The Unapologetic Mathematician

Mathematics for the interested outsider

Intertwiner Spaces

Another grading day, another straightforward post. It should come as no surprise that the collection of intertwining maps between any two representations forms a vector space.

Let’s fix representations \rho:A\rightarrow\mathrm{End}(V) and \sigma:A\rightarrow\mathrm{End}(W). We already know that \hom_\mathbb{F}(V,W) is a vector space. We also know that an intertwiner f\in\hom_{\mathbf{Rep}_\mathbb{F}(A)}(\rho,\sigma) can be identified with a linear map f\in\hom_\mathbb{F}(V,W). What I’m asserting is that f\in\hom_{\mathbf{Rep}_\mathbb{F}(A)}(\rho,\sigma) is actually a subspace of f\in\hom_\mathbb{F}(V,W) under this identification.

Indeed, all we really need to check is that this subset is closed under additions and under scalar multiplications. For the latter, let’s say that f is an intertwiner. That is, \left[\sigma(a)\right]\left(f(v)\right)=f\left(\left[\rho(a)\right](v)\right). Then given a constant c\in\mathbb{F} we consider the linear map cf we calculate


And so cf is an intertwiner as well.

Now if f and g are both intertwiners, satisfying conditions like the one above, we consider their sum f+g and calculate


Which shows that f+g is again an intertwiner.

Since composition of intertwiners is the same as composing their linear maps, it’s also bilinear. It immediately follows that the category \mathbf{Rep}_\mathbb{F}(A) is enriched over \mathbf{Vect}_\mathbb{F}.

December 11, 2008 - Posted by | Algebra, Representation Theory

1 Comment »

  1. […] of all, the intertwiners between any two representations form a vector space, which is really an abelian group plus stuff. Since the composition of intertwiners is bilinear, […]

    Pingback by The Category of Representations is Abelian « The Unapologetic Mathematician | December 15, 2008 | Reply

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