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4 votes
1 answer
302 views

Hattori-Stallings trace

Let $R$ be a (possibly non-commutative) unital ring and $M$ be a left $R$-module. If $M$ is finitely generated and projective, the natural map $$\iota:\mathrm{Hom}_R(M,R)\otimes_R M\to \mathrm{Hom}_R(...
Qwert Otto's user avatar
3 votes
1 answer
285 views

Extreme points of the set of all traces

Let $G$ be a finitely generated group with a bound on its complex unitary irreducible representations: That is assume all complex unitary irreducibles of $G$ have degrees at most $k$ for some integer $...
user3826143's user avatar
3 votes
0 answers
250 views

For which representations of a Lie algebra is the induced trace form basic?

Let $\mathfrak{g}$ be a simple Lie algebra. Let $\rho$ be a representation of $\mathfrak{g}$ on a finite-dimensional vector space $E$. Consider now the bilinear form on $\mathfrak{g}$: \begin{equation}...
Peter's user avatar
  • 556
8 votes
1 answer
249 views

Hecke algebra relation versus $\operatorname{SL}_2$ trace relation

The quadratic relation in the (type $A$) Hecke algebra is $(T-t)(T+t^{-1}) =0$, which can be rewritten as $$ T-T^{-1} = t-t^{-1}$$ Suppose $A \in \operatorname{SL}_2(\mathbb{Q})$ with eigenvalues $a,...
Peter Samuelson's user avatar
7 votes
1 answer
349 views

Is there a converse to the Brauer–Nesbitt theorem?

$\DeclareMathOperator\Tr{Tr}$Say that we have an algebra $R$ over $\mathbb{C}$. If, for two finitely generated (edit: and semisimple) $R$-modules $M, N$ we know that $\Tr_M(r)=\Tr_N(r)$ for all $r\in ...
Ioannis Zolas's user avatar