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 ...
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20 votes

Unifying Geometry for Characteristic Classes

I would compare at least 2 & 3, if not 4, using maps into classifying spaces. For a space $X$ homotopic to a finite CW-complex (at least), the pullback map $Map(X,Gr_n(\mathbb C^\infty)) \to \{$...
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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 ...
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  • 7,281
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, ...
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  • 33.4k
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 ...
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  • 29.2k
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 ...
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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 ...
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17 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 ...
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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 ...
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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 ...
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  • 40.8k
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 ...
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  • 19.6k
15 votes
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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}...
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15 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 ...
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  • 7,281
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 ...
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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 $...
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14 votes
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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 $w_2$ ...
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  • 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 ...
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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^...
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  • 29.2k
13 votes
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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 ...
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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 ...
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  • 7,672
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{...
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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 ...
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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 ...
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  • 7,901
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 $(\...
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12 votes
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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 ...
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  • 29.2k
12 votes
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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 $W(M) = 1 + w_1(M) + w_1(M)^2 +... .$ ...
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12 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-...
12 votes
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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 ...
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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 $n >4$, ...
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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 ...
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