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added pointer to the general definition in terms of equivariant sections of equivariant bundles of equivariant classifying spaces for equivariant K-theory:
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For $G$ a finite group (at least),
it is known that classifying $G$-space for rational $G$-equivariant KU-theory assigns the rationalized complex representation rings
$G/H \;\mapsto\; \underset{k}{\prod} B^{2k} \; \mathbb{Q} \otimes R_{\mathbb{C}}(H)$and that the equivariant fundamental group of $B_G PU(\mathcal{H})$ assigns the character group
$G/H \;\mapsto\; Hom\big(H, U(1)\big) \,.$Since the character group has a canonical action on the representation ring, it ought to be the case for the classifying coefficient $G$-bundle for 3-twisted equivariant K-theory, that the action of the equivariant fundamental group of the base on the homotopy fibers is, at each stage $G/H$, the canonical action of the $H$-character group on the $H$-representation ring.
Is this proven anywhere?
a reference item to add to twisted equivariant K-theory once the edit-functionality is back:
It is
well-known that the connected components of the $H$-fixed locus of the classifying space for equivariant K-theory is the representation ring $R(H)$
known that the possible twistings of equivariant K-theory over an $H$-fixed point include – in addition to the notorious “gerbe” – a complex line bundle with structure group $H^2(G; \mathbb{Z}) \simeq H^1\big(G; \mathrm{U}(1)\big) \,\subset \, \mathrm{U}(1)$ (aka “local system”).
But, in describing how this degree-1 twist actually acts on the classifying space, all authors I have seen (where “all” is no more than 2 or 3 groups, apparently) pass to the perspective of “delocalized” cohomology.
While the delocalized picture has some clear virtues, it does a fair bit of violence to the classifying picture of K-theory. In the latter picture, there is an evident guess for how the 1-twist acts: It ought to be the canonical operation of tensoring representations with group characters regarded as 1d reps:
$\array{ Hom(G; U(1)) \,\times\, R(G) && R(G) \\ (\kappa, \rho) &\mapsto& \kappa \otimes \rho }$Because what else can it be. But also because I think I have proven this now. (It follows from the observation mentioned in another thread, here.)
I keep wondering, though, if this has not been discussed elsewhere, before?
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