# Stiefel-Whitney total class with prescribed zeros

First things first, I am aware of the existence of this topic. It's related, but old and my question hasn't been discussed there. So I hope it's not wrong to start a new topic.

I'm currently searching for a vector bundle $E\to M$ with $M$ a manifold (no conditions on the rank of $E$) such that $$w(E) = 1 + w_2(E)+w_3(E)$$ with $w_2(E),w_3(E)$ both non zero.

I've shown that :

• If $w_2(E)=0$ then $w_3(E)=0$ (Wu's formula), so we can't really simplify the question.
• $M$ can not have a $\mathbf Z/2$ cohomology engendered by a single element of degree $1$. In fact one can not have $w_2(E)=x^2$ with $x$ of degree 1.
• $E\to M$ can not be the tangent bundle of $M$ (where $M$ is in this case a smooth manifold of dimension $3$ or $4$). (Wu's formula for tangent bundle and results on spin structures in dimension 4)

If you have any ideas, It would be much appreciated. Thanks!

edit: I'd like something less "trivial" than just the universal oriented vector bundle on something like an approximation of the grassmannian. The goal is to get something like a geometrical interpretation of such a total class. It's well understood that $w_1$ represents orientation and $w_2$ spin structure, but it's still a deep mystery to me the meaning of $w_3$.

edit2 : Mark Grant gave a first answer, and thanks to him. But it seems to me that it's not clear if such a manifold $M$ exists if we ask $M$ to be low dimensional : dimension $3$ or $4$ at most. Of course it get's uglier, mostly because we can't consider the tangent bundle as I pointed out before.

• Do you require $M$ to be a manifold? If not then the universal oriented bundle of rank 3 answers your question. – Mark Grant Apr 15 '17 at 19:41
• Yes I would like a manifold. Or at least a more geometric object. I edit in this way, thanks for the comment :-) – R. Alexandre Apr 15 '17 at 20:11
• But wouldn't then some finite approximation of the oriented grassmannian give a manifold with such a bundle? – Thomas Rot Apr 15 '17 at 22:40
• @ThomasRot Probably. But how does it look like ? And it's a bit "boring", I'd like something closer to usual manifolds. – R. Alexandre Apr 16 '17 at 8:42
• @MichaelAlbanese no there is another Wu's formula witch states in particular $Sq^1(w_2) = w_1w_2 + w_3$. And this formula is always true for any vector bundle. (It can be shown by a small recurrence and splitting principle.) – R. Alexandre Apr 16 '17 at 11:24

As Mark Grant pointed out, there is no such example when $E$ is the tangent bundle of a smooth four-dimensional manifold because orientable smooth four-manifolds are spin${}^c$, so $W_3 =0$ and therefore $w_3 = 0$.

The Wu manifold $X = SU(3)/SO(3)$ is a compact, smooth, five-dimensional manifold with total Stiefel-Whitney class $w(X) = 1 + w_2(X) + w_3(X)$ (in particular, it is an example of a non-spin${}^c$ manifold). In fact, $H^*(X; \mathbb{Z}_2) \cong \bigwedge(w_2(X), w_3(X))$.

On a smooth compact orientable five-dimensional manifold, the only potentially non-trivial Stiefel-Whitney number is $w_2w_3$. As there are no Pontryagin numbers in dimension five, we obtain an injective map $\Omega^{SO}_5 \to \mathbb{Z}_2$ given by the Stiefel-Whitney number $w_2w_3$. As $w_2(X)w_3(X) \neq 0$, this map is also surjective and hence an isomorphism. Therefore $\Omega^{SO}_5 \cong \mathbb{Z}_2$ with generator $[X]$.

