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This question is motivated by the calculation of the higher algebraic $K$-groups of a finite field.

Let $G$ be a finite group, the case I am most interested in is $G = \text{Gl}_n(\mathbb F_q)$, but the question makes sense for every $G$. Let us denote the representation ring of $G$ by $R_{\mathbb C} G$, this is the group completion of the semiring of all finite dimensional complex representations of $G$ (the addition comes from the direct sum, the multiplication from the tensor product). Then there is a ring homomorphsim ($K^0$ denotes topological $K$-theory) $$\Phi\colon R_{\mathbb C} G \to K^{0}(BG)$$ sending a representation $V$ to the class of the bundle $EG \times_G V \to BG$. But we can also define a map $$\Psi \colon R_{\mathbb C} G \to \tilde{K}^{0}(BG)$$ by choosing an invariant hermitian form on $V$ and an orthonormal basis which gives a map $G \to U(n)$. Composing with the inclusion $U(n) \to U$ and taking classifying spaces we get a map $BG \to BU$, hence after passing to homotopy classes we arrive at an element $$\Psi(V) \in \tilde{K}^{0}(BG),$$ and this should now be inependent of the choices we made since all possible inner products form a contractible space and choosing a different basis corresponds to an inner automorphism which is invisible in $BU$. The thus defined map should be a homomorphism of abelian groups. My question is: Is $\Psi$ just the composition of $\Phi$ with the projection $$K^{0}(BG) \to \tilde{K}^{0}(BG),$$ and if not, how can we understand $\Psi$ then? Or did I do a mistake and $\Psi$ does not make sense after all?

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  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ May 15, 2015 at 4:55
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    $\begingroup$ Could you elaborate a little? $\endgroup$ May 15, 2015 at 4:56
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    $\begingroup$ It follows more or less directly from the construction you've given. The way you construct a vector bundle out of a representation of $BG$ is that the map $G \to GL_n(\mathbb{C})$ induces a map $BG \to BGL_n(\mathbb{C})$. The map $U(n) \to GL_n(\mathbb{C})$ induces a homotopy equivalence $BU(n) \to BGL_n(\mathbb{C})$ so it doesn't matter which one you work with (this corresponds to your statement that the space of inner products is contractible). Then the way a vector bundle induces an element of K-theory is that the map $U(n) \to U$ induces a map $BU(n) \to BU$. $\endgroup$ May 15, 2015 at 5:01
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    $\begingroup$ None of this requires that $G$ is finite or even discrete and everything continues to makes perfect sense for, say, $G$ a compact Lie group. You can think of this map as the map from the genuinely $G$-equivariant K-theory of a point to the Borel-equivariant K-theory. $\endgroup$ May 15, 2015 at 5:03

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