Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with rt.representation-theory
Search options not deleted
user 37555
Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
5
votes
representations of $S_k \times S_j$
If all representations of a group can be realized over $\mathbb{R}$, then $S(G, \mathbb{C})$ equals the number of elements of order 2 in $G$. More generally, the number of solutions of the equation $x …
5
votes
Extending the discussion on "super Catalan" numbers
Let $p\neq 3$ be a prime. Then
\begin{eqnarray*}
\nu_p\left(\frac{(3x)!}{x!^3}\right) & = & \sum_k \left[\frac{3x}{p^k}\right]-3\left[\frac{x}{p^k}\right]\\
& = & \sum_k 3\left\{\frac{x}{p^k}\right\} …
84
votes
Feit-Thompson conjecture
It is not true anymore that a proof of this conjecture would lead to significant simplifications. Peterfalvi proved in 1984 a weaker version of this conjecture, which suffices to get rid of the chapte …