Hypotheses

• S1 · Invertible Euler characteristic.

  The order of the Weyl group is invertible in the coefficient field: $$|W|\in k^\times.$$

  Discussion

  Automatic for the paper, where \(k\) has characteristic zero. It is the only place the characteristic enters the key lemma; the geometric inputs (base change, projection formula, \(\pi^!=\pi^*[2d]\)) hold for any coefficients.

Proof

Step 1 reduces to a retraction of \(u\) on \(\mathbb BG\). Steps 2 and 5 compute \(q_*q^*\mathcal F\) from the square. Steps 3, 4 and 6 construct the retraction of \(u\) as a normalized trace and explain why the adjunction counit cannot play that role. Step 7 assembles \(\beta_{\mathcal F}\).

1. Purpose Reduce to the classifying stack

   If \(t:\mathcal A\to\omega_{\mathbb BG}\) satisfies \(t\circ u=\mathrm{id}\), and \(q_*q^*\mathcal F\simeq\gamma^!\mathcal A\otimes^!\mathcal F\) carries \(\eta_{\mathcal F}\) to \(\gamma^!u\otimes^!\mathrm{id}\), then \(\gamma^!t\otimes^!\mathrm{id}\) retracts \(\eta_{\mathcal F}\).

   Proof

   • \(\gamma^!\) is a functor, so \(\gamma^!t\circ\gamma^!u=\gamma^!(\mathrm{id})=\mathrm{id}_{\omega_{I\backslash\mathrm{Gr}}}\), and \(\omega_{I\backslash\mathrm{Gr}}\otimes^!\mathcal F=\mathcal F\).
2. Result The square is Cartesian

   \(I\backslash\mathrm{Fl}\simeq I\backslash\mathrm{Gr}\times_{\mathbb BG}\mathbb BB\), so \(q\) is the base change of \(\pi\) along \(\gamma\); it is proper and smooth of relative dimension \(d\).

   Proof

   • \(\mathrm{Fl}=LG/I=LG\times^K(K/I)\) is the bundle with fibre \(K/I\) associated with the \(K\)-torsor \(LG\to\mathrm{Gr}\); that is, \(\mathrm{Fl}=\mathrm{Gr}\times_{\mathbb BK}\mathbb BI\).
   • \(I=\mathrm{ev}^{-1}(B)=K\times_GB\), so \(K/I\simeq G/B\) and \(\mathbb BI=\mathbb BK\times_{\mathbb BG}\mathbb BB\). Hence \(\mathrm{Fl}=\mathrm{Gr}\times_{\mathbb BG}\mathbb BB\), with \(\gamma\) the composite \(\mathrm{Gr}\to\mathbb BK\to\mathbb BG\).
   • The left \(I\)-action is through the first factor and commutes with everything, so the same holds after the quotient.
   • Likewise \(I\backslash LX=\mathbb BI\times_{\mathbb BK}K\backslash LX=K\backslash LX\times_{\mathbb BG}\mathbb BB\), so \(p:I\backslash LX\to K\backslash LX\) is the base change of \(\pi\) along \(K\backslash LX\to\mathbb BK\to\mathbb BG\): the instance used in step 4 of the Proposition.
3. Result Lemma 1 · \(\mathcal A\) is an algebra with unit \(u\)

   \(\pi^!\simeq\pi^*[2d]\), so \(\pi^!\dashv\pi_*[-2d]\); the right adjoint of the symmetric monoidal functor \(\pi^!\) is lax symmetric monoidal, hence

   $$\mathcal A=\pi_*[-2d]\big(\omega_{\mathbb BB}\big)\ \text{is a commutative algebra in }(\mathcal D(\mathbb BG),\otimes^!),\qquad\text{unit }u:\omega_{\mathbb BG}\to\pi_*\pi^!\omega_{\mathbb BG}[-2d]=\mathcal A.$$
   Proof

   • \(\pi\) is smooth of relative dimension \(d\), so \(\pi^*=\pi^![-2d]\) is left adjoint to \(\pi_*\) (DG §6.1.9); shifting, \(\pi^!\dashv\pi_*[-2d]\).
   • \(\pi^!\) preserves \(\otimes^!\) and units. By HA, Corollary 7.3.2.7 its right adjoint is lax symmetric monoidal, so it carries the unit algebra \(\omega_{\mathbb BB}\) to an algebra whose unit is the adjunction unit at \(\omega_{\mathbb BG}\).
   • The note writes “relative dim \(2d\)” and then \(\pi^!\simeq\pi^*[2d]\); with \(d=\dim G/B\) the second line is the correct one.
4. Result The adjunction counit is not the retraction

   The counit of \(\pi_!\dashv\pi^!\) at \(\omega_{\mathbb BG}\) is \(c:\mathcal A\to\omega_{\mathbb BG}[-2d]\), and \(c\circ u=0\) for \(d\gt0\). The note's “counit map \(\operatorname{tr}\)” has to be the trace of the next steps.

