# Examples with zero first Stiefel-Whitney class and nonzero second Stiefel-Whitney class

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?

• Your second sentence makes it sound like you want to say "tangent bundle," not "vector bundle." Apr 19, 2015 at 4:57
• 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? Apr 19, 2015 at 5:15

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.

• 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.) Apr 19, 2015 at 5:03
• 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?) Apr 19, 2015 at 5:17
• @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. Apr 19, 2015 at 5:22
• @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.) 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 $$w_{2}(TM)\neq0$$