Understanding the canonical line bundle $H$, and the fact that $(H \otimes H)\oplus 1 \simeq H \oplus H$

Solution 1:

I'll work most of this answer with the example $H^{-1} \oplus H \cong \mathbb{C}^2$, because I find it easier to think about. This means that the clutching functions are $\left( \begin{smallmatrix} z & 0 \\ 0 & z^{-1} \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$. At the end, I'll make a minor comment about switching things up to do your example.

This is an interesting example because the above isomorphism is correct as topological, or smooth, vector bundles, but is NOT valid as holomorphic vector bundles. So the isomorphism will have to involve some non-holomorphic maps, even though the two vector bundles are holomorphic.

One quick note: The bundle you call "canonical" is not the one I learned to call canonical. I always called your $H$ the "tautological" line bundle.

As you say above, the fiber of $H$ above $(u:v)$ is $\mathbb{C} \cdot (u,v) \subseteq \mathbb{C}^2$. So we have an obvious inclusion $H \hookrightarrow \mathbb{C}^2$. The line bundle $H^{-1}$ is the dual to $H$. Fix a skew symmetric bilinear form $\omega $ on $\mathbb{C}^2$. Then there is a map $\mathbb{C}^2 \to H^{-1}$ taking the point $(x,y)$ to the linear functional $\omega( (x,y), \bullet)$. The kernel of this map is precisely the image of $H$. That is to say, the linear form $\omega( (x,y), \bullet)$ restricts to $0$ on $\mathbb{C} \cdot (u,v)$ precisely if $(x,y) \in \mathbb{C} \cdot (u,v)$. So we have a short exact sequence $$0 \to H \to \mathbb{C}^2 \to H^{-1} \to 0.$$ To be concrete, I'll take $\omega( (x_1, y_1), (x_2, y_2) ) = x_1 y_2 -x_2 y_1$.

Now, in the smooth category, every short exact sequence splits. So $\mathbb{C}^2 \cong H \oplus H^{-1}$ as claimed. Let's work out the isomorphism explicitly.

Recall the proof that every short exact sequence splits. Fix a positive definite Hermitian form on the middle term: We'll choose $\langle (x_1, y_1), (x_2, y_2) \rangle = x_1 \overline{x_2} + y_1 \overline{y_2}$. We are supposed to be split the sequence by lifting a vector $p$ in $H^{-1}$ to its unique preimage in $\mathbb{C}^2$ which is $\langle, \rangle$-orthogonal to $H$.

Let's suppose we are at the point $(x:y)$ in $\mathbb{P}^1$. Let our vector in $H^{-1}$, above the point $(x:y)$, be the linear functional $(x,y) \mapsto 1$. So the preimage of this point in $\mathbb{C}^2$ is $\{ (x', y') : x' y - y' x= 1\}$. The condition that $(x', y')$ be orthogonal to $H$ is that $x' \overline{x} + y' \overline{y} =0$. Solving these linear equations, we get $$(x', y') = \left( \frac{\overline{y}}{|x|^2+|y|^2}, \frac{- \overline{x}}{|x|^2+|y|^2} \right).$$

So, here is our isomorphism $H \oplus H^{-1} \to \mathbb{C}^2$. I'll describe it in the fiber over $(x:y)$. A basis of the fiber of $H \oplus H^{-1}$ is $e:=(x,y)$, in the first summand, and $f: = \left( (x,y) \mapsto 1 \right)$ in the second summand. Our trivialization sends $e \mapsto (x,y)$ and $f \mapsto (\overline{y}, - \overline{x}) /(|x|^2+|y|^2)$. Exercise: The map $H \oplus H^{-1} \to \mathbb{C}^2$ which I have described is independent of which representative $(x,y)$ we have chosen for $(x:y)$.

I'm running out of time to work on this. So I'll write down what this turns into in terms of clutching functions. There may be some minor errors here. We have $$\begin{pmatrix} z \vphantom{\frac{z}{1+|z|^{-2}}} & 1 \vphantom{\frac{z}{1+|z|^{-2}}} \\ \frac{-1}{1+|z|^2} & \frac{\overline{z}}{1+|z|^2} \end{pmatrix} \begin{pmatrix} 1 \vphantom{\frac{z}{1+|z|^{-2}}} & 0 \vphantom{\frac{z}{1+|z|^{-2}}} \\ 0 \vphantom{\frac{z}{1+|z|^{-2}}} & 1 \vphantom{\frac{z}{1+|z|^{-2}}} \end{pmatrix} \begin{pmatrix} \frac{1}{1+|z|^{-2}} & - z^{-1} \\ \frac{\overline{z}^{-1}}{1+|z|^{-2}} & 1 \end{pmatrix} = \begin{pmatrix} z \vphantom{\frac{z}{1+|z|^{-2}}}& 0 \\ 0 & z^{-1}\vphantom{\frac{z}{1+|z|^{-2}}} \end{pmatrix} $$

Note that the left matrix is smooth and invertible on the chart where $z \neq \infty$, and the right matrix is smooth and invertible on the chart where $z \neq 0$. So this shows that the two clutching functions give isomorphic vector bundles.

To return to your original request to equate $H^{\otimes 2} \oplus \mathbb{C}$ and $H^{\oplus 2}$, take the above equation and multiply both sides by $z$.