13 votes
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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 ...
Michael Albanese's user avatar
13 votes
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Kirby–Siebenmann class and the 4th Stiefel-Whitney class: $ \operatorname {ks} (M)$ v.s. $w_4(M)$

Initially, Stiefel-Whitney classes are only defined for smooth manifolds (if the manifold is not smooth, it doesn't have a tangent bundle). However, Wu's theorem states that $w = \operatorname{Sq}(\nu)...
Michael Albanese's user avatar
12 votes
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Is it possible to find a complex vector bundle on $S^2$ with nonzero first Chern class, which is trivial as a real vector bundle?

First of all, $c_1(E) \neq 0$ does not imply $w_2(E) \neq 0$. For example, thinking of $S^2$ as $\mathbb{CP}^1$, $TS^2$ is a complex vector bundle and $c_1(TS^2) = e(TS^2) \neq 0$ (in fact, $\langle e(...
Michael Albanese's user avatar
9 votes
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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 ...
Michael Albanese's user avatar
9 votes
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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 ...
Eric Wofsey's user avatar
9 votes
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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 ...
J Cameron's user avatar
  • 1,434
9 votes

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 ...
Rob Arthan's user avatar
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8 votes
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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 ...
Tony Blair's Witch Project's user avatar
8 votes
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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 ...
Najib Idrissi's user avatar
8 votes

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 ...
Michael Albanese's user avatar
8 votes
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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 ...
Michael Albanese's user avatar
8 votes
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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 ...
Quaere Verum's user avatar
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7 votes

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 ...
Thurmond's user avatar
  • 1,103
7 votes

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....
Francesco Polizzi's user avatar
7 votes

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 ...
Ted Shifrin's user avatar
7 votes

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 ...
Thomas Rot's user avatar
  • 9,983
7 votes

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 ...
Jason DeVito - on hiatus's user avatar
7 votes
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Pulling Back product of Characteristic Classes

Recall that for $x \in H^i(B_1)$ and $y \in H^j(B_2)$, the external product $a\times b \in H^{i+j}(B_1\times B_2)$ is defined as $p_1^*x\cup p_2^*y$ where $p_1 : B_1\times B_2 \to B_1$ and $p_2 : B_1\...
Michael Albanese's user avatar
7 votes
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What is the action of Steenrod squares on $BSO(n)$?

Surely Proposition 2 must be correct, as the argument is simplest and hardest to doubt. And indeed it is. One must be suspicious of a spectral sequence argument that gives you something on the nose. ...
user986876's user avatar
7 votes
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Classification of sphere bundles and vector bundles over a surface

The questions are not obviously equivalent, sphere bundles (whether smooth or topological) do not necessarily come from vector bundles. The first counterexample to this appears in dimension 4 thanks ...
Connor Malin's user avatar
  • 11.6k
6 votes
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Intuition behind the Thom Isomorphism.

$u$ is the Thom class, it is Kronecker dual to those classes "orthogonal" to $E_0$. That is, think of those simplices $\nu$ which intersect $E_0$ in one point, then $u(\nu)=1$. Now $E_0$ is ...
Rene Schipperus's user avatar
6 votes
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Relation between Pontryagin number and Euler number for a four-dimensional closed manifold

Let $M$ be a connected, closed, smooth, oriented four-dimensional manifold. By the Hirzebruch signature theorem, the first Pontryagin number is $$p_1(M) = \langle p_1(TM), [M]\rangle = 3\tau(M)$$ ...
Michael Albanese's user avatar
6 votes

Why are Stiefel-Whitney classes not represented in de Rham cohomology?

There is no (nonzero) map $f : H^*(M; \mathbb{Z}/2\mathbb{Z}) \to H^*(M; \mathbb{R})$ as you claimed. The field $\mathbb{Z}/2\mathbb{Z}$ has characteristic $2$, but the field $\mathbb{R}$ has ...
Najib Idrissi's user avatar
6 votes
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Classification of principal $G$-bundles over closed oriented surfaces with $G$ connected.

I don't know of a reference, but this follows from classifying space theory and a bit of homotopy theory. (Having written the answer, I now realise there is probably an easier answer using obstruction ...
Michael Albanese's user avatar
6 votes
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Chern number of a product of complex projective lines

First recall that $c(\mathbb{CP}^1) = c(T\mathbb{CP}^1) = (1 + a)^2 = 1 + 2a$, not $1 + 2a^2$. Keep in mind that $a \in H^2(\mathbb{CP}^1; \mathbb{Z})$ and is Poincaré dual to the fundamental class $[\...
Michael Albanese's user avatar
6 votes
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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}(\...
Michael Albanese's user avatar
6 votes

Applications of Chern class to gauge theories in physics

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 ...
Tyrone's user avatar
  • 15.9k
6 votes

Boil down the formal definition of Euler class

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 $(\...
Balarka Sen's user avatar
  • 13.9k

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