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One can construct the $d$-dimensional bordism category by declaring the objects to be the $(d-1)$-dimensional compact manifolds without boundary and the morphisms the $d$-dimensional bordisms between them. Call it $\mathcal{Cob}_d$. It is well known that the connected components of the geometric realization of this category are in one-to-one correspondence with $\pi_d(MO)$, where $MO$ is the Thom spectrum for the orthogonal group. This is the classical Thom-Pontryagin theorem.

One can think of constructing a similar category with supermanifolds. Namely, $\mathcal{Cob}_{(d|k)}$ is the category whose objects are $(d-1|k)$-dimensional supermanifolds and the morphisms the $(d|k)$-dimensional bordisms.

Does anyone know of a Thom-Pontryagin type result for this category? Is there a spectrum $MO_{|k}$, whose homotopy groups recover the connected components of the geometric realization of $\mathcal{Cob}_{(d|k)}$?

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    $\begingroup$ Just for completeness, there is a paper on that: Voronov and Zorich "Bordism theory and homotopy properties of supermanifolds" $\endgroup$
    – user_11437
    Commented Feb 21, 2015 at 15:19

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There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore these technicalities because I don't think it makes much of a difference to your question. Many of these technical issues have been addressed by other people, for example I would suggest looking at the work of Stolz-Teichner and the reference therein to see what sort of things people have tried to do.

In any case in the smooth category we have:

Batchelor's theorem: every $(d|k)$-dimensional supermanifold is (non-canonically!) of the form $\pi E$ for $E$ a rank $k$ vector bundle over a $d$-manifold. Moreover the isomorphism class of the vector bundle is uniquely determined by the super manifold.

Here $\pi E$ is the super manifold whose ring of functions is the global sections of the exterior algebra bundle $\wedge^* E^*$. So the morphisms from $\pi E$ to $\pi E'$ come from all algebra maps and are more than just the vector bundle maps (which correspond to homogeneous algebra maps).

If $(X, \mathcal{O}_X)$ is a super manifold, the vector bundle $E$ can be obtained by considering $\mathcal N / \mathcal N^2$ where $\mathcal N$ is the subsheaf of $\mathcal{O}_X$ of nilpotent elements.

So there are functors in both directions but there is no natural isomorphism from the identity functor on supermanifolds to $\pi(\mathcal N / \mathcal N^2)$. This is what the "non-canonical" means.

But we still get a bijection between isomorphism classes of supermanifolds and isomorphism classes of manifolds with vector bundles. Thus, unless you do something more fancy like work in families, when you pass to bordism classes you will just get the bordism group of $d$-manifolds equipped with rank $k$-vector bundles. This has two names:

$$ \pi_d( MO \wedge BO(k)) = MO_d(BO(k))$$

either as the d-th homotopy group of the smash of the spectrum $MO$ and space $BO(k)$ (which is also another Thom spectrum) or equivalently as the d-th $MO$-homology group of $BO(k)$. So that identifies the bordism group.

I think it is a very interesting question whether the whole Pontryagin-Thom construction can be carried out inside the world of supermanifolds, but that is a different question.

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    $\begingroup$ Thanks for the answer. I have a feeling that we are completely ignoring the automorphism information by this approach. If we thought of constructing some sort of a bordism $(\infty,n)$-category, it is interesting what's the resulting spectrum that comes from the geometric realization of this $(\infty,n)$-category. I assume these would be some sort of a filtration piece of $MO\wedge BO(k)$. $\endgroup$
    – Nerses Aramian
    Commented Feb 19, 2015 at 16:59
  • $\begingroup$ My guess is that you would just get the filtration by $MTO(n) \wedge BO(k)$, as $n$-varies. $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 20, 2015 at 10:17
  • $\begingroup$ How different are the automorphisms of $\pi E$ from those of $E$? $\endgroup$
    – Nerses Aramian
    Commented Feb 20, 2015 at 18:54
  • $\begingroup$ Also as a side question: are the characteristic numbers of the supermanifolds formed by pairing the products of Stiefel-Whitney classes of the tangent bundle and those of the defining vector bundle $E$ with the fundamental class? $\endgroup$
    – Nerses Aramian
    Commented Feb 22, 2015 at 19:57
  • $\begingroup$ Probably one can write something down, however, one will get something which depends on the bundles $E$ and $TM_0$ only and not really an invariant of the supermanifold. In fact, for any Lie supergroup $G$, isomorphism classes of principal $G$-bundles over the supermanifold $M$ are in one to one correspondence with isomorphism classes of principal $G_0$-bundles over $M_0$. This is so since $M_0 \rightarrow M$ and $G_0\rightarrow G$ are homotopy equivalences and the latter is in addition a morphism of groups. $\endgroup$
    – user_11437
    Commented Feb 23, 2015 at 10:52

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