Spherical Hecke algebra and classical Satake isomorphism

The Satake isomorphism is a way to identify the spherical Hecke algebra of a reductive group $G$ with the invariant part (under the action of the Weyl group) of cocharacters (also called one-parameter subgroups) and both are isomorphic to the Grothendieck ring of the category of the Langlands dual group $G^{\vee}$. This fits perfectly with the so-called Langlands duality philosophy, suggesting that algebraic objects associated with $G^{\vee}$ should be captured by topological objects associated with $G$.

More concretely, let $F$ be a non-archimedian local field, and $O$ the ring of integers, and let $G$ be a split reductive group over $\mathcal{O}$. Let $T \subset G$ be a maximal torus and $X_{\ast}(T) = \operatorname{Hom}(\mathbb{G}_m,T)$ the group of cocharacters. The group ring $\mathbb{Z}[X_{\ast}(T)]$ is endowed with an action of the Weyl group $W$. Let $q$ be the cardinality of the residue field of $K$, the Satake isomorphism reads

$$\mathbb{Z}_c[G(\mathcal{O}) \setminus G(K)/G(\mathcal{O})] \otimes_{\mathbb{Z}} \mathbb{Z}[q^{1/2},q^{-1/2}] \simeq \mathbb{Z}[X_{\ast}(T)]^W  \otimes_{\mathbb{Z}} \mathbb{Z}[q^{1/2},q^{-1/2}]$$ where the LHS is the spherical Hecke algebra, i.e., the ring of compactly support functions $G(F) \longrightarrow \mathbb{Z}$ that is left and right invariant under the action of $G(\mathcal{O})$. The Satake isomorphism nowadays has many offsprings, ranging from the complex, $\ell$-adic to motivic one. In this post, I'd like to stick to the classical one. At first, one might wonder why do we care about the Hecke algebras? One answer is dated back to automorphic representations. Shortly speaking, the spherical Hecke algebra is the algebra of spherical representations, in some sense, they are simplest smooth representations of the adeles. They are in bijection with classes of irreducible modules over the Hecke algebra. I devote this blog for some basic materials to understand the Satake isomorphism, mostly around spherical Hecke algebras.

Some facts on reductive groups 

As before, let $F$ be a local field with ring of integers $\mathcal{O}$ and a fixed uniformizer $\omega$. Let $G$ be a connected, reductive group over $\mathcal{O}$. We fix a split maximal torus $T$ contained in a Borel $B$, all of them are defined over $\mathcal{O}$. Let $W = N_G(T)/T$ be the Weyl group, where $N_G(T)$ is the normalization of $T$ in $G$. The groups 

$$X^{\ast}(T) = \operatorname{Hom}(T,\mathbb{G}_m) \ \ \ \ X_{\ast}(T) = \operatorname{Hom}(\mathbb{G}_m,T)$$ are groups of characters and cocharacters, respectively. They are free-abelian group and pair up to 

$$\left<\square,\square \right> \colon X^{\ast}(T) \times X_{\ast}(T) \longrightarrow \mathbb{Z}.$$ The group $X^{\ast}(T)$ contains roots $\Phi$, namely, those characters with non-trivial weight spaces in the adjoint representation of $\mathfrak{g}$, the Lie algebra of $G$. The choice of Borel $B$ leads to positive roots $\Phi^+$ and negative roots $\Phi^-$ such that $\Phi = \Phi^+ \cup \Phi^-$. We are also interested in 

$$X_{\ast}(T)^+ = \left \{\lambda \in X_{\ast}(T) \mid \left< \alpha,\lambda \right> \geq 0 \ \forall \ \alpha \in \Phi^+ \right \}$$ the set of positive coroots. There is a partial order on $X_{\ast}(T)$ by declaring that $\mu \leq \lambda$ if $\lambda - \mu \in \mathbb{Z}_{\geq 0}X_{\ast}(T)^+$.  

