The Unapologetic Mathematician

Mathematics for the interested outsider

The Coevaluation on Vector Spaces

Okay, I noticed that I never really gave the definition of the coevaluation when I introduced categories with duals, because you need some linear algebra. Well, now we have some linear algebra, so let’s do it.

Let V be a finite-dimensional vector space with dual space V^*. Then if we have a basis \left\{e_i\right\} of V we immediately get a dual basis \left\{\epsilon^i\right\} for V^* (yet another \epsilon to keep straight), defined by \epsilon^j(e_i)=\delta_i^j. We now define a map \eta:\mathbf{1}\rightarrow V^*\otimes V by setting \eta(1)=\epsilon^i\otimes e_i. That is, we take the tensor product of each dual basis element with its corresponding basis element, and add them all up (summation convention).

But this seems to depend on which basis \left\{e_i\right\} we started with. What if we used a different basis \left\{f_i\right\} and dual basis \left\{\phi^i\right\}? We know that there is a change of basis matrix f_i=t_i^je_j, so let’s see how this works on the dual basis.

The dual basis is defined by the fact that \phi^j(f_i)=\delta_i^j. So we use this new expression for f_i to write \phi^j(t_i^ke_k)=t_i^k\phi^j(e_k)=\delta_i^j. That is, \phi^j(e_k) must be the inverse matrix to t_i^k, which we’ll write as \left(t^{-1}\right)_k^j. But now we can check

\left[\left(t^{-1}\right)_i^j\epsilon^i\right](e_k)=\left(t^{-1}\right)_i^j\delta_k^i=\left(t^{-1}\right)_k^j

And so we find that \phi^i=\left(t^{-1}\right)_j^i\epsilon^j when we change bases.

Now we can use the same definition for \eta above with our new basis. We set \eta(1)=\phi^i\otimes f_i, and then substitute our expressions in terms of the old bases:

\eta(1)=\left(t^{-1}\right)_j^i\epsilon^j\otimes t_i^ke_k=\left(t^{-1}\right)_j^it_i^k\left(\epsilon^j\otimes e_k\right)=\delta_j^k\left(\epsilon^j\otimes e_k\right)=\epsilon^k\otimes e_k

which is what we got before. That is, this map actually doesn’t depend on the basis we chose!

Okay, now does this coevaluation — along with the evaluation map from before — actually satisfy the conditions for a duality? First, let’s start with a vector written out in terms of a basis: v=v^ie_i. Now we use the coevaluation to send it to v^ie_i\otimes\epsilon^j\otimes e_j. Next we evaluate on the first two tensorands to find v^i\delta_i^je_j=v^je_j. So we do indeed have the identity here. Verifying the other condition is almost the same, starting from an arbitrary covector \lambda=\lambda_i\epsilon^i.

So now we know that the category \mathbf{FinVect}(\mathbb{F}) has duals. Tomorrow we can promote this to a duality on the category of finite-dimensional representations of a Hopf algebra.

November 13, 2008 Posted by | Algebra, Linear Algebra | 2 Comments

   

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