1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – David Hill Apr 25 at 4:26
  • $\begingroup$ @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. $\endgroup$ – john Apr 25 at 10:55
  • $\begingroup$ and actually I did understand the other solution which was presented, it was clearer. $\endgroup$ – john Apr 25 at 14:56
  • $\begingroup$ I'm also interested in this question, can someone provide answer please! $\endgroup$ – Math Apr 29 at 12:31
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Tobias Kildetoft Apr 29 at 15:36
  • $\begingroup$ 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 $\rho_V : S_3 \to \operatorname{GL}(W)$ $\endgroup$ – Adam Higgins Apr 30 at 12:02
  • $\begingroup$ @john I’m not sure there is any substantial difference in our notations? What in particular do you take issue with? $\endgroup$ – Adam Higgins Apr 30 at 13:28
  • $\begingroup$ rereading this, can you tell me what is $\mathbb{C}^\times$? is it the space of complex numbers excluding zero? $\endgroup$ – john Jul 11 at 12:08
  • $\begingroup$ 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". $\endgroup$ – Adam Higgins Jul 11 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.