Example. The standard example is of course $\mathrm{GL}_n$. Then $T,B$ can be taken to be diagonal matrices and upper triangular matrices, respectively. Thus, 

$$X^{\ast}(T) \simeq \mathbb{Z}^n  = \mathbb{Z}e_1 \oplus \cdots \oplus \mathbb{Z}e_n$$ with $e_i$ the canonical basis and

$$X_{\ast}(T) \simeq \mathbb{Z}^n =   \mathbb{Z}e_1^{\vee} \oplus \cdots \oplus \mathbb{Z}e_n^{\vee}$$ with $e_i^{\vee}$ the dual basis. With this, one can easily see that

$$X_{\ast}(T)^+ = \left \{(\lambda_1,...,\lambda_n) \in \mathbb{Z}^n \mid \lambda_1 \geq \cdots \geq \lambda_n \right \}.$$

Any cocharacter $\lambda \in X_{\ast}(T)$ induces a morphism (by taking $F$-points) $\lambda \colon F^{\times} \longrightarrow T(F)$. Let $\lambda(\omega) \in T(F)$ be the image of $\omega$ under this morphism.  

Theorem (Cartan decomposition).  There is a decomposition 

$$G(F) = \coprod_{\lambda \in X_{\ast}(T)^+} G(\mathcal{O})\lambda(\omega) G(\mathcal{O})$$

 for any reductive group $G$ over $\mathcal{O}$. In other words, for any element $A \in G(F)$, there is a unique $\lambda \in X_{\ast}(T)^+$ and elements $X,Y \in G(\mathcal{O})$ so that $A = X\lambda(\omega)Y$.   

If you are only interested in classical groups, there is an elementary proof for this. The standard example is when $G = \mathrm{GL}_n$, the other classical groups are treated completely similar. In this case, the Cartan decomposition is a consequence of Gauss elimination. More concretely, we need Smith normal form

Theorem (Smith normal form). Let $R$ be a PID and $A \in \mathrm{M}_n(R)$ a matrix. There exists $X,Y \in \mathrm{GL}_n(R)$ such that

$$XAY =  \begin{pmatrix}
d_1 &  &  &  \\
 & d_2 &  &  \\
 &  & \cdots &  \\
 &  &  & d_n \\
\end{pmatrix}$$ where $d_i$ divides $d_{i+1}$. Moreover, $d_i$ are uniquely determined, up to units in $R$.  

Proof of Cartan decomposition for $\mathrm{GL}_n$. Let $A \in \mathrm{GL}_n(F) = \mathrm{GL}_n(\mathcal{O}[\omega^{-1}])$. Clearing denominators, there exists a $N > 0$ such that $\omega^N A \in \mathrm{GL}_n(\mathcal{O})$. Apply Smith normal form to $\omega^N A$, there exists $\overline{X},\overline{Y} \in \mathrm{GL}_n(\mathcal{O})$ so that $\overline{X}\omega^N A \overline{Y}$ is diagonal $\mathrm{diag}(d_1,...,d_n)$ with $d_i$ in $\mathrm{GL}_n(\mathcal{O})$ and $d_i$ dividing $d_{i+1}$. 

Since the determinant of $\omega^N A$ is not zero, diagonal entricies $d_i$ are non-zero. Therefore we can write $d_i = u_i \omega^{\lambda_i'}$ with $\lambda_i' \geq \lambda_{i+1}' \geq 0$. Finally, if we set $\lambda_i = \lambda_i' - N$, we get 

$$XAY =  \begin{pmatrix}\omega^{\lambda_1} &  &  &  \\
 & \omega^{\lambda_2} &  &  \\
 &  & \cdots &  \\
 &  &  & \omega^{\lambda_n} \\
\end{pmatrix}$$ with $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \in \mathbb{Z}$. The tuplet $(\lambda_1,...,\lambda_n)$ is clearly identified with a positive coroot as we discussed at the beginning. 

One more decomposition that we need is 

Proposition (Iwasawa decomposition). $G(F) = B(F)G(\mathcal{O})$. 

Unfortunately, I do not know any elementary proof of this without Bruhat-Tits theory. Even in the archimedean case $\mathrm{SL}_n(\mathbb{R})$, the elementary proof using linear algebra is already cubersome. 

Haar measures on locally compact groups

Let $G$ be a locally compact group, there exists a positive Radon measure $d_r g$ on $G$ that is right invariant under the action of $G$:

$$\int_G f(gx)d_r g = \int_G f(g)d_r g$$ for any $x \in G$ and $f \in \mathbb{C}_c(G)$, a compactly supported continuous function $f \colon G \longrightarrow \mathbb{C}$. Such a measure is uniquely determined up to a scalar in $\mathbb{R}_{>0}$, called right Haar measures. Similarly, there exists left Haar measures $d_l g$. If they coincide, we say $G$ is unimodular. 

