# An example needed of Schur's lemma

Reading Linear representations of finite groups by Serre, I need an example of the following:

Schur's lemma:

Let $$\rho^1\colon G \rightarrow GL(V_1)$$ and $$\rho^2\colon G \rightarrow GL(V_2)$$ be two irreducible representations of a group $$G$$, and let $$f$$ be a linear mapping of $$V_1$$ into $$V_2$$ such that $$\rho_s^2 \circ f = f \circ \rho_s^1$$ for all $$s \in G$$. Then:

(i) If $$\rho^1$$ and $$\rho^2$$ are not isomorphic, we have $$f=0$$

(ii) If $$V_1 = V_2$$ and $$\rho^1 = \rho^2$$, $$f$$ is scalar multiple of the identity.

Can someone form a concrete example so I can see what's going on?

• I've answered your questions before, and my sense is you are not ready for the answer. The map $f$ is a change of basis. Spend some time thinking about why the statement is true. – David Hill Apr 25 at 4:26
• @DavidHill I learn from examples as I see what is going on hence my questions on examples. it is rather obvious $f$ is a change of basis. – john Apr 25 at 10:55
• and actually I did understand the other solution which was presented, it was clearer. – john Apr 25 at 14:56
• I'm also interested in this question, can someone provide answer please! – Math Apr 29 at 12:31

If you translate Schur's Lemma into the language of representations of finite groups, you get the following.

Let $$G$$ be a finite group, $$\mathsf{k}$$ some field, and $$\rho_i : G \to \operatorname{GL}\left(V_i\right)$$ some irreducible representations of $$G$$ over $$\mathsf{k}$$. Then

1) If $$\rho_1$$ is not equivalent to $$\rho_2$$, then there are no non-trivial $$\mathsf{k}$$-linear maps $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$.

2) If $$\mathsf{k}$$ is algebraically closed, and $$\rho_1, \rho_2$$ are equivalent, then every non-trivial $$\mathsf{k}$$-linear map $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$ is an isomorphism.

By a linear map $$T : V_1 \to V_2$$ intertwining $$\rho_1, \rho_2$$, what I mean is that $$T\left( \rho_1(g) v \right) = \rho_2(g) T(v)$$ for all $$g \in G$$, $$v \in V$$, and two representations $$\rho_1, \rho_2$$ are called equivalent if there is a linear isomorphism intertwining them.

Schur's lemma is a direct result of the following observation

Let $$G$$ be a finite group, $$\mathsf{k}$$ some field, and $$\rho_i : G \to \operatorname{GL}\left(V_i\right)$$ some representations of $$G$$ over $$\mathsf{k}$$, and suppose that $$T : V_1 \to V_2$$ is a $$\mathsf{k}$$-linear map intertwining $$\rho_1$$, $$\rho_2$$. Then

1) $$(\rho_1, \operatorname{Ker}{T})$$ is a subrepresentation of $$(\rho_1, V_1)$$.

2) $$(\rho_2, \operatorname{Img}{T})$$ is a subrepresentation of $$(\rho_2, V_2)$$.

Now, if $$(\rho_i, V_i)$$ are irreducible, and $$T : V_1 \to V_2$$ is a $$\mathsf{k}$$-linear map intertwining $$\rho_1$$, $$\rho_2$$, then $$\ker{T}$$ is either $$\left\{0\right\}$$, or $$V_1$$. In the first case, $$T$$ is injective, whilst in the second case $$T$$ is the trivial map. Now suppose we have the first case, then since $$V_2$$ is irreducible, we have $$\operatorname{Img}T$$ is either $$\left\{0\right\}$$ or $$V_2$$. So our options are

1. $$V_1 = \left\{0\right\}$$, in which case $$T = 0$$,
2. $$V_1 \neq \left\{0\right\}$$ and $$T=0$$,
3. $$V_1 \neq \left\{0\right\}$$ and $$T \neq 0$$, in which case $$T$$ is injective and surjective, and so $$T$$ is a isomorphism intertwining $$V_1, V_2$$

Now suppose that we are in the third case, and $$V_1 = V_2$$, and that $$\mathsf{k}$$ is algebraically closed, and suppose that $$T : V_1 \to V_1$$ is a non-trivial intertwining map. Then by the above $$T$$ is an isomorphism, and since $$\mathsf{k}$$ is algebraically closed, $$T$$ has an eigenvector $$0 \neq v_1 \in V_1$$, with eigenvalue $$\lambda$$ say. Then it is easy to see that the $$\lambda$$-eigenspace of $$T$$ is a non-zero subrepresentation of $$V_1$$, and since $$V_1$$ is irreducible we have that $$V_1$$ is the $$\lambda$$-eigenspace of $$T$$, and so $$T(v) = \lambda v$$ for every $$v \in V_1$$, and so $$T = \lambda \operatorname{Id}_{V_1}$$, and moreover $$\lambda \neq 0$$. It follows that every non-trivial intertwining map $$T : V_1 \to V_1$$ is a scalar multiple of the identity.

