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 :)

  • $\begingroup$ Can we use the identity map on V? $\endgroup$ – Irreducible Jan 17 '13 at 10:52
  • $\begingroup$ 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) $\endgroup$ – user27126 Jan 17 '13 at 10:56
  • $\begingroup$ 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. $\endgroup$ – Irreducible Jan 17 '13 at 11:00
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jan 17 '13 at 11:05
  • $\begingroup$ But i work is such a algebraic fiel ;) is then the argumentation correct? $\endgroup$ – Irreducible Jan 17 '13 at 11:06

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.


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