Let $H < G$ be a subgroup of a finite group $G$. Let $X:=G/H$ and $\mathcal{F} \in Sh_G(X)$ be an equivariant sheaf on $X$ (w.r.t. left multiplication) associated to a finite dimensional representation $\pi$ of $H$ with character $\chi$. Let $Ind_H^G(\chi)$ be the character of $Ind_H^G(\pi)$ and for any $g \in G$ define $X^g=\{x\in X :gx=x \}$.

Recently I found out that the basic formula for the induced character $Ind_H^G(\chi)$ can be interpreted very naturally from the point of view equivariant sheaves giving the following elegant equation:

$$Ind_H^G(\chi)(g)=\Sigma_{x\in X^g} Tr(g^* ,\mathcal{F_x})$$

Where $Tr(g^*,\mathcal{F}_x)$ is the trace of the induced action of $g$ on stalk of $\mathcal{F}$ at $x \in X^g$.

The formula above is surprisingly elegant compared to the one I derived it from (which involved either choosing representatives for cosets or dividing by the order of $H$ while here we avoided both).

Question 1: Is there a reasonably geometric argument for why this formula is true?

Question 2: In what kind of generality does this formula hold? (continuous representations, locally compact groups, distributional characters etc...).

  • 1
    $\begingroup$ How do you define the sheaf $\mathcal F$? This kind of formula does exist more broadly and are called Lefschetz trace formulas. However there are of cohomogical nature, i.e. involve the spaces $H^k (X^g , {\mathcal F})$. $\endgroup$ – Paul Broussous Feb 5 '18 at 8:17
  • 4
    $\begingroup$ The classic reference for this kind of argument is the proof of the Weyl character formula (generalizing the Frobenius formula you quote) in Atiyah-Bott, A Lefschetz Fixed Point Formula for Elliptic Complexes (aka the Woods Hole fixed point theorem), in the setting of compact groups. Since then this approach to character formulas has been vastly generalized - see e.g. Schmid-Vilonen in the setting of real reductive groups. The basic idea is a simple geometric one, ie just that a character = a trace = sum/integral over diagonal, but intersection of a graph with the diagonal gives fixed points $\endgroup$ – David Ben-Zvi Feb 5 '18 at 16:46

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