Now, for a concrete example. Consider the representation of $$S_3$$ defined by $$\rho_V : S_3 \to \operatorname{GL}(\mathbb{C}^{3}) \ : \ \rho_V(\sigma)(ae_1 + be_2 + ce_3) = ae_{\sigma(1)} + be_{\sigma(2)} + ce_{\sigma(3)}.$$

Then $$\rho_V$$ is not irreducible, because the two subspaces

$$U = \left\{ae_1 + be_2 + ce_3 \mid a,b,c \in \mathbb{C} : a + b + c = 0 \right\}, \\ W= \left\{a(e_1 + e_2 + ce_3) \mid a \in \mathbb{C} \right\}$$

are both subrepresentations. Moreover, it is clear to see that $$U \cap W = \left\{ 0 \right\}$$, and so for dimension reasons we have $$V = \mathbb{C}^{3} = U \oplus W$$. Now, $$W$$ is one-dimensional and so must be irreducible, and take a moment to convince yourself that $$U$$ is irreducible too.

Now consider the representation $$\rho_t : S_3 \to \mathbb{C}^{\times}$$ such that $$\rho_t(\sigma) = t$$ for every $$\sigma$$, where $$t \neq 0$$. Suppose that $$T : \mathbb{C}^{\times} \to U$$ is an intertwining map, and suppose $$T(1) = ae_1 + be_2 + ce_3$$. Then since $$T$$ intertwines, we have $$t(ae_1 + be_2 + ce_3) = tT(1) = T(t) = T(\rho_t(\sigma)1) = \rho_V(\sigma)(T(1)) = \rho_V(\sigma)(ae_1 + be_2 + ce_3),$$

and so

$$tae_1 + tbe_2 + tce_3 = ae_{\sigma(1)} + be_{\sigma(2)} + ce_{\sigma(3)},$$

for every $$\sigma \in S_3$$. Taking $$\sigma = \operatorname{Id}$$ and $$\sigma = (12)$$ gives $$ta = a = b$$, and so $$a = b$$. Similarly you can show $$a = c$$ and $$c=b$$, but $$a + b + c =0$$, and so we get $$a = b = c =0$$. So every linear map intertwining $$\rho_t$$, and $$\rho_V$$ is zero, for whatever choice of $$\rho_t$$ we choose.

Now, suppose instead that $$T: \mathbb{C}^{\times} \to W$$ is an intertwining map, suppose that $$T(1) = a(e_1 + e_2 + e_3)$$. Then it is clear that $$T(x) = xa(e_1 + e_2 + e_3)$$, and so $$T$$ is a scalar multiple of the map $$I : \mathbb{C}^{\times} \to W$$ sending $$1$$ to $$e_1 + e_2 + e_3$$. This works regardless of the choice of $$t$$ (so long as $$t \neq 0$$ because $$(\rho_t, \mathbb{C}^{\times})$$ and $$(\rho_V, W)$$ are equivalent representations.

• For similar examples (in slightly different words, but only slightly), see also pure.au.dk/portal/files/120581284/intro_to_character_theory.pdf Example 4.3 and 4.4 (unfortunately, those are the old version of the notes as I don't have a good place to upload the newest version). – Tobias Kildetoft Apr 29 at 15:36
• What you write as $\rho^1_s$, I write as $\rho_1(s)$. I have two examples. In the first case $\rho_1$ is the map $\rho_V : S_3 \to \operatorname{GL}(U)$ and $\rho_2$ is the map $\rho_t : S_3 \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times}$. In the second case $\rho_2$ is the same, but now $\rho_1$ is the map $\rho_V : S_3 \to \operatorname{GL}(W)$ – Adam Higgins Apr 30 at 12:02
• @john I’m not sure there is any substantial difference in our notations? What in particular do you take issue with? – Adam Higgins Apr 30 at 13:28
• rereading this, can you tell me what is $\mathbb{C}^\times$? is it the space of complex numbers excluding zero? – john Jul 11 at 12:08
• For a general ring $R$ with a multiplicative identity $1_R$ say, $R^{\times}$ denotes the multiplicative group of units of $R$. Where $r \in R$ is said to be a unit if there exists $s \in R$ with $rs = sr = 1_R$. In this case the ring is $\mathbb{C}$ and in $\mathbb{C}$ every non-zero element is a unit (because it is a field), and so indeed $\mathbb{C}^{\times}$ is the complex numbers without zero, but when we write it we typically intend to emphasise that this a group. One last thing, the latex code for $\times$ is "\times". – Adam Higgins Jul 11 at 12:12