13
$\begingroup$

I have heard that one can do algebraic geometry internal to symmetric monoidal categories. Topological quantum field theories also exist internal to symmetric monoidal categories, and the usual definition is recovered by inserting the category of vector spaces.

I wonder whether this can, or has, been done for noncommutative geometry. I'm interested in Connes' approach via spectral triples. I guess it is feasible to define what a $*$-algebra internal to a symmetric monoidal category is, but how does the story go on?

Edit: As has been pointed out in the comments, this seems to hard in all generality. Let us restrict to finite-dimensional $*$-algebras and Hilbert spaces in order not to worry about topology and functional analysis.

Improve this question
$\endgroup$
7
  • $\begingroup$ I don't know much about $\ast$-algebras, but one can certainly do things like non-commutative Galois theory, descent theory and deformation theory by working with corings and comodules. $\endgroup$
    – Jonathan Beardsley
    Commented Apr 29, 2015 at 14:38
  • 4
    $\begingroup$ I'm pretty such that no such thing have been done, and I really don't think it is possible, but I have no concrete argument to bring to support this... the theory of spectral triple is definitely not 'algebraic' (in a naive sense of the word) but has a topological component (for example, the presence of an Hilbert space, and worst, of an unbounded dirac operator) which makes it very difficult to internalize. In fact, even the notion of *-algebra is not completely clear because the * has to be anti-linear so it already introduce some complication... $\endgroup$
    – Simon Henry
    Commented Apr 29, 2015 at 15:05
  • 1
    $\begingroup$ @SimonHenry, I agree that C-$*$-algebra with all the topology is a nasty to impossible business to internalise into a category, so thanks for that comment. But how about the finite dimensional case? Maybe it's like in topological quantum field theories, where the axiomatised framework only allows for finite dimensional vector spaces (or fully dualisable objects). $\endgroup$
    – Manuel Bärenz
    Commented Apr 29, 2015 at 16:18
  • 2
    $\begingroup$ Looks like an excellently provocative question that should certainly remain on the table. $\endgroup$
    – Paul Taylor
    Commented Apr 29, 2015 at 16:36
  • 1
    $\begingroup$ Here is a weird way to look at spectral triples: ncatlab.org/nlab/show/spectral+triple#As1DimensionalFQFTs There, they are some kind of nonstandard TQFT (I'm assuming that's the same as an FQFT) with values in super-Vectorspaces, somehow hinting that we could maybe replace $\mathrm{sVect}$ with an arbitrary symmetric monoidal category? $\endgroup$
    – Manuel Bärenz
    Commented Apr 29, 2015 at 17:07

1 Answer 1

Reset to default
6
$\begingroup$

I'll leave open how this might connect to "internal noncommutative geometry", but you can say something about internal $*$-algebras.

You can formulate $*$-algebras, anti-linear involution and all, internal to dagger monoidal categories, such as the category of Hilbert spaces. In fact, it turns out you can characterize finite-dimensional C*-algebras precisely as $*$-algebras internal to the category of finite-dimensional Hilbert spaces satisfying the so-called Frobenius law, see this paper.

This doesn't quite generalize to arbitrary dimension, because the carrier object still has to be a Hilbert space. So the best you can hope for is to characterize C*-algebras whose underlying Banach space is in fact a Hilbert space. But with this caveat, that is possible: these algebras are called H${}^*$-algebras, and are $*$-algebras internal to the category of Hilbert spaces that satisfy an axiom close to the Frobenius law, see this paper.

What's interesting is that you can also do this in other dagger monoidal categories. For example, in the category of sets and relations, the internal $*$-algebras as above (Frobenius algebras and H*-algebras) correspond precisely to groupoids! See this paper.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .