# Triviality of complexified tangent bundle of a closed surface

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.

• Maybe I'm being silly, but I don't understand your proof for $S$ orientable. – Michael Albanese Mar 28 at 14:31
• @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. – Yeah Mar 28 at 14:36
• I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TS\otimes_{\mathbb{R}}\mathbb{C}) = c_1(TS\oplus\overline{TS}) = c_1(TS) + c_1(\overline{TS}) = c_1(TS) - c_1(TS) = 0.$$ – Michael Albanese Mar 28 at 15:07
• @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 $L\to\mathbb{C}^n_{st}$. – 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.