# The Unapologetic Mathematician

## Representations of a Polynomial Algebra

Sorry for the delays, but tests are killing me this week.

Okay, so let’s take the algebra of polynomials, $\mathbb{F}[X]$ and consider its representation theory.

What is a representation of this algebra? It’s a homomorphism of $\mathbb{F}$-algebras $\rho:\mathbb{F}[X]\rightarrow\hom_\mathbb{F}(V,V)$. But the algebra of polynomials satisfies a universal property! A homomorphism of $\mathbb{F}$-algebras is uniquely determined by the image of the single element $X$, and we can pick this image freely. That is, once we pick a linear transformation $T:V\rightarrow V$ and set $\rho_T(X)=T$, then we are forced to use

$\displaystyle\rho_T\left(\sum\limits_{k=0}^na_nX^n\right)=\sum\limits_{k=0}^na_nT^n$

for all the other polynomials. That is, representations $\rho_T$ of $\mathbb{F}[X]$ are in bijection with the linear transformations $T\in\hom_\mathbb{F}(V,V)$.

But remember that these representations don’t live in a vacuum. No, they’re just the objects of a whole category of representations. We need to consider the morphisms between representations too!

So if $S:V\rightarrow V$ and $T:W\rightarrow W$ are linear transformations, what’s a morphism $\phi:\rho_S\rightarrow\rho_T$? It’s a linear map $\phi:V\rightarrow W$ such that $S\circ\phi=\phi\circ T:V\rightarrow W$. Notice that if $\phi$ intertwines the linear maps $S$ and $T$, then it will automatically intertwine the values of $\rho_S$ and $\rho_T$ for every polynomial.

Rather than try to examine this condition in detail (which leads to an interesting problem in the theory of quivers, if I recall), let’s just consider which representations are isomorphic. That is, let’s decategorify this category.

So we ask that the linear map $\phi:V\rightarrow W$ be an isomorphism, with inverse $\phi^{-1}:W\rightarrow V$. Then we can take the intertwining relation $S\circ\phi=\phi\circ T$ and compose on the left with $\phi^{-1}$ to find $T=\phi^{-1}\circ S\circ\phi$. But this uniquely specifies $T$ given $S$ and $\phi$. That is, given a representation $\rho_S:\mathbb{F}[X]\rightarrow\hom_\mathbb{F}(V,V)$ and an isomorphism $\phi:V\rightarrow W$, there is a unique representation $\rho_T:\mathbb{F}[X]\rightarrow\hom_\mathbb{F}(V,V)$ so that $\phi:\rho_S\rightarrow\rho_T$ is a natural isomorphism.

And we’re drawn again to consider the special case where $V=W$. Now an isomorphism is just a change of basis. Representations of $\mathbb{F}[X]$ are equivalent if they do “the same thing” to the vector space $V$, but just express it with different coordinates.

So here’s the upshot: the general linear group $\mathrm{GL}(V)$ acts on the hom-set $\hom_\mathbb{F}(V,V)$ by conjugation — basis changes. In fact, this is a representation of the group, but I’m not ready to go into that detail right now. What I can say is that the orbits of this action are in bijection with the equivalence classes of representations of $\mathbb{F}[X]$ on $V$.

October 30, 2008