Our lecturer gave us a hard exercice to go further in group theory (we stopped at group actions) :

Let G be a group, V and W complex vector spaces and $\rho_1 : G \mapsto GL(V) $ be a group homomorphism where GL(V) is the general linear group of V (i.e. the invertible linear maps V $\mapsto$ V, group under compositions of maps) and let $\rho_2 : G \mapsto GL(W) $.

$(\rho_1,V)$ and $(\rho_2,V)$ are called representation of G. I have to show that :

-For representations V, W of G, the direct sum $V\oplus W$ is a representation of G. i.e. I have to find a homomorphism $\omega_1 : G \mapsto GL(V\oplus W)$.

-Same again but for the tensor product $V\otimes W$ via $g(v\otimes w):=gv\otimes gw$ i.e. I have to find a homomorphism $\omega_2 : G \mapsto GL(V\otimes W)$.

The problem is : I struggle with the definitions of direct sum and tensor product... (never worked on this before).

Here is what I've done for the first question. I have to show that : let $v\in V$ and $w\in V$ and $(\rho, V\oplus W)$ be a representation of G. Then $\forall g\in G, \rho(g)(v+w)=\rho(g)(v)+\rho(g)(w):=\rho_1(g)(v)+\rho_2(g)(w) \in V\oplus W$. But I don't know how to prove it.

I tried to find lectures and understand as much as I could but there is usually no explanations about those two questions, it is considered trivial apparently !

Thanks for your help

  • 2
    $\begingroup$ Your task was to define a representation of $G$ on $V\oplus W$, and you have done so with the formula $\rho(v+w)=\rho_(v)+\rho_2(w)$. You still need to check that this $\rho$ is a representation. You do not need to prove that this is the only possible representation of $G$ on $V\oplus W$ (because it usually isn't). Similarly for $V\otimes W$. $\endgroup$ – Andreas Blass Nov 10 '13 at 0:42
  • $\begingroup$ I've shown that the map defined as :$\forall g\in G, \rho(g)(v+w)=\rho(g)(v)+\rho(g)(w):=\rho_1(g)(v)+\rho_2(g)(w) \in V\oplus W$ is a map from G to GL(V$\oplus$ W), i.e it is a representation. Is this not right ? $\endgroup$ – ALM Nov 10 '13 at 15:13
  • 2
    $\begingroup$ Representations are not arbitrary maps but homomorphisms. $\endgroup$ – Andreas Blass Nov 11 '13 at 1:06
  • $\begingroup$ I think I got it : $\forall g\in G, \rho(g)(v+w)=\rho(g)(v)+\rho(g)(w):=\rho_1(g)(v)+\rho_2(g)(w) \in V\oplus W$. Let $g_1,g_2 \in G$ then $\rho(g_1+g_2)(v+w)=\rho_1(g_1+g_2)(v)+\rho_2(g_1+g_2)(w)$ but $\rho_1$ and $\rho_2$ are group homomorphisms so we get $\rho(g_1+g_2)(v+w)=\rho_1(g_1)(v)+\rho_1(g_2)(v)+\rho_2(g_1)(w)+\rho_2(g_2)(w) =\rho_1(g_1)(v)+\rho_2(g_1)(w)+\rho_1(g_2)(v)+\rho_2(g_2)(w)$ which is by definition of$\rho$ equals to $\rho(g_1)(v+w)+\rho(g_2)(v+w)$ i.e $\rho$ is a group homomorphism, i.e a representation of G. $\endgroup$ – ALM Nov 11 '13 at 18:11
  • Notation

    $(G,\cdot_G)$ is a group with composition (or product) $\cdot_G$. The group of automorphisms of a vector space, let us say $V$, is denoted by $(\operatorname{Aut}(V),\circ)$, where $\circ$ is the composition of automorphisms.

The representations are defined in the OP; we use the following notation

$$\rho_1: (G,\cdot_G)\rightarrow (\operatorname{Aut}(V),\circ),$$ $$\rho_2:(G,\cdot_G)\rightarrow (\operatorname{Aut}(W),\circ).$$ We just recall that given any representation $\rho: (G,\cdot_G)\rightarrow (\operatorname{Aut}(T),\circ)$ we have

$$\rho(g_1\cdot_G g_2)=\rho(g_1)\circ\rho(g_2)$$

for all $g_1,g_2\in G$.

  • On direct sum.

$$ \rho_\oplus:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\oplus W),\circ)$$

is given by $\rho_\oplus:=\rho_1\oplus\rho_2$, i.e. $$\rho_\oplus(g)(v\oplus w)=\rho_1(g)(v)\oplus\rho_2(g)(w)\in V\oplus W$$ for all $v \in V$, $w\in W$ and $g\in G$.

