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The study of differentiable manifolds and differentiable maps. One fundamental problem is that of classifying manifolds up to diffeomorphism. Differential topology is what Poincaré understood as topology or “analysis situs”.
2
votes
Existence of a spin map from a standard sphere to any closed Riemaninan manifold with nonneg...
For now, let $M$ be any closed oriented $n$-dimensional manifold, and $f : S^m \to M$ a smooth map.
First suppose $k \geq 1$.
Since $m = n + 4k > 2$, we have $H^2(S^m; \mathbb{Z}_2) = 0$, and hence $f …
11
votes
A question about the existence of spin maps
If $M$ and $N$ are spin, then every map between them is a spin map. In particular, there exist spin maps $M \to N$ of degree zero.
If $M$ is spin and $N$ is not spin, then $f : M \to N$ is a spin map …
6
votes
Accepted
Does composition on the right by a volume-preserving diffeomorphism preserve homotopy class?
Let $X$ be a smooth, compact, orientable manifold and let $\omega$ be a choice of volume form. On $X\times X$, we have the natural volume form $\sigma = \pi_1^*\omega \wedge \pi_2^*\omega$ where $\pi_ …
7
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; …
18
votes
Accepted
Converse to Hopf degree theorem
See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property.
Note that $X$ having the Hopf property is equivalent to the injectivity of $\d …
10
votes
Accepted
The maximum number of vertical independent vector fields on the tangent bundle
I will address the first version of your question (i.e. no conditions on commuting flows).
A vector bundle $E \to B$ admits $k$ linearly independent vector fields if and only if $E$ has a subbundle is …
5
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 …
4
votes
Atiyah-Bott-Shapiro generalization to $U(n) \to ({Spin(2n) \times U(1)})/{\mathbf{Z}/4}$ for...
Let $\omega = e_1e_2\dots e_{2n-1}e_{2n}$.
For $n > 1$, the center of $Spin(2n)$ is $Z(Spin(2n)) = \{\pm 1, \pm\omega\}$. Note that $\omega^2 = (-1)^n$, so
$$Z(Spin(2n)) = \begin{cases}
\langle -1, \ …
32
votes
Accepted
If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundl...
The result you are hoping for is in fact false.
In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent b …
8
votes
Every _______ $d$-manifold has an $S$-structure
In this paper, Aleksandar Milivojevic and I prove that every orientable manifold of dimension $\leq 7$ is spin$^h$. We also construct, for every $d \geq 8$, infinitely many homotopy types of closed, s …
3
votes
A topological consequence of Riemann-Roch in the almost complex case
I just wanted to point out how this question is related to (spin${}^c$) Dirac operators and their indicies since this was alluded to in the comments to the question.
Let $(M, g)$ be an $2n$-dimension …
23
votes
If $M$ and $N$ are closed and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true th...
The manifolds $M$ and $N$ may not even be homotopy equivalent!
In Compact Flat Riemannian Manifolds: I, Charlap showed that there are two closed flat manifolds $M$ and $N$ of the same dimension which …
9
votes
Accepted
To what extent is a vector bundle on a manifold with boundary determined by its restriction ...
As I indicated in my comment, the inclusion $\iota : M_0 \to M$ is a homotopy equivalence. This can be shown using the fact that the boundary $\partial M$ has a collar neighbourhood; it then boils dow …
9
votes
Accepted
Index of Dirac operator and Chern character of symmetric product twisting bundle
Your first question can be answered by using the splitting principle.
If $V \to X$ is a complex vector bundle of rank two, then $c_1(S^3V) = 6c_1(V)$ and $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$.
Pr …
4
votes
Accepted
Under what condition is a fiber bundle cobordant to the trivial bundle?
Here are three special cases where it is always true:
Sphere bundles of vector bundles. If $E \to B$ is a vector bundle and $S(E) \to B$ is the associated sphere bundle, then $S(E)$ bounds the disc b …