 2 | I'm working on (yet) an(other) exercise from Mosher & Tangora's "Cohomology Operations and Applications to Homotopy Theory". This one is about the homotopy exact couple, which is defined for a complex $K$ by $D\_{p,q}=\pi\_{p+q}(K^p)$ and $E\_{p,q}=\pi\_{p+q}(K^p,K^{p-1})$. So that we have relative Hurewicz, we assume K to be simply connected. As stated in the title, the object of the exercise is to show that this is not a homotopy invariant but that its derived couple is.
The motivating example I've got in my head (let me know if you've got a better one) is $S^2$ realized either with 1 vertex, 1 edge, and 2 faces, or with 1 edge and 1 face. This already easily proves that the homotopy exact couple itself is not an invariant. For the harder part, I've drawn the (presumably standard) grid with rows like $\cdots \rightarrow D\_{p,q} \rightarrow E\_{p,q} \rightarrow D\_{p-1,q} \rightarrow \cdots$ connected by vertical inclusion maps $D\_{p,q} \rightarrow D\_{p+1,q-1}$, and I can see how these both give the same derived couple, but I'm having trouble figuring out exactly how to make this into a general argument. I begin with a homotopy equivalence $f:K \rightarrow L$, $g:L \rightarrow K$, and I can assume these maps are cellular so I get induced maps between all corresponding groups of the homotopy exact couples associated to $K$ and $L$. But what can I say about these maps? Clearly from my motivating example the restrictions to skeleta need not be homotopy equivalences, or even anything close. I'm pretty sure they commute with the intra-couple maps, but I haven't had any success pushing through the commutative algebra with that fact alone. It smells like obstruction theory should be involved here since in general you'll need to move $K^p$ through $K^{p+1}$ to realize the homotopy $gf\simeq 1\_K$ (consider it as a map $K\times I \rightarrow K$, which can be assumed to be cellular), but I don't think I understand it well enough to see how (or if that's even true, I guess). Am I headed in the right direction?
---
P.S. I'm camping right now so I typed all of this on my phone. Might this be a first for MO? Or have people been asking math questions from their phones since before I was born...
| https://mathoverflow.net/users/303 | How can I prove that the derived couple of the homotopy exact couple is an invariant? | Let me start by making a definition: an $n$-skeleton of a space $X$ is an
$n$-equivalence
$X\_n \to X$, where $X\_n$ is an $n$-dimensional (at most) CW complex
($X$ itself need not be a CW complex). Obviously, $n$-skeleta are not unique,
but any two $n$-skeleta for the same space factor through one another: there are
compositions
$X\_n' \to X\_n \to X$ and $X\_n \to X\_n' \to X$.
Let's concentrate on the $D$s. By definition,
$D^2\_{p,q} = \mathrm{im}( \pi\_{p+q} (K\_q) \to \pi\_{p+q}( K\_{q+1}))$. Any two $q$- and
$(q+1)$-skeleta $K\_q\to K\_{q+1}\to K$
and ${K\_q}'\to K\_{q+1}' \to K$ factor through one another, so
$\mathrm{im}( \pi\_{p+q} (K\_q) \to \pi\_{p+q}( K\_{q+1})) \cong
\mathrm{im}( \pi\_{p+q} ({K\_q}') \to \pi\_{p+q}( K\_{q+1}'))$.
This shows that $D^2\_{p,q}$ is independent of the choice of CW decomposition. The
isomorphism of the $E$-groups follows by the Five lemma.
EDIT: Of course the last bit of the second paragraph was ridiculous, and unnecessary;
fixed now.
| 3 | https://mathoverflow.net/users/3634 | 30617 | 19,933 |
https://mathoverflow.net/questions/20879 | 4 | A morphism of varieties over $\mathbb{C}$, $f:V\to W$ is proper if it is universally closed and separated. One way to check properness is the [valuative criterion](http://books.google.com/books?id=3rtX9t-nnvwC&lpg=PP1&dq=hartshorne&pg=PA101#v=onepage&q&f=false).
What other methods do we have for determining if a morphism is proper? Particularly, I'm interested in quasi-projective varieties, but ones that aren't actually projective. And while a completely algebraic, valid over all fields or for schemes answer would also be good, I'm looking at complex varieties, and may be able to assume that the singularities are all finite quotient singularities.
| https://mathoverflow.net/users/622 | When is a morphism proper? | Assume $V$ and $W$ are quasiprojective. Let $i:V\to X$ be a locally closed embedding with $X$ projective (for instance $X$ could be $P^n$). Consider the induced map $g:V\to X\times W$; this is also a locally closed embedding. Then $f$ is proper iff $g$ is a closed embedding, or equivalently if $g(V)$ is closed.
As for the topological approach, use the definition of properness given by Charles Staats. Let $f:X\to Y$ be a continuous map of Hausdorff second countable topological spaces. The base change $f':X'\to Y'$ of $f$ by a continuous map $g:Y'\to Y$ is defined by letting $X'$ be the set of pairs
$(x,y')$ in $X\times Y'$ such that $f(x)=g(y')$ (with the induced topology from $X\times Y'$), and $f':X'\to Y'$ the obvious projection. Then $f$ is proper if and only if all its base changes are closed. This may not be logically relevant, but I find it very comforting.
To connect the two cases note that, given a locally closed embedding of complex algebraic varieties, it is closed in the Zariski topology iff it is closed in the Euclidean topology.
| 9 | https://mathoverflow.net/users/4164 | 30627 | 19,941 |
https://mathoverflow.net/questions/30629 | 13 | I believe that I once saw a statement that every compact, smooth Calabi-Yau manifold in dimension at least 3 is algebraic, but I can remember neither the reference nor the proof (which would have been quite short) and I might just be confusing this with something else. Is it true?
| https://mathoverflow.net/users/7343 | Are Calabi-Yau manifolds in dimension >= 3 algebraic? | It depends a little bit on your definition of CY. If you're using a good one, it will imply that the Hodge numbers $h^{0,p} = 0$ for $p \neq 0,d$ (see, for example, Prop. 5.3 of Joyce's <http://arxiv.org/abs/math/0108088>). This implies that $H^2(X) \cong H^{1,1}(X)$. Since the Kaehler cone is an open set in $H^{1,1}(X)$, it contains an rational class, and we can scale that to be an integral class. So, by Kodaira and Chow, we're done.
| 15 | https://mathoverflow.net/users/947 | 30634 | 19,944 |
https://mathoverflow.net/questions/30633 | 15 | Background
----------
An ordinal $\alpha$ is called a *recursive ordinal* if there is a recursive well-order $R$ on $\mathbb{N}$ such that ordertype($\mathbb{N},R) = \alpha$. For example, $\omega\cdot 2$ is a recursive ordinal because the ordering of $\mathbb{N}$ as 0, 2, 4, 6, 8, ... 1, 3, 5, 7, ... is computable and has order type $\omega\cdot 2$.
Kleene encoded the recursive ordinals in the natural numbers in a nifty way which is described at [the Wikipedia page on Kleene's O](http://en.wikipedia.org/wiki/Kleene%27s_O). Now Kleene's $\mathcal{O}$ is a fairly powerful set -- given a Turing machine index for a linear order, $\mathcal{O}$ can decide whether that ordering is a well-ordering or not.
Using Kleene's $\mathcal{O}$, it is possible to describe how to iterate the Turing jump through the recursive ordinals. For each natural number $a\in\mathcal{O}$, we can define a set $H\_a$ recursively as follows:
1. $H\_a = \emptyset$ if $a=0$
2. $H\_a = {H\_b}'$ if $a=2^b$
3. $H\_a = \{\langle n, x \rangle | x \in H\_{\phi\_e(n)} \}$ if $a = {3\cdot 5^e}$
For each $a\in \mathcal{O}$, we have $H\_a$ <$\_T \ \mathcal{O}$ (strict inequality), and no $H\_a$ is powerful enough to decide which recursive orders are well-orders.
Question
--------
Among recursive non-well-orders, some hide their descending chains better than others do.
For example, if we only wanted to flag the non-well-orders sporting a *recursive* descending chain, the full power of $\mathcal{O}$ would not be necessary -- $\emptyset'''$ would do $(\exists e [ \phi\_e$ is total and $\forall n [\ \phi\_e(n+1)$ <$\_R\ \phi\_e(n)\ ]]$?). Thus there is a recursive linear non-well-order with no recursive descending chain.
In fact (by similar reasoning), for each $a \in \mathcal{O}$ there must be a recursive linear non-well-order with no recursive-in-$H\_a$ descending chain.
I wonder whether we could effectively construct these sneaky recursive non-well-orders.
>
> Is there a recursive function $f$ such that whenever $a\in\mathcal{O}$, $f(a)$ is a Turing index for a linear non-well-order with no $H\_a$ -computable descending chain?
>
>
>
| https://mathoverflow.net/users/6649 | Sneaky Recursive Non-Well-Orders | In the classic paper [Recursive pseudo-well-orderings](http://www.ams.org/journals/tran/1968-131-02/S0002-9947-1968-0244049-7/S0002-9947-1968-0244049-7.pdf), [TAMS 131 (1968), 526–543](http://www.ams.org/journals/tran/1968-131-02/S0002-9947-1968-0244049-7/home.html), Joe Harrison showed that one can in fact do much better: there are computable linear orderings which are not wellordered but have no [hyperarithmetic](http://en.wikipedia.org/wiki/Hyperarithmetical_theory) descending sequences. An index for such a linear ordering satisfies your requirements simultaneously for all $a \in \mathcal{O}$.
Here is a sketch of Harrison's argument. First note that the ordering of $\mathcal{O}$ has a c.e. extension ${\prec}$ to all of $\mathbb{N}$, namely the smallest relation such that
* $x \neq 0 \to 0 \prec x$,
* $x \prec 2^x$,
* $\phi\_e(n){\downarrow} \to \phi\_e(n) \prec 3\cdot5^e$, and
* $x \prec y \land y \prec z \to x \prec z$.
Harrison's $\mathcal{O}^\*$ is the intersection of all *hyperarithmetic* sets $X$ with the following properties.
* $0 \in X$.
* If $a \in X$ then $2^a \in X$.
* If $\phi\_e$ is total ${\prec}$-increasing and the range of $\phi\_e$ is contained in $X$,
then $3\cdot5^e \in X$.
Since Kleene's $\mathcal{O}$ is the intersection of *all* sets $X$ with the above properties, we have $\mathcal{O} \subseteq \mathcal{O}^\*$. Also, every property of $\mathcal{O}$ translates to a property of $\mathcal{O}^\*$ by restricting all second-order quantifiers to range over hyperarithmetic reals, provided that all axioms used in the proof of the property remain valid when all quantifiers are replaced in the same way. Since $\Sigma^1\_1$ dependent choice holds in the hyperarithmetic world, this includes a lot of properties of $\mathcal{O}$ including the fact that the initial intervals $I\_n = \{x : x \prec n\}$ are computable linear orderings that have no *hyperarithmetic* descending sequences.
Harrison shows that $\mathcal{O}^\*$ is (complete) $\Sigma^1\_1$. Since $\mathcal{O} \subseteq \mathcal{O}^\*$ is complete $\Pi^1\_1$, we must have some $n \in \mathcal{O}^\* \setminus \mathcal{O}$. Then the initial interval $I\_n = \{x : x \prec n\}$ is a computable non-wellfounded linear ordering without hyperarithmetic descending sequences. In fact, one can show that $\mathcal{O} \cap I\_n$ is a path through $\mathcal{O}$ and that $I\_n$ has order-type $\omega\_1^{CK}(1+\eta) + \alpha$ where $\eta$ is the order type of the rationals and $\alpha < \omega\_1^{CK}$.
| 15 | https://mathoverflow.net/users/2000 | 30644 | 19,950 |
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