# Computing a cobordism group of manifolds endowed with a real vector bundle with constraints on the Stiefel-Whitney classes

I am interested in computing the cobordism group of oriented manifolds $M$ of dimension 7 endowed with real vector bundles $N$ of rank 5 with the following conditions on the Siefel-Whitney classes:

$w_1(N) = 0 \;, \quad w_2(N) = w_2(TM) \;, \quad w_5(N) = 0 \quad (1)$

The first equation says that $N$ is an orientable bundle. The second says that $TM \oplus N$ is spin. The third is equivalent to the vanishing of the Euler class of $N$.

So to paraphrase the above, I want to know if given an oriented 7-manifold $M$ endowed with a bundle $N$ satisfying (1), it is always possible to find an oriented 8-manifold $M'$ bounded by $M$ together with a bundle $N'$ satisfying (1) and extending $N$ to $M'$.

If we ask the same question disregarding the bundle $N$, the obstruction is given by the stable homotopy group $\pi_7(MSO)$, which can be shown to vanish using the results of C.T.C. Wall. So given a 7-dimensional oriented manifold, one can always find an 8-dimensional manifold bounded by the latter.

Now taking into account $N$ and ignoring the last two conditions in (1), I believe that the obstruction to finding a bordism is given by $[S, \Sigma^{-7}MSO \wedge \Sigma^5 BSO]$, where $S$ is the sphere spectrum, $\wedge$ the smash product and $[.,.]$ denotes the homotopy classes of maps. Already at this level I am not sure how to compute this group.

Edit: As $N$ has to be an actual bundle and not only a stable one, the bordism group should be in this case $[S, \Sigma^{-7}MSO \wedge BSO(5)]$.

Finally, one has to take into account the last two constraints of (1). To this end, I imagine that one should use the fact that there is a map from $\Sigma^{-7}MSO \wedge \Sigma^5 BSO$ into $\Sigma^2H\mathbb{Z}_2 \wedge \Sigma^5H\mathbb{Z}_2$, determined by $(w_2(N) - w_2(TM), w_5(N))$ and that the relevant spectrum is in some appropriate sense the kernel of this map.

Any hint about how to compute this cobordism group, or reference to similar computations in the literature would be greatly appreciated.

For people curious about it, the motivation to compute this group comes from the physics of the M5-brane. The worldvolume of the M5-brane is oriented and 6-dimensional. It is embedded in an 11-dimensional manifold $X$ which is spin, with normal bundle $N$. The spin condition on $X$ and the orientability of $M$ account first two conditions in (1). To compute the global gravitational anomalies of the effective field theory on the worldvolume, one has to consider mapping tori of the worldvolume, endowed with a vector bundle $N$ satisfying (1). And it turns out that the best way to express the anomaly is in terms of an 8-dimensional manifold bounded by the mapping torus. This is why knowing if such bounded manifolds exist is crucial.

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Some of your suspensions have the wrong sign, perhaps they all do. Note that $\pi_7(MSO)=[S^7,MSO]=[S,\Sigma^{-7}MSO]$ (not $\Sigma^7MSO$). The group $[S,\Sigma^7MSO\wedge\Sigma^5BSO]$ is zero for trivial connectivity reasons, but you probably mean $[S,\Sigma^{-7}MSO\wedge\Sigma^{-5}BSO]$. –  Neil Strickland Jun 6 '13 at 22:48
Without the second and third assumptions on SW-classes, I believe you would be computing the bordism group $MSO_7(BSO(5))$. The advantage of viewing it as a generalized homology group is that it becomes readily computable using the bordism spectral sequence, see chapter II of Conner and Floyd's "Differential periodic maps". I'm not sure how to work the other conditions on SW-classes in, though. –  Mark Grant Jun 7 '13 at 7:01

This is similar to András's answer, but a bit different... Start with $BSO(5)$ and kill $w_5$, i.e take the fiber $F$ of $w_5:BSO(5)\to K(Z_2,5)$. Now consider the composition

$\xi: BSpin \times F \to BSO\times BSO(5)\to BO\times BO \to BO$,

where the first two arrows are products of the obvious map and the last one corresponds to addition of vector bundles. Finally take the Thom spectrum of the pullback of the universal bundle over BO along $\xi$.

