The extra multiplicative action

We wanted to construct this extension $$ 1 \to \mathbb{G}_ m \to \widehat{LG} \to LG \to 1. $$ Write $V = k((\varpi))^n = F^n$ and $\Lambda_0 = k[[\varpi]]^n = \mathcal{O}^n$. Given two $\mathcal{O}$-lattices $\Lambda_1, \Lambda_2 \subseteq V$, we may consider the relative determinant $$ \det(\Lambda_1 \mid \Lambda_2) = \det(\Lambda_1 / (\Lambda_1 \cap \Lambda_2)) \otimes \det(\Lambda_2 / (\Lambda_1 \cap \Lambda_2))^{-1}. $$ There is a canonical compatibility $$ \det(\Lambda_1 \mid \Lambda_2) \otimes \det(\Lambda_2 \mid \Lambda_3) \cong \det(\Lambda_1 \mid \Lambda_3), $$ and this also works in families. So for $G = \GL_n$, we have $$ \mathscr{L}_ \mathrm{det} / \mathrm{Gr}_ G = \lbrace \mathcal{O}\text{-lattices } \Lambda \subseteq F^n \rbrace $$ whose fiber over $\Lambda$ is $\det(\Lambda_0 \mid \Lambda)$. (This is the ample one.)

Now we have $LG$ acting on $\mathrm{Gr}_ G$. We want to make $\mathscr{L}_ \mathrm{det}$ equivariant with respect to this action, but this can’t be done. Basically, we need an isomorphism bewteen $$ \det(\Lambda_0 \mid g\Lambda) \cong \det(\Lambda_0 \mid \Lambda), $$ but this can’t be done canonically. To get this, we define an extension $$ 1 \to \mathbb{G}_ m \to \widehat{LG} \xrightarrow{\pi} LG \to 1 $$ so that we have $$ \pi^{-1}(g) = \det(g\Lambda_0 \mid \Lambda_0)^\times. $$ This can indeed be seen to be a nontrivial central extension.

Remark 1. From the construction we see that this canonically splits over $L^+G$.

For general $G$, we fix a faithful representation $\rho \colon G \hookrightarrow \GL_n$, and then we can pull back to get a central extension $$ 1 \to \mathbb{G}_ m \to (\widehat{LG})_ \rho \to LG \to 1. $$

Remark 2. This $(\widehat{LG})_ \rho$ is a nontrivial extension. For example, even after we restrict to the maximal torus $LT \to LG$, this is nontrivial. This is because we can compute the “commutator pairing” $LT \times LT \to \mathbb{G}_ m$ and see that this is nontrivial. This turns out to be essentially the tame symbol.

Remark 3. If $G$ is semisimple, there actually is a unique splitting $L^+G \to (\widehat{LG})_ \rho$, because there are no nontrivial characters $L^+G \to \mathbb{G}_ m$. On the other hand, the splitting over $I$ is not unique, but we can still use the one coming from $L^+G$, where this $L^+G$ comes from the pinning.

We now have a $\mathbb{G}_ m$-torsor over $I^+ \backslash LG / I^+$ which is $$ \tilde{T} \curvearrowright I^+ \backslash \widehat{LG} / I^+ \curvearrowleft \tilde{T}, $$ where we write $\tilde{T} = \hat{I} / I^+ \cong T \times \mathbb{G}_ m^\mathrm{cent}$. Now we can consider $$ \mathcal{H}_ \mathrm{mon} = \mathsf{Shv}_ \mathrm{mon}(I^+ \backslash \widehat{LG} / I^+). $$ Given two character sheaves $\chi$ and $\chi^\prime$ on $\tilde{T} \cong T \times \mathbb{G}_ m^\mathrm{cent}$, we have $$ {}_ {\chi\mathrm{-mon}} \mathcal{H}_ {\chi^\prime\mathrm{-mon}} = \mathsf{Shv}_ {\chi,\chi^\prime\mathrm{-mon}}(I^+ \backslash \hat{LG} / I^+). $$ These are naturally monoidal categories, where $m_\ast$ agrees with $m_!$ up to a shift as before.

