# A question about the complex conjugate bundle

If a complex vector bundle is constructed by the complexification of a real vector bundle, say $$E=F\otimes \mathbb{C}$$, then there's a conclusion that $$E$$ is isomorphic to its conjugate bundle by setting $$f: x+iy \to x-iy$$. However, usually a complex bundle is not isomorphic to its conjugate bundle. So what's the intrinsic difference between them? I'm reading the section on Pontrjagin classes in Milnor and Stasheff's "Characteristic Classes" and I am a little confused.  • Not every complex vector bundle is the complexification of a real vector bundle. I'm not sure what more you're looking for. Jan 16, 2021 at 15:09
• Thank you, I mean every complex vector space can set a map f like the above, but why the above proof does not hold for other complex bundles? Jan 16, 2021 at 15:18
• Let $V$ be a complex vector space. How do you define a conjugation map on $V$? Jan 16, 2021 at 15:19
• Is it true that all the complex vector space of the same dimensional is isomorphic? I think there must be some mistakes in my understanding. Jan 16, 2021 at 15:23
• Writing down a conjugation map on $V$ is equivalent to expressing $V$ as the complexification of a real vector space, see this answer. Jan 16, 2021 at 15:50

A complex conjugation on a complex vector space $$V$$ is an antilinear map $$\sigma : V \to V$$ such that $$\sigma\circ\sigma = \operatorname{id}_V$$; note that $$\sigma$$ induces an isomorphism $$V \to \overline{V}$$. Choosing a complex conjugation on $$V$$ is equivalent to choosing an isomorphism between $$V$$ and the complexification of a real vector space; see this answer. Every finite-dimensional complex vector space $$V$$ admits a complex conjugation because $$V \cong \mathbb{C}^n$$ for some $$n$$ and $$\mathbb{C}^n\cong\mathbb{R}^n\otimes_{\mathbb{R}}\mathbb{C}$$.
By the same argument as above, a complex vector bundle admits a complex conjugation (and hence is isomorphic to its conjugate bundle) if and only if it is the complexification of a real vector bundle. However, not every rank $$n$$ complex vector bundle is isomorphic to the trivial rank $$n$$ complex vector bundle (which is isomorphic to the complexification of the trivial rank $$n$$ real vector bundle). So a non-trivial bundle need not admit a complex conjugation map.
A natural question to ask is whether the existence of an isomorphism $$E\to \overline{E}$$ implies the existence of a complex conjugation. The answer is no! That is, there are complex vector bundles $$E$$ which do not admit a complex conjugation but nonetheless are isomorphic to $$\overline{E}$$.
Example: For any four-dimensional CW complex $$X$$, there is a bijection between isomorphism classes of principal $$SU(2)$$-bundles (or equivalently, rank two complex vector bundles with first Chern class zero) and $$H^4(X; \mathbb{Z})$$ given by $$E \mapsto c_2(E)$$; see this question.
Let $$E \to S^4$$ be a complex rank two vector bundle; note that $$c_1(E) = 0$$. As $$c_1(\overline{E}) = 0$$ and $$c_2(\overline{E}) = c_2(E)$$, we see that $$E$$ is isomorphic to $$\overline{E}$$. Suppose now that $$E$$ admits a complex conjugation. Then $$E$$ is isomorphic to $$E_0\otimes_{\mathbb{R}}\mathbb{C}$$ for some real rank two vector bundle $$E_0$$. As $$w_1(E_0) = 0$$, the bundle $$E_0$$ is determined up to isomorphism by its Euler class which is necessarily zero, so $$E_0$$ is trivial. Therefore, if $$E$$ admits a complex conjugation, then $$E$$ is trivial.
So for any $$\alpha \in H^4(S^4, \mathbb{Z}) \cong \mathbb{Z}$$ with $$\alpha \neq 0$$, the complex rank two vector bundle $$E \to S^4$$ with $$c_2(E) = \alpha$$ satisfies $$E \cong \overline{E}$$ but does not admit a complex conjugation.