# The Unapologetic Mathematician

## Some Representations of the General Linear Group

Sorry for the delays, but it’s the last week of class and everyone came back from the break in a panic.

Okay, let’s look at some examples of group representations. Specifically, let’s take a vector space $V$ and consider its general linear group $\mathrm{GL}(V)$.

This group comes equipped with a representation already, on the vector space $V$ itself! Just use the identity homomorphism $\mathrm{GL}(V)\rightarrow\mathrm{GL}(V)$ We often call this the “standard” or “defining” representation. In fact, it’s easy to forget that it’s a representation at all. But it is.

As with any other group, we have dual representations. That is, we immediately get an action of $\mathrm{GL}(V)$ on $V^*$. And we’ve seen it already! When we talked about the coevaluation on vector spaces we worked out how a change of basis affects linear functionals. What we found is that if $\rho(g)$ is our action on $V$, then the action on $V^*$ is by the transpose — the dual — of $\rho(g)^{-1}=\rho\left(g^{-1}\right)$. And this is exactly the dual representation.

Also, as with any other group, we have tensor representations — actions on the tensor power $V^{\otimes n}=V\otimes...\otimes V$ for any number $n$ of factors of $V$. How does this work? Well, every vector in $V\otimes...\otimes V$ is a linear combination of vectors of the form $v_1\otimes...\otimes v_n$, where each $v_k\in V$. And we know how to act on these: just act on each tensorand separately. That is, $\displaystyle\left[\rho(g)\right](v_1\otimes...\otimes v_n)=\left[\rho(g)\right]\left(v_1\right)\otimes...\otimes\left[\rho(g)\right]\left(v_n\right)$

Then we just extend this action by linearity to all of $V^{\otimes n}$.

December 2, 2008 -

## 7 Comments »

1. Out of curiosity: where are you heading with this? Were you planning to discuss Schur-Weyl duality? Comment by Todd Trimble | December 3, 2008 | Reply

2. Actually, I’m going to mention it tomorrow. But I’m not going to actually prove it, or use it as such. What I am going to do is introduce a particular representation that many readers may already be familiar with, but without picking a basis! Comment by John Armstrong | December 3, 2008 | Reply

3. […] Symmetric Group Representations Tuesday, we talked about tensor powers of the standard representation of . Today we’ll look at some representations of the symmetric […]

Pingback by Some Symmetric Group Representations « The Unapologetic Mathematician | December 4, 2008 | Reply

4. […] tomorrow I want to take last Friday’s symmetrizer and antisymmetrizer and apply them to the tensor representations of , which we know also carry symmetric group representations. Specifically, the th tensor power […]

Pingback by Symmetric Tensors « The Unapologetic Mathematician | December 22, 2008 | Reply

5. […] project by considering antisymmetric tensors today. Remember that we’re starting with a tensor representation of on the tensor power , which also carries an action of the symmetric group by permuting the […]

Pingback by Antisymmetric Tensors « The Unapologetic Mathematician | December 23, 2008 | Reply

6. […] vector space. But remember that this isn’t just a vector space. The tensor power is both a representation of and a representation of , which actions commute with each other. Our antisymmetric tensors are […]

Pingback by The Determinant « The Unapologetic Mathematician | December 31, 2008 | Reply

7. […] each degree. And if is invertible, so must be its image under each functor. These give exactly the tensor, symmetric, and antisymmetric representations of the group , if we consider how these functors act […]

Pingback by Functoriality of Tensor Algebras « The Unapologetic Mathematician | October 28, 2009 | Reply