Is $\mathcal{O}(-1)\cong\mathcal{O}(1)$?

Question

Disclaimer: I use some algebraic geometry notation below, out of convenience, but I know nothing about algebraic geometry.

As far as I can tell, the oriented 2-dimensional real vector bundles over $S^2\cong\mathbb{CP}^1$ are given by $\mathcal{O}(k)$ for $k\in\mathbb{Z}$. All of these are distinct: $\mathcal{O}(m)$ is not isomorphic to $\mathcal{O}(n)$ for $m\neq n$.

But what if we remove the orientation? Could it be true some of these become isomorphic as real vector bundles, like $\mathcal{O}(k)\cong\mathcal{O}(-k)$?

I tried to do this for the simplest example: $\mathcal{O}(1)$ and $\mathcal{O}(-1)$. Looking around on the internet, I found the following total spaces for these bundles:

\begin{align*} \mathcal{O}(-1) &= \{([\ell],v)\in\mathbb{CP}^1\times\mathbb{C}^2\mid v\in\ell\}\\ \mathcal{O}(1) &= \mathbb{CP}^2\setminus[0:0:1] \end{align*}

But I can't convince myself these spaces are homeomorphic, let alone bundle-isomorphic. And this is where I get confused: some answers on this site (like this one) seem to suggest these two total spaces are not homeomorphic. But other answers (like this one) seem to suggest they only differ by the per-fiber orientation.

So: is $\mathcal{O}(-1)\cong\mathcal{O}(1)$ as real non-oriented vector bundles over $S^2$? If not, why not? If so, is there an explicit bundle isomorphism?

Answer 1

In this case, you can give an explicit bundle isomorphism between $\mathcal{O}(1)$ and $\mathcal{O}(-1)$. I'll write $S$ and $T$ for the total spaces you listed in the question, so that $S$ is $\mathbb{CP}^2$ minus $[0:0:1]$ and $T$ is a subspace of $\mathbb{CP}^1\times\mathbb{C}^2$.

Since we want a bundle isomorphism, we need to send the point $[x:y:z]\in S$ to a point that looks like $\big([x:y]; v\big)\in T$. A naive choice of $v$ would be to scale the point $(x,y)\in\mathbb{C}^2$ by $z$: \begin{equation*} v = (xz, yz) \end{equation*}

But this isn't well-defined. A slightly better approach is to first normalize $(x,y)$: \begin{equation*} v = \frac{(xz, yz)}{|x|^2+|y|^2} \end{equation*}

This is better, but still not well-defined, since $[\lambda x:\lambda y:\lambda z]$ goes to different $v$ for different values of $\lambda\in\mathbb{C}^\ast$. That's where the complex conjugation comes in: \begin{equation*} v = \frac{(x\overline{z}, y\overline{z})}{|x|^2+|y|^2} \end{equation*}

This is now well-defined and gives a continuous map from $S$ to $T$, that covers the identity map on $\mathbb{CP}^1$: \begin{equation*} F([x:y:z]) = \bigg([x:y];\frac{\overline{z}(x, y)}{|x|^2+|y|^2}\bigg) \end{equation*}

Let $U_x=\{[x:y]\in\mathbb{CP}^1\mid x\neq0\}$, and similarly for $U_y$. Then the trivialization for $S$ is given by \begin{align*} U_x\times\mathbb{C} &\rightarrow \pi_S^{-1}(U_x)& U_y\times\mathbb{C} &\rightarrow \pi_S^{-1}(U_y)\\ \big([x:y],z\big)&\mapsto [x:y:xz] & \big([x:y],z\big)&\mapsto [x:y:yz] \end{align*} I'll abuse notation and write both of these as $\rho_S$. We can also write the trivialization for $\rho_T$: \begin{align*} U_x\times\mathbb{C} &\rightarrow \pi_T^{-1}(U_x)& U_y\times\mathbb{C} &\rightarrow \pi_T^{-1}(U_y)\\ \big([x:y],z\big)&\mapsto \big([x:y];(z,\tfrac{y}{x}z\big) & \big([x:y],z\big)&\mapsto \big([x:y];(\tfrac{x}{y}z,z\big) \end{align*}

These trivializations can then show what $F$ does to fibers, and show it really is a bundle isomorphism. For example, if $[x:y]\in U_x$ then \begin{equation*} \rho_T^{-1}F\rho_S(z) = \overline{z}\frac{|x|^2}{|x|^2+|y|^2} \end{equation*} which is certainly an invertible $\mathbb{R}$-linear map.

EDIT: It's possible to also give a definition for $F^{-1}$ \begin{equation*} F^{-1}([x,y]; v) = \begin{cases} [x:y:0] & v=0\\ [v:|v|^2] & v\neq0. \end{cases} \end{equation*}

Answer 2

One may prove that any complex vector bundle, viewed as a real, unoriented vector bundle is isomorphic to its complex dual as follows:

Let $\xi$ be a complex vector bundle over a base $B$ (assume this space to be paracompact). By a partition of unity argument, there exists a hermitian metric $h: \bar\xi\otimes \xi\to \underline{\mathbb{C}}$, where $\bar{\xi}$ denotes the conjugate bundle to $\xi$ and $\underline{\mathbb{C}}$ is the trivial complex line over $B$. One may view this as a map $\bar{\xi}\to \mathrm{Hom}(\xi, \mathbb{C}); v\mapsto(u\mapsto h(v,u))$ and the fact that this is a hermitian metric means that this is an isomorphism of complex vector bundles. Note that the underlying unoriented vector bundles $\xi$ and $\bar{\xi}$ are the same. These only differ in the complex structure given to them. This means that the map $(v\mapsto v)$ is an isomorphism of real vector bundles (this is real linear but complex antilinear). The composition $\xi\to \bar{\xi}\to \mathrm{Hom}(\xi, \mathbb{C})$ then gives a real vector bundle isomorphism between $\xi$ and $\xi^*=\mathrm{Hom}(\xi,\mathbb{C})$.

Just as a sanity check, we can see that this makes sense on the level of Stiefel-Whitney classes. The Stiefel-Whitney classes of a complex vector bundle are the mod 2 reduction of the Chern classes and the Chern classes of the conjugate bundle of $\xi$ are given by $c_i(\bar{\xi})=(-1)^{i}c_i(\xi)$ so the Stiefel-Whitney classes of $\xi$ match those of $\bar{\xi}$.