Some topological invariants that can be found e.g. in knot theory can be represented as integrals (Example: Integral for computing the Gauss linking number). Another example is the complex plane with poles: If the poles are defined by Terms like $\frac{1}{z-z_i}$ with some natural number $i$ that is indexing the pole then the number of poles (topological invariant) can be expressed as:

$N = \int_C \frac{f(z)dz}{2 \pi i}$.

In differential geometry the total curvature of a curve $c(t)$ parametrized by $t \in [0,1]$ is obtained by

$TC[c(t)]= \int_0^1 \kappa(t) dt$

with the curvature of the curve $\kappa(t)$. Also the Gauss-Bonnet Theorem computes the Euler characteristic from an integral.

Question: Can be expressed other topological invariants in Terms of integrals over (differential-)geometrical functions like curvature, Gauss curvature, etc.? What Basic ideas are used to link topological invariants with such integrals (how one can derive such identities)?

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    $\begingroup$ I think generally, the opposite question is asked. Namely, it is usually more natural to ask whether or not a differential-geometric construction is related to a topological invariant rather than the other way around, since differential structures come to us with a topology already on them. $\endgroup$ Apr 3 '15 at 22:45

I was hoping someone would give a more complete answer, but let me just indicate that the answer is yes, absolutely, a vast range of invariants can be so computed. This line of thought leads through Grothendieck-Riemann-Roch and the work of Hirzebruch to the Atiyah-Singer index theorem, which has been one of the impetus for the resurgence of interaction between mathematics and physics over the past thirty years.

  • $\begingroup$ -1 for being totally imprecise and containing no mathematics, but only talk about mathematics $\endgroup$ Apr 7 '15 at 0:04
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    $\begingroup$ @Herrmann I look forward to seeing your more detailed answer! $\endgroup$ Apr 7 '15 at 3:23
  • $\begingroup$ you don't even explicitly mention a topological invariant which admits an integral representation $\endgroup$ Apr 7 '15 at 11:40
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    $\begingroup$ @Herrmann you sound really stressed out about all this! Maybe you should take a break. $\endgroup$ Apr 7 '15 at 18:13
  • $\begingroup$ not so much. I think it is a comment, but not really an answer. $\endgroup$ Apr 7 '15 at 19:11

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