By definition, it follows that $\rho_\oplus(g_1\cdot_G g_2)=\rho_\oplus(g_1)\circ\rho_\oplus(g_2)$ and $\rho_\oplus(g^{-1})=\rho^{-1}_\oplus(g)$. This makes $\rho_\oplus$ a group homomorphism. Let us prove the first one as example. The first equation is proven by

$$(\rho_\oplus(g_1\cdot_G g_2))(v\oplus w)=(\text{def. of}~\rho_\oplus)= \rho_1(g_1\cdot_G g_2)(v)\oplus\rho_2(g_1\cdot_G g_2)(w)=(\text{def. of representations:})= \rho_1(g_1)(\rho_1(g_2)(v))\oplus\rho_2(g_1)(\rho_2(g_2)(w))= (\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w);$$

the last equality follows from the definition of composition $\circ$ in $\operatorname{Aut}(V\oplus W)$. In fact:

$$(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w):= \rho_\oplus(g_1)(\rho_\oplus(g_2)(v\oplus w))=(\text{def. of}~\rho_\oplus)= \rho_\oplus(g_1)(\underbrace{\rho_1(g_2)(v)}_{\in V}\oplus \underbrace{\rho_2(g_2)(w)}_{\in W})=(\text{again def. of}~\rho_\oplus)=\\ \underbrace{\rho_1(g_1)(\rho_1(g_2)(v))}_{\in V}\oplus \underbrace{\rho_2(g_1)(\rho_2(g_2)(w))}_{\in W},$$

as wished.

  • On tensor product

$$ \rho_\otimes:(G,\cdot_G)\rightarrow (\operatorname{Aut}(V\otimes W),\circ)$$

is given by $\rho_\otimes:=\rho_1\otimes\rho_2$, i.e. $$\rho_\otimes(g)(v\otimes w)=\rho_1(g)(v)\otimes\rho_2(g)(w)\in V\otimes W$$ for all $v \in V$, $w\in W$ and $g\in G$.

By definition, it follows that $\rho_\otimes(g_1\cdot_G g_2)=\rho_\otimes(g_1)\circ\rho_\otimes(g_2)$ and $\rho_\otimes(g^{-1})=\rho^{-1}_\otimes(g)$. This makes $\rho_\otimes$ a group homomorphism. The relations are proven in a similar way to the one used in the direct sum case.

  • $\begingroup$ Thanks very much, the notation helps a lot. But I don't understand the last step for the proof for the direct sum. Is it a property of the direct sum that: $\rho_1(g_1)(\rho_1(g_2))(v)) \oplus\rho_2(g_1)(\rho_2(g_2)(w))=(\rho_1(g_1)v \oplus\rho_2(g_1)w)\circ (\rho_1(g_2)v \oplus\rho_2(g_2)w$ which is indeed $(\rho(g_1)\circ \rho(g_2))(v\oplus w)$ ? $\endgroup$ – ALM Nov 13 '13 at 0:01
  • $\begingroup$ And similar for the tensor product I guess, is a property of the tensor product that : $\rho_1(g_1)(\rho_1(g_2))v\otimes \rho_2(g_1)(\rho_2(g_2))w=((\rho_1(g_1)v\otimes \rho_2(g_2)(w))\circ (\rho_1(g_2)v)\otimes \rho_2(g_2)(w))$, which is $(\rho(g_1)\circ(g_2))(v\otimes w)$, as required $\endgroup$ – ALM Nov 13 '13 at 0:43
  • 1
    $\begingroup$ I add some details to the proof in my answer. I edit it right now $\endgroup$ – Avitus Nov 13 '13 at 8:10
  • $\begingroup$ Thanks ! is it not $\rho_2$ instead of $\rho_1$ in the last two steps ? i.e. $(\rho_\oplus(g_1)\circ\rho_\oplus(g_2))(v\oplus w):= \rho_\oplus(g_1)(\rho_\oplus(g_2)(v\oplus w))=(\text{def. of}~\rho_\oplus)= \rho_\oplus(g_1)(\underbrace{\rho_1(g_2)(v)}_{\in V}\oplus \underbrace{\rho_2(g_2)(w)}_{\in W})=(\text{again def. of}~\rho_\oplus)=\\ \underbrace{\rho_1(g_1)(\rho_1(g_2)(v))}_{\in V}\oplus \underbrace{\rho_2(g_1)(\rho_2(g_2)(w))}_{\in W}$ $\endgroup$ – ALM Nov 13 '13 at 10:28
  • 1
    $\begingroup$ Yep, that's what I meant ! ALright I got it now, thanks very much ! $\endgroup$ – ALM Nov 13 '13 at 14:17

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