I have the following problem: Let $\Sigma$ be a compact and orientable surface. Show that $H^1(\Sigma-\{p\})=0$ for every $p\in \Sigma$.

Can anyone check my proof and give suggestions?

Sketch of proof: Let $B$ be a closed ball centered in $p\in \Sigma-\{p\}$ and consider the isomorphism $$\phi:H^1(\mathbb S^1)\rightarrow \mathbb R,\ [\omega]\mapsto \int_{\mathbb S^1}\omega.$$ If $\imath:\partial B\hookrightarrow \Sigma-\{p\}$ is the inclusion then $\imath^*:H^1(\Sigma-\{p\})\rightarrow H^1(\partial B)$ is symply the restriction. Since $H^1(\partial B)\simeq H^1(\mathbb S^1)$ we might consider the composition $$\phi\circ \imath^*:H^1(\Sigma-\{p\})\rightarrow \mathbb R.$$ Notice $\phi\circ \imath^*$ is given by, $$\phi\circ \imath^*([\omega])=\phi(\imath^*[\omega])=\phi([\imath^*\omega])=\int_{\mathbb S^1}\imath^*\omega=\int_{\partial B}\imath^*\omega.$$ I'm not pretty sure if I really can write the above equality.. Hence, by Stokes theorem, $$\int_{\partial B}\imath^*\omega=\int_{\partial(M-B^\circ)}\imath^*\omega=\int_{M-B^\circ}d\omega=0,$$ (here $B^\circ$ is the interior of $B$) for $d\omega=0$. Therefore $\phi\circ \imath^*([\omega])=0$ for all $[\omega]\in H^1(\Sigma-\{p\})$, so $\phi\circ \imath^*=0$ and since $\phi$ is invertible $\imath^*=0$ so that $\textrm{ker}(\imath^*)=H^1(\Sigma-\{p\})$. Since $\imath^*$ is injective $H^1(\Sigma-\{p\})=0$ (is $\imath^*$ really injective?).

  • 6
    $\begingroup$ A punctured compact surface deformation-retracts to a wedge of circles, so its $H^1$ is rarely zero. $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '13 at 20:26
  • $\begingroup$ @MarianoSuárez-Alvarez you mean the result is not true? $\endgroup$ – PtF Aug 9 '13 at 20:29
  • $\begingroup$ Indeed. Take for example $\Sigma$ to be a torus: it defirmation-retracts onto the union of one of its equators and one of its parallels. $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '13 at 20:36
  • $\begingroup$ Notice that the fact that this is not zero follows immediately from the answer you got here! math.stackexchange.com/questions/462129/… $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '13 at 20:58

Consider the one-form $dx$ on the punctured torus $T^{2*}=(\mathbb{R}^2 - \mathbb{Z}^2) / \mathbb{Z}^2$. Note that the image of the line $\gamma(t) = (t,\frac{1}{2})$ under the quotient is a circle, hence is a $1$-chain, and $$\int_{\gamma}dx = 1.$$ This implies that $dx$ is not a coboundary, for if it were and $dx = df$ for some function $f$, then by Stokes' theorem, $\int_\gamma df = f(\gamma(1)) - f(\gamma(0)) = 0$ since $\gamma(1) = \gamma(0)$.

Therefore $[dx]\in H^1_{dR}(T^{2*})$ is a nontrivial cohomology class.

To your proof, why should we believe $i^*$ is injective? Consider again $T^2=\mathbb{R}^2/\mathbb{Z}^2$. Let $\gamma$ be as above and $\delta = (\frac{1}{2},t)$. Now $dx(\gamma) = 1$, but $dx(\delta) = 0$; $dy(\gamma) = 0$ while $dy(\delta) = 1$. If $B$ is centered at $(0,0)$, then $dx(\partial B) = dy(\partial B) = 1 + 0 - 1 - 0 = 0$. This is an example of why a non-injective $i^*$ unless I am very much off my game today.

  • $\begingroup$ «is not a boundary» $\leadsto$ «is not exact». $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '13 at 21:33
  • $\begingroup$ @MarianoSuárez-Alvarez Bah. $\endgroup$ – Neal Aug 9 '13 at 21:39
  • $\begingroup$ :-) ${}{}{}{}{}$ $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '13 at 21:39
  • $\begingroup$ (I did fix it. :) $\endgroup$ – Neal Aug 9 '13 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.