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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”.
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; …
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 …
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 …
29
votes
1
answer
889
views
Is there an explicit description of a cobordism between $\mathbb{CP}^n$ and $\mathbb{RP}^n\t...
With a little bit of work, one can show that $\mathbb{CP}^n$ and $\mathbb{RP}^n\times\mathbb{RP}^n$ have the same Stiefel-Whitney numbers, so by a theorem of Thom, they are (unorientedly) cobordant.
…
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_ …
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 …
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 …
15
votes
1
answer
823
views
Are there non-smoothable homotopy/homology spheres?
A homotopy sphere is a topological $n$-manifold $M$ which is homotopy equivalent to $S^n$.
A homology sphere is a topological $n$-manifold $M$ such that $H_i(M) \cong H_i(S^n)$ for all $i$.
Note, by …
9
votes
Realization problem for Betti numbers
Suppose we are given non-negative integers $b_0, b_1, \dots, b_n$ with $b_k = b_{n-k}$. Is there a closed orientable manifold $M$ with $b_i(M) = b_i$? First we need $b_0 = b_n \geq 1$. It is enough to …
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 …
21
votes
1
answer
1k
views
Are homology spheres stably parallelisable?
A homology sphere is a closed smooth $n$-dimensional manifold with the same homology groups as $S^n$. Igor Belegradek's answer to a previous question of mine shows that the smoothness hypothesis is no …
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 …
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 …
3
votes
Topology of the blowup of a surface at a point (connected sum)
A much stronger statement is true. The following is from Huybrechts' Complex Geometry (note however that the emphasis is mine):
Let $x \in X$ be a point in a complex manifold $X$. Then the blow-up …
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 …