# In what way is $\chi(M)$ not an isometric invariant?

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:

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?

• 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 – Alessio Di Lorenzo Aug 14 '17 at 13:03
• 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 $2 \pi.$ – Narasimham Aug 14 '17 at 13:13

• @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? – Narasimham Aug 15 '17 at 12:37
• 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). – Henno Brandsma Aug 15 '17 at 13:27
• The sum of $k_g, K$ terms is a topological invariant, but not each term individually – Narasimham Aug 16 '17 at 16:24