Complexified bundle $\xi \otimes \mathbb{C}$ is isomorphic to its conjugate $\overline{\xi \otimes \mathbb{C}}$?

259 Views Asked by At

The following appears in Milnor's Characteristic classes, Chapter 15, and I don't fully understand what's going on.

Let $\xi$ be a real vector bundle. Milnor claims that the complexification $\xi \otimes \mathbb{C}$ is isomorphic to its conjugate bundle $\overline{\xi \otimes \mathbb{C}}$. Milnor introduces the (fiberwise) map \begin{align*} f: \xi \otimes \mathbb{C} & \to \overline{\xi \otimes \mathbb{C}} \\ x+iy & \mapsto x-iy, \end{align*} which certainly is an isomorphism of real bundles. He also notices the identity $$f(i(x + iy)) = -y -ix = -i f(x+iy),$$ but I think this exactly shows that $f$ is not an isomorphism of complex bundles, because it is not $\mathbb{C}$-linear, but conjugate-linear. And that is not even that special, because any complex bundle is conjugate linear to its conjugate bundle, via the identity map.

However on the next page he uses that isomorphism to conclude that $\xi \otimes \mathbb{C}$ and $\overline{\xi \otimes \mathbb{C}}$ have the same Chern classes.

What am I missing here?

1

There are 1 best solutions below

0
On BEST ANSWER

For notation's sake let $\kappa\colon \xi \otimes\mathbb{C} \to \xi \otimes \mathbb{C}$ and $\overline{\kappa}\colon \overline{\xi \otimes\mathbb{C}} \to \overline{\xi \otimes \mathbb{C}}$ be the names of the complex structures. Explicitly, for $z$ in some fibre $F\otimes \mathbb{C}$ we have $\kappa(z) = i\cdot z$ and $\overline{\kappa} = -i \cdot (z)$. Then for $f\colon \xi \otimes\mathbb{C} \to \overline{\xi \otimes\mathbb{C}}$ to be an isomorphism of vector bundles with complex structure it means $f(\kappa(v)) = \overline{\kappa}(f(v))$ for any $v$ in any fibre $F_x \otimes \mathbb{C}$. But if you unravel the definitions of $\kappa$ and $\overline{\kappa}$ this is precisely what the identity you wrote down is saying.