## Hot answers tagged vector-bundles

73

A much more general result is true.
Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma $ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$.
It says that exotic spheres cannot be distinguished by looking at the tangent bundle.
This result is one of the hidden gems of the golden age of topology and the proof invokes the whole ...

39

These answers look at bit complicated so maybe there is something obviously wrong with the following argument:
Every embedded two-sphere $\Sigma \subset S^2 \times {\mathbb R}^2$ is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from ...

35

This is more or less equivalent to Ryan's comment but with more details and a slightly different point of view.
Let $X$ be the total space of the tangent bundle, and put $Y=S^2\times\mathbb{R}^2$. If $X$ and $Y$ were homeomorphic, then their one-point compactifications would also be homeomorphic. We will show that this is impossible by considering their ...

33

You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N_{D/X}=\mathcal O_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O_D$. And the Todd class pops out right away.
Indeed from the exact sequence $0\to ...

31

The tangent bundle to a smooth structure on $S^7$ is classified by a map $S^7 \to G_7(R^{\infty})$. By the exact sequence for a fibration for the fiber bundle $O(7)\to V_7(R^\infty)\to G_7(R^\infty)$, we see that $\pi_7(G_7(R^\infty)) = \pi_6(O(7))$. But $\pi_6(O(7))=0$ (I found a table A1.1.3.2 of homotopy groups of orthogonal groups here(pdf), since this ...

30

Here's another way to answer the original question. There is a theorem of Bredon and Kosinski (Annals, 1966) which says that if a manifold $M^n$ is stably parallelizable, then either $M^n$ is parallelizable or the maximum number of linearly independent vector fields on $M^n$ is the same as on $S^n$. Since $S^7$ is parallelizable, this implies that exotic ...

29

Neither is more canonical than the other. The tangent bundle of $M$ represents the set of all possible derivatives of maps $R \rightarrow M$, and the cotangent bundle of $M$ represents the set of all possible derivatives of maps $M \rightarrow R$. They are dual to each other.
I hate to ruin such a nice terse answer, but I might as well describe how I think ...

27

If you want to differentiate functions from a manifold to (say) the real line R, then you want to use the cotangent bundle on the manifold.
If instead you want to to differentiate functions to the manifold from the real line (i.e. parameterised curves), then you want to use the tangent bundle on the manifold.
So the preference comes from whether you want ...

27

I don't think I can really give you the intuition that you seek because I don't think I quite have it yet either. But I think that understanding the relevance of Nigel Higson's comment might help, and I can try to provide some insight. (Full disclosure: most of my understanding of these matters has been heavily influenced by Nigel Higson and John Roe).
My ...

26

I shall show that the answer is no when $p=2$ (and it seems to me that a
somewhat more involved calculation will work for any $p$). We shall show that
there exists a vector bundle $\mathcal E$ such that $S^2\mathcal E$ is not
isomorphic to $\Gamma^2\mathcal E$ ($=(S^2\mathcal E^\ast)^\ast$).
Consider a vector bundle $\mathcal E$ which is an extension
...

25

If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.
This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial.
In ...

23

As Igor Belegradek showed in the comments, one could find an example by finding a CW-complex $X$ and a map $X \to BO(n)$ which is not nullhomotopic, but where the restriction to every finite subcomplex is nullhomotopic. Such a map is called a phantom map. The question "is this map nullhomotopic?" has the same answer whether or not we are asking our maps to ...

22

A few of the more obvious ones:
* Resolution of singularities in characteristic p
*Hodge conjecture
* Standard conjectures on algebraic cycles (though these are not so urgent since Deligne proved the Weil conjectures).
*Proving finite generation of the canonical ring for general type used to be open though I think it was recently solved; I'm not sure about ...

22

Dear Ila, the linear algebra result you mention is due to Dedekind-Weber and was published in Crelle's Journal dated 1882, in their article "Theorie der algebraischen Funktionen einer VerĂ¤nderlichen". Their motivation was proving Riemann-Roch on an arbitrary smooth projective curve $X$ by presenting the curve as a ramified covering of $\mathbb P^1$ and ...

20

This is true if $X$ satisfies Serre's condition $S_2$, i.e. $\mathcal O_X$ is $S_2$. Then a vector bundle is $S_2$ since locally it is isomorphic to $\mathcal O_X^n$.
More generally, a coherent sheaf $F$ on a Japanese scheme (for example: $X$ is of finite type over a field) which is $S_2$ has a unique extension from an open subset $U$ with ...

20

I'll use the definition of stack as a (weak) functor from the category of schemes to that of groupoids (as opposed to the definition as a fibered category over the category of schemes).
The prestack associated to the action of $GL_n$ on $\mathbb A_n$ is, by definition, given by
$$
X \mapsto \left\{\begin{matrix}\text{Objects: maps $s:X \to \mathbb ...

