I've heard this description of the first Stiefel-Whitney class of a vector bundle but I don't know why this is true. Can anyone help me with this, please?

The first Stiefel-Whitney class of a vector bundle is an element in the first cohomology group of the base space $H^1(B,Z_2)$. This element can be seen as a map $w_1 : \pi_1(B) \rightarrow Z_2$ (because of the universal coefficient theorem and the fact that $H_1$ is the abelianization of $\pi_1$), so it assigns to each loop based at a fixed point an element of $Z_2$. Now this number is $0$ if and only if that loop is orientation preserving (locally we have an orientation and when we go around a loop and come back to our first place, we can ask if orientation has changed or not)

  • $\begingroup$ Where did you hear this? $\endgroup$ – Adam Saltz May 3 '13 at 1:27
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    $\begingroup$ @Adam from one of PhD students $\endgroup$ – Mehdi May 3 '13 at 1:50
  • $\begingroup$ What definition(s) of $w_1$ do you know? If you have been studying Milnor-Stasheff, take a look at the obstruction theory chapter. $\endgroup$ – Sam Lisi May 3 '13 at 9:39
  • $\begingroup$ @Sam I've read Milnor-Stasheff, but , the obstruction chapter was somewhat brief, so I haven't learned it well. should I use the fact that stiefel-whitney classes are mod 2 reduction of obstruction classes? $\endgroup$ – Mehdi May 3 '13 at 16:06
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    $\begingroup$ @Mehdi : yes, that's what I was thinking about (to your first comment). I'll try to write something soon, but the short version of what I am thinking about is that the $\mathbb{Z}/2$ that shows up in the sequence $1 \to SO(n) \to O(n) \to \mathbb{Z}/2 \to 1$. $\endgroup$ – Sam Lisi May 3 '13 at 16:21

The answer is more straightforward than what I was saying in the comments above, though the general principle of what I was saying holds. Let $n$ denote the dimension of $M$.

Recall that the Stiefel-Whitney classes are characterized by a set of four axioms. The normalization axiom says that $w_1$ of the unorientable line bundle over $S^1$ is the generator of $H^1(S^1, \mathbb{Z}/2)$ and that $w_1$ of the orientable line bundle is $0 \in H^1(S^1, \mathbb{Z}/2)$.

Fix now a loop $\gamma \colon S^1 \to M$. The question of whether $TM$ is orientable over $S^1$ corresponds to whether the determinant bundle $\Lambda^n TM$ is trivial or not when restricted to $\gamma$. Naturality of the Stiefel-Whitney class gives \[ < w_1( \gamma^* \Lambda^n TM), [S^1] > = < \gamma^* w_1(\Lambda^n TM), [S^1]> \] where this pairing is between cohomology/homology of $S^1$. Now, the LHS is equal to $1$ iff $\gamma^*(\Lambda^n TM)$ is the non-trivial bundle, i.e. iff this loop is non-orientable. The RHS is equal to $ < w_1(\Lambda^n TM), \gamma_*[S^1]>$. This shows that $w_1$ exhibits the property you claim.

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    $\begingroup$ This works beyond $w_1$ too. For a 4-manifold $M$ with 2nd cohomology represented by surfaces $\Sigma\subset M$, you can view $w_2(M)$ as a map $\Sigma\to \mathbb{Z}_2$ given by trivializability of $TM|_\Sigma$ (and so if this is 0 on all surfaces, then you've got a spin structure). $\endgroup$ – Chris Gerig May 12 '13 at 21:07
  • $\begingroup$ Maybe you mean: "Now, the LHS is equal to 1 iff $\gamma^* \Lambda^n T M$ non-trivial bundle"? $\endgroup$ – kp9r4d Jun 30 '17 at 0:16
  • $\begingroup$ Thanks for catching that. $\endgroup$ – Sam Lisi Jul 7 '17 at 2:42

If you have a triangulated manifold then it is orientable when its n simplices can be given signs (+/-1) so that the signed sum is an integer cycle.

If the manifold is not orientable then any signed sum of the n simplices is not a cycle and it turns out that what goes wrong is that some of the n-1 simplices in the boundary get counted twice rather than cancelling each other out.

This boundary which is a sum of doubles of n-1 simplices can be divided by two to give a 2 torsion Z homology class. This n-1 dimensional cycle represents the obstruction to orienting the manifold. I would strongly suspect that the first Stiefel Whitney class of the tangent bundle is the Poincare dual to this homology class. Check it out.

  • $\begingroup$ This seems true: the Bockstein $\beta:H_n(M;\mathbb{Z}_2)\to H_{n-1}(M)$ maps the $\mathbb{Z}_2$-fundamental class $[X]_2$ to that 2-torsion generator. Reducing mod-2 and taking the dual, we have a map $H^0(M;\mathbb{Z}_2)=\mathbb{Z}_2\to H^1(M;\mathbb{Z}_2)$ which I can believe sends $PD[X]_2$ to $w_1(M)$. $\endgroup$ – Chris Gerig May 27 '13 at 21:49
  • $\begingroup$ +1. OK just worked it out and it is true; it follows from the definition of Stiefel-Whitney classes in terms of Steenrod squares of the Wu class (which involves the $\mathbb{Z}_2$-fundamental class). $\endgroup$ – Chris Gerig May 28 '13 at 0:29

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