We shall be interested in the modular quasi-character, which measures the failure of $G$ to be unimodular. Since $d_rg$ is defined up to a scalar, for any $h \in G$, there exists $\delta_G(h) \in \mathbb{R}_{>0}$ such that

$$d_r(hg) = \delta_G(h)d_r(g).$$ or alternatively, $d_l(h^{-1}) = \delta_G(h)d_l(h)$ since $d_l(h^{-1})$ is a right Haar measure. One sees immediately that $\delta_G \colon G \longrightarrow \mathbb{R}_{>0}$ is a homomorphism and it is trivial iff $G$ is unimodular.

Example. Let $F$ be a non-archimedean local field with ring of integers $\mathcal{O}$, a uniformizer $\omega$. There exists a Haar measure $dx$ on $F$. If we normalize $dx(\mathcal{O})=1$, we then can compute $dx(a + \omega^k \mathcal{O}) = dx(\omega^k\mathcal{O}) = 1/\left| \omega \right|^k$.  

Example. Let $F,\mathcal{O}$ be as above. Let $G$ be a smooth affine algebraic group over $\mathcal{O}$. If we endow $G(F)$ with the topology whose a basis of open neighborhoods of the identity is  

$$\operatorname{Ker}(G(\mathcal{O}) \longrightarrow G(\mathcal{O}/\omega^n \mathcal{O})).$$ This group is endowed with a unique Haar measure $\mu$ making $G(\mathcal{O})$ volume $1$. For integrals below, we implicitly understand that they are taken with respect to this measure. 

Example. Let us now turn to global field. It is sometimes convenient to explicitly describe Haar measures on $G(F_v)$ with $G$ a smooth affine algebraic group over $F$ a global field and $v$ a place. Let $\mathfrak{g} = \mathrm{Lie}(G)$ the Lie algebrai of $G$ and $n = \dim(G)$. There exists a top differential form

$$\omega_l \in \wedge^n g^{\vee}$$ uniquely defined up to an element of $F^{\times}$. There is a canonical inclusion $\wedge^n g^{\vee} \subset \wedge^n g^{\vee}_{F_v}$ and we still denote by $\omega_l$ the image of this differential form. Thus, we can define a Radon measure

$$\left| \omega_l \right|_v \colon \mathbb{C}_c(G(F_v)) \longrightarrow\mathbb{C}$$ given by

$$f \longmapsto \int_{G(F_v)} f(g)\left| \omega_l \right|_v(g).$$ Since $ \omega_l $ is left $G(F_v)$-invariant, the measure $\left| \omega_l \right|_v$ is left $G(F_v)$-invariant; thus a left Haar measure. 

Proposition. For the modular quasi-character $\delta$, one has

$$\delta_{G(F_v)}(g) = \left| \operatorname{det} \left( \operatorname{Ad}(g) \colon g \otimes_F F_v \longrightarrow g \otimes_F F_v\right) \right|_v.$$ Proof. The adjoint representation

$$\operatorname{Ad} \colon G \longrightarrow \mathrm{GL}_{\mathfrak{g}}$$ induces a representation (via the transpose of matricies)

$$\operatorname{Ad}^{\vee} \colon G \longrightarrow \mathrm{GL}_{\mathfrak{g}^{\vee}}.$$ Now for $g \in G$, the operator $\operatorname{Ad}^{\vee}(g^{-1})$ acts on $\mathfrak{g}^{\vee}$ and hence acts on $\wedge^n \mathfrak{g}^{\vee}$. We claim that

$$\operatorname{Ad}^{\vee}(g^{-1})(\omega_{\ell}) = c(g)\omega_{\ell}$$ where

$$c(g) = \left| \operatorname{det} \left( \operatorname{Ad}(g) \colon g \otimes_F F_v \longrightarrow g \otimes_F F_v\right) \right|_v.$$ This is elementary linear algebra, we leave the details for the reader. Now in terms of Haar measure

$$\left| \omega_l \right|_v(hg^{-1}) =  \left| \omega_l \right|_v(ghg^{-1}) = \left| \operatorname{Ad}^{\vee}(g^{-1})\omega_l \right|_v (h) = \left| c(g) \right|_v \left| \omega_l \right|_v (h).$$ Remind that by definition of the modular quasi-character

$$d_l(hg^{-1}) = \delta_{G(F_v)}(g)d_l(h)$$ for any left Haar measure on $G(F_v)$ and hence we win  

Using this, one can show that if $G$ is connected, reductive, then it is unimodular, i.e., the left and right Haar measures coincide. We need this in the sequel to prove that the spherical Hecke algebras of reductive groups are commutative.