Let us understand its geometry. We have for $w \in \tilde{W}$ the Schubert cell $$ i_w \colon I \backslash LG / I \hookrightarrow I \backslash IwI / I, $$ which is an affine locally closed embedding. We can restrict the $(\tilde{T} \times \tilde{T}) / \mathbb{G}_ m$-torsor $$ I^+ \backslash \widehat{LG} / I^+ \to I \backslash LG / I $$ and get the Schubert cell $$ i_w \colon I^+ \backslash \hat{I} w \hat{I} / I^+ \hookrightarrow I^+ \backslash \widehat{LG} / I^+. $$ Here, note that $\tilde{T}$ acts on both the left and the right. Now we can define $$ \mathcal{H}(w)_ \mathrm{mon} = \mathsf{Shv}_ \mathrm{mon}(I^+ \backslash \hat{I} w \hat{I} / I^+), $$ and similarly $$ {}_ {\chi\mathrm{-mon}} \mathcal{H}(w)_ {\chi\mathrm{-mon}}. $$

Lemma 4. Let $\dot{w}$ be a lift of $w$ to $\widehat{LG}$. Then we have an equivalence $$ \alpha_\dot{w} \colon \mathsf{Shv}_ \mathrm{mon}(\tilde{T}) \xrightarrow{\cong} \mathcal{H}(w)_ \mathrm{mon} , $$ given a choice of $\dot{w}$.

Definition 5. We define the functors $$ \Delta_{\dot{w}}^\mathrm{mon} = (i_w)_ ! \circ \alpha_{\dot{w}}, \quad \nabla_{\dot{w}}^\mathrm{mon} = (i_w)_ \ast \circ \alpha_{\dot{w}} \quad \colon \mathsf{Shv}_ \mathrm{mon}(\tilde{T}) \to \mathcal{H}_ \mathrm{mon}. $$

Then $\Delta_1^\mathrm{mon} = \nabla_1^\mathrm{mon}$, and the unit object of $\mathcal{H}_ \mathrm{mon}$ can be described as $\Delta_1^\mathrm{mon}(\tilde{\mathrm{Ch}})$.

There is $\tilde{W}$ acting on $\tilde{T}$, if we interpret $$ \tilde{W} = N_{LG}(LT) / L^+T = N_{\widehat{LG}}(\widehat{LT}) / L^+T $$ and now this acts on $\widehat{LT} / L^1 T = \tilde{T}$.

Claim 6. This action $\tilde{W} \hookrightarrow \Aut(\tilde{T})$ is faithful.

Essentially, if we calculate this action, we see that for $\tilde{W} = X_\ast(T) \rtimes W$ we already know that $W$ acts nontrivially. For the $X_\ast(T)$ part, if we consider $\lambda(\varpi) \in LT$ for $\lambda \in X_\ast(T)$, after choosing a lift we compute $$ \widetilde{\lambda(\varpi)} (t, c) \widetilde{\lambda(\varpi)}^{-1} = (t, c) (1, \operatorname{Comm}(\lambda(\varpi), t)). $$ But we have computed that this commutator pairing is $$ \operatorname{Comm}(\lambda(\varpi), \mu(a)) = (-1)^\cdots \lbrace \varpi, a \rbrace^{B(\lambda, \mu)} = (-1)^\cdots a^{-B(\lambda,\mu)} $$ for $a \in k^\times$. This we can see is nontrivial.

Lemma 7. If $v, w \in \tilde{W}$ and $\ell(w) + \ell(v) = \ell(wv)$, then 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 $$ \nabla_{\dot{w}}^\mathrm{mon}(\mathscr{L}) \star^u \nabla_{\dot{v}}^\mathrm{mon}(\mathscr{L}^\prime) \cong \nabla_{\dot{w}\dot{v}}^\mathrm{mon}(\mathscr{L} \star w(\mathscr{L}^\prime)) $$

Lemma 8. We have $$ \Delta_{\dot{w}}^\mathrm{mon}(\mathscr{L}) \star \nabla_{\dot{w}^{-1}}^\mathrm{mon}(\mathscr{L}^\prime) = \Delta_1^\mathrm{mon}(\mathscr{L} \star \dot{w}(\mathscr{L}^\prime)). $$