# Why is the definition of equivalent representations natural?

The definition of equivalent representations is given by:

Definition: Let $$G$$ be a group, $$\rho : G\rightarrow GL(V)$$ and $$\rho' : G\rightarrow GL(V')$$ be two representations of G. We say that $$\rho$$ and $$\rho'$$ are $$equivalent$$ (or isomorphic) if $$\exists \space T:V\rightarrow V'$$ linear isomorphism such that $$T{\rho_g}={\rho'_g}T\space \forall g\epsilon G$$.

What I would think a more natural definition would be:

Possible different definition: Let $$G$$ be a group, $$\rho : G\rightarrow GL(V)$$ and $$\rho' : G\rightarrow GL(V')$$ be two representations of G. We say that $$\rho$$ and $$\rho'$$ are $$equivalent$$ (or isomorphic) if $$\rho(G)$$ is isomorphic to $$\rho'(G)$$.

Then we would get the nice property that:

Given $$\rho : G \rightarrow GL_n(\mathbb{C})$$, $$\rho' : G \rightarrow GL_1(\mathbb{C})$$, such that $$\rho(g) = \text{Id}_n$$ and $$\rho'(g) = \text{Id}_1$$ for all $$g \in G$$. Then we have $$\rho \sim \rho'$$.

Can one say if there would be problems with the possible different definition?

I feel it would be more natural if a definition of equivalent representations would only consider the images of the representations, and not the underlying vector spaces.

• A representation of $G$ on $V$ can be thought of as a group action, where the bijections $V \to V$ induced by each $g \in G$ happen to be linear maps. The usual definition of equivalent representation is precisely saying that $V$ and $V’$ are isomorphic $G$-sets, and the isomorphism can be taken to be linear. So a similar question would be: why not consider two $G$-sets to be equivalent when the image of $G$ in their permutation groups are isomorphic? Commented Oct 21, 2018 at 15:06
• @Joppy indeed: Why not consider two G-sets to be equivalent when the image of G in their permutation groups are isomorphic? Commented Oct 21, 2018 at 15:41
• The moral reason I think is that $G$-sets or $G$-vector spaces are what we actually want to study, and hopefully having an action of $G$ helps introduce some nice symmetry into a problem. For example, in $G$-sets we get a decomposition into orbits, and in $G$-representations we get a decomposition into isotypic components. There is another reason to not use the different definition: it would mean that up to isomorphism, simple groups have precisely two representations, one trivial and one not. Commented Oct 22, 2018 at 0:46

The usual notion of isomorphism implies yours: Let $$\phi\colon V\to V'$$ be an intertwining operator that is a linear isomorphism. Then if $$\rho(g)=1$$, $$\rho'(g)\phi(v)=\phi(\rho(g)v)=\phi(v)$$ for all $$v\in V$$, and $$\rho'(g)=1$$. The same argument the other way round shows completes the proof that $$\ker\rho=\ker\rho'$$, so that $$\rho(G)\simeq \rho'(G)$$. However, your notion is isomorphism cannot distinguish between a representation $$V$$ and a $$V\oplus\mathbb{C}^n$$ for any $$n$$, where $$G$$ acts trivially on the second factor. The major task upon meeting a new representation is to decompose it into irreducibles. For finite groups, a great too for this is character theory, which won't work for your definition.