The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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October 30, 2008 - Posted by | Algebra, Linear Algebra, Representation Theory

1 Comment »

  1. [...] of Polynomials of Transformations When we considered the representation theory of the algebra of polynomials, we saw that all it takes to specify such a representation is choosing a single endomorphism . That [...]

    Pingback by Kernels of Polynomials of Transformations « The Unapologetic Mathematician | February 24, 2009 | Reply


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