# Tag Info

2

It seems that the notion of spectrum of vector bundles used in Rao's article is defined in Section 3 of C. Okonek and H. Spindler. Reflexive Garben vom Rang $r>2$ auf $\mathbb{P}^n$. Crelle Journal Reine Angew. Math. 344 (1983), pp. 38-64. If you are having trouble reading German, I could give the definition here, but refrain from it now because it ...

2

There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous equality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ ...

2

The easiest way to see that the norm of the curvature corresponds to the energy is to consider the special case of an abelian U(1)-Yang-Mills theory (i.e. electrodynamics). If you write out the norm squared of the curvature in terms of the $E$ and $B$ fields you get the expression $E^2 + B^2$. This is exactly the familiar energy density of an electrodynamic ...

1

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 ...

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