Proposition. If $G$ is connected, reductive, then $G(F_v)$ is unimodular (in fact, one only needs $G^{\circ}$ to be reductive). 

Proof.  Let $Z_G, G^{\mathrm{der}}$ be the center and the derived subgroup of $G$, respectively. By the general theory of reductive groups (my mind is in cases of groups over algebraically closed fields, so some technical details are possibly omitted), $(Z_G)_{\mathrm{red}}^{\circ} = R_G$ is the radical, which is also a torus by reductivity. Moreover, $R_G \cap G^{\mathrm{der}}$ is finite and $G = R_G G^{\mathrm{der}}$. 

There is a commutative algebra $$\require{AMScd}
\begin{CD}R_G \times G^{\mathrm{der}} @>{a}>> \mathbb{G}_m \\
@V{b}VV @VV{\operatorname{id}}V \\ G @>{c}>> \mathbb{G}_m;
\end{CD}$$ where on $R$-points, $a,c$ are defined as

$$a(r,g) = \operatorname{det}\left(\operatorname{Ad}(rg) \colon \mathfrak{g} \otimes_F R \longrightarrow \mathfrak{g} \otimes_F R \right) $$ $$c(g) =  \operatorname{det}\left(\operatorname{Ad}(g) \colon \mathfrak{g} \otimes_F R \longrightarrow \mathfrak{g} \otimes_F R \right)$$ and $b$ is simply the product map. The map $a$ is:

• trivial on $G^{\mathrm{der}}$ since $X^{\ast}(G^{\mathrm{der}})$ is already trivial. Indeed, for any character on $G^{\mathrm{der}}$, its kernel $N$ is a normal subgroup of $G^{\mathrm{der}}$ and $G^{\mathrm{der}}/N$ is commutative, implying that $N = G^{\mathrm{der}}$. 
• trivial on $R_G$ since it is a torus.

Since $b$ is the quotient morphism, $c$ must be trivial as desired.  

On the spherical Heck algebras

Let $F$ be a local field. Let $\omega \subset \mathcal{O}$ be a uniformizer of the ring of integers. Again, let $G$ be a reductive group over $\mathcal{O}$. The group $G(F)$ is a locally compact topological group as we already know. The spherical Heck algebra is the ring

$$\mathcal{H}_G = \mathbb{Z}_c[G(\mathcal{O}) \setminus G(F) / G(\mathcal{O})]$$ of locally constant, compactly support functions $G(F) \longrightarrow \mathbb{Z}$ that are left and right invariant under the action of $G(\mathcal{O})$. There is a convolution product given by the formula

$$(f \star g)(z) = \int_{G(F)}f(x)g(x^{-1}z)dx.$$ Now since our functions are locally, constant, compactly supported, they are of the form

$$f(x) = \sum_{i=1}^n f(x_i) \mathbf{1}_{x_i G(\mathcal{O})}$$ where $\mathbf{1}$ denotes the character function. The integral is simply

$$\int_{G(F)}f(x)dx = \sum_{i=1}^n f(x_i).$$ Thus, we obtain that

$$(f \star g)(z) = \sum_{x \in G(F)/G(\mathcal{O})} f(x)g(x^{-1}z).$$ Now since we have Cartan decomposition, we see that

Proposition. The spherical Hecke algebra $\mathcal{H}_G$ is generated by character functions $\left \{c_{\lambda} = \mathbf{1}_{G(\mathcal{O})\lambda(\omega)G(\mathcal{O})} \mid \lambda \in X_{\ast}(T)^+ \right \}$. 

The highly nontrivial result is that

Theorem. The algebra $\mathcal{H}_G$ is commutative. 

Proof.  The proof relies on the existence of an anti-convolution of $G$ fixing each double coset. Namely, there exists an automorphism $\iota \colon G(F) \longrightarrow G(F)$ such that $\iota^2 = \operatorname{id}$ and $\iota(x) \in G(\mathcal{O})x^{-1}G(\mathcal{O})$ for any $x \in G(F)$. In the case $G = \mathrm{GL}_n$, one might take $\iota(A) = {}^t(A^{-1})$ (transpose of inverse). 

