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The transposition action does not lift, at least, "typically". The easiest obstruction is in the case $k=1$; in general, I think it's the same, with a bit more characteristic classes (or, "projective space" replaced with other magic words like "Grassmannian"). So, restrict the bundle to matrices with a fixed image $V=\Bbb R\subset\Bbb R^n$. The bundle in ...


3

It looks like $n_3-n_1-n_2$, but double check the computation. Tensor everything by $L_1^{-1}$ to make $L_1$ trivial and recompute the classes to get $0$, $n_2-n_1$, and $n_3-2n_1$. Then project a section of (trivial now) $L_1$ to $E/L_2$: you are interested in the zeroes of this projection, which are counted by $c_1(E/L_2)=(n_3-2n_1)-(n_2-n_1)$.


2

On $M$ there is an exact sequence $$ 0 \to \mu^*\Omega_N \to \Omega_M \to i_*L^\vee \to 0, $$ where $i:D \to M$ is the embedding. This allows to understand the class of $D$ since $D = c_1(i_*L) = c_1(\Omega_M) - \mu^*c_1(N)$. In your case it is equal to $$ (-2,-2) - (-3d,-3k) = (3d-2,3k-2). $$ Thus $D$ is a curve of bidegree $(3d-2,3k-2)$. In particular, its ...


1

Yes: Denote by $E$ this family of subspaces. Choose $x\in X$ (with $\|x\|=1$) and $\phi_x\in Y$ with $\phi_x(x)=1$ and consider the weak star open set $U_x=\{\phi\in Y: \phi(x)\ne 0\}$. Then $$ E|_{U_x} \ni (y,\phi) \mapsto (y -\phi_x(y).x,\phi)\in \ker(\phi_x)\times U_x $$ is a trivializing vector bundle chart. The chart changes look like $$ ...


5

Consider the natural map $M \to\mathbb RP^n$ assigning $x\mapsto ker x$. Then obviously your line bundle arises as the pullback of the tautological bundle. But now you can write down a homotopically non-trivial map $S^1 \to M \to \mathbb RP^n$ (try to restrict your attention to hit $\mathbb RP^1\subset \mathbb RP^n$). No it is a matter of taste to say that ...


3

No, this bundle is not trivial (starting from dimension $2$). Introduce a metric, consider projector to a hyperplane, and rotate this hyperplane through $\pi$ about an axis. You get an orientation reversing loop.


1

A vector bundle $E\neq 0$ on $C$ is neither projective nor injective. Let $L$ be a line bundle; then $\mathrm{Ext}_C^1(E,L)\cong H^1(C,E^*\otimes L)$, which is nonzero by Serre duality when $\deg(L)\ll 0$. Similarly $\mathrm{Ext}_C^1(L,E)\cong H^1(C,E\otimes L^{-1})$ is nonzero for $\deg(L)\gg 0$. An argument along the same lines gives the same property ...


10

This is more of a comment than an answer, but it's too long for a comment, so I'm putting it here. It sounds as though you are asking what sorts of groups of diffeomorphisms there are acting transitively on spheres, so that for each such subgroup $\Gamma\subset \mathrm{Diff}(S^n)$, one can define a notion of $\Gamma$-bundles over manifolds, where a ...


0

Yes, because $F$ has at most 2 sub-line bundles of degree $c$, so you have plenty of choices. The reason is the following. Twisting by a line bundle of degree $-c$ you reduce to the case $c=0$. You can assume that $F$ contains at least one sub-line bundle of degree $0$, and twisting again that this is $\mathcal{O}_C$, so that you have an exact sequence ...



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