Anthony Zee's Proof of Schur's Lemma

Question

In Anthony Zee's proof of Schur's lemma (in his book Group Theory in a Nutshell for Physicists, page 102), he used the following fact (summarized by myself) without proof:

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Proposition: Let $G$ be a finite group, and $D(g)$ any unitary $n \times n$-matrix representation of $g \in G$. If $D$ is irreducible, then for any fixed $1 \le i,j \le n$, it cannot happen that $D_{ij}(g) = 0$ for all $g \in G$.

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Although Zee demonstrated a reducible representation can have $D_{ij}(g) = 0$ for all $g \in G$, he did not show the converse. So I want to ask how to prove that? (Please try avoiding general theory of representations, which may involving modules, rings, etc; I am not familiar with those more abstract math...)

Answer 1

From now on $|i\rangle$ denotes the i-th element of the normalized canonical basis of the finite-dimensional Hilbert space $\mathbb{C}^n$ where the unitary nxn matrices $D(g)$ are defined. Furthermore, $$D_{ij}(g):= \langle i| D(g) |j\rangle.$$

Suppose that $D_{ij}(g) =0$ for all $g$ and a choice of $i,j$.

Notice that $i\neq j$ necessarily, otherwise we would have $|||i\rangle||=0 \neq 1$ when choosing $g=e$ (the unit element of the group).

The subspace $S$ of the finite linear combinations $$\sum_{k=1}^N c_k D(g_k)|i\rangle\quad N=1,2,\ldots\:, \quad g_k \in H\:, c_k \in \mathbb{C}$$ is invariant under the action of $D$ trivially, because $$D(f)D(g_k)|j\rangle =D(g'_k)|i\rangle := D(f\circ g_k)|j\rangle$$ extended to the whole S by linearity. Regarding the space $S$,

(1) it does not coincide with the whole vector space since it is made of vectors orthogonal to $|i\rangle$;

(2) it is not made of the zero vector only, as it includes the vector $|j\rangle \neq 0$.

Hence $S$ is an invariant proper subspace of $\mathbb{C}^n$. In other words $D$ is reducible. In summary,

PROPOSITION. If a unitary finite-dimensional representation $D$ on $\mathbb{C}^n$ ($n>1$) of a group $G$ is irreducible, then $D_{ij}(g) =0$ for all $g$ and a choice of $i,j$ cannot hold, where $$D_{ij}(g):= \langle i| D(g) |j\rangle$$ and $|i\rangle$ is the generic element of the canonical basis of $\mathbb{C}^n$.

The thesis and the proof is still valid if the space is infinite dimensional the representation $D$ is unitary by referring to a fixed Hilbert basis. Just replace the subspace above with its closure.

What about the converse statement? What it is possible to prove is the following fact.

PROPOSITION. Let $G \ni g \mapsto D(g)$ be a representation of the group $G$ made of unitary operators $D(g): H \to H$ on the Hilbert space $H$ ($\dim(H)>1$, also infinite dimensional and non-separable). If $D$ is reducible, then there is a Hilbert basis $\{u_j\}_{j\in J} \subset H$ such that $$\langle u_i| D(g) u_j\rangle =0 \quad \mbox{for all $g\in G$ and a pair $i,j \in J$ with $i\neq j$.}$$

PROOF. let $S \subset H$ be a (closed) proper invariant subspace, i.e. $D(g)(S) \subset S$ for every $g\in G$. Since the representation is unitary $D(g)(S^\perp) \subset S^\perp$ for every $g\in G$ and also $S^\perp$ is proper because $H= S\oplus S^\perp$ and $S \neq H$, $S\neq \{0\}$. Let $B$ be a Hilbert basis of $S$ and $B'$ a Hilbert basis of $S^\perp$. $B\cup B'$ is a Hilbert basis of $H$ by construction. Any pair of vectors one in $B$ and the other in $B'$ satisfies the thesis. QED

Answer 2

Although Walter Moretti's answer is fine, I'd like to point out that if you strip it down it amounts to the following. In particular you don't need to assume the representation is unitary or even defined over a particular field.

Let $D$ be an irreducible representation of a finite group $G$ on a finite dimensional vector space $V$ over a field $F$. Then for any non-zero $v\in V$, the set $\{\sum_{g\in G}c_gD(g)v : c_g\in F\}\subseteq V$ is a non-zero subrepresentation, hence it is equal to $V$.

Now choose a basis $v_1,\ldots,v_n$ of $V$ so relative to this we can express each $D(g)$ as an $n$ by $n$ matrix $[D_{i,j}(g)]$. If for some fixed $i,j$, $D_{i,j}(g)=0$ for all $g\in G$ then $D(g)v_j$ is always in the subspace spanned by the $v_k$ with $k\neq i$ and it easily follows that this is a subrepresentation. By irreducibility this cannot happen.

Of course for a complex representation you can always make it unitary by choosing an invariant hermitian inner product.