   Proof

   • \(\pi_!=\pi_*\) and \(\pi^!\omega_{\mathbb BG}=\omega_{\mathbb BB}\), so the counit is \(\pi_*\omega_{\mathbb BB}=\mathcal A[2d]\to\omega_{\mathbb BG}\).
   • \(\operatorname{Hom}(\omega_{\mathbb BG},\omega_{\mathbb BG}[-2d])=\operatorname{Hom}(\mathbf 1,\mathbf 1[-2d])=H^{-2d}_{\mathrm{dR}}(\mathbb BG)=0\). On a point: \(k\to C^*(G/B)\to k[-2d]\) is the inclusion of \(H^0\) followed by integration, which is zero.
5. Result (2)–(3) · \(q_*q^*\mathcal F\simeq\gamma^!\mathcal A\otimes^!\mathcal F\)

   For \(\mathcal F\in\mathcal D_c(I\backslash\mathrm{Gr})\),

   $$q_*q^*\mathcal F\simeq\gamma^!\mathcal A\otimes^!\mathcal F\quad\text{(2)},\qquad\eta_{\mathcal F}=\gamma^!u\otimes^!\mathrm{id}_{\mathcal F}\ \text{under (2)}\quad\text{(3)}.$$
   Proof

   • \(q^*\mathcal F=q^!\mathcal F[-2d]=(q^!\mathcal F\otimes^!\omega_{I\backslash\mathrm{Fl}})[-2d]\), and the projection formula for the proper map \(q\) gives \(q_*(q^!\mathcal F\otimes^!\omega_{I\backslash\mathrm{Fl}})\simeq\mathcal F\otimes^!q_*\omega_{I\backslash\mathrm{Fl}}\).
   • Base change in the Cartesian square (DG, Proposition 7.6.8): \(q_*\omega_{I\backslash\mathrm{Fl}}=q_*\gamma'^!\omega_{\mathbb BB}\simeq\gamma^!\pi_*\omega_{\mathbb BB}\). Shift by \([-2d]\) to get (2).
   • (3): the identification is natural in \(\mathcal F\), and the unit of \(q^*\dashv q_*\) corresponds to \(\gamma^!\) of the unit of \(\pi^*\dashv\pi_*\), by compatibility of the base change isomorphism with adjunction units (Beck–Chevalley).
6. Method (1) · The normalized trace retracts \(u\)

   \(\mathcal A\) is dualizable with \(\mathcal A^\vee=\mathcal A[2d]\), and

   $$\operatorname{tr}_{\mathcal A}\circ u=\dim\mathcal A=\chi(G/B)=|W|=n,\qquad\tfrac{1}{n}\gamma^!\operatorname{tr}_{\mathcal A}\circ\gamma^!u=\mathrm{id}_{\omega_{I\backslash\mathrm{Gr}}}\quad\text{(1)}.$$
   Proof

   • Dualizable. For \(\mathcal G\in\mathcal D(\mathbb BG)\), the projection formula and the adjunctions \(\pi_!\dashv\pi^!\), \(\pi^*\dashv\pi_*\) give \(\underline{\operatorname{Hom}}^!(\pi_*\omega_{\mathbb BB},\mathcal G)\simeq\pi_*\pi^!\mathcal G[-2d]\simeq\pi_*\omega_{\mathbb BB}[-2d]\otimes^!\mathcal G\), naturally in \(\mathcal G\); so \(\pi_*\omega_{\mathbb BB}\) is dualizable with dual \(\pi_*\omega_{\mathbb BB}[-2d]\), and \(\mathcal A^\vee=\mathcal A[2d]\).
   • Trace of the unit. \(m\circ(u\otimes\mathrm{id})=\mathrm{id}_{\mathcal A}\), so \(\operatorname{tr}_{\mathcal A}\circ u=\mathrm{ev}\circ s\circ\mathrm{coev}=\operatorname{tr}(\mathrm{id}_{\mathcal A})=\dim\mathcal A\in\operatorname{End}(\omega_{\mathbb BG})=k\) (Ponto–Shulman, Definition 2.2).
   • Value. \(i^!\) for \(i:\mathrm{pt}\to\mathbb BG\) is symmetric monoidal, preserves traces, and is the identity on \(\operatorname{End}(\mathbf 1)=k\). By base change \(i^!\mathcal A=C^*(G/B)\), a perfect complex whose cohomology has one Schubert class per \(w\in W\), all in even degrees (Brion, Proposition 1.2.1 and §1.3). Its Euler characteristic is \(|W|\) (Ponto–Shulman, Example 3.3).
   • By S1, \(n=|W|\) is invertible; apply \(\gamma^!\) to \(\tfrac{1}{n}\operatorname{tr}_{\mathcal A}\circ u=\mathrm{id}\) to get (1).
7. Conclusion (4) · Define \(\beta_{\mathcal F}\) and conclude

   Let \(\beta_{\mathcal F}\) be (2) followed by \(\tfrac{1}{n}\gamma^!\operatorname{tr}_{\mathcal A}\otimes^!\mathrm{id}_{\mathcal F}\). Then by (3) and (1),

   $$\beta_{\mathcal F}\circ\eta_{\mathcal F}=\big(\tfrac{1}{n}\gamma^!\operatorname{tr}_{\mathcal A}\circ\gamma^!u\big)\otimes^!\mathrm{id}_{\mathcal F}=\mathrm{id}_{\mathcal F}.$$
   Proof

   • All maps are natural in \(\mathcal F\), so \(\beta\) is a natural transformation \(q_*q^*\Rightarrow\mathrm{id}\) retracting \(\eta\).
   • \(q^*\mathcal F\) is coherent because \(q\) is smooth; this is what the Proposition's step 7 uses.