Last time we define the category $$ \mathcal{H}_ \mathrm{mon} = \mathsf{Shv}_ \mathrm{mon}(I^+ \backslash \widetilde{LG} / I^+). $$ Given $w \in \tilde{W}$, we also had $$ i_w \colon I^+ \backslash \widetilde{LG} / I^+ \xrightarrow{j_w} I^+ \backslash \widetilde{LG} / I^+ $$ and then for a choice of lift $\dot{w}$ we had $$ \alpha_{\dot{w}} \colon \mathsf{Shv}_ \mathrm{mon}(\tilde{T}) \xrightarrow{\cong} \mathsf{Shv}_ \mathrm{mon}(I^+ \backslash \widetilde{LG}_ w / I^+). $$ Then we defined $$ \Delta_{\dot{w}}^\mathrm{mon} = (i_w)_ ! [-\ell(w)] \circ \alpha_{\dot{w}}, \quad \nabla_{\dot{w}}^\mathrm{mon} = (i_w)_ \ast [-\ell(w)] \circ \alpha_{\dot{w}}. $$
Proposition 1. 1. If $\ell(wv) = \ell(w) + \ell(v)$ we have $$ \Delta_{\dot{w}}^\mathrm{mon}(\mathscr{L}) \star^u \Delta_{\dot{v}}^\mathrm{mon}(\mathscr{L}^\prime) \cong \Delta_{\dot{w}\dot{v}}^\mathrm{mon}(\mathscr{L} \star w(\mathscr{L}^\prime)), $$ and similarly for $\nabla$.
- We have $$ \Delta_{\dot{w}}^\mathrm{mon}(\mathscr{L}) \star^u \nabla_{\dot{w}}^\mathrm{mon}(\mathscr{L}^\prime) \cong \Delta_e^\mathrm{mon}(\mathscr{L} \star w(\mathscr{L}^\prime)). $$
- For $s$ a simple reflection there is a fiber sequence (not canonically because $\dot{s}^2 \neq e$) $$ \begin{align} \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L} \star s(\mathscr{L}^\prime)) \oplus \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L} \star s(\mathscr{L}^\prime))[-1] &\to \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L}) \star^u \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L}^\prime) \br &\to \Delta_e^\mathrm{mon}(\mathscr{L} \star s(\mathscr{L}^\prime)). \end{align} $$
Where does the third part come from? Write $$ \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L}^\prime) \to \nabla_{\dot{s}}^\mathrm{mon}(\mathscr{L}^\prime) \to \mathscr{F}. $$ Then we have $$ \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L}) \star^u \Delta_{\dot{s}}^\mathrm{mon}(\mathscr{L}^\prime) \to \Delta_e^\mathrm{mon}(\mathscr{L} \star s(\mathscr{L}^\prime)) \to \Delta_s^\mathrm{mon}(\mathscr{L}) \star^u \mathscr{F}. $$ Here, we note that $\mathscr{F}$ is supported on the closed stratum $e$, so the right term is supported on the $s$ stratum. So actually, this right term is just the restriction of $\Delta_s^\mathrm{mon}(\mathscr{L}) \star^u \Delta_s^\mathrm{mon}(\mathscr{L}^\prime)$ to the $s$-stratum. We can now directly compute that this is some compactly supported cohomology of $\mathbb{G}_ m$.
Block decomposition
For simplicity, we assume that $\Lambda$ is an algebraically closed field. We recall that we have $$ \mathsf{Shv}_ \mathrm{mon}(\tilde{T}) \cong \operatorname{IndCoh}(\mathcal{R}_ {I_F^t, \hat{\tilde{T}}}). $$ On the other hand, for $\chi \in \hat{\tilde{T}}$ of order prime to $p$, we had this component $\hat{\chi} \hookrightarrow \mathcal{R}_ {I_F^t,\hat{\tilde{T}}}$ and then we had $$ \mathsf{Shv}_ \mathrm{mon}(\tilde{T}) = \bigoplus_\chi \mathsf{Shv}_ {\chi\mathrm{-mon}}(\tilde{T}), \quad \mathsf{Shv}_ {\chi\mathrm{-mon}}(\tilde{T}) \cong \mathsf{IndCoh}(\hat{\chi}). $$
Now because $\tilde{W}$ acts on $\hat{\tilde{T}}$, given $\chi, \chi^\prime \in \hat{\tilde{T}}$ we can define $$ {}_ \chi \tilde{W}_ {\chi^\prime} = \lbrace w \in \tilde{W} : \chi = w \chi^\prime \rbrace. $$ We have a Coxeter subgroup $$ {}_ \chi \tilde{W}^0_\chi = \langle s_\alpha : \alpha^\vee \colon \mathbb{G}_ m \to \tilde{T} \text{ affine coroot}, (\alpha^\vee)^\ast \mathrm{Ch}_ \chi \text{ is trivial} \rangle \subseteq {}_ \chi \tilde{W}_ \chi \to \Omega_\chi. $$ We then have a similar combinatorics $$ \tilde{W}_ \chi^0 \backslash {}_ \chi \tilde{W}_ {\chi^\prime} \cong {}_ \chi \tilde{W}_ {\chi^\prime} / \tilde{W}^0_{\chi^\prime} \cong {}_ \chi \Omega_{\chi^\prime} \hookrightarrow {}_ \chi \tilde{W}_ {\chi^\prime}, $$ where this can be thought of as the subset of minimal length elements in both types of cosets. When $\chi, \chi^\prime$ are trivial on the central $\mathbb{G}_ m^\mathrm{cen} \subseteq \tilde{T} = \mathbb{G}_ m^\mathrm{cen} \times T$, this reduces to the previous combinatorics.