Given such automorphism $\iota$, for any $f \in \mathbb{Z}_c[G(F)]$, we set $\overline{f}(x) = f(\iota(x))$ and $\widetilde{f}(x) = f(x^{-1})$. If moreover, $f \in \mathcal{H}_G$, then $\overline{f} = \widetilde{f}$. The measure $dx \circ \iota$ is left invariant, implying the existence of a constant $c>0$

$$\int_{G(F)}\overline{f}(x) dx = c\int_{G(F)}f(x)dx $$ but since $\iota^2 = \operatorname{id}$, $c=1$. Thus

$$\overline{f \star g} = \overline{f} \star \overline{g}$$ for $f,g \in \mathbb{Z}_c[G(F)]$. On the other hand, the unimodularity of $G$ implies that

$$\int_{G(F)} \widetilde{f}(x) dx = \int_{G(F)} f(x)dx$$ and an easy computation shows that

$$\widetilde{f \star g} = \widetilde{g} \star \widetilde{f}$$ for $f,g \in \mathbb{Z}_c[G(F)]$. For $f,g \in \mathcal{H}_G$, we combine all to get

$$\overline{f \star g} = \overline{f} \star \overline{g} = \widetilde{f} \star \widetilde{g} = \widetilde{g \star f} = \overline{g \star f}$$ and thus $f \star g = g \star f$ as desired.  

On the Satake transformation

Let's keep the setup as in the previous part and let $q = \left| \mathcal{O}/(\omega) \right|$ the cardinality of the residue field. Let $B \subset G$ be a Borel. Let $B = T.U$ be the decomposition with $U$ the unipotent radical of $B$ and $T$ a maximal torus. Let $du$ be the unique Haar measure on $U(F)$ giving $N(\mathcal{O})$ volume $1$. As we know, the modular quasi-character $\delta \colon B(F) \longrightarrow \mathbb{R}_{>0}$ 

$$\delta(b) = \left| \operatorname{det}(\operatorname{Ad}(b) \colon  \operatorname{Lie}(B_F) \longrightarrow \operatorname{Lie}(B_F)\right|.$$ It is also easy to see that $\delta$ is trivial on $U(F)$ and hence it reduces to $\delta \colon T(F) \longrightarrow \mathbb{R}_{>0}$. Counting on weight spaces, we see that

$$  \left| \operatorname{det}(\operatorname{Ad}(b) \colon  \operatorname{Lie}(B_F) \longrightarrow \operatorname{Lie}(B_F)\right| = \text{sums of positive roots}$$ and hence if $t = \mu(\omega)$ with $\mu \in X_{\ast}(T)$

$$\delta^{1/2}(t)  = \left| \omega^{\left<\mu,2\rho \right>}\right|^{1/2} =   q^{-\left<\mu,\rho \right>}$$ with $\rho$ the half-sum of positive roots. We define the Satake transform as

$$\operatorname{Sat} \colon \mathcal{H}_G \longrightarrow \mathcal{H}_T \otimes \mathbb{Z}[q^{\pm 1/2}] \ \ \ \ \ \ \ \ f \longmapsto \operatorname{Sat}(f)$$ with

$$\operatorname{Sat}(f)(t) = \delta(t)^{1/2} \int_{N(F)}f(tu)du.$$ This integral is inspried by the theory of Harish-Chandra. Now let $G^{\vee}$ be the Langlands dual group with maximal torus $T^{\vee}$. Obviously, $X_{\ast}(T) = X^{\ast}(T^{\vee})$ and hence an isomorphism $\mathcal{H}_T \simeq X^{\ast}(T^{\vee})$. We can state the full version of Satake's theorem

Theorem.  The Satake transform induces isomorphisms of rings

$$\mathcal{H}_G \otimes_{\mathbb{Z}} \mathbb{Z}[q^{\pm 1/2}] \simeq \mathbb{Z}[X^{\ast}(T^{\vee})]^W \otimes_{\mathbb{Z}} \mathbb{Z}[q^{\pm 1/2}] \simeq K_0(\operatorname{Rep}(G^{\vee})) \otimes_{\mathbb{Z}} \mathbb{Z}[q^{\pm 1/2}]$$ where $\operatorname{Rep}$ denotes the category of representations. 

