# Tag Info

## Hot answers tagged characteristic-classes

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### Is there a name given to a space whose first three Stiefel-Whitney classes vanish?

No there isn't because if $w_1 = 0$ and $w_2 = 0$, then $w_3 = 0$. More generally, the smallest positive $k$ such that $w_k \neq 0$ is always a power of two. This general fact follows from Wu's ...
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### Formula for the Stiefel-Whitney classes of a tensor product

I provided a solution to this exercise in my note here; I have copied the proof below. Note, this may not be the exact proof Milnor had in mind. Lemma: Let $L_1$ and $L_2$ be real line bundles over a ...
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### Chern classes, cohomology classes with real/integer coefficients

Chern classes can be defined in multiple different but roughly equivalent ways. As you have seen, they can be defined to be certain classes in $H^{2i}(X;\mathbb{Z})$, or they can be defined to be ...
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### Characteristic classes, Möbius strip, and the cylinder

You can use the fact that if you cut the open Möbius strip around center the resulting space is connected. Any homeomorphism from the Möbius strip to a cylinder will induce an isomorphism on ...
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### Characteristic classes, Möbius strip, and the cylinder

As an alternative to James's answer, you can look at the one-point compactifications. For the cylinder, you get a space homeomorphic to a sphere with two points identified, which has $\Bbb{Z}$ for its ...
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### Has anyone seen this combinatorial identity involving the Bernoulli and Stirling numbers?

The coefficients $B_j^{(r)}$ defined by$$\sum_{j = 0}^\infty B_j^{(r)} {{x^j}\over{j!}} = \left({x\over{e^x - 1}}\right)^r$$are usually called higher order Bernoulli numbers, so your identity is a ...
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### Is Hatcher's proof of thom isomorphism theorem flawed?: I don't believe that $H^n(E,E_0)\cong H^n(R^n,R^n-0)$

Your mistake is in the equality $\tilde{H}^2(S^2 \times (D^2, S^1)) = \tilde{H}^2(S^2 \times S^2)$ you wrote. It might be a bit easier to see why this is false in one dimension lower. Consider the ...
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### Chern classes of tangent bundle over the Grassmannian G(2,4)

Let $E$ and $F$ be complex vector bundles over $B$ of ranks $r$ and $s$ respectively (or locally free sheaves if you prefer). One can compute the Chern classes of the tensor product $E\otimes F$ in ...
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### Real $2n$-plane bundle with a complex structure is a complex $n$-plane bundle

Not every trivialisation is fiberwise complex-linear, however, there is a trivialisation which is. This is the bundle version of the statement that for an $n$-dimensional complex vector space $V$, not ...
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### Allendoerfer and Weil's generalization of Gauss-Bonnet Theorem

Every Riemannian manifold $(X,g)$ can be isometrically embedded into $\mathbb{R}^N$ for $N$ sufficiently large. So in principle, when one wants to prove something about Riemannian manifolds, it ...
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### Motivating Characteristic Classes Using $S^2$

While $S^2$ is perhaps the most convenient space for thinking about actual topology, some special features of $S^2$ seem to have thrown you for a loop. In particular, you're confusing the so-called ...
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### Book on characteristic classes

The following is a celebrated classic. J. Milnor is a Fields medalist, famous for the power of his mathematical thinking and the clarity and precision of his style. Milnor, John W.; Stasheff, James D....
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### Book on characteristic classes

For someone coming from the complex geometry perspective, I would suggest reading some combination of Chern (Complex Manifolds without Potential Theory), Wells (Differential Analysis on Complex ...
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### How does one introduce characteristic classes

I was always confused by characteristic classes until I understood the definition of characteristic classes via the classifying map. Corresponding to a vector bundle with structure group $G$ there ...
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### Splitting of the tangent bundle and Euler characteristic of surfaces

To complement Connor's answer: Classifing spaces argument: Because $M$ is orientable, its tangent bundle is classified by a map $\phi:M\rightarrow BSO(2)=\mathbb{C}P^\infty$. The Euler class of the ...
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### Is there a description of the Grassmannian as a homogeneous space where the principal bundle is one associated to the universal vector bundle?

Yes there is. \begin{align*} \operatorname{Gr}_{n,k}(\mathbb{R}) &= O(n)/(O(k)\times O(n-k))\\ \operatorname{Gr}_{n,k}(\mathbb{C}) &= U(n)/(U(k)\times U(n-k))\\ \operatorname{Gr}_{n,k}(\...
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1) The second Chern class is an invariant associated to a complex vector bundle $E\rightarrow M$ over a manifold (spacetime) $M$ (or, equivalently, to a principal $U(n)$-bundle over $M$). It is for ...
A choice of a local orientation of $\Bbb R^n$ at the origin is equivalent to choosing a vector space orientation of $T_0\Bbb R^n\simeq \Bbb R^n$ which is in turn equivalent to choosing a basis \$(\...