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Why is the Euler characteristic of a boundary even? How can one prove this and is there an geometric way to think about it?

pki
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2 Answers2

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Let $M$ be your (compact) manifold. You can glue two copies $M_1$, $M_2$ of $M$ along their boundary, getting a closed manifold $2M$. Using the Mayer-Vietoris long exact sequence for the triad $(2M;M_1,M_2)$. It gives us the relation $\chi(2M)=2\chi(M)-\chi(\partial M)$, because $M_1$ and $M_2$ intersect along $\partial M$.

Now, if $\dim M$ is odd, then $\dim 2M$ is also odd and $\chi(2M)=0$, so $\chi(\partial M)=2\chi(M)$ is even. If $\dim M$ is even, then $\dim \partial M$ is odd and therefore $\chi(\partial M)=0$ is also even.

Mariano Suárez-Álvarez
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  • @Marino, for using Mayer-Vietoris we need an open cover from $2M$, how will the boundary show up in the sequence? – PtF Jul 20 '13 at 19:32
  • The boundary has a collar. Use that. – Mariano Suárez-Álvarez Jul 20 '13 at 19:36
  • should I use the collar neighbourhood theorem? – PtF Jul 20 '13 at 19:45
  • Well, that theorem is the one that proves that the boundary has a collar, so if I am suggesting that yu usse a collar, you are surely going to need it! – Mariano Suárez-Álvarez Jul 20 '13 at 19:49
  • well thanks, I'll try =D What would be a good book that has the proof of this theorem? I'll need to reference it because we didn't see it in class.. – PtF Jul 20 '13 at 19:50
  • I got up to this point: Let $X$ and $Y$ be two copies of $Z$ ($Z$ is my manifold with compact boundary $M$). With this notation $\partial X=\partial Y=M$. Since $X$ and $Y$ have compact border we can get a manifold without boundary $N=X\cup Y$ (a theorem assure this). By the collar neighbourhood theorem there is a neighbourhood $U$ of $\partial X$ and a neighbourhood $V$ of $\partial Y$ diffeomorphic to $\partial M\times [0, 1)$ ( $U=V$). Consider the open cover ${(X-M)\cup V, (Y-M)\cup U}$ of $N$. I can't see how I can apply Mayer-Vietoris to get $\chi(N)=2\chi(X)-\chi(M)$. – PtF Jul 21 '13 at 14:42
  • Let $S^2$ be the two spheres, and let $H_+$ and $H_-$ be the closed north and south hemispheres, respectively. How do you apply M-V to the triad $(S,H_+,H_-)$? – Mariano Suárez-Álvarez Jul 21 '13 at 19:06
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Another perspective: Triangulate $M$; we want to compute $\chi(\partial M) \bmod 2$. Let $T$ be the set of all faces (of all dimensions) in the triangulation of $M$. We will count, modulo $2$, the number of pairs $(\sigma, \tau)$ of elements in $T$ with $\sigma \subsetneq \tau$.

Fixing $\sigma$ If $\sigma$ meets the interior of $M$, I claim that there are an even number of $\tau$ containing $\sigma$. Proof sketch: Working in a neighborhood of $x$, let $x$ be a point in the interior of $\sigma$, let $V$ be a linear space through $x$ transverse to $S$, and let $S$ be a sphere around $x$ in $V$. Then the intersection of $S$ with the triangulation $T$ gives a triangulation of the sphere $S$, with one face for each $\tau$ with $\sigma \subsetneq \tau$. A triangulation of a sphere has an even number of faces.

Similarly, if $\sigma$ is in the boundary of $M$, then $S$ meets $M$ in a closed hemisphere and the triangulation of $S$ has an odd number of faces.

So the number of pairs $(\sigma, \tau)$ as above is congruent $\bmod 2$ to the number of faces in $\partial M$, and thus congruent $\bmod 2$ to $\chi(\partial M)$.

Fixing $\tau$ Let $\tau$ have dimension $d$. Then there are $2^{d+1}-2$ faces $\sigma$ with $\sigma \subsetneq \tau$. In particular, this number is even, so the count is even.

David E Speyer
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