# Proof of Schur's lemma

Can someone give me a simplified proof of Schur's lemma in group theory. Sorry if the question looks a standard textbook proof. But I find the proof complicated in books. It would be helpful if someone can provide a link that proves Schur's lemma in a simpler way.

## 1 Answer

Using and extending notations on the Wiki page.

$\phi: M\to N$ is a linear map and there is a pair of linear representations of the same group $G$: One representation, say $U$, acting on $M$ and the other, say $U$, acting on $N$. The most elementary statement of the lemma established that,

Schur's lemma.

Under the said hypotheses if both the following conditions hold:

(a) $\phi U_g = V_g \phi$ for all $g\in G$

and

(b) $M$ is irreducible for $U$, $N$ is irreducible for $V$,

then $\phi$ is bijective or it is the null map.

Proof. Suppose that $\phi$ is not bijective, our final goal is obviously to prove that $\phi$ is the null map.

If $\phi$ is not bijective, it may happen for two reasons only:

(1) $Ran(\phi)$ is smaller than $N$ or

(2) $Ker(\phi)$ is larger than $\{0\} \subset M$ (or both).

Let us examine separately these cases.

Suppose (1) is valid. $y\in Ran(\phi)$ is equivalent to say that $y= \phi x$ for some $x\in M$. For such $y$, in view of (a) we have $V_gy = V_g\phi x = \phi U_g x = \phi U_gx' \in Ran(\phi)$. All that entails $$V_g Ran(\phi) \subset Ran(\phi)\quad \mbox{for all g \in G.}$$ As $V$ is irreducible for (b), it must be (i) $Ran(\phi)=\{0\}$ or (ii) $Ran(\phi)= N$. In the case (i) $\phi$ is the null map and the proof concludes, in the case (ii) case, since $\phi$ is not bijective by hypotheses,we must have that $Ker(\phi)$ is larger than $\{0\}$ and so we can restrict to the other case treated below.

Suppose (2) is valid. In this case (a) implies: $$U_g Ker(\phi) \subset Ker(\phi)\quad \mbox{for all g \in G.}$$

Indeed: If $x \in Ker(\phi)$ then $\phi x=0$ and so, by (a), $\phi U_gx =V_g\phi x = V_g 0 =0$, so $U_gx \in Ker(\phi)$ as wanted.

Since $U$ is irreducible for (b), it must be $Ker(\phi)=M$ and this means that $\phi$ is the null map and the proof stops, or $Ker(\phi)=\{0\}$ that is impossible by hypotheses ($Ker(\phi)$ is larger than $\{0\}$).

In all cases $\phi$ is the null map if it is not bijective. QED

• See the amusing comment here about van Vleck's and George Mackey's proof here: maybe the OP was reading something akin to the former! – Selene Routley Feb 1 '14 at 13:33
• Indeed it is possible to state all the proof in few lines: since $Ran(\phi)$ and $Ker(\phi)$ are evidently invariant subspaces under the representations in the corresponding spaces, they must coincide either to the whole space or to $\{0\}$. This is equivalently to say that $\phi$ is bijectve or the null map. – V. Moretti Feb 1 '14 at 13:41
• However this is not the most interesting version of S. lemma. The physically fundamental statements reads: A unitary group representation on a Hilbert space is irreducible if and only if the bounded operators commuting with it are all of the form $cI$ with $c\in \mathbb C$. – V. Moretti Feb 1 '14 at 13:45