2.1 Some representation theory

对照

Recall the definition of a group representation:

Definition · Representation

Definition (Representation). A representation $(\pi ,V)$ of a group $G$ on a complex vector space $V$ (with a chosen basis identifying $V\simeq {\mathbf{C}}^n)$ is a homomorphism

$$\pi :G\to GL(n,\mathbf{C})$$

This is just a set of $n$ by $n$ matrices, one for each group element, satisfying the multiplication rules of the group elements. $n$ is called the dimension of the representation.

We are mainly interested in the case of $G$ a Lie group, where $G$ is a differentiable manifold of some dimension. In such a case we will restrict attention to representations given by differentiable maps $\pi$. As a space, $GL(n,\mathbf{C})$ is the space ${\mathbf{C}}^{n^2}$ of all $n$ by $n$ complex matrices, with the locus of non-invertible (zero determinant) elements removed. Choosing local coordinates on $G,\pi$ will be given by $2n^2$ real functions on $G,$ and the condition that $G$ is a differentiable manifold means that the derivative of $\pi$ is consistently defined. Our focus will be not on the general case, but on the study of certain specific Lie groups and representations $\pi$ which are of central interest in quantum mechanics. For these representations one will be able to readily see that the maps $\pi$ are differentiable.

To understand the representations of a group $G,$ , one proceeds by first identifying the irreducible ones:

Definition · Irreducible representation

Definition (Irreducible representation). A representation $\pi$ is called irreducible if it is has no subrepresentations, meaning non-zero proper subspaces $W\subset V$ such that $({\pi }_{|W},W)$ is a representation. A representation that does have such a subrepresentation is called reducible.

Given two representations, their direct sum is defined as:

Definition · Direct sum representation

Definition (Direct sum representation). Given representations ${\pi }_1$ and ${\pi }_2$ of dimensions $n_1$ and $n_2$ , there is a representation of dimension $n_1+n_2$ called the direct sum of the two representations, denoted by ${\pi }_1\oplus {\pi }_2$ . This representation is given by the homomorphism

$$({\pi }_1\oplus {\pi }_2):g\in G\to \left(\begin{array}{cc}{\pi }_1(g)& 0\\ 0& {\pi }_2(g)\end{array}\right)$$

In other words, representation matrices for the direct sum are block diagonal matrices with ${\pi }_1$ and ${\pi }_2$ giving the blocks. For unitary representations

Theorem

Theorem 2.1. Any unitary representation $\pi$ can be written as a direct sum

$$\pi ={\pi }_1\oplus {\pi }_2\oplus \cdots \oplus {\pi }_m$$

where the ${\pi }_j$ are irreducible.

Proof

Proof. If $(\pi ,V)$ is not irreducible there exists a non-zero $W\subset V$ such that $({\pi }_{|W},W)$ is a representation, and

$$(\pi ,V)=({\pi }_{|W},W)\oplus ({\pi }_{|W^{\perp }},W^{\perp })$$

Here $W^{\perp }$ is the orthogonal complement of $W$ in $V$ (with respect to the Hermitian inner product on $V$). $({\pi }_{|W^{\perp }},W^{\perp })$ is a subrepresentation since, by unitarity, the representation matrices preserve the Hermitian inner product. The same argument can be applied to $W$ and $W^{\perp }$ , and continue until $(\pi ,V)$ is decomposed into a direct sum of irreducibles. □

Note that non-unitary representations may not be decomposable in this way. For a simple example, consider the group of upper triangular 2 by 2 matrices, acting on $V={\mathbf{C}}^2$ . The subspace $W\subset V$ of vectors proportional to $\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ is a subrepresentation, but there is no complement to $W$ in $V$ that is also a subrepresentation (the representation is not unitary, so there is no orthogonal complement subrepresentation).

Finding the decomposition of an arbitrary unitary representation into irreducible components can be a very non-trivial problem. Recall that one gets explicit matrices for the $\pi (g)$ of a representation $(\pi ,V)$ only when a basis for $V$ is chosen. To see if the representation is reducible, one can’t just look to see if the $\pi (g)$ are all in block-diagonal form. One needs to find out whether there is some basis for $V$ for which they are all in such form, something very non-obvious from just looking at the matrices themselves.

The following theorem provides a criterion that must be satisfied for a representation to be irreducible:

Theorem

Theorem (Schur’s lemma). If a complex representation $(\pi ,V)$ is irreducible, then the only linear maps $M:V\to V$ commuting with all the $\pi (g)$ are $\lambda 1$, multiplication by a scalar $\lambda \in \mathbf{C}$.

Proof

Proof. Assume that $M$ commutes with all the $\pi (g)$ . We want to show that $(\pi ,V)$ irreducible implies $M=\lambda 1$ . Since we are working over the field $\mathbf{C}$ (this doesn’t work for $\mathbf{R}$), we can always solve the eigenvalue equation

$$\det (M-\lambda 1)=0$$

to find the eigenvalues $\lambda$ of $M$. The eigenspaces

$$V_{\lambda }=\{v\in V:Mv=\lambda v\}$$

are non-zero vector subspaces of $V$ and can also be described as $\ker (M-\lambda 1)$ the kernel of the operator $M-\lambda 1$ . Since this operator and all the $\pi (g)$ commute, we have

$$v\in \ker (M-\lambda 1)⟹\pi (g)v\in \ker (M-\lambda 1)$$

so $\ker (M-\lambda 1)\subset V$ is a representation of $G$. If $V$ is irreducible, we must have either $\ker (M-\lambda 1)=V$ or $\ker (M-\lambda 1)=0$ . Since $\lambda$ is an eigenvalue, $\ker (M-\lambda 1)\ne 0$ , so $\ker (M-\lambda 1)=V$ and thus $M=\lambda 1$ as a linear operator on $V$. □

More concretely Schur’s lemma says that for an irreducible representation, if a matrix $M$ commutes with all the representation matrices $\pi (g)$ , then $M$ must be a scalar multiple of the unit matrix. Note that the proof crucially uses the fact that eigenvalues exist. This will only be true in general if one works with $\mathbf{C}$ and thus with complex representations. For the theory of representations on real vector spaces, Schur’s lemma is no longer true.

An important corollary of Schur’s lemma is the following characterization of irreducible representations of $G$ when $G$ is commutative.

Theorem

Theorem 2.2. If $G$ is commutative, all of its irreducible representations are one dimensional.

Proof

Proof. For $G$ commutative, $g\in G$ , any representation will satisfy

$$\pi (g)\pi (h)=\pi (h)\pi (g)$$

for all $h\in G$ . If $\pi$ is irreducible, Schur’s lemma implies that, since they commute with all the $\pi (g)$ , the matrices $\pi (h)$ are all scalar matrices, i.e., $\pi (h)={\lambda }_{h}1$ for some ${\lambda }_h\in \mathbf{C}.$ . $\pi$ is then irreducible when it is the one dimensional representation given by $\pi (h)={\lambda }_h$ □

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正文：英文 · 原著转录

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原书 PDF · 印刷页 14、15、16

来源版本：2025-10-20

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