17 votes

Tensor product of vector bundles

Perhaps the fundamental issue here is an unintended consequence of the orthodox set-theoretic foundations of mathematics, which can give the mistaken impression that mathematical objects need be ...
Terry Tao's user avatar
  • 109k
12 votes

Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$. The answer is "basically no" ...
Sam Nead's user avatar
  • 26.3k
9 votes
Accepted

Pullback of a vector bundle extension class

For a counterexample, one can take $X$ an elliptic curve over a field of characteristic not $2$, $\phi$ the involution sending each point to its inverse under the group law, $F$ and $G$ both $\mathcal ...
Will Sawin's user avatar
  • 137k
9 votes

Tensor product of vector bundles

I guess we should start by speaking somewhat philosophically about what it means to think geometrically about bundles. For me, this usually means looking at their sections: either constructing ...
Paul Siegel's user avatar
  • 28.8k
8 votes

Obstructions to the existence of a flat connection on a vector bundle

A $d$-dimensional flat real vector bundle $E→M$ is classified by a map $\def\B{{\sf B}}\def\GL{{\rm GL}}M→\B\GL(d)_δ$, where $\GL(d)_δ$ is the orthogonal group equipped with the discrete topology. ...
7 votes
Accepted

Is the subscheme parametrizing the k-th degeneracy loci Cohen-Macaulay?

First, let $$ M_k \subset \mathrm{Mat}(e,f) $$ be the variety of matrices of size $e \times f$ and rank at most $k$. This variety is well-known to be Cohen--Macaulay. Now consider the projection $\pi \...
Sasha's user avatar
  • 37.2k
6 votes

Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

The answer is no. This is a corrected version of Nicolast's comment. Let $E$ be an elliptic curve, let $f: E \to E$ be an endomorphism and let $H_1(f) : H_1(E) \to H_1(E)$ be the induced map on $H_1$. ...
David E Speyer's user avatar
6 votes
Accepted

Identifying the circle bundle of the canonical line bundle $\mathcal{O}(-n-1)$ over a projective space $\mathbb{CP}^n$

Write $X:=\mathbb{CP}^{n}$, the space of lines in $\mathbb C^{n+1}$. For each $k\ge1$ the total space of circle bundle $\mathcal O(-k)\to X$ is $\{\ell\in X,v\in(\mathbb C^{n+1})^{\otimes k}:v\in\ell^{...
Kenta Suzuki's user avatar
  • 1,897
6 votes
Accepted

Converging paths implies converging parallel transports along those paths?

The following is a precise version of the "convergence of equations implies convergence of solutions" claim. Theorem. Let $\Omega\subseteq \mathbb{R}^n$ be open, and $Q\subseteq \mathbb{R}$. ...
Willie Wong's user avatar
  • 37.6k
5 votes

Proving that $H^1(X,\mathcal{Hom}(\mathcal{G},\mathcal{E})) \cong \text{Ext}^1(\mathcal{G},\mathcal{E})$ holds for locally free sheaves

This isomorphism holds for any degree of cohomology and it only requires that the first sheaf $\mathcal{G}$ is locally free. This relies on the fact that $\mathcal{H}om(\mathcal{G},-)$ is exact. There ...
Ben C's user avatar
  • 3,301
5 votes
Accepted

Learning roadmap for holonomy theory

Dominic Joyce has two relevant books: Compact Manifolds with Special Holonomy and Riemannian Holonomy Groups and Calibrated Geometry for the study of holonomy groups of Riemannian manifolds, the ...
Ben McKay's user avatar
  • 25.6k
5 votes
Accepted

A smooth family of lattices on the tangent bundle?

A choice of a (smoothly varying) lattice in each tangent space is the same as a reduction of structure group of the tangent bundle from $GL_n(\mathbb R)$ to $GL_n(\mathbb Z)$. For instance, it implies ...
Bertram Arnold's user avatar
5 votes

A smooth family of lattices on the tangent bundle?

