30
votes
What are the possible Stiefel-Whitney numbers of a five-manifold?
Recall that on a closed $n$-manifold $M$, there is a unique class $\nu_k$ such that $\operatorname{Sq}^k(x) = \nu_kx$ for all $x \in H^{n-k}(M; \mathbb{Z}_2)$; this is called the $k^{\text{th}}$ Wu ...
- 18.2k
25
votes
Accepted
A difficult integral for the Chern number
If Stokes' theorem counts as a standard technique, then here's an answer:
Introduce a "vector potential"
\begin{equation}
A_i = \frac{1-\hat n_z}{\hat n_x^2 + \hat n_y^2}(\hat n_x \partial_i ...
- 556
19
votes
Accepted
Is a 4-dimensional submanifold of a spin manifold always spin?
Let $i$ denote an immersion $N \to M$. There is an exact sequence of vector bundles on $N$ given by
$$0 \to TN \to i^*TM \to \nu \to 0$$
where $\nu$ is the normal bundle. As total Stiefel-Whitney ...
- 18.2k
18
votes
Accepted
Is the cohomology ring $H^*(BG,\mathbb{Z})$ generated by Euler classes?
Let $G = \mathbb{Z}/p \times \mathbb{Z}/p$ for $p$ odd. Then $H^3(BG; \mathbb{Z}) \cong \mathbb{Z}/p$ is not in the subring generated by Euler classes, since the non-trivial irreducible ...
- 7,984
18
votes
Accepted
Wu formula for manifolds with boundary
A relative Wu formula for manifolds with boundary is discussed in Section 7 of
Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.
In particular, ...
- 34.7k
17
votes
Intuition/idea behind a proof of the splitting principle?
Perhaps my very short (4 pages plus bibliography) paper ``A note on the splitting principle'' http://www.math.uchicago.edu/~may/PAPERS/Split.pdf
may be illuminating. It shows that the splitting ...
- 29.7k
17
votes
Accepted
Who discovered this definition of Stiefel-Whitney classes?
For the relation $Sq(U) = \Phi(w)$, where $Sq$ is the total Steenrod squaring operation, $U$ is the Thom class, $\Phi$ is the Thom isomorphism and $w$ is the total Stiefel-Whitney class, I would cite ...
- 7,984
17
votes
Betti numbers as characteristic numbers?
No. The Stiefel-Whitney and Pontryagin numbers of a closed oriented manifold are cobordism invariants, but the Betti numbers are not.
More explicitly, all closed oriented $3$-manifolds are frameable ...
- 113k
17
votes
Accepted
A 4-manifold with Stiefel-Whitney classes $w_1\neq 0$, $w_3 \neq 0$, and $w_1^2=0$?
Let's begin by reformulating the question a bit. Note that any orientable 4-manifold is spin-c, so in particular has $w_3 = 0$. The condition that $w_1 \not= 0 $ is thus redundant.
Wu's theorem gives ...
- 1,886
16
votes
Accepted
Nice things that can be proved easily with characteristic classes
In this blog post you'll find a computation of the cohomology ring of a hypersurface of degree $d$ in $\mathbb{CP}^3$ using characteristic classes. This turns out to be a weirdly good exercise in ...
- 113k
16
votes
Accepted
A Compact Manifold with odd Euler characteristic whose tangent bundle admits a field of lines
I believe there is no example satisfying all your constraints. If I recall (my memory is a little foggy on this) the result likely goes back to Hopf, and one of his variations on the Poincare-Hopf ...
- 42.1k
15
votes
Four-dimensional vector bundles over $S^4$, intuition?
It probably helps if you notice that the unit quaternions are isomorphic to $SU(2)$, and the unit quaternions act as rotations on the quaternions by both left and right multiplication. So, we get a ...
- 21k
15
votes
Accepted
Whitney sum formula for Pontryagin classes I
I do not see why the difference $c_1(E_\mathbb{C})c_1(F_\mathbb{C})$ between these two expressions is necessarily 2-torsion. What am I missing?
Real line bundles are classified by $H^1(X, \mathbb{Z}...
- 113k
15
votes
Pin$^+$ and Pin$^−$ structure for manifolds in any dimensions
The condition for having a $Pin^+$-structure is the vanishing of $w_2$, and for having a $Pin^-$-structure is the vanishing of $w_2 +w_1^2$, for manifolds of any dimension. This is because the Lie ...
