Projection Operator

Motivation

Let $k$ be a field and $R$ be a $k$-algebra. If we have $E$ as a $R$-module, can we break $E$ as a non-isomoprhic simple spaces? So, it means can we find the projection operator as a bicommutant?

Lemma

Density Theorem

Statement

Let $k$ be a field, $R$ be a semi simple $k$-algebra, and $V_1,\dots ,V_m$ finite dimensional $k$-spaces which are also simple $R$-modules, and such that $V_i$ are not $R$-isomorphic to $V_j$ for $i\ne j$. Then there exists elements $e_i\in R$ such that $e_i$ acts as the identity on $V_i$ and $e_i{V}_j=0$ if $j\ne i$.

Proof

Let $E=V\oplus V^{\prime }$ where $V$ one of $V_i$ and $V^{\prime }$ is the direct sum of the others. Let $\left\{e_1,\dots ,e_n\right\}=(k\text{-basis of }V)\cup (k\text{-basis of }{V}^{\prime })$ Let ${\pi }_{V_0}:E\to E$ be a projection in $V$. Then ${\pi }_{V_0}\in R^{\prime \prime }$ since $\phi \in R^{\prime \prime }$, $\phi (V)\subset V$ and $\phi (V^{\prime })\subset V^{\prime }$.

By the density theorem, there exists $r\in R$ such that $r{e}_i={\pi }_V(e_i)$.

Corollary

Let $k$ be a field with $char(k)=0$, and $R$ is a $k$-algebra. Let $E,F$ be a semi simple $R$-module and finite dimensional vector space over $k$. For each $r\in R$, let $f{r}^E\in En{d}_k(E)$ and $g_r^F\in En{d}_k(F)$. If $tr(f_r^E)=tr(g_r^F)$, then $E{\cong }_{R-mod}F$.¹

Proof

Let $V$ be a simple $R$-module.² Let $E=V^n\oplus \cdots V^{\prime }$ and $F=V^m\oplus V^{\prime \prime }$, where $V^{\prime }$ and $V^{\prime \prime }$ are direct sum of simple moudles that are not isomorphic to $V$.

Let $r_V\in R$ as a main statement. then

$tr(f_r^E)=tr(r_V:E\to E)=di{m}_k{V}^n=ndi{m}_{k}V.$

Similarly, $g_r^F=mdi{m}_{k}V$.

Since the characteristic of $k$ is 0, $n=m$.

Application

Characters determine Representations over characteristic 0 field $k$.

Let $G$ be a finite group with representation $\rho :G\to G{L}_n(k)=Aut(k^n)$ and $\rho :G\to G{L}_m(k)=Aut(k^m)$ where $k^n$ and $k^m$ are $kG$-modules.

If $tr(\rho (g))=tr({\rho }^{\prime }(g))$ for all $g$, then $E{\cong }_{R-mod}F$ as $kG$-modules.

Footnotes

1. Make clear for myself. Trace operator on $k$-endomorphism, gives $R$-mod isomorphism. ↩

2. For the semi simple case use block decomposition. ↩