# The Unapologetic Mathematician

## Algebra Representations

We’ve defined a representation of the group $G$ as a homomorphism $\rho:G\rightarrow\mathrm{GL}(V)$ for some vector space $V$. But where did we really use the fact that $G$ is a group?

This leads us to the more general idea of representing a monoid $M$. Of course, now we don’t need the image of a monoid element to be invertible, so we may as well just consider a homomorphism of monoids $\rho:M\rightarrow\hom_\mathbb{F}(V,V)$, where we consider this endomorphism algebra as a monoid under composition.

And, of course, once we’ve got monoids and $\mathbb{F}$-linearity floating around, we’re inexorably drawn — Serge would way we have an irresistable compulsion — to consider monoid objects in the category of $\mathbb{F}$-modules. That is: $\mathbb{F}$-algebras.

And, indeed, things work nicely for $\mathbb{F}$-algebras. We say a representation of an $\mathbb{F}$-algebra $A$ is a homomorphism $\rho:A\rightarrow\hom_\mathbb{F}(V,V)$ for some vector space $V$ over $\mathbb{F}$. How else can we view such a homomorphism?

Well, it turns an algebra element into an endomorphism. And the most important thing about an endomorphism is that it does something to vectors. So given an algebra element $a\in A$, and a vector $v\in V$, we get a new vector $\left[\rho(a)\right](v)$. And this operation is $\mathbb{F}$-linear in both of its variables. So we have a linear map $\mathrm{ev}\circ(\rho\otimes1_V):A\otimes V\rightarrow V$, built from the representation $\rho$ and the evaluation map $\mathrm{ev}$. But this is just a left $A$-module!

In fact, the evaluation above is the counit of the adjunction between $\underline{\hphantom{X}}\otimes V$ and the internal $\hom$ functor $\hom_\mathbb{F}(V,\underline{\hphantom{X}})$. This adjunction is a natural isomorphism of $\hom$ sets: $\hom_\mathbb{F}(A\otimes V,V)\cong\hom_\mathbb{F}(A,\hom_\mathbb{F}(V,V))$. That is, left $A$-modules are in natural bijection with representations of $A$. In practice, we just consider the two structures to be the same, and we talk interchangeably about modules and representations.

As it would happen, the notion of an algebra representation properly extends that of a group representation. Given any group $G$ we can build the group algebra $\mathbb{F}[G]$. As a vector space, this has a basis vector $e_g$ for each group element $g\in G$. We then define a multiplication on pairs of basis elements by $e_{g_1}e_{g_2}=e_{g_1g_2}$, and extend by bilinearity.

Now it turns out that representations of the group $G$ and representations of the group algebra $\mathbb{F}[G]$ are in bijection. Indeed, the basis vectors $e_g$ are invertible in the algebra $\mathbb{F}[G]$. Thus, given a homomorphism $\alpha:\mathbb{F}[G]\rightarrow\hom_\mathbb{F}(V,V)$, the linear maps $\rho(g)=\alpha(e_g)$ must be invertible. And so we have a group representation $\rho:G\rightarrow\mathrm{GL}(V)$. Conversely, if $\rho:G\rightarrow\mathrm{GL}(V)$ is a representation of the group $G$, then we can define $\alpha(e_g)=\rho(g)\in\mathrm{GL}(V)\subset\hom_\mathbb{F}(V,V)$ and extend by linearity to get an algebra representation $\alpha:\mathbb{F}[G]\rightarrow\hom_\mathbb{F}(V,V)$.

So we have representations of algebras. Within that we have the special cases of representations of groups. These allow us to cast abstract algebraic structures into concrete forms, acting as transformations of vector spaces.

October 24, 2008 -

## 3 Comments »

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