# 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 ...
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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" A_i = \frac{1-\hat n_z}{\hat n_x^2 + \hat n_y^2}(\hat n_x \partial_i ...
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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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### 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, ...
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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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### 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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### 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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### 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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### 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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### 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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• 55.7k

### 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" ...
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### 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 ...
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### rational cohomology of finite real grassmannian

I could not find the explicit formulas in the Algebraic models book (they seem to only do infinite Grassmannians and Stiefel varieties) or Mimura-Toda (they do the complex and symplectic case but not ...
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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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### Steenrod powers of the Thom class

I don't know a reference, but you can proceed as follows. By the splitting principle, it suffices to give the formula for vector bundles which are sums of complex line bundles, and we may as well then ...
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### Vector bundle over an oriented manifold with non-vanishing w_2w_3

As far as I know the Wu manifold $X=SU(3)/SO(3)$ is orientable and has mod 2 cohomology ring $H^*(X;\mathbb{Z}_2)=\Lambda(\omega_2(X),\omega_3(X))$. Thus $\omega_2(X)\cdot\omega_3(X)\neq 0$, and in ...
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### What is the first Pontryagin class of the $n$-dimensional representation of $S_n$?
If I am not confused, $p_1(V)=-c_2(V\otimes {\mathbb C})$. According to Theorem 7.1 in the book Characteristic Classes and the Cohomology of Finite Groups by Charles Thomas, $c_2$ of the standard ...