• That's a nice example in high dimensions, thanks! Do you have any reference for this manifold ? I've never encountered this before ... – R. Alexandre Apr 16 '17 at 12:11
• I don't know of a reference. I learnt about Wu's manifold from Appendix D of Lawson and Michelsohn's Spin Geometry, this answer, and this page on the Manifold Atlas Project. – Michael Albanese Apr 16 '17 at 12:31

Expanding on my and Thomas Rot's comments above, let $\tilde{G}_3(\mathbb{R}^{3+k})$ denote the Grassman manifold of oriented $3$-planes in $\mathbb{R}^{3+k}$. This is a perfectly natural closed manifold of dimension $3k$, and it has a tautological rank $3$ bundle with $w_1=0$ and $w_2,w_3\neq 0$ for $k$ sufficiently large.

In response to your edit, recall that a vector bundle has a $\operatorname{Spin}^c$ structure iff $w_1=0$ and $w_2$ is the reduction of an integral class. So $W_3=\beta(w_2)$ is the obstruction to an orientable bundle admitting a $\operatorname{Spin}^c$ structure (here $\beta:H^*(-;\mathbb{Z}/2)\to H^{*+1}(-;\mathbb{Z})$ denotes the Bockstein).

Since the Wu formulae imply that $Sq^1(w_2)=w_3$ when $w_1=0$, and $Sq^1=\rho\circ\beta$ where $\rho$ denotes reduction mod 2, the conditions you impose ensure that the vector bundle is orientable but not $\operatorname{Spin}^c$.

• Thanks! I'm not very familiar with Spin (and Spin^c) structures. But as a I can see, there is no manifold in dimension less than 4 that doesn't admit a Spin^c structure. Does it mean we can't find any example in dimension less than 4? – R. Alexandre Apr 16 '17 at 11:01
• @R.Alexandre: That's right, that every orientable 4-manifold admits a Spin^c structure. I don't see why that would imply that every oriented bundle over a 4-manifold admits a Spin^c structure. – Mark Grant Apr 16 '17 at 11:13
• oh yeah I put it wrong. Maybe you have an idea about finding such a bundle over a $4$-manifold ? – R. Alexandre Apr 16 '17 at 11:25

In your second edit, you ask whether there exists an example of such a bundle over a lower-dimensional manifold.

Four-dimensional example

Let $$M = (\mathbb{RP}^2\times\mathbb{RP}^2)\#(S^1\times S^3)$$. Note that

$$H^1(M; \mathbb{Z}_2) \cong H^1(\mathbb{RP}^2\times\mathbb{RP}^2; \mathbb{Z}_2)\oplus H^1(S^1\times S^3;\mathbb{Z}_2).$$ Let $$a$$ and $$b$$ denote elements of $$H^1(M; \mathbb{Z}_2)$$ corresponding to generators of $$H^1(\mathbb{RP}^2\times\mathbb{RP}^2; \mathbb{Z}_2)$$, and let $$c$$ denote the element of $$H^1(M; \mathbb{Z}_2)$$ corresponding to the generator of $$H^1(S^1\times S^3; \mathbb{Z}_2)$$.

Consider the rank four vector bundle $$E = L_a \oplus L_b \oplus L_c\oplus L_{a + b + c}$$ where $$L_x$$ is the unique real line bundle over $$M$$ with $$w_1(L_x) = x$$; note that $$L_{a+b+c} \cong L_a\otimes L_b\otimes L_c$$. We have

\begin{align*} w_1(E) =&\ w_1(L_a) + w_1(L_b) + w_1(L_c) + w_1(L_{a + b + c})\\ =&\ a + b + c + (a + b + c) = 0\\ &\\ w_2(E) =&\ w_1(L_a)w_1(L_b) + w_1(L_a)w_1(L_c) + w_1(L_a)w_1(L_{a + b + c})\\ &+ w_1(L_b)w_1(L_c) + w_1(L_b)w_1(L_{a + b + c}) + w_1(L_c)w_1(L_{a + b + c})\\ =&\ ab + ac + a(a + b + c) + bc + b(a + b + c) + c(a + b + c)\\ =&\ ab + a^2 + b^2 \neq 0\\ &\\ w_3(E) =&\ w_1(L_a)w_1(L_b)w_1(L_c) + w_1(L_a)w_1(L_b)w_1(L_{a + b + c})\\ &+ w_1(L_a)w_1(L_c)w_1(L_{a + b + c}) + w_1(L_b)w_1(L_c)w_1(L_{a + b + c})\\ =&\ abc + ab(a + b + c) + ac(a + b + c) + bc(a + b + c)\\ =&\ a^2b + ab^2 \neq 0\\ &\\ w_4(E) =&\ w_1(L_a)w_1(L_b)w_1(L_c)w_1(L_{a + b + c})\\ =&\ abc(a + b + c) = 0. \end{align*}

So $$E$$ is a rank four vector bundle over a four-manifold $$M$$ with $$w(E) = 1 + w_2(E) + w_3(E)$$.

In fact, we can do better. As $$H^4(M; \mathbb{Z}) \cong \mathbb{Z}_2$$, reduction mod $$2$$ defines an isomorphism $$H^4(M; \mathbb{Z}) \to H^4(M; \mathbb{Z}_2)$$. Under this isomorphism, $$e(E)$$ is mapped to $$w_4(E) = 0$$, so $$e(E) = 0$$ and hence $$E \cong F\oplus\varepsilon^1$$. Note that $$F \to M$$ is a rank three vector bundle with $$w(F) = 1 + w_2(F) + w_3(F)$$.

Three-dimensional characterisation

Let $$X$$ be a three-dimensional CW complex. Recall that there is a bijection between isomorphism classes of orientable rank three bundles on $$X$$ and homotopy classes of maps $$X \to BSO(3)$$. As $$X$$ is three-dimensional, we can instead map to $$BSO(3)[3]$$, the third stage of the Postnikov tower for $$BSO(3)$$. As $$\pi_1(BSO(3)) = 0$$, $$\pi_2(BSO(3)) = \mathbb{Z}_2$$, and $$\pi_3(BSO(3)) = 0$$, we see that $$BSO(3)[3]$$ is a $$K(\mathbb{Z}_2, 2)$$. Moreover, as the map $$BSO(3) \to BSO(3)[3]$$ induces an isomorphism on $$\pi_1$$ and $$\pi_2$$, the map $$H^2(BSO(3)[3]; \mathbb{Z}_2) \to H^2(BSO(3); \mathbb{Z}_2)$$ is also an isomorphism. It follows that there is a bijection between orientable rank three bundles on $$X$$ and $$H^2(X; \mathbb{Z}_2)$$ given by the second Stiefel-Whitney class of the bundle.

Now suppose that $$X$$ is a connected three-dimensional manifold. In order for $$w_3(E) \in H^3(X; \mathbb{Z}_2)$$ to be non-zero, we need $$X$$ to be closed. Furthermore, if $$X$$ is closed,

$$w_3(E) = \operatorname{Sq}^1(w_2(E)) = \nu_1(X)w_2(E) = w_1(X)w_2(E)$$

so $$X$$ must be non-orientable. By Poincaré duality, there is at least one $$\alpha \in H^2(X; \mathbb{Z}_2)$$ such that $$w_1(X)\alpha \neq 0$$. For each such $$\alpha$$, there is a unique $$SO(3)$$-bundle $$E \to X$$ with $$w(E) = 1 + \alpha + w_1(X)\alpha$$.

In conclusion, we have the following statement:

Let $$X$$ be a connected, closed three-manifold. There is a real rank three vector bundle $$E \to X$$ with $$w(E) = 1 + w_2(E) + w_3(E)$$ if and only if $$X$$ is non-orientable. Moreover, on any non-orientable $$X$$, for every choice of $$\alpha \in H^2(X; \mathbb{Z}_2)$$ satisfying $$w_1(X)\alpha\neq 0$$, there is a unique real rank three bundle $$E$$ with $$w(E) = 1 + \alpha + w_1(X)\alpha$$.

• Thank you for this example in dimension 4! – R. Alexandre Dec 9 '18 at 15:31