- 17.5k
14
votes
Accepted
Infinite Grassmannian does not have the homotopy type of a finite-dimensional complex
We have $H^*(BO(k); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_1, \dots, w_k]$ where $\deg w_i = i$. In particular, $H^n(BO(k); \mathbb{Z}_2) \neq 0$ for every $n$ as $w_1^n$ is a non-zero element. Therefore $...
- 18.2k
14
votes
Accepted
Second Stiefel-Whitney class is a square
At least there are quite a lot of such manifolds: up to multiplying by powers of 2, any oriented bordism class contains such a manifold.
Proof: Let $f: X \to BSO$ be the universal map such that $w_2$ ...
- 751
14
votes
Accepted
Fourth obstruction, Pontryagin and Euler class
Geometric generators for $\pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces ...
- 17k
14
votes
Analogy between Stiefel-Whitney and Chern classes
Here is one way I like to think of the analogy.
The maximal torus of diagonal matrices $T^{n} \subset U(n)$ gives a map $BT^n \to BU(n)$ which on integral cohomology gives an isomorphism from $H^...
- 29.7k
13
votes
Accepted
characteristic classes of homotopy equivalent manifolds
Wu's formula for the Stiefel-Whitney classes implies that they are invariants of homotopy type. See for example here.
Chern classes are not even diffeomorphism invariant, and it is possible to have ...
- 1,673
13
votes
A dictionary of Characteristic classes and obstructions
The following classes are of a slightly different flavour because they depend on the additional choice of a connection.
Assume that $E\to B$ carries a flat connection $\nabla$. Then the Kamber-...
13
votes
Chern class on a symplectic manifold
The answer is no.
First of all, if you want $f\omega$ to define a cohomology class, you should ask that $d(f\omega)=0$, and this is equivalent to $df\wedge\omega=0$, since $d\omega=0$. Using Darboux ...
- 7,782
13
votes
Accepted
Is there a closed 5-manifold $M$ with $w_1(M)w_2(M)\ne 0$?
Note that the third Wu class is $\nu_3 = w_1w_2$, so on a closed connected smooth $n$-manifold $M$, $\operatorname{Sq}^3 : H^{n-3}(M; \mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$ is given by $\operatorname{...
- 18.2k
13
votes
Why is the first integral Pontryagin class a homeomorphism invariant?
For spin manifolds this is proved in Corollary 1.22 (p.17) of Kammeyer's Diploma. Kreck's claim that the "spin" assumption can be dropped is mentioned after the corollary, with the caveat that "the ...
- 28.1k
13
votes
Accepted
Conversion formula between "generalized" Stiefel-Whitney class of real vector bundles: O(n) and SO(n)
First I will write up what your question is asking in terms of Arun Debray's comment. I strongly suggest that when discussing questions like this, you use precise notation as in the following; I found ...
- 8,988
13
votes
Why do Chern classes and Stiefel-Whitney classes satisfy the "same" Whitney sum formula?
I prefer this approach which I believe is due to Grothendieck. (I haven't checked how this compares with the sources cited by Nick Kuhn.)
Let $(\mathbb{K},R,d)$ be $(\mathbb{R},\mathbb{Z}/2,1)$ or $(\...
- 53.9k
13
votes
What is the Todd class *really*?
To my knowledge, currently the best "motivation" for the Todd class comes from the so called "orientation theory" and the formal group laws associated to "oriented" ...
- 2,711
13
votes
Computation on characteristic classes
I wrote a blog post that turned into quite a nice exercise in characteristic classes. The goal was to compute the cohomology of a smooth hypersurface of degree $d$ in $\mathbb{CP}^3$, as a ring. This ...
- 113k
12
votes
Accepted
Examples of Stiefel-Whitney classes of manifolds
It goes back to Wu in the 1950's that if one can compute the mod 2 cohomology of a manifold, with its Steenrod operations, then one can explicitly compute its Stiefel-Whitney classes,
via the Wu ...
- 29.7k
12
votes
Accepted
Immersing spaces in $\mathbb{R}^{n+1}$, Stiefel-Whitney classes
Read Milnor Stasheff: Characteristic classes.
For the first question:
The total normal Stiefel Whitney class of $M$ is $1-w_1(M)$, hence the inverse is
$W(M) = 1 + w_1(M) + w_1(M)^2 +... .$
...
- 2,309
12
votes
Accepted
Differential characters, Chern-Simons forms, and differential cohomology
The simple beginning of this story is that the curvature of a $\mathrm{U}(1)$ connection does not tell you the bundle it's a connection on — not even up to isomorphism. Differential cohomology ...
- 21k
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