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 ...
Michael Albanese's user avatar
27 votes
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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 ...
Michał Jan's user avatar
19 votes
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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 ...
Michael Albanese's user avatar
18 votes
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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 ...
John Rognes's user avatar
  • 8,692
18 votes
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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, ...
Mark Grant's user avatar
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 ...
Peter May's user avatar
  • 30.2k
17 votes
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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 ...
John Rognes's user avatar
  • 8,692
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 ...
Qiaochu Yuan's user avatar
17 votes
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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 ...
Johannes Nordström's user avatar
16 votes
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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 ...
Qiaochu Yuan's user avatar
16 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 ...
Oscar Randal-Williams's user avatar
16 votes
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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 ...
Ryan Budney's user avatar
  • 42.9k
14 votes
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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 $...
Michael Albanese's user avatar
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 $w_2$ ...
user80296's user avatar
  • 751
14 votes
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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 ...
Matthias Wendt's user avatar
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^...
Peter May's user avatar
  • 30.2k
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
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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 ...
John Baez's user avatar
  • 21.3k
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 ...
diverietti's user avatar
  • 7,852
13 votes
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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{...
Michael Albanese's user avatar
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 ...
Igor Belegradek's user avatar
13 votes
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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 ...
mme's user avatar
  • 9,263
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 $(\...
Neil Strickland's user avatar
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" ...
Tintin's user avatar
  • 2,741
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 ...
Qiaochu Yuan's user avatar
12 votes
Accepted

Vector bundles with exactly one nonzero SW-class

For $i = 2^k$ you can get an example with $X = \Bbb{RP}^{m}$ for any $m \geq 2^k$. If $L$ is the canonical line bundle on $\Bbb{RP}^{m}$ then let $E = \bigoplus_{i=1}^{2^k} L$ be the a sum of several ...
Tyler Lawson's user avatar
  • 51.1k
12 votes
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Generalize Wu formula to integral cohomology classes

I do not think such a cohomology operation can exist, as it would descend to a rational cohomology class in $H^{n+4}(K(\mathbb Z,n);\mathbb Q)$. But for any odd $n$ and also for all even $n >4$, ...
Jens Reinhold's user avatar
12 votes
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Oriented Bordism Group and Un-Oriented Bordism Group of points $pt$

Unoriented cobordism: can be read off from the structure of the unoriented cobordism ring (calculated in Thom's thesis): $\Omega_6^O = (\mathbb Z/2)^3$, $\Omega_7^O = \mathbb Z/2$, $\Omega_8^O = (\...
Arun Debray's user avatar
  • 6,766
12 votes

Mathematical/Physical uses of $SO(8)$ and Spin(8) triality

Seiberg and Witten showed that the $\mathcal{N}=2$ supersymmetric SU(2) gauge theory with $N_f=4$ flavor is endowed with SO(8) flavor symmetry, and it enjoys SO(8) triality. Later, Gaiotto's ...
Satoshi  Nawata's user avatar
12 votes

Chern classes of generators of $K(S^{2n})$

Note that $K(S^{2n}) = \mathbb{Z}\times\mathbb{Z}$ has many different sets of generators and these can have different Chern classes, so the question doesn't have a well-defined answer. For one ...
Michael Albanese's user avatar

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