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.

  • 4
    $\begingroup$ Do you require $M$ to be a manifold? If not then the universal oriented bundle of rank 3 answers your question. $\endgroup$ – Mark Grant Apr 15 '17 at 19:41
  • $\begingroup$ Yes I would like a manifold. Or at least a more geometric object. I edit in this way, thanks for the comment :-) $\endgroup$ – R. Alexandre Apr 15 '17 at 20:11
  • $\begingroup$ But wouldn't then some finite approximation of the oriented grassmannian give a manifold with such a bundle? $\endgroup$ – Thomas Rot Apr 15 '17 at 22:40
  • $\begingroup$ @ThomasRot Probably. But how does it look like ? And it's a bit "boring", I'd like something closer to usual manifolds. $\endgroup$ – R. Alexandre Apr 16 '17 at 8:42
  • 1
    $\begingroup$ @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.) $\endgroup$ – 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 $[X]$.

| cite | improve this answer | |
  • $\begingroup$ That's a nice example in high dimensions, thanks! Do you have any reference for this manifold ? I've never encountered this before ... $\endgroup$ – R. Alexandre Apr 16 '17 at 12:11
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $\operatorname{Spin}^c$.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – R. Alexandre Apr 16 '17 at 11:01
  • 1
    $\begingroup$ @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. $\endgroup$ – Mark Grant Apr 16 '17 at 11:13
  • $\begingroup$ oh yeah I put it wrong. Maybe you have an idea about finding such a bundle over a $4$-manifold ? $\endgroup$ – 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 $w(E) = 1 + \alpha + w_1(X)\alpha$.

| cite | improve this answer | |
  • $\begingroup$ Thank you for this example in dimension 4! $\endgroup$ – R. Alexandre Dec 9 '18 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.