# Spin manifold and the second Stiefel-Whitney class

We know that:

Spin structures will exist if and only if the second Stiefel-Whitney class $w_2(M)\in H^2(M,\mathbb Z/2)$ of $M$ vanishes.

Can someone use simple words and logic to show why the above is true?

Note. More precisely, from Wikipedia: André Haefliger found necessary and sufficient conditions for the existence of a spin structure on an oriented Riemannian manifold (M,g). The obstruction to having a spin structure is certain element [k] of $H^2(M,\mathbb{Z}/2)$. For a spin structure the class [k] is the second Stiefel-Whitney class $w_2(M)\in H^2(M,\mathbb{Z}/2)$ of M. Hence, a spin structure exists if and only if the second Stiefel-Whitney class $w_2(M)\in H^2(M,\mathbb Z/2)$ of M vanishes.

A. Haefliger (1956). "Sur l’extension du groupe structural d’un espace fibré". C. R. Acad. Sci. Paris 243: 558–560.

Spin structures and the second Stiefel-Whitney class are themselves not particularly simple, so I don't know what kind of an answer you're expecting. Here is an answer which at least has the benefit of being fairly conceptual.

First some preliminaries. Recall that a real vector bundle of rank $n$ on a space is the same thing as a principal $\text{GL}_n(\mathbb{R})$-bundle (namely its frame bundle) and that principal $G$-bundles are classified by maps into the classifying space $BG$. In particular, a smooth manifold $M$ of dimension $n$ has a tangent bundle which has a classifying map $M \to B\text{GL}_n(\mathbb{R})$. Additional information allows us to reduce the structure group of this classifying map as follows:

• If $M$ is equipped with a Riemannian metric then the transition maps for the tangent bundle can be chosen to lie in $\text{O}(n) \subseteq \text{GL}_n(\mathbb{R})$, so we get a classifying map $M \to B \text{O}(n)$.
• If $M$ is in addition equipped with an orientation then (possibly by definition, depending on your definition of an orientation) the transition maps for the tangent bundle can in addition be chosen to lie in $\text{SO}(n) \subseteq \text{O}(n)$, so we get a classifying map $M \to B \text{SO}(n)$.
• If $M$ is in addition equipped with a spin structure then (probably by definition, depending on your definition of a spin structure) the above classifying map can be lifted to a classifying map $M \to B \text{Spin}(n)$.

Now, what does this have to do with the second Stiefel-Whitney class? First let me tell a simpler story about the first Stiefel-Whitney class. The first Stiefel-Whitney class is a cohomology class $w_1 \in H^1(B\text{O}(n), \mathbb{F}_2)$ giving a characteristic class for $\text{O}(n)$-bundles which vanishes iff those bundles can be reduced to $\text{SO}(n)$-bundles. Why?

One reason is the following. $w_1$ can be regarded as a homotopy class of maps $B \text{O}(n) \to B \mathbb{Z}_2$ (where I use $\mathbb{Z}_2$ to mean the cyclic group of order $2$). Now, it's known that any such map comes from a homotopy class of maps $\text{O}(n) \to \mathbb{Z}_2$, and there's an obvious candidate for such a map, namely the determinant. This gives an exact sequence

$$1 \to \text{SO}(n) \to \text{O}(n) \to \mathbb{Z}_2 \to 1$$

which, after applying the classifying space functor, gives a homotopy fibration

$$B \text{SO}(n) \to B \text{O}(n) \xrightarrow{w_1} B \mathbb{Z}_2$$

exhibiting $B \text{SO}(n)$ as the homotopy fiber of the first Stiefel-Whitney class.

The homotopy fiber of a map between (pointed) spaces is analogous to the kernel of a map between groups; in particular, if $w : B \to C$ is a map of groups, then a map $f : A \to B$ satisfies $w \circ f = 0$ if and only if $f$ factors through a map $A \to \text{ker}(w)$. The same kind of thing is happening here: a classifying map $f : M \to B \text{O}(n)$ satisfies that $w_1 \circ f$ is homotopic to a constant map if and only if it factors up to homotopy through the homotopy fiber $M \to B \text{SO}(n)$.

