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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.

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Michael Albanese
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Sunhf
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    Not every complex vector bundle is the complexification of a real vector bundle. I'm not sure what more you're looking for. – Michael Albanese Jan 16 '21 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? – Sunhf Jan 16 '21 at 15:18
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    Let $V$ be a complex vector space. How do you define a conjugation map on $V$? – Michael Albanese Jan 16 '21 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. – Sunhf Jan 16 '21 at 15:23
  • Yes, two finite-dimensional vector spaces over the same field are isomorphic if and only if they have the same dimension. – Michael Albanese Jan 16 '21 at 15:31
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    Writing down a conjugation map on $V$ is equivalent to expressing $V$ as the complexification of a real vector space, see this answer. – Michael Albanese Jan 16 '21 at 15:50
  • Thank you very much ! I think I've got the main problem in it. Although we can define conjugation on every complex space, there isn't a canonical way to define a continuous conjugation map like above on the whole vector bundle. For example, a conjugation conjugation map like above means there's a n-dimensional real vector bundle embedded in the underlying 2n-dimensional real vector bundle, but we can not always find a n-dimensional real vector bundle embedded in a 2n-dimensional vector bundle. – Sunhf Jan 17 '21 at 15:34
  • Sounds like you've got it. I wrote an answer below to summarise our discussion. – Michael Albanese Jan 17 '21 at 23:16

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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.

Michael Albanese
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