Conjugation on Vector Spaces and on Vector Bundles

I recently read a very interesting answer, that is about the conjugation structure on a complex vector bundle.

Just as what we do in complex numbers, a conjugation on complex vector space $V$ ( $\dim_{\mathbb{C}}V=n$ ) is an anti-linear isomorphism $\sigma: V\longrightarrow V$ satisfying $\sigma^2=\mathrm{id}_V$ . If we identify $V$ with $\Bbb{C}^n$ by choosing a priori basis, namely $\phi:V\longrightarrow \Bbb{C}^n$ , then there is a standard conjuagtion on $V$ defined by $\phi\circ\sigma\circ\phi^{-1}$ , where $\sigma(z_1,...,z_n):=\left(\bar{z}_1,...,\bar{z}_n\right)$ is the conjuagtion on $\mathbb{C}^n$ .

However, the isomorphism between $V$ and $\mathbb{C}^n$ is not natural (depends on the choice of the basis), thus the existence of a conjuagtion means something more. In fact, if $V$ admits a conjuagtion $\sigma$ , then there is a subspace $W$ which was fixed by $\sigma$ , i.e. $\sigma(W)=W$ , and it has real dimension $n$ , then we can write $V=W\otimes_{\mathbb{R}}\mathbb{C}$ , then the $\sigma$ can be read as $\sigma(w\otimes z)=w\otimes\bar{z}$ . Conversely, it is obviously that if $V$ is the complexification of a real vector space $W$ , then there is a natrual conjugation $\sigma$ read as above. So, we obtained:

A complex vector space admits a conjuagtion if and only if it is the complexification of some real vector space.

The above result seems likes an abstract nonsense, since every complex vector space can be chose by a basis so that the conjugation can be defined. But things are different if we globalize it to a complex vector bundle $E\longrightarrow X$ .

First of all, on a rank $n$ complex vector bundle $E$ , we can define its conjuagte bundle $\overline{E}$ by taking the conjugation of the transition functions $\bar{g}_{\alpha\beta}$ , where $g_{\alpha\beta}$ ‘s are the transition functions on $E$ . If we endow $E$ with an Hermitian metric, then $g_{\alpha\beta}$ ‘s are taking values in $U(n)$ , so we obatined that $\overline{E}\cong E^*$ .

Now, if $E$ admits a complex conjuagtion $\sigma$ , then $\sigma$ induces the isomorphism $E\cong\overline{E}\cong E^*$ , and by our previously result, $E$ is a complexification a real vector bundle $W$ . However, this is not always the case, since we can choose a connection $d+\omega$ on $E$ with the curvature provided by $\Omega=d\omega+\omega\wedge\omega$ , since the structure group is $U(n)$ , $\Omega$ is an $\mathfrak{u}(n)$ -valued 2-form, hence it is skew-Hermitian. By Chern-Weil theory, we have

$\begin{aligned}c_1\left(\overline{E}\right)&=\left[\frac{i}{2\pi}\mathrm{Tr}\left(\overline{\Omega}\right)\right]\\&=-\left[\frac{i}{2\pi}\mathrm{Tr}(\Omega)\right]\\&=-c_1(E)\end{aligned}$

Hence one must have $c_1(E)=0$ . But the second Chern class is not necessarily vanishing, since $c_2(E^*)=c_2(E)$ for $\mathrm{rank}E=2$ . If $W$ is a real vector bundle, the second Chern class $c_2(W\otimes\mathbb{C})$ is also known as the 1st Pontryagin class of $W$ , denoted by $p_1(W)$ .

One shall also be very careful that even if $E\cong \overline{E}$ , the conjugation on $E$ still can be not necessarily existed. One example is as following.

Recall that on a compact 4-manifold, the rank 2 Hermitian vector bundles with vanishing 1st Chern classes were classified by $H^4(X;\pi_3(SU(2)))=H^4(X;\mathbb{Z})$ , that is they were classified by the 2nd Chern classes. Hence if we let $X=S^4$ , then $H^4(S^4;\Bbb{Z})\cong \Bbb{Z}$ , hence we can pick a non-trivial rank 2 complex vector bundle $E$ with vanishing 1st Chern classes. Since

$c_1(E)=-c_1(\overline{E})=-c_1(E^*)=0$ and $c_2(E)=c_2(E^*)=c_2(\overline{E})$

we deduce that $E=E^*=\overline{E}$ . Now, if $E$ exists a conjuagtion, then $E=W\otimes\mathbb{C}$ for some real-rank 2 vector bundle $W$ , and $W$ has the structure group $SO(2)$ , but on $S^4$ we know that the $SO(2)-$ bundle were classified by $\pi_3(SO(2))=0$ hence $W$ must be trivial, which leads that $E$ is also trivial, a beautiful contradiction.