18
votes
Accepted
Splitting of tangent bundle
To expand on my comment above, suppose that $TS^{2k}\cong\xi\oplus\eta$ for some non-trivial vector bundles $\xi$ and $\eta$ over $S^{2k}$ of dimensions $m$ and $\ell$, respectively. Hence $0<m,\...
12
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 ...
9
votes
Accepted
Pairing of cotangent and tangent bundles
For (1), recall that if $R$ is a ring, then a derivation $D: R \to R$ satisfies the Leibniz rule, which by induction on $n$ implies that if $D^n$ denotes the $n$-fold iterate of $D$, then
$$D^n(fg) = \...
8
votes
Splitting of tangent bundle
I like to play around with projective spaces for these type of questions. The tangent bundles of real projective spaces have stiefel whitney class $(1+a)^{n+1}$. So for example for $n=4$, we have ...
5
votes
Accepted
Kähler metric on the projective space
Every Kähler manifold admits an infinite dimensional family of Kähler metrics, by adding a Kähler potential with small enough second derivatives. One can even do this in any open set, using compactly ...
4
votes
Accepted
tangent bundle of Hilbert schemes of points on a projective surface
The tangent bundle on $S^{[n]}$ is not quite isomorphic to $(T_S)^{[n]}$, rather by Theorem B of Stapleton's paper listed below there is an injection of the former into the latter.
Stapleton, David, ...
4
votes
Accepted
Tangent bundle for orthogonal and isotropic Grassmannians
The tangent bundle to the orthogonal Grassmannian fits into an exact sequence
$$
0 \to T_{\mathrm{OG}(k,V)} \to \mathcal{S}^\vee \otimes \mathcal{Q} \to S^2\mathcal{S}^\vee \to 0.
$$
Taking into ...
4
votes
Splitting of tangent bundle
A condition that prevents a vector bundle from splitting a Whitney sum
is that the projection of the associated sphere bundle induces a non-surjective map on some homotopy group. This is proved by L....
3
votes
Can cotangent bundles see exotic smooth structures?
The questions at hand are much easier to answer for open manifolds than for closed manifolds. Here is a large family of simple examples: There are uncountably many diffeomorphism types of exotic $\...
3
votes
Accepted
Normal bundle of veronese as iteration extension of symmetric powers
Your third exact sequence is incorrect --- the correct form is
$$
0 \to S^{d-1}V \otimes \mathcal{O}(d-1)
\to S^{d}V \otimes \mathcal{O}(d)
\to S^dT
\to 0.
$$
3
votes
Accepted
Natural extension homomorphism and wrong-way maps in K-theory
The natural extension homomorphism is defined in footnote 4 on page 7 of the expository paper by Gregory Landweber you are looking at.
In that reference, K-theory with compact supports is defined for ...
1
vote
Accepted
Derivative of the symplectomorphism evaluated at a point of the zero section of the cotangent bundle
On Question 1, $\phi$ can be defined on the whole of $T^*L$ so long as the exponential map can be defined on the normal bundle $NL$, which will be the case for example if $M$ is geodesically complete. ...
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