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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?

Ben Webster
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Qiaochu Yuan
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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 02 '10 at 01:30

7 Answers7

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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.

Chris Heunen
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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 $W \to V$? – LSpice May 09 '11 at 04:15
  • This is super cool! I think it's worth emphasizing that a linear map can be recovered up to a scalar from the associated functor (part iv of Theorem 1 in the paper cited). – Vectornaut Nov 14 '13 at 00:00
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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.

Ben Webster
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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.

Kevin Walker
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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.

Mike Shulman
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  • I think your remark is actually evidence in favor of such a categorification being possible! You see... – Vectornaut Nov 13 '13 at 23:47
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    ... in a setting where a functor is the categorification of a linear map, a natural transormation should be the categorification of something like a homotopy of linear maps---let's say a constant-rank path through the space of linear maps. In this setting, an equivalence of categories would be a pair of linear maps whose compositions are isotopic to the identity. Your remark is the categorification of the fact that any isomorphism between inner product spaces can be improved, by the Gram-Schmidt process, to a pair of linear maps whose compositions are isotopic to the identity. – Vectornaut Nov 13 '13 at 23:48
  • But that sort of "natural transformation" would not yield the usual notion of "adjoint linear map" as a category-theoretic adjunction, would it? – Mike Shulman Nov 14 '13 at 16:33
  • I'm not sure what you mean. Are you saying that the notion of isotopy of linear maps somehow yields the notion of adjunction of linear maps, so that choosing how to categorify isotopy of linear maps will force a particular choice of how to categorify adjunction of linear maps? I don't see how that would work. – Vectornaut Nov 15 '13 at 20:10
  • The question was whether there is "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". Does your proposal of isotopies as natural transformations have this property? – Mike Shulman Nov 17 '13 at 05:54
  • I haven't proposed a way to realize vector spaces as categories in the way the OP wanted. I'm just claiming that your argument does not rule out such a realization. – Vectornaut Nov 18 '13 at 05:32
  • Your argument, if I understand it correctly, is that in any such realization, the statement "every equivalence of categories can be improved to an adjoint equivalence" would translate into a categorification of the statement "every isomorphism between inner product spaces is unitary/orthogonal," which is false. I claim, however, that in any such realization, the statement above would translate into a categorification the statement "any isomorphism between inner product spaces can be improved to a pair of linear maps whose compositions are isotopic to the identity," which is true. – Vectornaut Nov 18 '13 at 05:33
  • I think your claim is wrong. I concluded that isomorphisms could be improved to orthogonal ones because if adjunctions were adjoint linear maps, then an adjoint equivalence would be an isomorphism that consists of adjoint linear maps, hence orthogonal. This depends only on the OP's desiderata that adjunctions are adjunctions, and has nothing to do with what we choose the natural transformations to be. – Mike Shulman Nov 19 '13 at 06:19
  • Also, what distinguishes an equivalence from an adjoint equivalence is not the existence of natural transformations, but rather the fact that they satisfy the triangle laws. So if natural transformations were isotopies, then "a pair of linear maps whose compositions are isotopic to the identity" would be just the notion of equivalence; an adjoint equivalence would then require a coherence condition on those isotopies. – Mike Shulman Nov 19 '13 at 06:20
  • Aaack, now I see the confusion! There's a typo in my first comment, which I repeated later: in the clause "any isomorphism between inner product spaces can be improved, by the Gram-Schmidt process, to a pair of linear maps whose compositions are isotopic to the identity," the phrase "a pair of linear maps" should have been "a pair of adjoint linear maps." Thanks for pointing that out! I wish I could edit the original comment... – Vectornaut Nov 19 '13 at 08:03
  • Unfortunately, I still don't understand your argument, maybe because I've misunderstood what you mean when you say, "Every equivalence of categories can be improved to an adjoint equivalence." Following Wikipedia, I'm assuming you mean that if $F \colon \mathcal{D} \to \mathcal{C}$ and $G \colon \mathcal{C} \to \mathcal{D}$ form an equivalence, then there exists a functor $G' \colon \mathcal{C} \to \mathcal{D}$ such that $F$ and $G'$ form an adjoint equivalence. Is that right? – Vectornaut Nov 19 '13 at 08:45
  • Here's another place I might be misreading your argument. You seem to be working under the assumption that if functors were linear maps, then an equivalence of categories would be an isomorphism of vector spaces. Is that right? – Vectornaut Nov 19 '13 at 08:56
  • No, I mean that there exist natural transformations $\eta$ and $\epsilon$ exhibiting $F$ and $G$ themselves as forming an adjoint equivalence. – Mike Shulman Nov 20 '13 at 14:58
  • You're right, though, that there is an extra assumption that equivalences correspond to isomorphisms. – Mike Shulman Nov 20 '13 at 14:59
  • Okay, I think it's just the assumption that equivalences correspond to isomorphisms that's tripping me up. I'm more used to the strict point of view, where isomorphism of categories is analogous to isomorphism of spaces, and equivalence of categories is analogous to homotopy-equivalence of spaces. – Vectornaut Nov 20 '13 at 22:57
  • Ah, I see. I live in higher-category-theory land where one never even thinks about "isomorphisms of categories". Moreover, vector spaces (over an arbitrary field) aren't necessarily topological, so "homotopy equivalence" doesn't always make sense. But I think the same argument applies to homotopy equivalence: it's not true that every linear homotopy equivalence of topological vector spaces can be improved to an orthogonal one. – Mike Shulman Nov 21 '13 at 23:19
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Not exactly an answer to the question as posed, but it's worth noting that adjoint linear maps and adjoint functors can both be realized as instances of the same thing, namely morphisms in a Chu construction.

On one hand, a vector space $V$ with an inner product (or indeed any bilinear form) $B_V:V\otimes V \to \Bbbk$ can be regarded as an object of $\mathrm{Chu}(\mathrm{Vect},\Bbbk)$, call it $V'$. A morphism $V'\to W'$ in $\mathrm{Chu}(\mathrm{Vect},\Bbbk)$ is then a pair of linear maps $f:V\to W$ and $g:W\to V$ such that $B_V(v,g w) = B_W(f v,w)$, i.e. an adjoint pair of transformations in the linear sense.

On the other hand, any category $C$ has a hom-functor $\hom_C : C^{\mathrm{op}}\times C\to \mathrm{Set}$ and can thereby be regarded as an object of the 2-categorical Chu construction $\mathrm{Chu}(\mathrm{Cat},\mathrm{Set})$, as discussed here; call this object $C'$. Then a morphism $C'\to D'$ in $\mathrm{Chu}(\mathrm{Cat},\mathrm{Set})$ is a pair of functors $f:C^{\mathrm{op}}\to D^{\mathrm{op}}$ (or equivalently $C\to D$) and $g:D\to C$ together with an isomorphism (this is why the Chu construction has to be 2-categorical here) $\hom_C(c,g d) \cong \hom_D(f c,d)$, i.e. an adjoint pair of functors in the categorical sense.

Mike Shulman
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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."

Martin Brandenburg
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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).

Andrew Critch
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