The Unapologetic Mathematician

Mathematics for the interested outsider

Categories with Duals

Now we’ve got monoidal categories to categorify the notion of a monoid, we should consider what the proper analogue of an inverse is. What we’ll do is define a “dual object”, which is sort of halfway a left inverse and halfway a right inverse.

So, we say a monoidal category \mathcal{M} “has duals” if for every object M there is a “dual object” M^* and natural transformations \eta_M:\mathbf{1}\rightarrow M^*\otimes M and \epsilon_M:M\otimes M^*\rightarrow\mathbf{1}. We do not ask that these be isomorphisms, since for the best examples they’re actually not.

Instead we ask that the transformations be compatible in the sense that the following equations hold:
\lambda_M\circ(\epsilon_M\otimes1_M)\circ\alpha_{M,M^*,M}^{-1}\circ(1_M\otimes\eta_M)\circ\rho_M^{-1}=1_M
\rho_{M^*}\circ(1_{M^*}\otimes\epsilon_M)\circ\alpha_{M^*,M,M^*}\circ(\eta_M\otimes1_{M^*})\circ\lambda_M^{-1}=1_{M^*}
The first one means (reading right to left) that if we create a copy of the identity on the right with \rho^{-1}, then hit it with \eta to create M^*\otimes M, then associate, then hit the M\otimes M^* on the left with \epsilon to get the identity, then remove it with \lambda, it’s just the same as doing nothing at all! And there’s a similar interpretation of the other equation. I can hear your furiously scratching your heads and saying “huh?” right now, but I’ll come back to this with another viewpoint later.

Anyhow, we also want this duality to be a functor — a contravariant functor even — which you should have guessed when I called \eta and \epsilon natural transformations. So we need for every f:M\rightarrow N and arrow f^*:N^*\rightarrow M^* so that (f\circ g)^*=g^*\circ f^* and 1_M^*=1_{M^*}. There are also a naturality conditions for \eta and \epsilon that you should be able to write down if you think of this as a covariant functor (\underline{\hphantom{X}})^*:\mathcal{M}^\mathrm{op}\rightarrow\mathcal{M}. The functor should also be monoidal, in the sense that (M\otimes N)^*\cong N^*\otimes M^* (why this form? write out the condition explicitly).

The motivating example here is the category \mathbf{FinVect}_k of finite-dimensional vector spaces over a field k. Here we know that tensor product over k gives a monoidal structure, and we’ll use the dual module of linear functionals as our functor. Indeed, if we set M^*=\hom_{\mathbf{Vect}_k}(M,k) we clearly have a contravariant functor. We can verify that it’s monoidal in the proper sense, and that the pairing gives a natural transformation M\otimes M^*\rightarrow\mathbf{1}=k. The other natural transformation required is also there, but we don’t yet have the tools needed (unless you’ve taken some linear algebra on your own). Still, you can keep this example in mind.

Now one thing that should be bugging you is that we have arrows M\otimes M^*\rightarrow\mathbf{1} and \mathbf{1}\rightarrow M^*\otimes M\rightarrow, but there are two more it seems we should have. To handle this, we’ll insist that (M^*)^*\cong M — that duality “is its own inverse”, so that we can just use \eta_{M^*} and \epsilon_{M^*} as the other natural transformations we need. This does indeed hold in our motivating example, but we again won’t prove it completely until later.

Note that there’s no reason to assume that \epsilon_{M^*}\circ\eta_M=1_\mathbf{1} or \epsilon_{M}\circ\eta_{M_*}=1_\mathbf{1}, so these still aren’t isomorphisms. If they are, though, then we really do have natural isomorphisms to replace left and right inverse rules in the definition of a group. If you’re feeling up to it, try to state (prove?) a coherence theorem that says what diagrams must commute to ensure they all do.

About these ads

July 7, 2007 - Posted by | Category theory

10 Comments »

  1. It looks like you’ve got some misplaced dollar signs in the penultimate paragraph, after “so that we can just use \eta_M“.

    Comment by Blake Stacey | July 8, 2007 | Reply

  2. Thanks, fixed.

    Comment by John Armstrong | July 8, 2007 | Reply

  3. [...] this sort of naturality before, though. If we read the diagram in , consider a monoidal category with duals, and use the functor , then this is exactly the sort of naturality we find in the duality arrows [...]

    Pingback by Extraordinary Naturality « The Unapologetic Mathematician | August 30, 2007 | Reply

  4. [...] Another thing vector spaces come with is duals. That is, given a vector space we have the dual vector space of “linear functionals” [...]

    Pingback by Dual Spaces « The Unapologetic Mathematician | May 27, 2008 | Reply

  5. [...] category of matrices also has duals. In fact, each object is self-dual! That is, we set . We then need our arrows and [...]

    Pingback by The Category of Matrices II « The Unapologetic Mathematician | June 3, 2008 | Reply

  6. typoes in par 4: for every f:N->M an arrow f^*:M*->N*

    Comment by Avery Andrews | July 19, 2008 | Reply

  7. Thanks.

    Comment by John Armstrong | July 19, 2008 | Reply

  8. [...] of a bialgebra is monoidal. What do we get for Hopf algebras? What does an antipode buy us? Duals! At least when we restrict to finite-dimensional [...]

    Pingback by Representations of Hopf Algebras I « The Unapologetic Mathematician | November 12, 2008 | Reply

  9. [...] 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 [...]

    Pingback by The Coevaluation on Vector Spaces « The Unapologetic Mathematician | November 13, 2008 | Reply

  10. [...] if that weren’t enough, has duals! Indeed, we have a cone in the dual vector space defined by if and only if for all . Or in other [...]

    Pingback by Ordered Linear Spaces II « The Unapologetic Mathematician | May 15, 2009 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: