Schur's Lemma

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 22 January 2010

Schur's lemma

Let $A$ be an algebra and let $M$ be an $A$-module. Define $${\mathrm{End}}_A\left(M\right)=\left\{T\in \mathrm{End}\left(M\right)|Ta=aT\text{for all}a\in A\right\}\text{.}$$

(Schur's lemma). Let $A$ be a be a finite dimensional algebra over an algebraically closed field $\overline{𝔽}$.

1. Let $A^{\lambda }$ be a simple $A$-module. Then ${\mathrm{End}}_A\left(A^{\lambda }\right)=\overline{𝔽}·{\mathrm{Id}}_{A^{\lambda }}\text{.}$
2. If $A^{\lambda }$ and $A^{\mu }$ are nonisomorphic simple $A$-modules then ${\mathrm{Hom}}_A\left(A^{\lambda },A^{\mu }\right)=0$.

		Proof.
	(b): Let $T:A^{\lambda }\to A^{\mu }$ be a nonzero $A$-module homomorphism. Since $A^{\lambda }$ is simple, $\ker T=0$ so $T$ is injective. Since $A^{\mu }$ is simple, $\mathrm{im}T=A^{\mu }$ and so $T$ is surjective. So $T$ is an isomorphism. Thus we may assume that $T:A^{\lambda }\to A^{\lambda }$. (a): Since $\overline{𝔽}$ is algebraically closed $T$ has an eigenvector and a corresponding eigenvalue $\alpha \in \overline{𝔽}$. Then $T-\alpha ·\mathrm{Id}$ is not invertible. So $T-\alpha ·\mathrm{Id}=0$. So $T=\alpha ·\mathrm{Id}$. So ${\mathrm{End}}_A\left(A^{\lambda }\right)=\overline{𝔽}·\mathrm{Id}$. $\square$

(The centraliser theorem). Let $A$ be a finite dimensional dimensional algebra over an algebraically closed field $\overline{𝔽}$. Let $M$ be a simple $A$-module and set $Z={\mathrm{End}}_A\left(M\right)$. Suppose that $$M\cong \underset{\lambda \in \hat{M}}{⨁}{\left(A^{\lambda }\right)}^{\oplus m_{\lambda }}\text{,}$$ where $\hat{M}$ is the index set for the reducible $A$-modules $A^{\lambda }$ which appear in $M$ and the $m_{\lambda }$ are positive integers.

1. $M\cong {\oplus }_{\lambda \in \hat{M}}{M}_{m_{\lambda }}\left(\overline{𝔽}\right)$.
2. As an $\left(A,Z\right)$-module $$M\cong \underset{\lambda \in \hat{M}}{⨁}{A}^{\lambda }\otimes Z^{\lambda }\text{,}$$ where the $Z^{\lambda }$, $\lambda \in \hat{M}$, are the simple $Z$-modules.

		Proof.
	(a): Index the components in the decomposition of $M$ by dummy variables ${\epsilon }_i^{\lambda }$ so that we may write $$M\cong \underset{\lambda \in \hat{M}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{A}^{\lambda }\otimes {\epsilon }_i^{\lambda }\text{.}$$ For each $\lambda \in \hat{M}$, $1\le i,j\le m_{\lambda }$, let ${\phi }_{ij}^{\lambda }:A^{\lambda }\otimes {\epsilon }_j\to A^{\lambda }\otimes {\epsilon }_i$ be the $A$-module isomorphism given by $${\phi }_{ij}^{\lambda }\left(m\otimes {\epsilon }_j^{\lambda }\right)=m\otimes {\epsilon }_i^{\lambda }\text{,}\text{for}m\in A^{\lambda }\text{.}$$ By Schur's lemma, $$\begin{array}{rcl}{\mathrm{End}}_A\left(M\right)& =& {\mathrm{Hom}}_A\left(M,M\right)\\ & \cong & {\mathrm{Hom}}_A\left(\underset{\lambda }{⨁}\underset{j}{⨁}{A}^{\lambda }\otimes {\epsilon }_j^{\lambda },\underset{\mu }{⨁}\underset{i}{⨁}{A}^{\mu }\otimes {\epsilon }_i^{\mu }\right)\\ & \cong & \underset{\lambda ,\mu }{⨁}\underset{i,j}{⨁}{\delta }_{\lambda \mu }{\mathrm{Hom}}_A\left(A^{\lambda }\otimes {\epsilon }_j^{\lambda },A^{\mu }\otimes {\epsilon }_i^{\mu }\right)\\ & \cong & \underset{\lambda }{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}\overline{𝔽}{\phi }_{ij}^{\lambda }\text{.}\end{array}$$ Thus each element $z\in {\mathrm{End}}_A\left(M\right)$ can be written as $$z=\sum _{\lambda \in \hat{M}}\sum _{i,j=1}^{m_{\lambda }}{z}_{ij}^{\lambda }{\phi }_i^j\text{,}\text{for some}{z}_{ij}^{\lambda }\in \overline{𝔽}\text{,}$$ and can be identified with an element of ${⨁}_{\lambda }{M}_{m\lambda }\left(\overline{𝔽}\right)$. Since ${\phi }_{ij}^{\lambda }{\phi }_{ij}^{\mu }={\delta }_{\lambda \mu }{\delta }_{jk}{\phi }_{ij}^{\lambda }$ it follows that $${\mathrm{End}}_A\left(M\right)\cong \underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(\overline{𝔽}\right)\text{.}$$ (b): As a vector space, $Z^{\mu }=\mathrm{span}\left\{{\epsilon }_i^{\mu }|1\le i\le i_{\mu }\right\}$ is isomorphic to the simple $\underset{\lambda }{\oplus }{M}_{m_{\lambda }}\left(\overline{𝔽}\right)$-module of column vectors of length $m_{\mu }$. The decomposition of $M$ as $A\oplus Z$-modules follows since $$\left(a\otimes {\phi }_{ij}^{\lambda }\right)\left(m\otimes {\epsilon }_k^{\mu }\right)={\delta }_{\lambda \mu }{\delta }_{ij}\left(a\otimes {\epsilon }_i^{\mu }\right)\text{,}\text{for all}m\in A^{\mu }\text{,}a\in A\text{.}$$ $\square$

Reference

[HA] T. Halverson and A. Ram, Partition algebras, European Journal of Combinatorics 26, (2005), 869-921; arXiv:math/040131v2.

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