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The third in the "Loop groups and twisted K-theory" series by FHT treats compact Lie groups without any connectedness assumptions. I am trying to unwind what Theorem 2 of that paper (available here, page 5) says for a finite group.

Theorem 2: "For regular $\tau$, there is a natural isomorphism $R^\tau(LG_s) \cong K_G^\tau(G)$ ..."

For $G$ finite, I think that the regularity condition (page 7) is empty, as it has something to do with a maximal torus of $G$. So we can take $\tau = 0$.

On the right-hand side, that means untwisted $K$-theory, which is naturally identified with $\bigoplus_g R'(Z_G(g))$, where $R'$ denotes the Grothendieck group of finite-dimensional complex representations, and $g$ runs over conjugacy class representatives. ($R'$ to avoid conflict with $R^\tau$, whose meaning I don't understand.)

What is on the left-hand side, when $\tau$ is zero and $G$ is finite? It is a Grothendieck group of representations of the group of free loops in $G$, modified somehow by $\tau$ ($\tau = 0$, so not modified at all). But I don't understand what it means. For finite $G$, loops are constant and the Lie algebra is trivial, so from the definitions on page 5, $LG = LG_s = G$. But it can't simply be the Grothendieck group of complex representations of $G$, that is much smaller than the right-hand side.

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  • $\begingroup$ I would think $\tau$ is more naturally expressed as a based map $BG \to K(\mathbb{Z},4)$, so it corresponds to a group cohomology class for finite $G$. Then the twisted K-theory is the K-theory of a (Dijkgraaf-Pasquier-Roche) twisted quantum double. $\endgroup$
    – S. Carnahan ♦
    Commented Jul 6, 2014 at 23:52
  • $\begingroup$ In the FHT paper, $\tau$ on the right is a parameter that twists $K$-theory, for instance a $G$-equivariant Clifford algebra on $G$. It has a characteristic class in some combination of $H^1_G(G;\mathbf{Z}/2)$ and $H^3_G(G;\mathbf{Z})$. Do you see a way to get from here to a class in $H^4(BG;\mathbf{Z})$? $\endgroup$
    – ya-tayr
    Commented Jul 7, 2014 at 0:36
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    $\begingroup$ I believe it goes the other way. Observe that $EG\times_G BG \simeq LBG$, so the idea is to transgress your class in $H^4(BG;\mathbb{Z})$ over to $H^3(LBG;\mathbb{Z}) \cong H^3_G(G;\mathbb{Z})$. Take a look at David Ben-Zvi's answer here (mathoverflow.net/questions/20671/…) for more. $\endgroup$
    – Dan Kneezel
    Commented Jul 7, 2014 at 4:14
  • $\begingroup$ Rats! I meant to write $EG\times_G G \simeq LBG$ (where $G$ is acting on itself by conjugation). $\endgroup$
    – Dan Kneezel
    Commented Jul 7, 2014 at 4:43

2 Answers 2

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Take a look at the Annals version of FHT III (2011). There, Theorem 2 is more carefully stated only for compact connected Lie $G$. I think what you want is Theorem 3 (whose statement and surrounding context are essentially the same as the '08 version of the paper you linked to):

Theorem 3. Let $G$ be a compact Lie group, $f \in G$. For a regular twisting $\tau$, there is a natural isomorphism $R^{\tau - \underline{\sigma}-\underline{d}}(L_f G) \stackrel{\sim}{\to} K_G^\tau([fG_1])$.

If I'm not mistaken, for $G$ finite and $\tau = 0$, Theorem 3 reduces to an isomorphism $R(Z_G(f))\cong K_G(G/Z_G(f))$. This isomorphism is, of course, an elementary observation in equivariant $K$-theory (cf. Example (ii), p. 132, Segal 1968).

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This is maybe more of a comment. Some time ago there was a paper by Simon Willerton titled "The twisted Drinfeld double of a finite group via gerbes and finite groupoids" which tried to make sense of the statement of Freed-Hopkins-Teleman in the case of finite groups (i.e. Dijkgraaf-Witten theory).

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