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3d Chern-Simons gauge theories based on a Lie group $G$ are classified by an element $k_{CS}\in H^4(BG,\mathbb{Z})$, its level. Via the CS/WZW correspondence the theory is related with a 2d non-linear sigma model with target space $G$, and with a WZW term classified by an element $k_{WZW}\in H^3(G,\mathbb{Z})$, its level. The relation between the two levels is given by a transgression map

$$ \tau : H^4(BG,\mathbb{Z}) \rightarrow H^3(G,\mathbb{Z}) $$

See for instance

  • Robbert Dijkgraaf, Edward Witten, Topological gauge theories and group cohomology, Comm. Math. Phys. 129(2) (1990) 393-429, doi:10.1007/BF02096988, Project Euclid

for an early discussion relating CS/WZW correspondence with transgression.

More generally a CS level for a (d+1)-dimensional Chern-Simons theory transgresses to a WZW level for a d-dimensional WZW level through a transgression map (for $d>2$ this is related with anomalies in QFT)

$$ \tau : H^{d+2}(BG,\mathbb{Z}) \rightarrow H^{d+1}(G,\mathbb{Z}) $$

I think that for simple, simply connected and compact $G$ this map is an isomorphism. Is this correct?

Second, dropping the assumption that $G$ is simple and simply connected, the map can fail to be surjective or injective. For instance if $G=SO(3)$, then $\tau : H^4(B SO(3),\mathbb{Z})=\mathbb{Z} \rightarrow H^3(SO(3),\mathbb{Z})=\mathbb{Z}$ is the multiplication by $2$, that is injective but not surjective. As an opposite case, if $G=U(1)$, for any even $d\geq 2$ $H^{d+2}(BU(1),\mathbb{Z})=\mathbb{Z}$ while $H^{d+1}(U(1),\mathbb{Z})=0$ so $\tau$ is the zero map, and it's not injective.

I'd like to know how can I characterize in general kernel and image of $\tau$ in terms of $G$. I would expect there are mixed situations, and cases where the map is non-trivial but not injective (I would expect so for instance for $G=U(n)$). Is there also some long exact sequence from where I can read these pieces of information?

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  • $\begingroup$ People usually consider the compact semisimple case for $H^4(BG)$, that I've seen, and I think that is possible to get a general result. For the even more special compact simple case, this degree is known completely. $\endgroup$
    – David Roberts ♦
    Commented Jun 7 at 8:40
  • $\begingroup$ Can you elaborate on this? $\endgroup$
    – Andrea Antinucci
    Commented Jun 7 at 10:10
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    $\begingroup$ See eg mathoverflow.net/a/390077/4177 The induced map on cohomology $H^4(BG) \to H^4(B\widetilde{G})$ is injective for $G$ compact, simple and connected, and where $\widetilde{G}$ is the simply-connected cover, and transgression is an isomorphism $\tau\colon H^4(B\widetilde{G})\xrightarrow{\simeq} H^3(\widetilde{G})\simeq \mathbb{Z}$. The table at the above link doesn't have $G_2$, $F_4$ or $E_8$, but these cases all have trivial centre $\endgroup$
    – David Roberts ♦
    Commented Jun 7 at 11:00
  • $\begingroup$ Note that for any connected Lie groups $G$, it is known that $H^4(BG,\mathbb{Z})$ is torsion free. (in the previous comment, the cohomology is all integral). See also the comments below Theorem 6 in Henriques' paper arxiv.org/abs/1602.02968 linked at that question. $\endgroup$
    – David Roberts ♦
    Commented Jun 7 at 11:11
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    $\begingroup$ Note that any compact connected Lie group is a finite quotient of one of the form $K_1\times \cdots \times K_n\times U(1)^d$, where each $K_i$ is connected, compact, and simply connected. So if transgression is natural (I can't think why not, but I don't know a reference off the top of my head), then the problem is reduced to understanding transgression for Lie groups of the above form, and what the pullback map does from $G$ to this finite cover. $\endgroup$
    – David Roberts ♦
    Commented Jun 7 at 13:18

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