Definitions

The action correspondence is \(m:LG\times^I LX\to LX\), \([g,y]\mapsto gy\), and \(\mathcal G*_IM=m_!(\mathcal G\widetilde\boxtimes M)\). A trace is the alternating Frobenius trace on stalk cohomology. Write \(\mathbb F_q\) for the residue field and \(F=\mathbb F_q(\!(t)\!)\).

Hypotheses

Use (Lev) and (Dict). The bounded kernel descends to a finite-type model and the Schubert support is proper. The relevant Iwahori congruence quotients are connected, so Lang's theorem identifies rational flag points with \(G(F)/I\).

Lemma IIb · Trace of convolution. Let \(\mathcal G\in\mathcal H_{c,I}\) have support in \(\mathrm{Fl}_{\le w}\), and let \(M\in\mathcal C_I\) be a Weil object. Use (Dict)(d), the induced Weil kernel and Haar normalization \(\operatorname{vol}I=1\).

$$\operatorname{tr}(\mathcal G\ast_IM)=\operatorname{tr}\mathcal G\ast\operatorname{tr}M,\qquad(\operatorname{tr}\mathcal G\ast\operatorname{tr}M)(x)=\sum_{gI\in G(F)/I}\operatorname{tr}(\mathcal G)(g)\,\operatorname{tr}(M)(g^{-1}x),\qquad\operatorname{tr}\mathrm{IC}_0=\varphi_0 .$$

Steps 1–3 · Finite correspondence · Steps 4–6 · Frobenius traces

Finite correspondence

  1. Rewrite the action map

    Shearing identifies the correspondence with a projection.

    $$[g,y]\longmapsto(gI,gy),\qquad LG\times^ILX\simeq\mathrm{Fl}\times LX,\qquad m=\mathrm{pr}_2.$$
    Proof
    1. The relation \((g,y)\sim(gi,i^{-1}y)\) leaves both \(gI\) and \(gy\) unchanged.
    2. The inverse sends \((gI,x)\) to \([g,g^{-1}x]\), independently of the representative.
    3. Both maps are defined over the residue field and commute with Frobenius.
    Sources

  2. Identify the kernel stalk

    The descended twisted external product has stalk

    $$K_{(gI,x)}\simeq\mathcal G_g\otimes M_{g^{-1}x}.$$
    Proof
    1. Pull back to \(LG\times LX\), where the twisted product becomes the ordinary external product.
    2. Apply the inverse shearing map. Equivariance identifies the stalks when \(g\) is replaced by \(gi\).
    3. Descent preserves this tensor description and its Frobenius action.
    Sources

  3. Choose a finite model

    Choose \(N\) supporting \(M\); (Dict)(d) supplies \(N'\) and \(M''\).

    $$\operatorname{supp}K\subseteq\mathrm{Fl}_{\le w}\times Y_{N'},\qquad K=\operatorname{pullback}(K_{N',M''}).$$
    Proof
    1. The support bound places every output in \(Y_{N'}\).
    2. The factorization requirement in (Dict)(d) makes the kernel depend on finitely many output coefficients.
    3. The descended projection is proper because \(\mathrm{Fl}_{\le w}\) is projective. Proper base change now computes the stalk at \(x\) on this finite-type model.
    Sources

Frobenius traces

  1. Apply the trace formula

    $$\operatorname{tr}(\mathcal G*_IM)(x)=\sum_{gI\in\mathrm{Fl}_{\le w}(\mathbb F_q)}\operatorname{tr}(\mathcal G_g)\operatorname{tr}(M_{g^{-1}x}).$$
    Proof
    1. Proper base change identifies the output stalk with the cohomology of the fibre.
    2. Grothendieck–Lefschetz replaces its alternating Frobenius trace by the sum of stalk traces over rational fibre points.
    3. The alternating trace of a tensor product is the product of alternating traces. The support has finitely many rational points.
    Sources

  2. Identify rational cosets

    $$\mathrm{Fl}(\mathbb F_q)=G(F)/I,\qquad\operatorname{tr}(\mathcal G*_IM)=\operatorname{tr}\mathcal G*\operatorname{tr}M.$$
    Proof
    1. The obstruction to representing a rational flag by a rational loop lies in \(H^1(\mathbb F_q,I)\).
    2. Lang's theorem kills this obstruction on the connected finite congruence quotients; compatible lifts through the pro-unipotent kernels give the same conclusion for \(I\).
    3. Each right coset has measure one under \(\operatorname{vol}I=1\). The sum in Step 4 is therefore the usual convolution integral.
    Sources

  3. Check a valuation translate

    For \(X=\mathbb A^1\), \(G=\mathrm{GL}_1\), the flag space is discrete.

    $$I=\mathcal O^\times,\quad\mathrm{Fl}(\mathbb F_q)=\{t^jI:j\in\mathbb Z\},\quad\mathcal G=\delta_{t^j},\quad M=\overline{\mathbb Q}_{\ell,\mathcal O}.$$
    Proof
    1. Only the coset \(t^jI\) contributes.
    2. The action is the isomorphism \(y\mapsto t^jy\) from \(\mathcal O\) onto \(t^j\mathcal O\).
    3. With the unshifted constant Weil structures,
    $$\delta_{t^j}*_IM=\overline{\mathbb Q}_{\ell,t^j\mathcal O},\qquad\mathbf1_{t^jI}*\mathbf1_{\mathcal O}(x)=\mathbf1_{\mathcal O}(t^{-j}x)=\mathbf1_{t^j\mathcal O}(x).$$

    For \(j=1\), the trace is one precisely when the constant coefficient of \(x\in\mathcal O\) vanishes. This is the trace calculation used in III.

    Sources