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Let $G$ be a finite group. There exists a decomposition, as a $\mathbb{C}[G]$-module, $\mathbb{C}[G]=n_{1}V_{1}\oplus \cdots\oplus n_{r}V_{r}$, where the $V_{i}$ form a complete set of irreducible submodules of $\mathbb{C}[G]$, these submodules corresponding to the irreducible representations of $G$.

Why is the dimension of the representation $V_{i}$ equal to $n_{i}$, the number of copies of $V_{i}$ in the decomposition? I can prove that the space of $\mathbb{C}[G]$-module homomorphisms from $V_{1}$ to $V$ has dimension $n_{1}$, but I don't quite see how this implies that the dimension of the representation is also $n_{1}$.

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This should be implicit in the way the decomposition is constructed! That said, if you can prove that $\dim V_i\leqslant n_i$ for all $i$ then you can conclude via the fact that $\dim\mathbb{C}[G]=|G|=\sum_i\dim V_i^2$. –  Alex Youcis Jan 19 '12 at 17:42
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If you know some character theory, the result is quite simple. Check out Theorem 3.19 on page 46 of these nice concise notes on characters for groups. –  Bill Cook Jan 19 '12 at 19:21
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For any $\mathbb{C}[G]$-module $V$ there is a natural isomorphism of vector spaces $$Hom_{\mathbb{C}[G]}(\mathbb{C}[G],V)\to V$$ given by $f\mapsto f(1)$ (the inverse of the isomorphism is $v\mapsto (\alpha\mapsto\alpha\cdot v)$). If $V$ is irreducible then, as you noticed, $\dim Hom_{\mathbb{C}[G]}(\mathbb{C}[G],V)$ is the multiplicity of $V$ in $\mathbb{C}[G]$, so this multiplicity is $\dim V$.

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Another way to see this result is to note that $\mathbb{C}[G]$ naturally acts by vector space endomorphisms of its irreducible representations; that is, there is a canonical algebra homomorphism $$\mathbb{C}[G] \to \bigoplus \text{End}(V_i).$$

Since $\mathbb{C}[G]$ is semisimple and has a faithful representation (namely itself), this map is injective. By the Jacobson density theorem, it is also surjective, hence bijective. In fact it is an isomorphism of $\mathbb{C}[G]\text{-}\mathbb{C}[G]$-bimodules, whence $$\mathbb{C}[G] \cong \bigoplus V_i \otimes V_i^{\ast}$$

(where the first factor is acted on by the first copy of $\mathbb{C}[G]$ and the second factor is acted on by the second) and the conclusion follows.

Of course, there is also the standard argument using characters, but I always found this result rather mysterious from the character-theoretic point of view.

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