A smooth family of lattices cannot in general be chosen in the tangent bundle of a manifold, even in such a simple case as $S^2$. Indeed, if one could do this it would imply that one can also choose ...
Mikhail Katz's user avatar
  • 15.2k
5 votes
Accepted

Isomorphism between tangent bundle of $S^2$ and the kernel of a bundle homomorphism

Since $\pi\circ\iota$ is constant, we have $d\pi\circ d\iota = 0$, so the image of $d\iota$ is contained in the kernel of $d\pi$, i.e. $(d\iota)(T\mathbb{CP}^1) \subseteq L_n$. Since $\iota$ is an ...
Michael Albanese's user avatar
5 votes
Accepted

Bochner Laplacian in coordinates

Example 10.1.32 (which starts on page 456) does not consider $\nabla$ the Levi-Civita for a Riemannian metric. It is considering a general vector bundle $E$ equipped with a Hermitian metric $\langle,\...
Willie Wong's user avatar
  • 37.6k
4 votes

Vector bundles on $\mathbb{P}^1$

Consider the exact sequence, $0\to E(-1)\to E\to E_p\to 0$, where the first map is just multiplication by $x$. Then $E_p$ is just the skyscraper sheaf at $p=(0,1)$. Taking cohomologies you get $0\to \...
Mohan's user avatar
  • 6,117
4 votes

Derived flat bundles

Maybe not exactly what you are looking for, but perhaps of relevance are Flat super-connections. That let $E\to B$ be a $\mathbb{Z}_2$ graded vector bundle, and $A=d_0+d_1+d_2+\ldots$ be an odd ...
Ryan Mickler's user avatar
4 votes
Accepted

Self-intersection of zero section of line bundle over elliptic base curve

The result in question is a special case of the self-intersection formula (Fulton, intersection theory, p. 103) which states that the intersection of a smooth subvariety $X$ of a smooth variety $Y$ ...
Will Sawin's user avatar
  • 137k
4 votes

Singularities of fibrations in conics

It might be that $X$ is singular along some fiber that is a double line. To arrange this, start with $Y \rightarrow \mathbb{P}^1$ with such a fiber over the point $P$ and then pull back by a double ...
inkspot's user avatar
  • 3,052
4 votes

Tensor product of vector bundles

A rank-$n$ vector bundle over $X$ consists of a bunch of trivial vector bundles over open subsets $U_i$ covering $X$, patched together on the overlaps. On each overlap $U_{ij}=U_i\cap U_j$, there is ...
Steven Landsburg's user avatar
3 votes
Accepted

Orientation bundle and its flat connection

There is a different construction of orientation bundles. One considers the $\{\pm1\}$-principal bundle $o(TM)$ of fibrewise orientations of $TM$. The associated real line bundle $o(TM)\times_{\pm 1}\...
Sebastian Goette's user avatar
3 votes

Obstructions to the existence of a flat connection on a vector bundle

A slightly different point of view for answering this question is the following one: First, if $M$ is simply connected, then $E\to M$ admits a flat connection if and only if $E$ is trivial, so in this ...
3 votes
Accepted

Equivariant sheaves on $\mathbb P^1$

Let me explain why the line bundle $\mathcal{O}(1)$ does not admit a $\mathrm{PGL}(2)$-equivariant structure. Indeed, if it does, then the vector space $$ \mathrm{Hom}(\mathcal{O}, \mathcal{O}(1)) $$ ...
Sasha's user avatar
  • 37.2k
3 votes

Deriving the definition of vector bundle morphisms from Cartan geometry (a.k.a. why are they linear?)

One obvious fact: Cartan geometries are rigid, i.e. the automorphisms of a Cartan geometry, on a manifold with finitely many connected components, form a finite dimensional Lie group acting smoothly. ...
Ben McKay's user avatar
  • 25.6k
2 votes
Accepted

Factorization systems for vector bundles

Normally, we consider vector bundles over some base space $X$. The simplest case is if $X$ is a point. Then the category of vector bundles over $X$ is just the category of vector spaces (over whatever ...
David White's user avatar
  • 29.8k
2 votes
Accepted

Normal bundle of veronese as iteration extension of symmetric powers

Your third exact sequence is incorrect --- the correct form is $$ 0 \to S^{d-1}V \otimes \mathcal{O}(d-1) \to S^{d}V \otimes \mathcal{O}(d) \to S^dT \to 0. $$
Sasha's user avatar
  • 37.2k
1 vote

Integral mean value property

This is a comment but will be too long for that. The equation given is a (very simple) example of a difference-differential equation (in fact, a difference equation but there are several related ...
memorial's user avatar
  • 396
1 vote

Isomorphism between $\operatorname{End}_0(E)$ and $\operatorname{End}_0(E')$ as Lie algebra bundles

With this type of question, it is better to write to one of the authors directly. $\DeclareMathOperator\Fl{Fl}\DeclareMathOperator\End{End}$First, a vertical vector field is given by a global vector ...
Vicente Muñoz's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible