# Tag Info

### 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 ...

### 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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### 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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### 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, ...

### 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 ...

### 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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### 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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### 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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### 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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### 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 ...

### 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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### 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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### 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 ...

### 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 ...