# The Unapologetic Mathematician

## Direct Sums of Representations

We know that we can take direct sums of vector spaces. Can we take representations $\rho:A\rightarrow\mathrm{End}(V)$ and $\sigma:A\rightarrow\mathrm{End}(W)$ and use them to put a representation on $V\oplus W$? Of course we can, or I wouldn’t be making this post!

This is even easier than tensor products were, and we don’t even need $A$ to be a bialgebra. An element of $V\oplus W$ is just a pair $(v,w)$ with $v\in V$ and $w\in W$. We simply follow our noses to define $\displaystyle\left[\left[\rho\oplus\sigma\right](a)\right](v,w)=\left(\left[\rho(a)\right](v),\left[\sigma(a)\right](w)\right)$

The important thing to notice here is that the direct summands $V$ and $W$ do not interact with each other in the direct sum $V\oplus W$. This is very different from tensor products, where the tensorands $V$ and $W$ are very closely related in the tensor product $V\otimes W$. If you’ve seen a bit of pop quantum mechanics, this is exactly the reason quantum system exhibit entanglement while classical systems don’t.

Okay, so we have a direct sum of representations. Is it a biproduct? Luckily, we don’t have to bother with universal conditions here, because a biproduct can be defined purely in terms of the morphisms $\pi_i$ and $\iota_i$. And we automatically have the candidates for the proper morphisms sitting around: the inclusion and projection morphisms on the underlying vector spaces! All we need to do is check that they intertwine representations, and we’re done. And we really only need to check that the first inclusion and projection morphisms work, because all the others are pretty much the same.

So, we’ve got $\iota_1:V\rightarrow V\oplus W$ defined by $\iota_1(v)=(v,0)$. Following this with the action on $V\oplus W$ we get $\displaystyle\left(\left[\rho(a)\right](v),\left[\sigma(a)\right](0)\right)=\left(\left[\rho(a)\right](v),0\right)$

But this is the same as if we applied $\iota_1$ to $\left[\rho(a)\right](v)$. Thus, $\iota_1$ is an intertwiner.

On the other hand, we have $\pi_1:V\oplus W\rightarrow V$, defined by $\pi_1(v,w)=v$. Acting now by $\rho$ we get $\left[\rho(a)\right](v)$, while if we acted by $\rho\oplus\sigma$ beforehand we’d get $\displaystyle\pi_1\left(\left[\rho(a)\right](v),\left[\sigma(a)\right](w)\right)=\left[\rho(a)\right](v)$

Just as we want.

The upshot is that taking the direct sum of two representations in this manner is a biproduct on the category of representations.

December 9, 2008