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Isometric invariants relating curvature/integrals of $ K,k_g$ along with the topological invariant $\chi(M)=( 2-2g) $ or$\,(2-r) $ the Euler characteristic appear on either side of the equation of the Gauss-Bonnet Theorem:

GBthm

So $\chi(M) $ can be expressed in terms derived from isometric invariant Christoffel symbols of first fundamental form.

If so, is $\chi(M) $ an isometric invariant and topological invariant at the same time?

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  • $\begingroup$ Can you state (or give a reference to) what is the relation between $\chi(M)$ and the Christoffel symbols? I didn't manage to find it $\endgroup$ – Alessio Di Lorenzo Aug 14 '17 at 13:03
  • $\begingroup$ I have not calculated it myself but $ \int\int K dA,\, \int k_g ds $ are isometric invariants. We can add them and divide by $2 \pi.$ $\endgroup$ – Narasimham Aug 14 '17 at 13:13
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Yes. Every isometry is a homeomorphism, then the Euler characteristic is an isometric invariant because it is a topological invariant.

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  • $\begingroup$ However in general every homeomorphism is not an isometry. So to me it appears Euler characteristic is not automatically an isometric invariant in full bijective sense. $\endgroup$ – Narasimham Aug 14 '17 at 17:33
  • $\begingroup$ @Narasimham what do you mean by "in the full bijective sense". It is a topological invariant, so a fortiori an isometric invariant. $\endgroup$ – Henno Brandsma Aug 14 '17 at 18:31
  • $\begingroup$ @Henno Brandma Please help me understand. Since topological mapping condition is stronger, it goes without saying the special cases of $ K, \chi(M)$ are invariant in topological maps, Right? $\endgroup$ – Narasimham Aug 15 '17 at 12:37
  • $\begingroup$ If $M$ and $M'$ are homeomorphic, $\chi(M) = \chi(M')$. So in particular we get the same conclusion if $M$ and $M'$ are even isometric (this is the stronger condition, in my terminology; it imposes more restrictions). $\endgroup$ – Henno Brandsma Aug 15 '17 at 13:27
  • $\begingroup$ The sum of $k_g, K$ terms is a topological invariant, but not each term individually $\endgroup$ – Narasimham Aug 16 '17 at 16:24

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