What happens when we compactify every fibre of a holomorphic $\Bbb C$ bundle over a compact Riemann surface?

Question

I am a beginner in learning about holomorphic line bundles over Riemann surfaces. In particular, I am a complete naïf as regards sheaf cohomology (so I hope the answer(s) won't involve that too deeply).

Let $\Bbb C\to E\to M$ be a holomorphic line bundle over the compact Riemann surface $M$.

Let $\pi:E\to M$ denote this original bundle.

If we include each fibre in its one-point compactification $\Bbb CP^1\cong S^2$, then I believe we end up with a holomorphic $S^2$ bundle over $M$ (by using the same transition functions). (Is that correct?)

Assuming it is correct:

Denote the resulting bundle by $\xi(\pi)$.

We then have a holomorphic $S^2$-bundle $\xi(\pi)$ over $M$ with two distinguished sections: the $0$-section and the $\infty$-section.

Questions:

1. Suppose that we now delete the image of the $0$-section. The result is a holomorphic $\Bbb C$-bundle over $M$ with one distinguished section, the $\infty$-section. Which bundle is it in terms of the original one?

The most obvious guesses are a) the original bundle and b) its inverse $\pi^\ast$ in the Picard group, isomorphic to the canonical bundle of $M$.

2. Which original bundles $\pi$ end up with holomorphically isomorphic $\Bbb CP^1$-bundles $\xi(\pi)$ as a result?

3. Which $\Bbb CP^1$-bundles over $M$ can be obtained in this way?

Answer 1

Given a holomorphic line bundle $E \to M$, we obtain a holomorphic $\mathbb{CP}^1$-bundle $\mathbb{P}(E\oplus\mathcal{O}) \to M$ with structure group $PGL(2, \mathbb{C})$ where $\mathcal{O}$ is the trivial holomorphic line bundle. Note that $\mathbb{P}(E\oplus\mathcal{O})$ is precisely the space obtained by forming the one-point compactification of the fibers of $E$. In the special case where $M = \mathbb{CP}^1$ and $E = \mathcal{O}(n)$, the total space $\mathbb{P}(E\oplus\mathcal{O})$ is called the $n^{\text{th}}$ Hirzebruch surface.

There are two natural sections of $\mathbb{P}(E\oplus\mathcal{O}) \to M$:

• the zero section $s_0$ where $s_0(m) = [0, z]$ where $z \in \mathbb{C}\setminus\{0\}$, and
• the infinity section $s_{\infty}$ where $s_{\infty}(m) = [e, 0]$ where $e \in E_m\setminus\{0\}$.

Given any holomorphic vector bundle $V \to M$ of rank $r + 1$, and any holomorphic line bundle $L \to M$, there is an isomorphism of holomorphic $\mathbb{CP}^r$-bundles $\mathbb{P}(V) \to \mathbb{P}(V\otimes L)$ given by $[v] \mapsto [v\otimes\ell]$ where $\ell$ is any non-zero element of $L$ (over the same point of $M$ that $v$ lies over). In our case, we see there is an isomorphism

\begin{align*} \mathbb{P}(E\oplus\mathcal{O}) &\to \mathbb{P}((E\oplus\mathcal{O})\otimes E^*)\\ [e, z] &\mapsto [(e, z)\otimes\alpha].\ \end{align*}

Note that there is an isomorphism of vector bundles $(E\otimes\mathcal{O})\otimes E^* \to \mathcal{O}\oplus E^*$ which descends to an isomorphism of the corresponding projective bundles

\begin{align*} \mathbb{P}((E\otimes\mathcal{O})\otimes E^*) &\to \mathbb{P}(\mathcal{O}\oplus E^*)\\ [(e, z)\otimes\alpha] &\mapsto [(\alpha(e), z\alpha)]. \end{align*}

Furthermore, the isomorphism $\mathcal{O}\oplus E^* \to E^*\oplus\mathcal{O}$ descends to an isomorphism of projective bundles

\begin{align*} \mathbb{P}(\mathcal{O}\oplus E^*) &\to \mathbb{P}(E^*\oplus\mathcal{O})\\ [z, \alpha] &\mapsto [\alpha, z]. \end{align*}

The composition of these three isomorphisms gives an isomorphism $\Phi : \mathbb{P}(E\oplus\mathcal{O}) \to \mathbb{P}(E^*\oplus\mathcal{O})$ given by $\Phi([e, z]) = [z\alpha, \alpha(e)]$. Note that $\Phi([0, z]) = [z\alpha, 0]$, so the isomorphism $\Phi$ maps the zero section of $\mathbb{P}(E\oplus\mathcal{O})$ to the infinity section of $\mathbb{P}(E^*\oplus\mathcal{O})$. Therefore, removing the image of the zero section of $\mathbb{P}(E\oplus\mathcal{O})$ is equivalent to removing the image of the infinity section of $\mathbb{P}(E^*\oplus\mathcal{O})$ which gives $E^*$. This answers question 1.

To answer the remaining questions, note that

• the set of isomorphism classes of holomorphic line bundles over $M$ is given by $H^1(M, \mathcal{O}^*)$ where $\mathcal{O}^*$ denotes the sheaf of germs of holomorphic maps $M \to \mathbb{C}^*$,
• the set of isomorphism classes of holomorphic rank $r + 1$ vector bundles over $M$ is given by $H^1(M, \mathcal{GL}(r+1, \mathbb{C}))$ where $\mathcal{GL}(r + 1, \mathbb{C})$ denotes the sheaf of germs of holomorphic maps $M \to GL(r + 1, \mathbb{C})$, and
• the set of isomorphism classes of holomorphic $\mathbb{CP}^r$-bundles over $M$ with structure group $PGL(r + 1, \mathbb{C})$ is given by $H^1(M, \mathcal{PGL}(r + 1, \mathbb{C}))$ where $\mathcal{PGL}(r + 1, \mathbb{C})$ denotes the sheaf of germs of holomorphic maps $M \to PGL(r + 1, \mathbb{C})$.

The short exact sequence of groups $0 \to \mathbb{C}^* \to GL(r + 1, \mathbb{C}) \to PGL(r + 1, \mathbb{C}) \to 0$ induces a short exact sequence of sheaves $0 \to \mathcal{O}^* \to \mathcal{GL}(r + 1, \mathbb{C}) \to \mathcal{PGL}(r + 1, \mathbb{C}) \to 0$. This induces an exact sequence of cohomology sets

$$0 \to H^1(M, \mathcal{O}^*) \to H^1(M, \mathcal{GL}(r + 1, \mathbb{C})) \to H^1(M, \mathcal{PGL}(r + 1, \mathbb{C})) \to H^2(M, \mathcal{O}^*).$$

Note that $H^1(M, \mathcal{O}^*)$ acts on $H^1(M, \mathcal{GL}(r + 1, \mathbb{C}))$ via the tensor product, i.e. $[L]\ast[V] = [L\otimes V]$. The map $H^1(M, \mathcal{O}^*) \to H^1(M, \mathcal{GL}(r + 1, \mathbb{C}))$ is precisely restricting the action to the trivial bundle $[\mathcal{O}^{r+1}]$, i.e. $[L] \mapsto [L\otimes(\mathcal{O}^{r+1})] = [L^{r+1}]$.

On the other hand, the map $H^1(M, \mathcal{GL}(r + 1, \mathbb{C})) \to H^1(M, \mathcal{PGL}(r + 1, \mathbb{C}))$ is given by $[V] \mapsto [\mathbb{P}(V)]$. It follows from exactness that $[V_1], [V_2] \in H^1(M, \mathcal{GL}(r+1, \mathbb{C}))$ have the same image in $H^1(M, \mathcal{PGL}(r+1, \mathbb{C}))$ (i.e. $[P(V_1)] = [P(V_2)]$ if and only if $[V_1]$ and $[V_2]$ are in the same orbit of $H^1(M, \mathcal{O}^*)$. That is, $\mathbb{P}(V_1)$ and $\mathbb{P}(V_2)$ are isomorphic if and only if there is $[L] \in H^1(M; \mathcal{O}^*)$ with $[L]\ast[V_1] = [V_2]$, i.e. $V_2 \cong L\otimes V_1$ - note, we already saw above that $V_2\cong L\otimes V_1$ implies $\mathbb{P}(V_1)$ and $\mathbb{P}(V_2)$ are isomorphic, and this argument shows the converse is also true.

Returning to question 2, we see that two holomorphic line bundles $E_1$ and $E_2$ give rise to isomorphic projective bundles if and only if there is a holomorphic line bundle $L$ such that $E_2\oplus\mathcal{O} \cong (E_1\oplus\mathcal{O})\otimes L$. I don't know if this condition can be further simplified. In the case of Hirzebruch surfaces, we have $\mathbb{P}(\mathcal{O}(n)\oplus\mathcal{O})\cong\mathbb{P}(\mathcal{O}(m)\oplus\mathcal{O})$ if and only if $m = \pm n$.

It follows from the previous discussion that $\mathbb{P}(V)$ arises from the construction you outline if and only if $V$ splits as the sum of two line bundles: if $\mathbb{P}(V) \cong \mathbb{P}(E\oplus\mathcal{O})$, then $V \cong (E\oplus\mathcal{O})\otimes L \cong E\otimes L \oplus L$, and if $V \cong L_1\oplus L_2$, then $V\cong ((L_1\otimes L_2^*)\oplus\mathcal{O})\otimes L_2$, so $\mathbb{P}(V) \cong \mathbb{P}(E\oplus\mathcal{O})$ where $E \cong L_1\otimes L_2^*$. So we obtain an answer to question 3: the construction gives rise to those projective bundles which are projectivisations of the sum of two line bundles.