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Section 2.5 Schur's lemmas

The two lemmas of Schur are fundamental in representation theory. The ultimate goal of the next few sections is to develop tools to determine whether a given representation is irreducible or not. The fundamental result in this direction is called the “great orthogonality theorem”. But to be able to prove this theorem, we first need to introduce two lemmas that are interesting in their own right, known as “Schur's lemmas”.

Schur's lemmas are concerned with the type of matrices that commute with all matrices of irreducible representations of a finite group \(G\text{.}\) This will be important to prove the great orthogonality theorem.

Subsection 2.5.1 Schur's first lemma

Schur's first lemma gives us a criterion to determine when two representations are irreducible.

Suppose that \(T\) is \(n\)-dimensional, and \(S\) is \(m\)-dimensional. First we note that \(A\) is a \(m \times n\) matrix. Suppose that there is a subspace \(W \subset V\) such that \(A w = 0\) if and only if \(w \in W\text{.}\) Let \(P\) be the projection matrix on the subspace \(W \subset V\text{.}\) Then \(A P = 0\text{.}\) It follows that

\begin{equation*} A T(g) P = S(g) A P = 0. \end{equation*}

But then, this implies that \(T(g) P \in W\text{,}\) or, in other words, \(T(g) w \in W\) for any \(w \in W\text{.}\) This means that \(W\) is a \(T\)-invariant subspace. Since \(W\) is irreducible, the only possible invariant subspaces are the trivial ones, namely either \(W = V\) or \(W = \{0 \}\text{.}\) In the first case, this implies that \(A=0\text{.}\) In the second case, this means that \(A v\) is never zero for all non-zero \(v \in V\text{.}\)

We can redo the same argument by starting with a subspace \(W' \in U\) such that \(u A = 0\) if and only if \(u \in W'\text{.}\) Following the same steps as above, we end up with two possibilities; either \(A=0\) or \(u A\) is never zero for all non-zero \(u \in U\text{.}\)

So we are left with two cases: either \(A=0\text{,}\) or \(A v\) and \(u A\) are never zero for non-zero \(v \in V\) and \(u \in U\text{.}\) We now show that the second case implies that \(A\) is an invertible square matrix. First, if the number of rows is less than the number of columns, than there must exist a \(v \in V\) such that \(A v = 0\text{,}\) which is a contradiction. Similarly, if the number of columns is less than the number of rows, than there must exist a \(u \in U\) such that \(u A =0\text{,}\) which is a contradiction. Therefore \(A\) must be square, that is, \(m=n\text{.}\) Finally, we know that a square matrix has a non-trivial kernel if and only its determinant is zero. Therefore \(A\) must have non-zero determinant, or equivalently, it must be invertible.

Therefore, we conclude that either \(A=0\) or \(A\) is an invertible square matrix. In the latter case, if then follows that

\begin{equation*} A T(g) A^{-1} = S(g), \end{equation*}

that is, \(T\) and \(S\) are equivalent representations.

Subsection 2.5.2 Schur's second lemma

The second lemma studies what kind of matrices commute with all matrices of a given irreducible representation.

This is a rather powerful statement. In general, if you were given a bunch of matrices \(T_1, \ldots, T_n\text{,}\) there would potentially be many matrices that commute with all of those. However, if the matrices \(T_i\) are not arbitrary but actually form an irreducible representation of a finite group, things are much more constrained: only multiples of the identity matrix commute with the \(T_i\text{.}\)

The second lemma is a direct consequence of the first one. \(A\) must have at least one eigenvalue, which we call \(\lambda\text{.}\) Then \((A - \lambda I)\) has determinant zero, and hence is not invertible. But since \(A T(g) = T(g) A\text{,}\) it follows that:

\begin{equation*} (A - \lambda I) T(g) = T(g) (A - \lambda I). \end{equation*}

From Schur's first lemma Lemma 2.5.1, it then follows that \(A - \lambda I = 0\text{,}\) since \(A - \lambda I\) is not invertible. Therefore \(A = \lambda I\) for some \(\lambda \in \mathbb{C}\text{.}\)

Subsection 2.5.3 Irreducible representations of abelian groups

A direct consequence of Schur's second lemma is the following theorem:

Let \(T:G \to GL(V)\) be an irreducible representation of a finite abelian group \(G\text{.}\) Then for any \(a \in G\text{,}\) \(T(a)\) commutes with all \(T(g)\text{,}\) \(g \in G\text{.}\) Thus, by Schur's second lemma, \(T(a) = \lambda I\) for some \(\lambda \in \mathbb{C}\text{.}\) This is true for all \(a \in G\text{,}\) and hence all \(T(a)\) are scalar multiples of the identity matrix. It then follows that the only way that \(T\) can be irreducible is if it is one-dimensional.

Note that as usual, while we stated the result only for finite groups, it holds whenever Schur's second lemma holds. In particular, it is true for compact groups: all complex irreducible representations of compact abelian groups are one-dimensional.

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