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3

Maybe the simplest example of the phenomenon mentioned by Jason is the case of rank 2 vector bundles on $\mathbb{P}^3$ with $c_1$ even, studied in this paper of Atiyah and Rees. Besides $c_1$ and $c_2$ which encode the $K$-theory class, there is another topological invariant $\alpha $ with values in $\mathbb{Z}/2$, which can be $0$ or $1$ for holomorphic ...


3

As always, you should specify that $X$ is projective (or at least proper). Even so, this really depends on what you mean by a "connected family". Certainly this is false if you want the base of your family to be irreducible. Here is the simplest counterexample I see. On $X=\mathbb{P}^2$, let $V_1$ be $\mathcal{O}(-7)\oplus \mathcal{O}(-7)\oplus ...


2

Yes. If this map $\pi_{8t+1}U\to \pi_{8t+1}O$ were trivial then $\pi_{8t+1}(O/U)$ would have an element of order $2$. But $O/U$ is homotopy equivalent to $\Omega O$, and $\pi_{8t+2}O$ is trivial. I do not know a reference offhand.


1

Orientability of $TM|_\Sigma$ is equivalent to non-triviality of normal bundle $N\Sigma$. Denote generator of $\pi_1(\Sigma)$ by $\gamma$; if we have $N\Sigma$ is trivial, then $TM|_{i(\gamma)}$ is non-orientable, so $\gamma$ cannot be nullhomotopic.


12

Yes, there is such a twistor fibration over each $S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic $n$-plane field transverse to the fibers of the mapping. Namely, one writes $S^{2n} = \mathrm{SO}(2n{+}1)/\mathrm{SO}(2n)$ and then, using the inclusion $\mathrm{U}(n)\subset\mathrm{SO}(2n)$, one has the coset fibration $$ ...


6

One possible generalization of Penrose twistor fibration is the fibration $\pi:\mathcal{J}_M\to M$ over any oriented $2n$-dimensional conformal manifold $(M,[g])$, whose fiber $\pi^{-1}(x)$ is the space of orthogonal almost complex structures on $(T_xM,[g_x])$, namely the orthogonal $J_x:T_xM\to T_xM$ squaring to $-\mathrm{id}$ such that the complex ...



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