What's the simplest/most concrete vector bundle you can think of that has zero first Stiefel-Whitney class but non-zero second? That would be the simplest space that doesn't have spinors. (See Spin manifold and the second Stiefel-Whitney class) (Which we can interpret as...fermions can't exist?)

It can't be the tangent bundle of a two or three-manifold. (See Vanishing of the second Stiefel–Whitney classes of orientable surfaces and Second Stiefel-Whitney Class of a 3 Manifold ) How complicated do they have to be? What are some examples?

  • $\begingroup$ Your second sentence makes it sound like you want to say "tangent bundle," not "vector bundle." $\endgroup$ Apr 19, 2015 at 4:57
  • $\begingroup$ Ah! I want to say vector bundle, because a lower rank vector bundle over a weird CW complex seems more interesting. Especially if the bundle is rank 2 because then you're talking about electrons! Is that possible? $\endgroup$
    – spitespike
    Apr 19, 2015 at 5:15

3 Answers 3


Consider $T\mathbb{CP}^k$, the tangent bundle of $\mathbb{CP}^k$. I claim it satisfies the desired conditions if and only if $k$ is even, in particular, $T\mathbb{CP}^2$ is an example.

There are several ways to see that $w_1(T\mathbb{CP}^k) = 0$ for every $k$:

  • as $T\mathbb{CP}^k$ is a complex bundle, its odd Stiefel-Whitney numbers are zero, in particular $w_1(T\mathbb{CP}^k) = 0$;
  • every complex vector bundle is orientable so $w_1(T\mathbb{CP}^k) = 0$; and
  • $\mathbb{CP}^k$ is simply connected so $H^1(\mathbb{CP}^k; \mathbb{Z}_2) = 0$.

As for the condition that the second Stiefel-Whitney class be non-zero, we will need the restriction that $k$ is even. To see this, recall that $c(T\mathbb{CP}^k) = (1 + \alpha)^{k+1}$, so $c_1(T\mathbb{CP}^k) = \binom{k+1}{1}\alpha = (k + 1)\alpha$, and hence $w_2(T\mathbb{CP}^k) = (k + 1)\bar{\alpha}$ where $\bar{\alpha}$ is the image of $\alpha$ under the natural map $H^2(\mathbb{CP}^k; \mathbb{Z}) \to H^2(\mathbb{CP}^k; \mathbb{Z}_2)$ induced by the reduction modulo $2$ map $\mathbb{Z} \to \mathbb{Z}_2$. As $\alpha$ is a generator for $H^2(\mathbb{CP}^k; \mathbb{Z})$, it is not divisible by two, so $\bar{\alpha} \neq 0$. Therefore, $w_2(T\mathbb{CP}^k) = (k + 1)\bar{a}$ is non-zero if and only if $k$ is even.

  • 6
    $\begingroup$ I was recently made fun of at a conference for saying that $\mathbb{CP}^2$ is my favorite manifold, but this is a great example of why I like it so much. It's the simplest example of a closed orientable manifold with lots of generic features that you don't see in lower dimensions: 1) it's not stably parallelizable, 2) it has a nonzero Pontryagin class, 3) it's not a boundary, 4) it doesn't have a spin structure... (of course these properties are related.) $\endgroup$ Apr 19, 2015 at 5:03
  • $\begingroup$ This is great! Thanks. Conceptually: does that mean that fermions couldn't exist at all or that wavefunctions necessarily vanish at some points (sort of like the hairy-ball theorem?) $\endgroup$
    – spitespike
    Apr 19, 2015 at 5:17
  • 2
    $\begingroup$ @spitespike: one way of saying what the wavefunction of a fermion is is that it's a section of a particular bundle, namely a spinor bundle associated to a spin structure on a manifold. If a manifold doesn't have a spin structure then you can't define this bundle, so you can't define its sections. You can try to remove part of the manifold to kill $w_2$ and hence define a spin structure and a spinor bundle on what's left, and then I guess you'll get a notion of fermions where the wavefunctions have singularities on the thing you removed. $\endgroup$ Apr 19, 2015 at 5:22
  • 1
    $\begingroup$ @QiaochuYuan: It's OK; $\mathbb{CP}^2$ is one of my favorite manifolds too. (My background is in $4$-manifold topology, where it's also a common example or counterexample.) $\endgroup$
    – anomaly
    Apr 22, 2015 at 16:55

Since you said in comments that you were interested in general vector bundles, we can lower the dimension from Michael's answer.

Consider the tautological complex line bundle over $S^2 = \mathbb{C}P^1$. That is, thinking of $\mathbb{C}P^1$ as the set of complex lines through the origin in $\mathbb{C}^2$, the vector bundle is $E = \{(p,v)\in \mathbb{C}P^1\times \mathbb{C}^2: v\in p\}$. The projection map is simply projection onto the first factor.

(This is the analogue of the Mobious bundle over $S^1\cong\mathbb{R}P^1$. Alternatively, one can think about it by taking the universal complex line bundle over $\mathbb{C}P^\infty$ and then pulling back via the inclusion of the $2$-skeleton $S^2\rightarrow \mathbb{C}P^\infty$.)

Of course, $w_1(E) = 0$ since $H^1(S^2;\mathbb{Z}/2\mathbb{Z}) = 0$. For $w_2$, we note that the first Chern class, which is well known to equal the Euler class, is given by a generator of $H^2(S^2;\mathbb{Z})$. But the mod 2 reduction of the Euler class is the top Stiefel-Whitney class, so $w_2(E)\neq 0 \in \mathbb{Z}/2\mathbb{Z}\cong H^2(S^2;\mathbb{Z}/2\mathbb{Z})$.


I am restricting to the case of smooth four-manifolds. Rokhlin's theorem tells us that- if a manifold is Spin, then the signature should be multiple of 16. Spin structure exists iff $w_{2}(TM)=0$. Consider the manifold $\mathbb{C}P^{2}\#\mathbb{C}P^{2}\#\overline{\mathbb{C}P^{2}}$. This manifold is orientable, so $w_{1}(TM)=0$. This manifold has signature 1, hence not Spin by Rokhlin's theorem. So $w_{2}(TM)\neq0$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .