Sorry for the delays, but tests are killing me this week.
What is a representation of this algebra? It’s a homomorphism of -algebras . But the algebra of polynomials satisfies a universal property! A homomorphism of -algebras is uniquely determined by the image of the single element , and we can pick this image freely. That is, once we pick a linear transformation and set , then we are forced to use
for all the other polynomials. That is, representations of are in bijection with the linear transformations .
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 and are linear transformations, what’s a morphism ? It’s a linear map such that . Notice that if intertwines the linear maps and , then it will automatically intertwine the values of and 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 be an isomorphism, with inverse . Then we can take the intertwining relation and compose on the left with to find . But this uniquely specifies given and . That is, given a representation and an isomorphism , there is a unique representation so that is a natural isomorphism.
And we’re drawn again to consider the special case where . Now an isomorphism is just a change of basis. Representations of are equivalent if they do “the same thing” to the vector space , but just express it with different coordinates.
So here’s the upshot: the general linear group acts on the hom-set 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 on .