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The vast majority of physical theories are formulated on a spacetime that is mathematically represented by a pseudo-Riemannian manifold, i.e. a smooth manifold with a metric tensor structure. The nature of this pseudo-Riemannian manifold is very different between different theories: for example, in Newtonian mechanics it is Euclidean space (with a flat positive-definite metric); in special relativity, classical E&M, and particle physics it is usually taken to be Minkowski space (with a flat but indefinite metric), and in general relativity and cosmology it is a generic curved Lorentzian manifold. But in almost every case, the metric structure is critical to the theory.

But topological quantum field theory is an important exception to this pattern: a TQFT can be formulated on an arbitrary smooth manifold, whether or not it has a metric, because it is only sensitive to the manifold's topology (which can be defined independently of any metric).

Are there any other examples of useful/realistic physical theories that can be defined on a smooth spacetime manifold without a metric structure? In particular, any "simpler" theories, such as a theory of classical point particles or fields, whether relativistic or not? (That one might be challenging, since kinetic energy terms like $\frac{1}{2} m g_{\mu \nu} v^\mu v^\nu$ or $\frac{1}{2} g^{\mu \nu} \partial_\mu\varphi \partial_\nu \varphi$, etc. typically contain an inner product.)

(When I talk about the manifold that the theory is formulated on, I mean the physical spacetime itself, not some more abstract configuration/phase/state space.)

tparker
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  • Nice question, Note If it was purely topological, then homeomorphic manifolds would be physically equivalent, which is a somewhat looser statement than saying diffeomorphic manifolds are physically equivalent (especially in dimension 4!) – R. Rankin Feb 08 '23 at 03:06
  • @R.Rankin Yes, I'm not requiring that the theory be purely topological (i.e. homeomorphism alone gives physical equivalence). I'm allowing the theory to depend on the manifold's smooth structure, and I'm only requiring that diffeomorphic smooth manifolds are physically equivalent. (Diffeomorphic in the mathematicians' sense of the word, not the physicists' sense.) – tparker Feb 08 '23 at 04:16
  • @R.Rankin Although I would say that "Homeomorphic manifolds are physically equivalent" is a stricter statement than "Diffeomorphic manifolds are physically equivalent", not a looser statement, since the former statement automatically implies the latter but not vice versa. – tparker Feb 08 '23 at 04:21
  • Lol my bad, I actually mean that the other way, thinking of course of the infinite set of exotic smooth structures on some (or possibly still all!) 4 manifolds – R. Rankin Feb 08 '23 at 06:04
  • There's always finsler geometry, which still has a metric structure, but is a generalization of riemannian metric. Not to Segway you – R. Rankin Feb 08 '23 at 06:13
  • Wait. aren't diffeomorphisms a subset of homeomorphisms? meaning diffeomorphism invariance is a stricter requirement than homeomorphism invariance. last comment I swear – R. Rankin Feb 08 '23 at 07:41
  • @R.Rankin Yes, every diffeomorphism is a homeomorphism but not vice versa, so the requirement that two smooth manifolds be diffeomorphic is stricter than the requirement that they be homeomorphic. But that means that the requirement that “homeomorphic manifolds must be physically equivalent” is stricter than the requirement that “diffeomorphic manifolds must be physically equivalent”, because the former requirement covers more pairs of manifolds. – tparker Feb 08 '23 at 12:40

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