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In Chapter 19 of Homotopical Topology by Fomenko and Fuchs, there is an exercise asserting that a complex vector bundle $\xi$ is the complexification of a real vector bundle if and only if there is an isomorphism $\xi\cong\bar\xi$. I have been trying to do this exercise, but getting caught up on some of the details.

I know that if $V$ is a complex vector space, then specifying a real structure on $V$ (i.e. a subspace $W\subset V$ with $V=W\oplus iW$) is equivalent to specifying a conjugate-linear involution on $V$. However, I have been unable to go from a general conjugate-linear isomorphism $\xi\rightarrow\xi$ to one that equals its own inverse.

Any suggestions?

Nikhil Sahoo
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1 Answers1

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It turns out that the exercise is wrong. The following answer is based on a homework problem assigned by Alexander Givental and multiple conversations with Tong Zhou.

For the quaternionic projective plane $\mathbb H\mathbb P^1$, we take the set of right-lines in $\mathbb H^2$ (i.e. let $\mathbb H^\times$ act on $\mathbb H^2$ by right-multiplication and define $\mathbb H\mathbb P^1$ to be the quotient of $\mathbb H^2-\{0\}$ under this action). Every $\ell\in \mathbb H\mathbb P^1$ is a quaternionic line under right-multiplication. This makes the tautological bundle $\xi=\{(\ell,v)\in \mathbb H\mathbb P^1\times \mathbb H^2\mid v\in \ell\}$ into a quaternionic line bundle. We can forget the quaternionic structure and consider the complex vector bundle $\xi|\mathbb C$ over $\mathbb H\mathbb P^1$. Right-multiplication by $j$ preserves each fiber and anti-commutes with $i$, so we get a conjugate-linear automorphism of $\xi|\mathbb C$.

On the other hand, one can show that $c_2(\xi|\mathbb{C})$ is a generator of $H^4(\mathbb{HP}^1)=\mathbb{Z}$. But if there were a real vector bundle $\eta$ over $\mathbb{HP}^1$ of rank 2 such that $\eta\otimes\mathbb{C}\cong \xi|\mathbb{C}$, then we would have $e(\eta)=0$ (because $H^2(\mathbb{HP}^1)=0$) and thus $$c_2(\xi|\mathbb{C})=e(\eta\oplus \eta)=e(\eta)\cup e(\eta)=0.$$

Nikhil Sahoo
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  • The $\eta$ in my last paragraph has rank 2, because then the complexification $\eta\otimes \mathbb C$ has rank 4. I was using the notation $\mathbb C\eta=\eta\otimes \mathbb C$ for the complexification and $\mathbb C\xi=\xi|\mathbb C$ for the restriction of structure. This isn't the best choice of notation, so I've tried to modify it for clarity. Hopefully, that helps. – Nikhil Sahoo Apr 05 '20 at 17:07
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    "left-multiplication by $j$ is [...] obviously $\mathbb H$-linear." How so? $j$ is not central in $\mathbb H$. Or are you viewing $\mathbb H$ as an $\mathbb H$-vector space from the right but as $\mathbb C$- vector space from the left? Possible, and I don't doubt your ultimate conclusion, but this feels fishy. – Torsten Schoeneberg Jan 25 '21 at 02:45
  • @TorstenSchoeneberg Good point! I was a little careless with my justification (and even my definition of the tautological bundle over $\mathbb H\mathbb P^1$). Does the new fix help? – Nikhil Sahoo Jan 25 '21 at 19:42
  • I guess so, although to be honest I don't feel competent enough to "get" what is happening here. It does make sense to me though. – Torsten Schoeneberg Jan 28 '21 at 05:43
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    More generally, any complex rank 2 bundle over $S^4$ with non-zero $c_2$ provides a counterexample, see here. – Michael Albanese Feb 02 '24 at 14:08