Definition 2. We let ${}_ \chi \mathscr{H}_ {\chi^\prime} \subseteq \mathscr{H}_ \mathrm{mon}$ be the full subcategory generated by $\Delta_w^\mathrm{mon}(\mathrm{Ch}_ \chi)$ for $w \in {}_ \chi \tilde{W}_ {\chi^\prime}$. For $\beta \in {}_ \chi \Omega_{\chi^\prime}$, we also let $$ {}_ \chi \mathcal{H}_ {\chi^\prime}^\beta \subseteq {}_ \chi \mathcal{H}_ {\chi^\prime} $$ be the full subcategory generated by $\Delta_w^\mathrm{mon}(\mathrm{Ch}_ \chi)$ for $w \in {}_ \chi \tilde{W}_ {\chi^\prime}^\beta$.
Theorem 3. We have a decomposition $$ \mathcal{H}_ \mathrm{mon} = \bigoplus_{\chi,\chi^\prime,\beta} {}_ \chi \mathcal{H}_ {\chi^\prime}^\beta. $$
Remark 4. When $\Lambda$ is not a field, and something like $\bar{\mathbb{Z}}_ \ell$, the blocks will be something like characters over $\bar{\mathbb{F}}_ \ell$. But we will still have a similar picture.
Lemma 5. Fix $\chi$ and let $s$ be a simple reflection. If $s \notin {}_ \chi \tilde{W}^0_\chi$ then we have $$ \Delta_{\dot{s}}^\mathrm{mon}(\mathrm{Ch}_ \chi) \cong \nabla_{\dot{s}}^\mathrm{mon}(\mathrm{Ch}_ \chi). $$
We can reduce to the case of $\mathrm{SL}_ 2 \hookrightarrow \widetilde{LG}$, which sends $$ \begin{pmatrix} t \br & t^{-1} \end{pmatrix} \to \tilde{T}, \quad \begin{pmatrix} 1 & \ast \br & 1 \end{pmatrix} \to U_\alpha. $$ What this reduces to is, given the diagram $$ \begin{CD} \mathbb{A}^1 \times \mathbb{G}_ m @>>> \mathbb{A}^2 - \lbrace (0,0) \rbrace \br @VVV @VVV \br \mathbb{A}^1 @>{j_s}>> \mathbb{P}^1, \end{CD} $$ if $\mathrm{Ch}_ \chi$ is a nontrivial Kummer sheaf then $$ (j_s)_ ! (\Lambda_ {\mathbb{A}^1} \boxtimes \mathrm{Ch}_ \chi) = (j_s)_ \ast (\Lambda_ {\mathbb{A}^1} \boxtimes \mathrm{Ch}_ \chi). $$ The difference really happens around $0 \in \mathbb{P}^1$, so we can just understand what is happening here. If we throw out $\infty \in \mathbb{P}^1$ and look at what is happening, this boils down to the fact that if we look at $$ \mathbb{G}_ m \times \mathbb{G}_ m \hookrightarrow \mathbb{A}^1 \times \mathbb{G}_ m, $$ the extension of $\mathrm{Ch}_ \chi \boxtimes \mathrm{Ch}_ \chi$ is clean.
Lemma 6. The convolution $(-) \star^u \Delta_{\dot{s}}^\mathrm{mon}(\tilde{\mathrm{Ch}})$ sends ${}_ \chi \mathcal{H}_ {\chi^\prime}^\beta \to {}_ \chi \mathcal{H}_ {s\chi^\prime}^{\beta^\prime}$ where $[\beta^\prime] = [\beta] s$.
