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Stream: learning: questions

Topic: Modules with an inner product

Paolo Perrone (Jun 23 2025 at 08:49):

Does anyone know if people in algebraic geometry/commutative algebra/K-theory have studied modules equipped with a bilinear form?
Under the Serre-Swan theorem these should correspond to vector bundles with an inner product, like for example the tangent bundle of a Riemannian manifold.

Ryo Suzuki (Jun 23 2025 at 13:16):

Hermitian K-theory could be relevant
https://ncatlab.org/nlab/show/Hermitian+K-theory

Simon Burton (Jun 23 2025 at 19:50):

How about lattices with an inner product? Those seem to be pretty common.. I guess you might want integral lattices.

John Baez (Jun 23 2025 at 20:38):

There's a lot of work on modules with bilinear forms. It's pretty easy to see that if $R = C(X)$ is the ring of real continuous functions on a compact Hausdorff space $X$, a module $M$ over $R$ equipped with a nondegenerate bilinear form $g: M \otimes_R M \to R$ is the same as a real vector bundle over $X$ equipped with an inner product. The only clever part is that nondegenerate here should mean that the map

$\flat : M \to [M,R]$

that arises from currying $g$ should be an isomorphism. Here the brackets mean the internal hom in the category of $R$-modules, and $[M,R]$ is usually called the dual of $M$. There's a theorem that any module over any commutative ring that's isomorphic to its dual must be finitely generated and projective! Thus, we can apply Swan's theorem and conclude $M$ is the module of sections of some vector bundle over $X$. The final step is to show that $g$ comes from an inner product on $M$.

If this was too terse, the following equation may help:

$\flat (s) (t) = g(s \otimes t)$

for any $s,t \in M$.

John Baez (Jun 23 2025 at 20:43):

I should just warn you that if we let $R$ be the ring of complex continuous functions on $X$, the story changes, because an inner product on a complex vector bundle is not the same as a nondegenerate symmetric bilinear form on its module of sections: it corresponds to a nondegenerate sesquilinear form.

Paolo Perrone (Jun 23 2025 at 22:01):

Thank you! Where can I find more on this?

John Baez (Jun 23 2025 at 22:02):

I don't know. I looked around a bit, but didn't see anything. It's just "obvious".

John Baez (Jun 23 2025 at 22:02):

I'd be happy to explain this stuff further....

Paolo Perrone (Jun 24 2025 at 07:10):

My question is, do people use this to do Riemannian geometry in an algebraic way?

John Baez (Jun 24 2025 at 08:18):

There's a kind of offshoot of Connes' approach to noncommutative geometry, which is some people (physicists mainly) using Swan's theorem to "algebraicize" differential geometry in the way we've been discussing. I can easily imagine them doing Riemannian geometry in the way I was just explaining. But I don't actually remember seeing this stuff. I worked on this back in the late 1980s but never published anything on it.

John Baez (Jun 24 2025 at 08:20):

Algebraic geometers are more interested in Kahler geometry than Riemannian geometry: that's where you have a complex variety, with its complex tangent bundle, and you put a nondegenerate sesquilinear form on that complex tangent bundle, which called a hermitian form. If it's very nice you call it a Kahler structure.

John Baez (Jun 24 2025 at 08:21):

I poked around for 3 minutes looking for work on what you're wondering about, but I didn't see it. It's so easy to do - with the proper training - that it's hard to believe it doesn't exist.

David Corfield (Jun 24 2025 at 08:40):

Did you follow up on links at the nLab page you first referenced, Serre-Swan theorem? Like smooth Serre-Swan theorem:

image.png

From there derivations of smooth functions are vector fields:

image.png

And thence to the thickets of 'geometry of physics'.

Paolo Perrone (Jun 24 2025 at 08:44):

Yes, that's what I'm interested in. How much of the 'metric' aspects have people made algebraic?

David Corfield (Jun 24 2025 at 08:58):

Perhaps some pointers from embedding of smooth manifolds into formal duals of R-algebras

image.png

This is all Urs's stuff, so you might ask at a suitable thread on the nForum.

Paolo Perrone (Jun 24 2025 at 08:59):

Good idea, thanks!