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10

In the context of geometry, a distinction between "soft" and "hard" was introduced by Gromov, as explained here and applied to Soft and Hard Symplectic Geometry. In Gromov's words, 'hard' refers to a strong and rigid structure of a given object, while 'soft' suggests some weak general property of a vast class of objects. Riemannian geometry is hard, while ...

2

No. Namely, the rank of $\Phi:E\to F$ cannot drop locally (some minor, in local frames, has $\det\ne 0$). But it can increase locally. So the rank of $ker(\Phi)$ can drop locally. But the rank of $\Psi:G\to E$ cannot drop locally.

0

As pointed out to me by Guangbo Xu, if the $S^1$-action gives $V$ the structure of an $S^1$-bundle $\pi:V\to B$, then such connection exists if and only if $\pi_*(c_1(L))=0 \in H_1(B)$, where $\pi_*$ is the integration map along the fibers of the Gysin sequence. However, one still has to think what happens if we choose another loop of diffeomoerphisms ...

12

W.Sutherland. A note on the parallelizability of sphere bundles over sphere. J. London Math. Soc. 39 (1964), 55--62. The answer is yes.

0

What about Mehta and Seshadri (Math. Ann. 248, 1980)? Their paper handles the genus 2 or more cases (with a focus on one marked point), but my understanding is that their result and proof (generically) generalizes to genus 1 and 0 (and arbitrary many marked points). Perhaps a good project for your student would be to read their paper, and explain how the ...

5

This discussion seems to involve two holomorphic structures on the smooth manifold $T S^2$. Structure 1 Identify $S^2$ with $\mathbb{CP}^1$. Place a complex structure on the real tangent bundle to $\mathbb{CP}^1$ by using the complex structure on $\mathbb{CP}^1$. This is the structure Alex Degtyarev's answer refers to. As Alex says, this is the line bundle ...

1

$TS^2\to S^2$ has Euler class $2$; ergo, the line bundle in question is $\mathcal{O}_{\Bbb P^1}(2)$. It's holomorphic sections form a vector space of dimension $3$; they can be thought of as degree $\le2$ polynomials on $\Bbb C$. Added in proof: All this is independent of the holomorphic structure on $TS^2$ assuming that there is a fibration $TS^2\to S^2$. ...

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