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 ...
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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 (...
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  • 32k
12 votes
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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) ...
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  • 2,894
11 votes
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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 ...
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  • 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 ...
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10 votes
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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 $R$. ...
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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 ...
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7 votes
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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, ...
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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 \...
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  • 32k
7 votes
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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 ...
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7 votes
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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 ...
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  • 32k
7 votes
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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 ...
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6 votes
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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 ...
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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. ...
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  • 50.6k
6 votes
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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 ...
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  • 1,680
6 votes
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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; \...
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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 \...
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  • 118k
5 votes
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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 ...
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  • 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 ...
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4 votes
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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 $*$-...
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4 votes
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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 ...
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  • 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 $n=3$....
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  • 5,756
4 votes
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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 ...
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4 votes
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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 $M$ ...
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3 votes
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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 ...
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  • 7,501
3 votes
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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 ...
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  • 7,672
3 votes
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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 $Z$ ...
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  • 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 ...
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  • 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\...
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  • 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 ...
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  • 38.1k

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