First assume that $s \in {}_ {\chi^\prime} \tilde{W}^0_ {\chi^\prime}$. Then for $w \in {}_ \chi \tilde{W}^\beta_ {\chi^\prime}$ we also have $ws \in {}_ \chi \tilde{W}^\beta_ {\chi^\prime}$ and so we compute $$ \Delta_w^\mathrm{mon}(\mathrm{Ch}_ \chi) \star^u \Delta_s^\mathrm{mon}(\tilde{\mathrm{Ch}}) = \begin{cases} \Delta_{ws}^\mathrm{mon}(\mathrm{Ch}_ \chi), & \ell(ws) = \ell(w) + 1 \br \Delta_{ws}^\mathrm{mon}(\mathrm{Ch}_ \chi) \star^u \Delta_s^\mathrm{mon}(\tilde{\mathrm{Ch}}) \star^u \Delta_s^\mathrm{mon}(\tilde{\mathrm{Ch}}), & \ell(ws) = \ell(w) - 1 \end{cases} $$ and we can check that in both cases this is in ${}_ \chi \mathcal{H}^\beta_ {\chi^\prime}$.
In the second case, we consider the case when $s \notin \tilde{}_ {\chi^\prime} \tilde{W}^0_ {\chi^\prime}$. Then we have $\Delta_s^\mathrm{mon}(\mathrm{Ch}_ {\chi^\prime\mathrm{-mon}}) = \Delta_s^\mathrm{mon}(\mathrm{Ch}_ {\chi^\prime\mathrm{-mon}})$ and so we compute $$ \Delta_w^\mathrm{mon}(\mathrm{Ch}_ \chi) \star^u \nabla_s^\mathrm{mon}(\mathrm{Ch}_ {\chi^\prime\mathrm{-mon}}) = \begin{cases} \Delta_{ws}^\mathrm{mon}(\mathrm{Ch}_ \chi \star \mathrm{Ch}_ {w\chi^\prime\mathrm{-mon}}) & \ell(ws) = \ell(w) + 1 \br \Delta_{ws}^\mathrm{mon}(\mathrm{Ch}_ \chi) & \ell(ws) = \ell(w) - 1. \end{cases} $$ In both cases we land in ${}_ \chi \mathcal{H}^{\beta^\prime}_ {s\chi^\prime}$.
We can now prove the block decomposition theorem. It suffices to show that for $v \in {}_ {\chi_1} \tilde{W}^{\beta_1}_ {\chi_1^\prime}$ and $w \in {}_ {\chi_2} \tilde{W}^{\beta_2}_ {\chi_2^\prime}$ with ${}_ {\chi_1} \tilde{W}^{\beta_1}_ {\chi_1^\prime} \neq {}_ {\chi_2} \tilde{W}^{\beta_2}_ {\chi_2^\prime}$ we have $$ R\Hom(\Delta_v^\mathrm{mon}(\mathrm{Ch}_ {\chi_1}), \Delta_w^\mathrm{mon}(\mathrm{Ch}_ {\chi_2})) = 0. $$ We prove this by induction on $\ell(w)$. When $\ell(w) = 0$ we have $$ R\Hom(\Delta_v^\mathrm{mon}(\mathrm{Ch}_ {\chi_1}), \Delta_w^\mathrm{mon}(\mathrm{Ch}_ {\chi_2})) = R\Hom(i_w^\ast(\Delta_v^\mathrm{mon}(\mathrm{Ch}_ {\chi_1})), \mathrm{Ch}_ {\chi_2}), $$ which is zero unless $w = v$ and $\chi_1 = \chi_2$, in which case we are in the same block.
For the inductive step, we write $w = xs$ for $\ell(w) = \ell(x) + 1$. Then we have $$ \Delta_w^\mathrm{mon}(\mathrm{Ch}_ {\chi_2}) = \Delta_x^\mathrm{mon}(\mathrm{Ch}_ {\chi_2}) \star^u \Delta_s^\mathrm{mon}(\tilde{\mathrm{Ch}}) $$ and so what we need to show is $$ R\Hom(\Delta_w^\mathrm{mon}(\mathrm{Ch}_ {\chi_1}) \star \nabla_s^\mathrm{mon}(\tilde{\mathrm{Ch}}), \Delta_x^\mathrm{mon}(\mathrm{Ch}_ {\chi_2})) = 0. $$ This follows from the induction hypothesis, because the blocks of the right side doesn’t intersect the blocks of the left side.