19

The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example ...

19

If your space is a manifold, knowing the vector bundles over that space amounts to knowing all of its tubular neighbourhoods when you embed the space in another manifold. This frequently allows you to find many relationships between the two manifolds.
One classical application would be the proof that all smooth embeddings $S^n \to S^{n+2}$ (co-dimension two ...

19

The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory of vector bundles on $\mathbb{P}^2$ (that is, very complicated).
Here is a simple example of an indecomposable rank 2 bundle. By the KĂ¼nneth formula, we see ...

18

I would say that the importance of the classification of vector bundles comes first from the fact that it leads naturally to the "characteristic classes" and their complete description. Characteristic classes are computable and powerful invariants of vector bundles.
Look at the book Milnor-Stasheff "Characteristic classes" which is the wonderful classic in ...

18

Let me mention a couple of problems related to vector bundles on projective spaces.
The Hartshorne conjecture. In its weak form it says that any rank 2 vector bundle on $\mathbf{P}^n_{\mathbf{C}},n>6$ is a direct sum of line bundles, which implies that any codimension 2 smooth subvariety whose canonical class is a multiple of the hyperplane sectionis a ...

18

Let $U$ and $V$ be two copies of the real line and make a space $X$ by gluing them by the identity along the strictly positive half-line: $x\in U$ equals $x\in V$ for $x>0$. Now make a rank one vector bundle over this space by taking a trivial bundle over each of the lines: glue $U\times\mathbb R$ to $V\times \mathbb R$ by identifying $(x,y)\in ...

18

Well, in algebraic geometry, here's a couple of reasons:
1) Subvarieties: Take a vector bundle, look at a section, where is it zero? Lots of subvarieties show up this way (not all, see this question) but generally, we can get lots of information out of vector bundles regarding subvarieties.
2) Invariants of spaces: The Picard group of Line bundles and more ...

18

Any curve of large enough degree will do. Set $F:= E'\otimes E^{\vee}$; if $d$ is a very large integer, then $\mathrm H^1(F(-d)) = 0$. Take any curve $C$ of degree $d$, and suppose that $E\mid_C$ and $E'\mid_C$ are isomorphic; this isomorphism is given by a section of $F\mid_C$. Since $\mathrm H^1(F(-d)) = 0$, this section extends to a global section of $F$, ...

17

This may be overkill, but to elaborate on Ryan's answer in another way:
Without mentioning either boundaries or any other compactifications, we can define the intersection number of $x\in H_p$ and $y\in H_q$ for homology classes in an oriented $(p+q)$-manifold. First turn them into compactly supported cohomology classes by duality, then cup these to get ...

17

To your first question, "function on a space" $X$ usually means a morphism from $X$ to one of several "ground spaces" of choice, for example the reals if you work with smooth manifolds, Spec(A) if you work with schemes over a ring, etc. (This is a fairly selective use of the word "function" which used to confuse me.) A section $\gamma$ of a (some-kind-of) ...

17

The answer is positive. Let $P$ be the principal $\mathrm{GL}_n$-bundle associated with $E$; then the space of flags is the quotient $P/B$, where $B$ is the Borel subgroup of $\mathrm{GL}_n$ consisting of upper triangular matrices. Set $Z = P/T$, where $T$ is the maximal torus consisting of diagonal matrices. A point of $Z$ is a point of $X$, plus $n$ ...

16

Yes, it is true that (over an algebraically closed field) the only positive-dimensional smooth projective variety on which every algebraic vector bundle splits as a sum of line bundles is $\mathbb P^1$.
More generally Ballico has proved that if every algebraic vector bundle splits as a sum of line bundles on a reduced but maybe reducible ...

16

The answer for projective spaces is negative. I think the simplest example are 2-bundles on $\mathbb{P}^3(\mathbb{C})$. In that case the Schwarzenberger condition is that $c_1c_2$ should be even. Atiyah and Rees have proved that for any pair $(c_1,c_2)$ satisfying this there are holomorphic vector bundles $\xi$ with $c_1(\xi)=c_1,c_2(\xi)=c_2$ (see Atiyah, ...

16

Many (compact orientable) hyperbolic 3-manifolds have non-trivial $SU(2)$ representations.
By Mostow rigidity, the representation of the fundamental group $\Gamma$ of a closed hyperbolic 3-manifold into $SL(2,\mathbb{C})$ (lifted from $PSL(2,\mathbb{C})$) may be conjugated so that it lies in $SL(2,K)$, for $K$ a number field (because transcendental ...

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