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.

Suppose $A$ is an algebra over a field $k$. I look to the following exercise:

  • Show that if $V$ is an irreducible finite dimensional representation of A then any element $z\in Z(A)$ (with $Z(A)$ the center of $A$) acts in V by multiplication by some scalar $\chi_V(z)$ Show that $\chi_V(z)$ is a homomorphism.

I think you have to use the lemma van Schur. I want to construct an intertwining map from $V$ to $V$. Is this the good way? Thank you for hints and solutions :)

share|improve this question
    
Can we use the identity map on V? –  Irreducible Jan 17 '13 at 10:52
    
You can try to cook up some intertwining maps, for example, $z - c \cdot \mathrm{id}$ for suitable constant $c$. (Assuming that your field is algebraically closed) –  user27126 Jan 17 '13 at 10:56
    
I think you can it do as follows: define $\psi:V\rightarrow V$ by $\psi(v)=\rho(z)v$ (with $\rho$ the representationmap from $V$). Then we get: $\psi(\rho(a)v)=\rho(z)\rho(a)v=\rho(a)\rho(z)v$ because $z$ is in $Z(A)$. But then $\psi$ is an intertwijner. –  Irreducible Jan 17 '13 at 11:00
    
For Schur's lemma to work you need to assume that $k$ is algebraically closed. Otherwise the result is false. Think about the complex numbers acting on $\mathbb{R}^2$ by multiplication to find a counterexample. –  Jyrki Lahtonen Jan 17 '13 at 11:05
    
But i work is such a algebraic fiel ;) is then the argumentation correct? –  Irreducible Jan 17 '13 at 11:06
show 1 more comment

1 Answer

Let me denote the representation in the question by $\rho$. Schur's lemma states that an intertwining operator $T$ from $\rho$ to $\rho$ must be either bijective or zero. Now, if $z \in Z(A)$ then also $\rho(z) \in Z(End(V))$ and so also $A_{\lambda} := \rho(z) - \lambda {\rm Id} \in Z(End(V))$.

Now, suppose $k$ is algebraically closed. Then the equation $\det A_{\lambda} = 0$ has a root, say $\alpha$. But then $A_{\alpha}$ is a non-bijective intertwining operator and hence zero and so $\rho(z) = \alpha {\rm Id}$ acts by multiplication.

share|improve this answer
add comment

Your Answer

 
discard

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.