The Unapologetic Mathematician

Mathematics for the interested outsider

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 - Posted by | Algebra, Linear Algebra, Representation Theory


  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

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