Does anybody know how to prove the following statement:

The complexified tangent bundle $TS\otimes\mathbb{C}$ of a closed surface $S$ is topologically trivial iff the Euler characteristic $\chi(S)$ is even.

Note that if $S$ is orientable, then $\chi(S)=2-2g(S)$ is always even. And in fact it is easy to see $TS\otimes\mathbb{C}$ is trivial in the orientable case because $TS$ is stably trivialized by normal bundle in $\mathbb{R}^3$.

I have trouble dealing with the non-orientable case. I can’t even see an idea for $\mathbb{RP}^2$ or the Klein bottle.

  • $\begingroup$ Maybe I'm being silly, but I don't understand your proof for $S$ orientable. $\endgroup$ – Michael Albanese Mar 28 at 14:31
  • $\begingroup$ @MichaelAlbanese Let $\nu$ be normal bundle of $S$ in $\mathbb{R}^3$, then $TS\oplus(S\times\mathbb{R})\cong TS\oplus\nu\cong S\times\mathbb{R}^3$. Then complexify these bundles, and take first Chern classes. $\endgroup$ – Yeah Mar 28 at 14:36
  • $\begingroup$ I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TS\otimes_{\mathbb{R}}\mathbb{C}) = c_1(TS\oplus\overline{TS}) = c_1(TS) + c_1(\overline{TS}) = c_1(TS) - c_1(TS) = 0.$$ $\endgroup$ – Michael Albanese Mar 28 at 15:07
  • $\begingroup$ @MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TL\otimes\mathbb{C}$ is equivalent to the existence of a Lagrangian immersion $L\to\mathbb{C}^n_{st}$. $\endgroup$ – Yeah Mar 28 at 19:08

If $E \to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E \cong F\oplus\varepsilon_{\mathbb{C}}^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TS\otimes_{\mathbb{R}}\mathbb{C}$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TS\otimes\mathbb{C} \cong L\oplus\varepsilon_{\mathbb{C}}^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TS\otimes_{\mathbb{R}}\mathbb{C}$ is trivial if and only if $c_1(L) = c_1(TS\otimes_{\mathbb{R}}\mathbb{C}) \in H^2(S; \mathbb{Z})$ is zero.

If $S$ is not orientable, then $H^2(S; \mathbb{Z}) \cong \mathbb{Z}_2$ and the reduction modulo $2$ map $H^2(S; \mathbb{Z}) \to H^2(S; \mathbb{Z}_2)$ is an isomorphism. Under this map, $c_1(TS\otimes_{\mathbb{R}}\mathbb{C}) \mapsto w_2(TS\otimes_{\mathbb{R}}\mathbb{C})$. Now, as a real bundles, $TS\otimes_{\mathbb{R}}\mathbb{C} \cong TS\oplus TS$, so

$$w_2(TS\otimes_{\mathbb{R}}\mathbb{C}) = w_2(TS\oplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$

On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $\nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TS\otimes_{\mathbb{R}}\mathbb{C}$ is trivial if and only if $w_2(TS) = 0$. Now note that $\langle w_2(TS), [S]\rangle \equiv \chi(S) \bmod 2$, so $TS\otimes_{\mathbb{R}}\mathbb{C}$ is trivial if and only if $\chi(S)$ is even.

  • $\begingroup$ Thank you Michael! $\endgroup$ – Yeah Mar 27 at 20:11

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