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Let $V$ be a vector space over $\mathbb C$ and $W$ a $\operatorname{End}(V)$-module. I'm having difficulty seeing why the map $$ \operatorname{Hom}_{\operatorname{End}(V)}(V,W) \otimes V \to W $$ $$ \phi \otimes v \mapsto \phi(v) $$ is surjective. I'm having trouble because I can't construct elements of $\operatorname{Hom}_{\operatorname{End}(V)} (V,W)$. It seems hard because I know that the above map is injective which means showing surjectiveness is not as simple as showing that for any $w \in W$ there exists a module map and $v \in V$ such that $\phi(v) = w$.

Thanks!

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2 Answers 2

In the case that $V$ and $W$ are finite-dimensional, here is a "non-intrinsic" way to see this isomorphism. If $V$ is finite-dimensional, then every $\operatorname{End}(V)$-module is semisimple, and moreover every simple $\operatorname{End}(V)$-module is isomorphic to $V$:

Choose a basis $\{e_i\}$ for $V$ and define $P_i(e_j)=\delta_{ij}e_i$, so $P_1,\ldots,P_n\in\operatorname{End}(V)$ with $P_iP_j=\delta_{ij}P_i$ and $\sum_iP_i=1$. Each of the left modules $\operatorname{End}(V)\cdot P_i$ is then simple and isomorphic to $V$, via the map $A\cdot P_i\mapsto Ae_i$. This is easier to see if you think of this choice of basis as giving an isomorphism between $\operatorname{End}(V)$ and the ring of $n\times n$ matrices over your field, in which case $\operatorname{End}(V)\cdot P_i$ is the submodule of matrices whose only nonzero column is the $i$th one.

Let $M$ be a simple $\operatorname{End}(V)$-module and let $x\in M\setminus\{0\}$. Since $\sum_iP_i=1$, there must be some $i$ for which $P_ix\neq 0$, hence $M=\operatorname{End}(V)\cdot P_ix$. The map of modules $\operatorname{End}(V)\cdot P_i\to M$ sending $aP_i$ to $aP_ix$ is then an isomorphism, by Schur's lemma.

Combining this with semisimplicity, if $W$ is finite-dimensional then it is isomorphic as an $\operatorname{End}(V)$-module to some direct sum $V^{\oplus n}$.

If we unravel the chain of isomorphisms $\operatorname{Hom}_{\operatorname{End}(V)}(V,V^{\oplus n})\otimes V\cong \operatorname{Hom}_{\operatorname{End}(V)}(V,V)^{\oplus n}\otimes V\cong (\operatorname{Hom}_{\operatorname{End}(V)}(V,V)\otimes V)^{\oplus n}\cong (\mathbb{C}\otimes V)^{\oplus n}\cong V^{\oplus n}$, we see that it is the same as the evaluation map you define.

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Thanks. Is there an easy way to see that any simple $End(V)$-mod is isomorphic to $V$? –  Eric O. Korman Nov 6 '11 at 1:29
    
I've added an explanation of this fact to the answer. –  Brad Nov 7 '11 at 1:54
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I just thought of the following argument. Suppose first that $W$ is finite dimensional. Then the map I gave is a map of $\operatorname{End}(V)$ mods (with the trivial action on $Hom_{\operatorname{End}(V)}(V,W)$) and so must be an isomorphism by Schur's lemma. In the general case we'll have $W = \bigoplus_i W_i$ with $W_i$ irreducible. Then $$ \operatorname{Hom_{\operatorname{End}(V)}} (V, W) \simeq \bigoplus \operatorname{Hom_{\operatorname{End}(V)}} (V, W_i) \simeq \bigoplus W_i \simeq W. $$

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The first isom. is unclear if $V$ is infinte-dim. –  Martin Brandenburg Nov 6 '11 at 9:53

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