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I am reading the proof of the following proposition.

Proposition. As algebras, $\mathbb{C} G \cong \bigoplus \mathrm{End}(W_i),$ where $G$ is a finite group and $W_i$ are irreducible representation of $G$.

The proof goes as follows.

A representation $G \to \mathrm{Aut}(W_i)$ extends to a homomorphism $\mathbb{C}G \to \mathrm{End}(W_i)$. Thus we have $\phi: \mathbb{C} G \to \bigoplus \mathrm{End}(W_i)$.

I don't understand the next step where we want to show $\phi$ is injective.

The text book says that "This is injective since the representation on the regular representation is faithful".

What is the representation on the regular representation? And why is it faithful?

I appreciate any help.

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1 Answer 1

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The "representation on the regular representation" is probably just a typo for "the regular representation". This is the representation on the vector space $\mathbb CG$ in which each element of $G$ acts by multiplication from the left (on the standard basis vectors of $\mathbb CG$). That representation is faithful because every non-identity element of $G$ moves some vectors in $\mathbb CG$, for example any element of $G$ (considered as a basis vector in $\mathbb CG$). Finally, if we decompose the regular representation into irreducible summands, then each non-identity element must act nontrivially in at least one of those irreducible representations.

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Ok, that is a typo. Then what is the relation between the regular representation and $\bigoplus \mathrm{End}(W_i)$? –  Snow Jul 19 '13 at 2:14
    
The regular representation is isomorphic the direct sum of all the irreducible representations, each occurring with multiplicity equal to its dimension. But for the purpose at hand, all that is needed is that the regular representation can be decomposed as a sum of irreducible ones. If an element $x$ of $\mathbb CG$ were sent to zero by the $\phi$ in your question, it would annihilate all elements in every irreducible representation and therefore also in the regular representation. But to annihilate the identity of $G$ in the regular representation, $x$ has to be $0$, so $\phi$ is injective. –  Andreas Blass Jul 19 '13 at 3:57

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