Are there any papers on the cohomology of the classifying space of the general linear supergroup $GL(n, m)$ or unitary supergroup $U(n, m)$?

I know basically nothing about supergeometry. It seems that in my situation I just need to know the cohomology of the classifying spaces of the supergroups mentioned above. When I google for this topic, I basically found nothing, but other information seems to imply that ordinary cohomology theory (for example, singular or de Rham) cannot detect the superdifferentiable structure of supermanifolds, so some cohomology theories for supermanifolds were invented (I know there are different classes of supermanifolds, but I don't the groups or its classifying spaces belong to which class). Thus, as a start, I think I need to know the ordinary cohomology of $BGL(n, m)$ and $BU(n, m)$ (if it exists).


  • $\begingroup$ The usual definition of B first takes the underlying space of GL(n,m), which only sees the underlying ordinary manifold of GL(n,m). Thus the classifying space of the super Lie group GL(n,m) is the same as the classifying space of its underlying ordinary Lie group. $\endgroup$ – Dmitri Pavlov Sep 10 '15 at 11:02
  • $\begingroup$ I see. So can I say classifying complex supervector bundle is just the same as classifying its underlying complex vector bundle?If this is true, then it's interesting that the super structure is lost when doing classification. $\endgroup$ – GRR Sep 10 '15 at 11:14
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    $\begingroup$ If you use the traditional definition of a classifying space, you don't see the odd direction. But one could consider many other notions of classifying spaces, with different universal properties. For example, one could argue that the right notion of a classifying space in supergeometry is a stack of spaces on the site of superpoints (i.e., R^{0|n}), whose value on R^{0|n} is the classifying space of Hom(R^{0|n},G). One can then extract nontrivial spaces from such a stack by taking various homotopy colimits, which will detect a significant part of the odd direction. $\endgroup$ – Dmitri Pavlov Sep 10 '15 at 11:21
  • $\begingroup$ A category where one can build $BG$ for a Lie supergroup $G$ would be the site of simplicial presheaves on the category of supermanifolds (with the Grothendieck topology coming from open covers). However, if one localizes this category at $\mathbb{R}^{1|0}$ one loses all the odd directions and gets the category of simplicial presheaves on manifolds localized at $\mathbb{R}$. This category is equivalent to the homotopy category of topological spaces. (cont.) $\endgroup$ – user_11437 Oct 20 '15 at 12:41
  • $\begingroup$ @DmitriPavlov Can you elaborate on how to produce nontrivial spaces? $\mathrm{Hom}(\mathbb{R}^{0|n}, G)_0$ is still homotopy equivalent to $G_0$. I can only think of extracting the Thom space of $(TG)_1 \rightarrow G_0$ by restricting to some sort of proper morphisms. Thanks in advance! $\endgroup$ – user_11437 Oct 20 '15 at 12:41

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