Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

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.

share|improve this answer
An orthonormal basis would be nice. From what I read so far, that sort of property motivates the definition of Young symmetrizers, or primitive central idempotents in a group algebra. Unfortunately I need to prove this statement pretending that I know none of these. –  galileilei Sep 24 '12 at 5:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.