The second isomorphism is in fact follows more generally by the well-known result below. Let $T$ be a torus, we make a convention that the group ring $\mathbb{Z}[X^{\ast}(T)]$ is generated by symbols $e^{\mu}$ with $\mu \in X^{\ast}(T)$ subjecting to the relation $e^{\mu}e^{\lambda} = e^{\mu+\lambda}$

Theorem. Let $G$ be a reductive group over a field with a split maximal torus $T$. Let $W$ be the Weyl group. For any $V \in \operatorname{Rep}(G)$, we define the character map

$$\operatorname{ch}(V) = \sum_{\mu \in X^{\ast}(T)} \dim(V_{\mu}) e^{\mu} \in \mathbb{Z}[X^{\ast}(T)]$$ where $V_{\mu} = \left \{v \in V \mid \mu(t)v = t \cdot v \right \}$ the weight space of $\mu$, then $\operatorname{ch}$ induces an isomorphism of rings

$$K_0(\operatorname{Rep}(G)) \simeq \mathbb{Z}[X^{\ast}(T)]^W.$$  Example. A trivial example is when $G= T$ is a torus (hence $W=1$ trivial). The ring $\mathcal{H}_T$ is obviously commutative. Moreover, since $T$ is commutative

$$T(\mathcal{O}) \setminus T(F) / T(\mathcal{O}) = T(F)/ T(\mathcal{O})$$ and hence we get

$$\begin{align*} (c_{\lambda} \star c_{\mu})(z) & = \int_{T(F)} c_{\lambda}(x)c_{\mu}(x^{-1}z) dx \\ & = \int_{T(F)} \mathbf{1}_{x \in \lambda(\omega)T(\mathcal{O}), x^{-1}z \in \mu(\omega)T(\mathcal{O})}dx \\ & = c_{\lambda+\mu}(z) \end{align*}$$ which means that there is an isomorphism $\mathcal{H}_T \simeq \mathbb{Z}[X_{\ast}(T)]$ via $\mu \longmapsto e^{\mu}$. 

We will not provide a full proof but will illustrate the idea of proof. Let $\lambda,\mu \in X_{\ast}(T)^+$ be two positive coroots. Let $c_{\lambda}$ be the characteristic function $\mathbf{1}_{G(\mathcal{O})\lambda(\omega)G(\mathcal{O})}$ and $\chi_{\lambda}$ be the characteristic function $\mathbf{1}_{T(\mathcal{O})\lambda(\omega)T(\mathcal{O})}$.  Write

$$\operatorname{Sat}(c_{\lambda}) = \sum_{\mu \in X_{\ast}(T)} a_{\lambda}(\mu)\chi_{\mu}.$$ For each $\lambda \in X_{\ast}(T)^+$, we can write

$$G(\mathcal{O})\lambda(\omega)G(\mathcal{O}) = \coprod x_i G(\mathcal{O})$$ with $x_i \in B(F)$. By the Iwasawa decomposition, we can also write $x_i = t_i u_i$ with $t_i \in T(F)$ and $u_i \in U(F)$ (the unipotent radical of $B$).

Let us compute

$$\begin{align*}a_{\lambda}(\mu) &=  \operatorname{Sat}(c_{\lambda})(\mu(\omega))  \\ & = \delta(\mu)^{1/2} \int_U \mathbf{1}_{G(\mathcal{O})\lambda(\omega)G(\mathcal{O})}(\mu(\omega u)du \\ & = q^{-\left<\mu,\rho \right>} \sum_i \int_{U \cap \mu(\omega)^{-1}x_i G(\mathcal{O})} du \\ & = q^{-\left<\mu,\rho \right>} \left| \left\{ i \mid t_i \equiv \mu(\omega) \ \operatorname{mod} \ G(\mathcal{O}) \right\} \right|\end{align*}$$ Clearly, one sees that $\operatorname{Sat}(c_{\lambda})(\lambda(\omega)) = q^{\left<\lambda,\rho \right>}$ non-zero. Moreover, (this is HARD), if $\mu \in X_{\ast}(T)^+$ such that $\operatorname{Sat}(c_{\lambda})(\mu(\omega)) \neq 0$, it is true that $\mu \leq \lambda$. In the end, we have

$$\operatorname{Sat}(c_{\lambda}) = a_{\lambda}(\lambda)\chi_{\lambda} + \sum_{\mu < \lambda}a_{\lambda}(\mu)\chi_{\mu}.$$ Thus, once we can prove that $\operatorname{Sat}$ takes values in $\mathbb{Z}[X_{\ast}(T)]^W$, the above shows that $\operatorname{Sat}$ is represented by a triangular, invertible matrix, hence an isomorphism.