# Is this the right categorical generalisation of dual space

### Motivatation

I am presently looking at a structure that I am trying to pin down- my strategy being to pull the thing up into the greatest possible generality (based on the bits I'm sure about) and narrow it down from there.

The situation I have is somewhat similar to the dual space structure in vector spaces, though almost certainly less well-behaved and vector spaces alone will not cut it.

### The Construction

Consider a vector space $V$ over a field $k$- its dual space $V^*$ appears naturally as the set linear of maps

$$w^* : V \to k$$

which, coincidentally, form a vector space in themselves. We can generalise this to arbitrary categories $A$, $B$, $C$ by setting $B=hom(A,C)$. Then, at least in some sense, $B=A^*$. So far so standard, but I want more: a nice property of dual spaces is that an element of $V \otimes V^*$ can be canonically seen as an element of $hom(V,V)$- this is because of the way that $k$ acts on $V$ by multiplication. We can mimic this by letting $C$ be a monoid acting on $A$.

In summary: A category $A$ acted on by a monoid $C$ and a dual A $A^*:=hom(A,C)$

I am particularly interested in when $A$ is also a monoid, especially so when $A$ is a space of stochastic matrices.

### Questions:

So this isn't too unlikely a construction, in fact it's probably forehead-slappingly well known, so:

• What is it called, if anything?
• In what cases can we have $A=B=C$? Is that necessarily a permutation group for example?
• Are there any useful canonical examples, besides vector spaces?
• Better yet, theorems??? Papers???

As you can probably tell, I am no category theorist, so any help would be awesome.

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I believe the right generalization of dual spaces is that of a dual object in a tensor category, which I will assume symmetric for convenience.

Recall what makes a dual space of a vector space work: We have a map $V \times V^* \to k$ (for $k$ the ground field). The problem is, this isn't a homomorphism in the category of vector spaces; it is rather a bilinear map. So you can think of it as a map $V \otimes V^* \to k$ instead. This is why you need a tensor structure to think of duals.

This isn't enough, though, because we need to know that the pairing is nondegenerate. One way to express this is that there's a map $k \to V \times V^*$ mapping 1 to the "Casimir element" (which is the sum $\sum e_i \otimes e_i^{\vee}$ where $e_i$ ranges over a basis of $V$ and $e_i^{\vee}$ the dual basis; it is independent of the choice of $e_i$ as a quick computation shows). The Casimir morphism satisfies the condition that $V \to (k) \otimes V \to (V \otimes V^*) \otimes V \to V \otimes (V^* \otimes V)$ is just the identity.
Conversely, this is enough to show that the pairing is nondegenerate.

So, anyway, how does this make sense in a symmetric tensor category? Basically, $V$ is the object, $V^*$ the putative dual, and $k$ replaced by the unital object. This definition is entirely arrow-theoretic, and it all goes through as usual. It is an exercise to check that the dual is unique.

Some examples:

1. This coincides with the usual dual in the category of vector spaces

2. This coincides with the dual sheaf if one is working in the category of locally free sheaves on a scheme

3. This corresponds to the dual (contragredient) representation in the (tensor) category of representations of any Hopf algebra (so this includes representations of finite groups and Lie algebras)

Oh, and what happens if you don't have a symmetric tensor category? Then you have to worry about "left" and "right" duals, respectively. For more about all this, I recommend the notes of Pavel Etingof on tensor categories.

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+1 for a Nice explanation! I am still unsure as to whether this construction guarantees the tensor product of an element and a dual element is necessarily a hom from elements to elements though... – Tom Boardman Jul 25 '10 at 0:55
Actually, since we are doing things categorically, we do not have the notion of element. It is, however, still true that $V \otimes V^*$ is isomorphic to something called "internal hom" $\hom(V, V)$, but this is not the same as external hom (in fact, we have that the category-theoretic morphisms $1 \to \hom(V,V)$ correspond precisely to the morphisms $V \to V$). Note that internal hom is an object of the category, while externel hom is a set. – Akhil Mathew Jul 25 '10 at 1:12
Incidentally, for why it doesn't make sense to consider elements in tensor categories, there are examples where the objects of the tensor category have no readily definable structure as vector spaces or anything but are purely formal symbols. See this paper of Deligne (math.ias.edu/files/deligne/Symetrique.pdf) for several such examples (actually, families of tensor categories). – Akhil Mathew Jul 25 '10 at 1:16
There is a blog post of David Speyer explaining this paper, sbseminar.wordpress.com/2009/02/25/…. (Sorry for the repeated comments.) – Akhil Mathew Jul 25 '10 at 5:03

You can discuss duals in a monoidal category (which may not be symmetric). This has been mentioned by Akhil.

Let $V$ and $W$ be objects (in your monoidal category) with $K$ the identity for tensor product. Then you require morphisms $K\rightarrow V\otimes W$ and $W\otimes V\rightarrow K$ which satisfy the zig-zag identities (so-called because this becomes clear if you draw string diagrams).

Formally the zig-zag identities are $$V=K\otimes V\rightarrow V\otimes W\otimes V\rightarrow V\otimes K=V$$ is the identity map and $$W=W\otimes K\rightarrow W\otimes V\otimes W\rightarrow K\otimes W=k$$ is the identity map.

This is equivalent to $Hom(W\otimes X,Y)=Hom(X,V\otimes Y)$ and $Hom(X\otimes V,Y)=Hom(V,Y\otimes W)$ (both natural in $X$ and in $Y$.

Then you say $V$ is left/right dual to $W$ and $W$ is right/left dual to $V$ (I can never remember which). Then we can define $V$ to be dual to $W$ if it is left dual and right dual.

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