Now the reason I gave such a sophisticated description of orientations is that the story for spin structures is completely parallel. Namely, the second Stiefel-Whitney class is a cohomology class $w_2 \in H^2(B\text{SO}(n), \mathbb{F}_2)$ which can be regarded as a homotopy class of maps $B \text{SO}(n) \to B^2 \mathbb{Z}_2$. You can produce such classes by applying the classifying space functor to a homotopy class of maps $\text{SO}(n) \to B \mathbb{Z}_2$, or equivalently a cohomology class in $H^1(\text{SO}(n), \mathbb{F}_2)$, and there's a natural candidate for such a class, namely the cohomology class classifying the nontrivial double cover $\text{Spin}(n) \to \text{SO}(n)$. This also turns out to imply that we get a homotopy fibration

$$B \text{Spin}(n) \to B \text{SO}(n) \xrightarrow{w_2} B^2 \mathbb{Z}_2$$

exhibiting $B \text{Spin}(n)$ as the homotopy fiber of the second Stiefel-Whitney class, and if you believe this then it again follows from the universal property of the homotopy fiber that a map $f : M \to B \text{SO}(n)$ lifts to a map $M \to B \text{Spin}(n)$ iff $w_2 \circ f$ is homotopic to a constant map.

(This homotopy fibration is a "delooping" of the more obvious homotopy fibration $B \mathbb{Z}_2 \to B \text{Spin}(n) \to B \text{SO}(n)$ coming from the short exact sequence $1 \to \mathbb{Z}_2 \to \text{Spin}(n) \to \text{SO}(n) \to 1$.)

This argument can be continued all the way up the Whitehead tower of $B \text{O}(n)$; the next step is a string structure, etc.

• Nice answer. Notation question: What is $B^2G$? – Cheerful Parsnip May 25 '14 at 3:49
• @Grumpy: $B^2 A$ ($A$ has to be abelian for this to make sense) is the double delooping of $A$. When $A$ is discrete this is the Eilenberg-MacLane space $K(A, 2)$. – Qiaochu Yuan May 25 '14 at 3:51
• An excellent answer from a real expert. Thanks! :) – wonderich May 25 '14 at 4:04
• It's worth mentioning that the homotopy fiber of a classifying map $M \to BG$ is, up to homotopy, the bundle $E \to M$ it represents. This is how we get a homotopy fibration $\text{Spin}(n) \to \text{SO}(n) \to B \mathbb{Z}_2$ which then deloops to the second homotopy fibration above. – Qiaochu Yuan May 25 '14 at 5:17
• – Grigory M May 26 '14 at 18:42

The situation is very simple, actually.

Recall that $Spin(n)$ is the 2-fold cover of $SO(n)$, i.e. we have a short exact sequence $1\to\mathbb Z/2\to Spin(n)\to SO(n)\to1$. It induces a cohomology exact sequence $$\require{AMScd} \begin{CD} H^1(X,\underline{Spin}(n)) @>>> H^1(X,\underline{SO}(n)) @>>> H^2(X;\mathbb Z/2)\\ @| @| @|\\ \operatorname{Vect}_{Spin}(X) @>>> \operatorname{Vect}_{SO}(X) @>{w_2}>> H^2(X;\mathbb Z/2) \end{CD}$$ (where $\underline G$ is the sheaf of $G$-valued functions on $X$ and cohomology are Čech cohomology) — i.e. an $SO$-structure lifts to a $Spin$-structure iff some class in $H^2(X;\mathbb Z/2)$ is zero (and as a free bonus we see that any two liftings differ by an element of $H^1(X;\mathbb Z/2)$).

(Of course, we can get even more down to earth by working explicitly with cocycles and not even mentioning sheaves and Čech cohomology.)

Now this obstruction is some mod 2 characteristic class. But actually there is only one such class in $H^2(BSO)$, so it coincides with usual $w_2$.

• thanks Grigory M, +1, also a very nice answer. – wonderich May 26 '14 at 20:31
• To make it clear that Grigory and I aren't saying different things, exactness here is precisely a reflection of the statement that $B \text{Spin}(n) \to B \text{SO}(n) \to B^2 \mathbb{Z}_2$ is a homotopy fibration. – Qiaochu Yuan May 26 '14 at 22:04
• @QiaochuYuan Sure (the only difference is the first sentences, essentially ;-) – Grigory M May 27 '14 at 6:43
• I see you write $H^2(X;\Bbb Z/2)$ without the underline, so do you mean standard simplicial cohomology? I ask because I think I can prove $H^2(X;\underline{\Bbb Z/2})\cong H^2(X;\Bbb Z/2)$ when $X$ is a manifold (I need a good cover), but in general I'm not sure how it works. Bredon seems to say I need the space to be homologically locally connected. – Ryan Unger Jul 2 '18 at 19:57
• @0celo7 well, yes, OP explicitly asked about manifolds and tangent bundles (and I'm afraid I know almost nothing about cohomology of spaces not homotopy equivalent to CW complexes…) – Grigory M Jul 2 '18 at 20:41