## Can adjoint linear transformations be naturally realized as adjoint functors?

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Last week Yan Zhang asked me the following: is there a way to realize vector spaces as categories so that adjoint functors between pairs of vector spaces become adjoint linear operators in the usual sense?

It seems as if one needs to declare an inner product by fiat for this to work out. An obvious approach is to take the objects to be vectors and hom(v, w) to be the inner product (so the category should be enriched over C). But I don't see how composition works out here, and Yan says he tried this and it didn't work out as cleanly as he wanted. In this setup I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens. I think John Baez's ideas about categorified linear algebra, especially categorified Hilbert spaces, are relevant here but I don't understand them well enough to see how they work out.

Anyone who actually knows some category theory care to clear things up?

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" I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens."  Make it happen!  Mod out the the messy part so you get what you want.  This works remarkably often.  Of course, sometimes it gives you the trivial object... – SixWingedSeraph Jan 2 2010 at 1:30

## 6 Answers

There's a canonical way of going the other way, starting with two linear categories with nice finiteness properties, with adjoint functors between them and getting a pair of vector spaces with adjoint linear transformations. The vector spaces are generated by formal symbols for each object in the category, and the inner product between any objects is the dimension of the Hom space (so Hom spaces had better be finite dimensional). Note that this doesn't have to be symmetric.

Functors give linear transformations, and adjoint functors are adjoint in the usual sense.

You can soup up this construction when you have some more structures on your category. For example, if you have a direct sum, then you can impose the relation $[A+B]=[A]+[B]$, and everything will work fine.

If your category is abelian, you can take Grothendieck group, where $[A]+[C]=[B]$ for every short exact sequence $0\to A \to B \to C\to 0$, but then you have to be much more careful about the fact that lots of functors (including Hom with objects in the category!) aren't exact: they don't send short exact sequences to short exact sequences. You need to use derived functors to fix this.

There's no canonical way of going the direction you asked, though in practice we have a very good record of being able to and I don't know of any really good examples of there being two equal natural seeming but different such constructions.

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I think it's more natural to take advantage of the monoidal structure and regard the vector spaces as functors rather than objects. For simplicity, consider only finite dimensional vector spaces. Given V, we have a functor $F_V: Vect \to Vect$ which sends $W$ to $W \otimes V$. The familiar identification $Hom(U\otimes V, W) = Hom(U, W\otimes V^*)$ shows that the (category theory) adjoint of $F_V$ is `$F_{V^*}$`. (That's $F$ sub $V^*$, in case the font is too small to read.) Chaining together two of these adjunctive identifications of Hom sets, we have

`$Hom(V, X) = Hom(1, X\otimes V^*) = Hom(X^*, V^*)$`.

The above identification sends a linear transformation $g:V\to X$ to the (linear algebra) adjoint `$g^*: X^*\to V^*$`. If $V$ and $X$ are inner product spaces then we can of course identify `$V^*$` with $V$ and `$X^*$` with $X$.

Maybe that's too elementary and not the answer you were looking for. But it seems to me it's the most simple and obvious way to relate linear algebra adjoints to category theory adjoints.

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It just occurred to me that there may be a certain sense in which this is impossible in principle. Every equivalence of categories can be improved to an adjoint equivalence, by modifying either the unit or the counit. This is true for all sorts of categories (internal, enriched, fibered, etc.). So if there were a way to realize all vector spaces (or, say, inner product spaces) as some kind of category such that adjoint linear transformations became adjoint functors, we would expect that any isomorphism of vector spaces would give an equivalence of such categories, and hence could be improved to an adjoint equivalence, i.e. an isomorphism whose adjoint is its inverse. But this is false; not every isomorphism between inner product spaces is unitary/orthogonal.

I can't decide whether this is deep or nonsensical, but I thought I'd throw it out there.

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A neat correspondence between adjoint functions and adjoint functors is possible, if you relax your understanding of what it means for a category to "realize" a Hilbert space a bit. (The adjoint of a linear function only exists if the vector spaces are Hilbert spaces and the function is continuous, so I'll take the question to be about Hilbert space instead of vector spaces.)

Given a Hilbert space $H$, "realize" it as the partially ordered set of closed subspaces $S(H)$, regarded as a category. Then a continuous linear function $f \colon H \to K$ induces a contravariant functor $S(f) \colon S(H)^{\text{op}} \to S(K)$. Now, denoting the adjoint function of $f$ by $f^\dagger \colon K \to H$, we get an adjunction between $S(f)$ and $S(f^\dagger)$. In fact, up to a scalar, any contravariant adjunction between $S(H)$ and $S(K)$ comes from an adjoint pair of functions between $H$ and $K$!

All this comes from a 1974 paper by Paul H. Palmquist, a student of Mac Lane, called "Adjoint functors induced by adjoint linear transformations" in Proceedings of the AMS 44(2):251--254.

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“The adjoint of a linear function only exists if the vector spaces are Hilbert spaces and the function is continuous”—I guess you mean the only situation in which we may naturally view the adjoint of a continuous map $V \to W$ as a map $W \to V$? – L Spice May 9 2011 at 4:15

There a simple way to make this work:

Say `T:V->X` is a map of inner-product vector spaces. You can view `V` as a category, where `Hom(v,w)` is a singleton set containing one real number, the inner product `<v,w>`, and similarly for `X`.

Composition, a binary operation, is defined (stupidly, as in any category with singleton hom-sets) as follows:

`Comp_{uvw} : Hom(u,v)xHom(v,w) -> Hom(u,w) `

by `(<u,v>,<v,w>) |-> <u,w>`

Then the adjoint T*:X->V satisfies

`<Tv,x>=<v,T*x>`, i.e. `Hom(Tv,x)=Hom(v,T*x)`

, meaning it is a right adjoint to T (in a very strong sense: we have equality of these hom-sets instead of just natural isomorphism).

The triviality of this example reflects the fact that that T and T* are called "adjoint" simply because they belong on opposite sides of a comma :)

In general, if H is any function of two variables, we can say that g is right adjoint to f "with respect to H" if H(f(a),b)=H(a,g(b)), and say that "adjoint functors" are "adjoint with respect to Hom" (up to natural isomorphism, of course).

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Okay, but in this setup are the functors between such categories precisely the linear transformations?  I don't have much intuition for whether this Hom makes sense. – Qiaochu Yuan Oct 14 2009 at 23:36
this approach forgets the whole vector space structure... – Martin Brandenburg Dec 28 2009 at 12:40

check out

John C. Baez, Higher-Dimensional Algebra II: 2-Hilbert Spaces, online.

"The analogy to adjoints of operators between Hilbert spaces is clear. Our main point here is that that this analogy relies on the more fundamental analogy between the inner product and the hom functor."

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