# Is there an “arithmetic cobordism category”?

This question is a clumsy attempt to apply a certain analogy. I hope that if the answer is negative it comes with a clarification of the scope and limitations of the analogy.

Arithmetic topology is based on an analogy between number fields and 3-manifolds where primes are something like knots, the Legendre symbol is something like a linking number, etc. In quantum topology, on the other hand, one way to study 3-manifolds is to study 3d TQFTs, e.g. functors $Z : 3\text{Cob} \to \text{Vect}$. These functors assign to every 3-manifold, interpreted as a cobordism from the empty 2-manifold to itself, a morphism $k \to k$ where $k$ is the base field, and therefore give $k$-valued invariants of 3-manifolds.

If the analogy between number fields and 3-manifolds is strong enough, there might conceivably exist an "arithmetic cobordism category" whose morphisms are number fields and whose objects are... whatever boundaries of number fields are in arithmetic topology. (One might need to adapt this construction depending on whether number fields are considered to have "boundaries" at all.) It might conceivably be possible to adapt constructions of 3d TQFTs to the arithmetic case and therefore to find "quantum invariants" of number fields.

So is any construction like this possible, or am I just talking nonsense?

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Something TQFT-like in number theory cropped up here - londonnumbertheory.wordpress.com/2010/05/10/… –  David Corfield May 26 '10 at 9:03
This goes into that direction: math.uiuc.edu/K-theory/0547 –  Thomas Riepe May 26 '10 at 9:28
In the analogy 3manifolds = numberfields, 1-manifolds = finitefields, you're asking 2manifolds = ?. You could also ask 0manifolds = ?. –  André Henriques May 26 '10 at 12:45
Number fields are not (compact) 3-manifolds. Rings of integers and S-integers are. Local fields are 2-manifolds, the boundary around a knot. Rings of integers in them are the tubular neighborhood of the knot. That doesn't give a lot of 2-manifolds to work with. Not enough for Heegaard splittings. But you could try to approach the Casson invariant some other way, without mentioning Heegaard splittings or the TQFT more generally. –  Ben Wieland Jul 10 '10 at 3:31