The reason is that $N$ is classified by a map to $F$, and the difference $\nu-N$ is classified by a map to $BSpin$. (I assume you want the spin structures on your manifolds and bordisms too. Spin structures on $\nu-N$ are in bijection with spin structures on $TM+N$.)

Then the Thom spectrum is $MSpin\wedge \Sigma^{-5}TF$, where $TF$ is the Thom space for $F\to BSO(5)$, so the bordism group is $\Omega_{12}^{Spin}(TF)$. One can try to compute this via the Atiyah-Hirzebruch spectral sequence, or also by using the Adams spectral sequence.

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Thanks, this helps a lot. But there is one point I don't understand in your construction. Where is the suspension operator in the expression for the Thom spectrum coming from? It seems to me that the bundle over $BSpin \times F$ is the universal bundle over $BSpin$ times the pull back to $F$ of the universal bundle over $BSO(5)$, whose Thom spectrum is $MSpin \wedge TF$. –  Samuel Monnier Aug 4 '13 at 16:27
To me the best way to formulate it is that maps to $BO$ correspond to K-theory classes or equivalently stable vector bundles or equivalently virtual vector bundles of dimension 0, that means formal differences $E-F$, where $E,F$ are honest vector bundles and $dim(E-F):=dim(E)-dim(F)$. Strictly speaking, this equivalence is only correct for compact base spaces. For example the universal "virtual vector bundle" over $BO$ is the class which restricts to the actual virtual bundle $\eta_n-n\mathbb R$ to each $BO(n)$, where $\eta_n$ is the universal $O(n)$-bundle. –  nsrt Aug 5 '13 at 7:59
And if you pull this back to $BSO(5)$, you have $\tilde{\eta}_5 -5\mathbb R$, where $\tilde{\eta}_5$ is the universal $SO(5)$-bundle. A virtual vector bundle $E-n\mathbb R$ has Thom spectrum $\Sigma^{-n} TE$, where $TE$ is the Thom space of the vector bundle $E$. –  nsrt Aug 5 '13 at 8:02
Ok... but what is the reason for using virtual bundles of dimension $0$? $N$ is the pull back from $F$ of $\tilde{\eta}_5$, which is why I would have naively thought that the Thom spectrum should be $MSpin \wedge TF$. There is some basic point I must be missing... –  Samuel Monnier Aug 5 '13 at 12:31
Maybe the point you are missing is that the Pontryagin-Thom isomorphism relates an $n$-dimensional bordism group to the $(n+k)$-th homotopy group of a Thom space, when the fiber of the vector bundle is $k$-dimensional. So you can work with the Thom spectrum $MSpin\wedge TF$, but you want the 12-th homotopy group then. –  nsrt Aug 5 '13 at 13:01

Here is the space whose stable homotopy group will be the cobordism group you are looking for.

Start with $BSpin(5)$ and kill the cohomology class $w_5$ in it. (That is consider the map $w_5: BSpin(5) \to K(Z_2,5)$, pull back by this map the tautological (path) fibration. Denote by $B \to BSpin(5)$ the obtained fibration. $B$ is the resulting space, which is obtained from $BSpin(5)$ after killing $w_5.)$

Pull back the 5-dimensional universal spin vector bundle from $BSpin(5)$ to $B.$ Let $T$ be the Thom space of this bundle. This is the space whose stable homotopy group is your cobordism group.

Namely $\pi_{q+12}(S^qT)$ is your cobordism group.

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This answers the question for framed $M$, but the questions asked for $M$ merely oriented. –  Oscar Randal-Williams Jul 17 '13 at 13:57
No, these manifolds will not be framed, They will be exactly what you wanted. –  András Szűcs Jul 18 '13 at 6:19
My apologies, they are not framed. But the homotopy group you suggest classifies tuples $(M^7, N \to M, \varphi : TM \oplus N \oplus \epsilon^q \cong \epsilon^{q+12})$, where $N$ is a 5-dimensional vector bundle with $B$-structure (in your notation). This implies the conditions on Stiefel--Whitney classes that was asked for, but also implies other conditions (such as $p_1(TM) = - p_1(N)$) which were not. –  Oscar Randal-Williams Jul 18 '13 at 8:42
You are right. My construction identified the normal bundle of the manifold with the required 5-bundle, so it was wrong. I think the correct way is how nsrt did it above. –  András Szűcs Jul 18 '13 at 9:16
Thanks for your answer and for the correction. –  Samuel Monnier Aug 4 '13 at 16:27