# Idempotent and Hermitian vectors in Group Algebra

Let $C$ be the field of complex number and $G$ a finite group, then define $C[G]$ be a vector space over $C$, with elemnts of $G$ as the basis. Then any element in $C[G]$ can be written as $\sum_{g \in G} a_g e_g$ where $a_g \in C$ and $g \in G$. We also have a multiplication structure on $C[G]$, carried naturally from the group structure of $G$.

My question here concerns with a hint to Exercise 4.4 in Representation Theory, by William Fulton and Joe Harris, page 518:" More generally, if $A = C[G]$ is a group algebra, call an element $a = \sum a_g e_g$ Hermitian if $a_{g^{-1}} = \overline{a_{g}}$ for all $g$ in the summation. If $a$ and $b$ are idempotent and Hermitian, then $Aab \equiv Aba$." $a$ and $b$ in the original questions are $a_{\lambda}$ and $b_{\lambda}$ defined in Young Symmetrizer. In that case we could use the fact that Young symmetrizer is idempotent.

If Hermitian here means the same thing for matrices, I imagine $a$ and $b$ here as orthogonal projections of $C[G]$, therefore they commute. But here $Aab$ and $Aba$ are really sub-modules, or left ideals of $C[G]$, and an explanation in term of these concepts would be nice.

EDIT: clarify the definition of Hermitian.

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When you say that $a = \sum a_g e_g$ is hermitian in the group algebra do you mean to say that $a_{g^{-1}} = \overline{a_g}$ for all $g \in G$? – user38268 Sep 23 '12 at 22:41
Yes it is for all $g$ in the summation. I will add that. – galileilei Sep 24 '12 at 5:01

I think, it is really about the commutativity of your 'orthorgonal projections'. Consider the two $A\to A$ linear maps: $s\mapsto s\cdot a$ and $s\mapsto s\cdot b$.
Endow $A$ with scalar products, taking the given basis $(g)_{g\in G}$ as an orthonormal basis, then the matrix of $s\mapsto s\cdot a$ is given by $((g\cdot a)_h)_{g,h\in G} = \dots = (a_{g^{-1}h})_{g,h}$, so 'Hermitian' and idempotency for $a$ means that this matrix is indeed an orthogonal projection.