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Stream: deprecated: algebra & CT

Topic: Finitely generated projective modules

Tobias Fritz (Jul 23 2020 at 02:25):

When studying compact closed categories, it's very common to present the category of finite-dimensional vector spaces over a field as one of the primary examples. Less well-known seems to be the more general fact that the category of finitely generated projective modules over a commutative ring is also compact closed. This is relatively easy to prove from the dual basis lemma. (Which has an unfortunate name, since the relevant families of module elements are typically not bases! Incidentally, the finitely generated projective modules are also precisely the compact objects in the category of modules.)

Question: Does anyone know of a reference for the fact that finitely generated projective modules form a compact closed category?

I'm hoping that someone will already have written a pedagogical exposition of this. I know of references such as Lemma 3.60 in the thesis of Chris Schommer-Pries where a more general statement is proven, namely that the dualizable 1-morphisms in the bicategory of rings, bimodules and bimodule maps are the finitely generated projective bimodules. But this is overkill for my purposes.

Tim Hosgood (Jul 23 2020 at 09:34):

i’m guessing there might be a proof in terms of vector bundles on affine schemes, right?

Tobias Fritz (Jul 23 2020 at 12:34):

I admittedly don't know about vector bundles on schemes (even affine ones), and one also doesn't need to in order to prove that the category of fg projective modules is compact closed. The argument goes like this: let $M$ be such a module and $M^* := \mathrm{Hom}_R(M,R)$. Then the dual basis lemma says that there are finitely many $x_i \in M$ and $f_i \in M^*$ such that
$\sum_i f_i(y) x_i = y$
for all $y \in M$. (This holds in particular when $(x_i)$ is a basis and $(f_i)$ is its dual basis, but of course when $M$ is not free then neither can actually be a basis!) Thus upon using the obvious evaluation map $M \otimes M^* \to R$ and the coevalution map $R \to M^* \otimes M$ corresponding to the element $\sum_i f_i \otimes x_i \in M^* \otimes M$, the zig-zag equations follow from the formula above.

In fact, it must be possible to run this argument in reverse and to show that if a module $M$ has a dual object in the monoidal category sense, then $M$ is fg projective and the duality is conjugate to one as above, but right now I'm not sure how that goes.

Paolo Perrone (Jul 23 2020 at 13:52):

I think Tim is referring to the Gelfand-type duality given by the Serre-Swan theorem. It could be that a proof in the literature has been (equivalently) given in the opposite category.

Tim Hosgood (Jul 23 2020 at 13:56):

yeah, this is what i was thinking, especially as you already seem to be familiar with the proof for vector spaces, which are just vector bundles over a point :slight_smile:

Tobias Fritz (Jul 23 2020 at 16:51):

That's a good question. Based on that, I've found that the statement is also a special case of Proposition 4.7.5 of Tensor categorical foundations of algebraic geometry. Thanks!

If anyone knows of a reference that would be more accessible to category theorists without an algebraic geometry background, I'd still appreciate that. (Not for myself, but to point my readers to.)

Simon Burton (Jul 24 2020 at 16:45):

Thanks for the link @Tobias Fritz , i'm really enjoying this PhD thesis by Martin Brandenburg. My vague hope is to learn enough category theory that understanding algebraic geometry becomes a side effect of that.

Jens Hemelaer (Jul 25 2020 at 22:33):

Does someone here know if the idempotent completion of a compact closed category is again compact closed?
Then you get that the category of f.g. projectives is compact closed from the category of f.g. free modules being compact closed. I only found a reference for dagger categories, and I don't think that helps here.

Tobias Fritz (Jul 25 2020 at 22:44):

Interesting question! I bet that this is true, and that it can be proven by taking the dual of $(X,e : X \to X)$ to be given by $(X^*,e^* : X^* \to X^*)$, but I haven't checked the details.

Cole Comfort (Sep 11 2020 at 04:43):

I have completely necro'd this thread, but I'll take a shot.

If I recall correctly, the cagtegory of finitely generated projective modules over a ring is (bimonoidally) equivalent to the category of matrices over a semiring.

This can be alternatively characterized as a decorated cospan category where the associated monoid corresponds to the multiplication of the ring. And because every decorated cospan category is just a fancy category of cospans, it becomes equipped with a frobenius algebra structure compatible with the tensor product. This frobenius algebra actually comes from the graph of the commutative monoid corresponding to a copy of the presentation of $({\sf FinSet},+)$; and its transpose, considered as a span; the distributive law between which is a special commutative frobenius algebra.

So given some object $n$, the monoid of the Frobenius algebra is in the image of $n^{(\_)}:({\sf FinSet},+) \to ({\sf Mat}(R),\otimes)$ and the comonoid is in the image of the functor $(n^{(\_}))^T$

So you might expect that matrices are more like spans of sets, where pullback is defined as the bialgebra law between a commutative monoid and a cocomutative comonoid. However, when you are looking at the multilinear tensor product side of things, the contravariance of this functor turns monoids into comonoids and comonoids into monoids, so that pullback in the opposite category is pushout, of which the corresponding distributive law is a special commutative frobenius algebra.

Tobias Fritz (Sep 11 2020 at 10:23):

Cole Comfort said:

If I recall correctly, the cagtegory of finitely generated projective modules over a ring is (bimonoidally) equivalent to the category of matrices over a semiring.

That doesn't sound right—this would mean in particular that the monoid of isomorphism classes of finitely generated projective modules is isomorphic to $(\mathbb{N},\cdot)$, which is typically not the case. For example if $k$ is a field, then $k^n$ (with componentwise operations) is a ring, the modules of which can be identified with $n$-tuples of vector spaces over $k$, with one component for each component of $k^n$. It's easy to see that all of these modules are projective, e.g. by virtue of being direct summands of a free module. It follows that the monoid of isomorphism classes is isomorphic to $\mathbb{N}^n$ with componentwise multiplication.

Cole Comfort (Sep 11 2020 at 13:56):

Oops I meant so say equivalent to matrices over a ring. I can't find a reference, but is this not true?

Morgan Rogers (he/him) (Sep 11 2020 at 14:23):

Are you thinking of Morita equivalence? Where the category of modules over a ring $R$ is monoidally equivalent to the category of modules over the $n \times n$ matrices in $R$? I don't know where the projective part would have come from in that case, though.

Cole Comfort (Sep 11 2020 at 14:24):

Like is the category of finitely generated projective modules over R equivalent to matrices over R?

Reid Barton (Sep 11 2020 at 14:27):

What do you mean by "the category of matrices over a ring"? The category whose Hom-sets are given by sets of matrices? In algebra this would go by the name "the category of free modules over a ring".

Cole Comfort (Sep 11 2020 at 14:32):

Hmm, I guess I proved it for the wrong category :sweat_smile: