13
votes
Accepted
Possible mistake in Cohen notes "Immersions of manifolds and homotopy theory" (version 27 March 2022)
The statement is false in two ways. First, two immersions might not even be homotopic even though their normal bundles are both, say, trivial. Second, even if $M=\mathbb R^m$, regular homotopy classes ...
- 48.7k
12
votes
A tale of two maps into a Grassmannian
I guess, when you say "in the sense of Hartshorne" you mean the projective spectrum of $\oplus S^kE$.
Yes, the morphisms are the same, and to see this just note that there is a natural (...
- 32k
12
votes
Accepted
Embedding of a bundle with connection into a bundle with flat connection?
The paper “Existence of universal connections” by Narasimhan, M. S.; Ramanan, S. proves that the Grassmanian is universal for connections not just bundles. That is any connection in a U(n) or O(n) ...
- 2,894
11
votes
Accepted
Derivations on the continuous functions of a manifold
More is true: if $X$ is a topological manifold, then in fact $\operatorname{Der}(C(X)) = 0$, where $C(X)$ denotes the $\mathbb{R}$-algebra of $\mathbb{R}$-valued continuous functions on $X$. In ...
- 5,681
10
votes
Complex vector bundles on compact complex manifolds
This is an explanation of my comment above, namely: "Complex vector bundles over a CW complex of dimension $\leq 4$ are classified by their Chern classes and rank. Moreover, every possible choice ...
- 17.3k
10
votes
Accepted
Triviality of vector bundles on affine open subsets of affine space
For your final question, the answer is that all vector bundles over $U$ are trivial.
Sketch of proof: Let $R=k[x_1,\ldots,x_n]$.
Let $L_1, L_2,\ldots, L_m$ be the equations of hyperplanes in $R$. ...
- 20.9k
8
votes
For each $k$, is there a vector bundle $E$ such that $E\oplus\varepsilon^k$ is trivial but $E\oplus\varepsilon^{k-1}$ is not?
This is more of an extended comment on ways to approach this problem than an answer.
If we have a vector bundle $E$ on $X$ and have a trivialization $E ⊕ \epsilon^k \to \epsilon^{n+k}$, then this ...
- 48k
7
votes
Accepted
When are two resolutions of a coherent sheaf homotopic
If this were true, then any short exact sequence of vector bundles would split. Indeed, if $0 \to \mathscr E_1 \to \mathscr E_2 \to \mathscr E_3 \to 0$ is a short exact sequence of vector bundles, ...
- 20.4k
7
votes
When are two resolutions of a coherent sheaf homotopic
No. The simplest example is given by the following two resolutions of the structure sheaf of a point $P \in \mathbb{P}^1$:
$$
0 \to \mathcal{O}_{\mathbb{P}^1}(-1)
\to \mathcal{O}_{\mathbb{P}^1}
\to \...
- 32k
7
votes
Accepted
Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors?
The answer is no.
At least when $G$ and $H$ are semisimple, the quotient $G/H$ is diffeomorphic to the normal bundle of $K_G/K_H$ inside $G/H$ (where $K_G$ and $K_H$ denote respectively maximal ...
- 1,823
7
votes
Accepted
Square root of a line bundle up to a finite surjective morphism
Assume $\mathcal{L}$ is associated with an effective Cartier divisor $D$. Let $D'$ be another Cartier divisor such that $D + D'$ is divisible by 2 in $\mathrm{Pic}(X)$. Let
$$
g \colon X' \to X
$$
be ...
- 32k
7
votes
Accepted
Examples and properties of spaces with only trivial vector bundles
Let $B$ be a closed manifold with such that every vector bundle is trivial. Then $H^1(B; \mathbb{Z}_2) = 0$, otherwise there would be a non-trivial line bundle. Therefore every bundle over $B$ is ...
- 17.3k
6
votes
Accepted
Global choice of eigenvectors on an open surface
Not necessarily. To construct a counter-example, start from the other direction. Suppose that the tangent bundle of $M$ can be split as the direct sum $TM = L_1\oplus L_2$ where $L_1$ and $L_2$ are ...
- 98.4k
6
votes
Examples and properties of spaces with only trivial vector bundles
Here is one constraint, which seems relevant in light of Michael Albanese's answer:
Claim: Let $B$ be a closed orientable odd-dimensional manifold with no stably nontrivial complex vector bundles. ...
- 50.6k
6
votes
Accepted
Different definitions of "charged spinors": "bundle splicing" vs. "twisted spinor bundles"
I'll assume that the vector space "$V$" occuring in constructions (1) and (2) doesn't have to be the same. In that case I'll rename vector space in construction (2) to "$W$."
Then ...
- 1,680
6
votes
Accepted
Stable normal bundle and immersions
This follows from obstruction theory; also see this answer.
If $E \to X$ is a rank $r$ real vector bundle over a CW complex $X$, then the obstructions to finding a nowhere-zero section lie in $H^i(X; \...
- 17.3k
5
votes
Surfaces of general type such that $\operatorname{Sym}^n \Omega_X$ is globally generated (but $\Omega_X$ is not)
Let $E$ be an elliptic curve, $A$ an abelian surface, and $x\in A$ a point that's not 2-torsion.
Let $Y \subset A \times E$ be a surface of high degree, stable under inversion, and containing $x \...
- 118k
5
votes
Accepted
Is there a stable reduction for a family of vector bundles?
Consider the Harder-Narasimhan filtration of $E_0$. Assume for simplicity it has length 2:
$$
0 \to F_0 \to E_0 \to E_0/F_0 \to 0,
$$
where $F_0$ and $E_0/F_0$ are semistable and the slope of $F_0$ is ...
- 32k
5
votes
Unsplitting sequence of vector bundles
$\def\CC{\mathbb{C}}$A splitting would be a global map $f : G(k,n) \times \CC^n \to \CC^n$ such that $f(L,v) \in L$ for all $L \in G(k,n)$ and $v \in \CC^n$. But, since $G(k,n)$ is projective and ...
- 140k
4
votes
Accepted
Hermitian vector bundles and Hilbert $C^*$-modules
In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:
A construction of Kaplansky (Rings of operators, Thm 26) shows that if $\mathcal{A}$ is a $*$-...
- 7,509
4
votes
Accepted
Do we know anything about Harder-Narasimhan filtrations of tensor products of vector bundles?
It is a result of Narasimhan and Seshadri that if $V$ is semistable and $W$ is semistable then $V \otimes W$ is semistable.
If $E$ has a filtration with associated graded $E_i/ E_{i-1}$, and $F$ has a ...
- 118k
4
votes
Examples and properties of spaces with only trivial vector bundles
Here is another obstruction.
Suppose $M^n$ is a closed simply connected manifold which admits only trivial vector bundles. Then $M$ cannot be a $\mathbb{Z}/2\mathbb{Z}$-homology sphere, unless $n=3$....
- 5,756
4
votes
Accepted
Quantitative results for stabilizing tangent bundles of homology spheres
If $E \to X$ is a rank $r$ real vector bundle, then it is classified by a map $X \to BO(r)$. The existence of an isomorphism $E \cong E_0\oplus\underline{\mathbb{R}}$ (equivalently, the existence of a ...
- 17.3k
4
votes
Accepted
Tangent bundle of a compact two-dimensional manifold
I have an almost complete answer. I start with a summary of the different cases. In all the cases $M$ is assumed to be a closed 2-manifold.
$M$ is orientable. Then $TM$ is trivial if and only if $M$ ...
- 182
3
votes
Accepted
Identification of tangent spaces by parallel transport along geodesics
Ok, given the comments, what you are really asking for, is for a class of connected Riemannian manifolds for which the following construction (or the map $\Phi$) is a (smooth) trivialization of the ...
- 7,501
3
votes
Accepted
Semi-stability of $S^n\Omega_S$ with respect to $K_S$
Ciao Francesco!
The answer to your question is yes, and the work of Bogomolov you are looking for about semistability of the tangent space (for minimal surfaces of general type indeed, no need of ...
- 7,672
3
votes
Accepted
Embedding of a blow-up
The map $Z \to \mathbb{P}^1$ is a conic bundle, so to understand the vector bundle $\mathbb{E}$ it is enough to compute the pushforward of the anticanonical class.
Now, the anticanonical class of $Z$ ...
- 32k
3
votes
Sheafification of presheaf of trivial vector bundles is the stack of vector bundles
If $G$ is an affine groupe scheme over some base $S$, you can consider the groupoid $G\rightrightarrows S$. The corresponding prestack $[G\rightrightarrows S]^{pre}$ is (equivalent to) the prestack of ...
- 3,733
3
votes
Grassmannian of line subbundle of a stable rank 2 vector bundle on a smooth projective curve
For $g=2$,$d=0,$ this is studied in M.S. NARASIMHAN, S. RAMANAN: Moduli of vector bundles on a compact Riemann surface. Ann. of Math. 89, 19-51. It is shown that the divisor is linear equivalent to $2\...
- 6,335
3
votes
Hermitian vector bundles and Hilbert $C^*$-modules
Yes, every finitely generated Hilbert module comes from a hermitian complex vector bundle in this way. In fact more is true: arbitrary Hilbert modules over $C(X)$ correspond to continuous (in an ...
- 38.1k
Only top scored, non community-wiki answers of a minimum length are eligible