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16 votes
2 answers
818 views

Decomposing $(\mathbb C^n)^{\otimes m}$ as a representation of $S_n\times S_m$

$V=\mathbb C^n$ is a $\mathbb CS_n$-module, where $S_n$ is the symmetric group of degree $n$, via the representation sending a permutation to the corresponding permutation matrix. The tensor power $V^...
Benjamin Steinberg's user avatar
9 votes
3 answers
947 views

Where was it first stated that there are no 4-transitive finite groups other than symmetric, alternating and Mathieu groups?

It seems to be well-known that the six-transitive finite groups are the symmetric and alternating groups, and the only other four-transitive finite groups are the Mathieu groups (the statement can be ...
Igor Rivin's user avatar
  • 96.4k
5 votes
1 answer
889 views

A generalized Burnside's lemma

Let $G$ be a finite group acting on a set $X$, and let $S\subseteq G$ be a union of conjugacy classes. Then I believe I can prove: $$ \sum_{[x]\in X/G} \frac{|G_x \cap S|}{|G_x|} = \sum_{g\in S} \...
Mike Shulman's user avatar
  • 66.8k
3 votes
0 answers
233 views

A bridge between the algebraic / differential geometry of $\frak{sl}_2(\mathbb{C})$ and the Sheffer-Appell calculus and combinatorics

In "Four examples of Beilinson-Bernstein localization", Anna Romanov introduces the basis $m_k = \frac{(-1)^k}{k!} \partial^k \delta $ on p. 9, where $\partial$ is a partial derivative and $\...
Tom Copeland's user avatar
  • 10.5k
3 votes
4 answers
654 views

A generalization of Landau's function

For a given $n > 0$ Landau's function is defined as $$g(n) := \max\{ \operatorname{lcm}(n_1, \ldots, n_k) \mid n = n_1 + \ldots + n_k \mbox{ for some $k$}\},$$ the least common multiple of all ...
StefanH's user avatar
  • 798
2 votes
1 answer
846 views

Why Triangle of Mahonian numbers T(n,k) forms the rank of the vector space?

I am looking for an explanation of why Triangle of Mahonian numbers T(n,k) form the rank of the vector space $H^k(GL_n/B)$? With respect to the property of Kendall-Mann numbers where the statement ...
Mikhail Gaichenkov's user avatar