[proofplan] We compare all lattices through the maximal torus of the simply connected compact semisimple group with the same [Lie algebra](/page/Lie%20Algebra). First, roots are actual torus characters because they occur as weights of the adjoint representation, so their integral span lies in the character lattice. Second, every torus character has integral coroot pairings, hence belongs to the full weight lattice. Finally, central quotients are classified by which simply connected torus characters are trivial on the finite central kernel; the perfect finite pairing between the centre and $P/Q$ identifies exactly the intermediate lattices. [/proofplan] [step:Place the three lattices in a common Cartan dual] Let $\widetilde G$ be the simply connected compact semisimple Lie group with Lie algebra $\mathfrak g$, and let \begin{align*} p:\widetilde G\to G \end{align*} be the covering homomorphism with finite central kernel. Let $\widetilde T$ be the identity component of $p^{-1}(T)$. Then $\widetilde T$ is a maximal torus of $\widetilde G$, and the restricted map \begin{align*} p_T:\widetilde T\to T \end{align*} is a surjective homomorphism with finite kernel. Define the pullback map on character lattices \begin{align*} p_T^*:X^*(T)&\to X^*(\widetilde T) \end{align*} \begin{align*} \lambda&\mapsto \lambda\circ p_T. \end{align*} Since $p_T$ is surjective, $p_T^*$ is injective. We therefore identify $X^*(T)$ with its image in $X^*(\widetilde T)$. By the definition of the full weight lattice for the simply connected compact semisimple group, and equivalently by the highest-weight lattice convention in [citetheorem:9752], we identify \begin{align*} X^*(\widetilde T)=P. \end{align*} Thus it remains to prove that $Q\subset p_T^*X^*(T)\subset P$, where the second inclusion is automatic from the displayed identification once the first embedding has been fixed. [guided] The point of this first step is to remove an ambiguity in the theorem statement. The torus $T$ depends on the particular compact group $G$, while the full weight lattice $P$ is most naturally attached to the simply connected group with the same Lie algebra. We therefore compare both lattices by pulling characters back to the maximal torus $\widetilde T$ of the simply connected cover. Let $\widetilde G$ be the simply connected compact semisimple Lie group with Lie algebra $\mathfrak g$, and let \begin{align*} p:\widetilde G\to G \end{align*} be the covering homomorphism. Its kernel is finite and central because $p$ is a covering homomorphism between connected Lie groups with the same Lie algebra. Define $\widetilde T$ to be the identity component of $p^{-1}(T)$. Since $p$ is locally an isomorphism and $T$ is a maximal connected abelian subgroup of $G$, the subgroup $\widetilde T$ is a maximal torus of $\widetilde G$, and the restricted map \begin{align*} p_T:\widetilde T\to T \end{align*} is surjective with finite kernel. Now define the character pullback \begin{align*} p_T^*:X^*(T)&\to X^*(\widetilde T) \end{align*} \begin{align*} \lambda&\mapsto \lambda\circ p_T. \end{align*} This is a [group homomorphism](/page/Group%20Homomorphism) between abelian groups. It is injective because if $\lambda\circ p_T=1$ as a character of $\widetilde T$, then surjectivity of $p_T$ implies $\lambda(t)=1$ for every $t\in T$, so $\lambda$ is the trivial character. Thus $X^*(T)$ may be regarded as a sublattice of $X^*(\widetilde T)$. For the simply connected compact semisimple group, the character lattice of $\widetilde T$ is the full weight lattice $P$ of the root system. This is the lattice convention encoded by [citetheorem:9752], where the centre of the simply connected group is described using the quotient $P/Q$. Hence all three lattices are now being compared inside the single lattice \begin{align*} X^*(\widetilde T)=P. \end{align*} [/guided] [/step] [step:Show that the root lattice is contained in the torus character lattice] Write $\operatorname{Ad}$ for the adjoint representation of $G$ on the complexified Lie algebra $\mathfrak g_{\mathbb C}$. For each root $\alpha\in R$, the root space \begin{align*} \mathfrak g_{\mathbb C,\alpha}:=\{X\in\mathfrak g_{\mathbb C}:\operatorname{Ad}(t)X=\alpha(t)X\text{ for every }t\in T\} \end{align*} is non-zero by the definition of the root system. Thus $\alpha:T\to S^1$ is a character occurring in the restriction of the adjoint representation of $G$ to $T$. Hence $\alpha\in X^*(T)$ for every $\alpha\in R$. Since the root lattice is \begin{align*} Q:=\operatorname{span}_{\mathbb Z}(R), \end{align*} and since $X^*(T)$ is an [abelian group](/page/Abelian%20Group) closed under products and inverses of characters, every integral product of roots is again a character of $T$. Therefore \begin{align*} Q\subset X^*(T). \end{align*} After applying the injective pullback $p_T^*$ from the previous step, this is the asserted inclusion of $Q$ into the common lattice. [/step] [step:Use coroot integrality to bound the character lattice by the full weight lattice] Let $\lambda\in X^*(T)$. By the [integrality criterion for torus characters](/theorems/9729) [citetheorem:9729], the differential of $\lambda$ is integral on the integral lattice of $T$. For each root $\alpha\in R$, write $\alpha^\vee$ for the associated coroot, viewed in the integral lattice of $T$ through the coroot circle subgroup. The associated coroot circle in $T$ has derivative $\alpha^\vee$, so evaluating $\lambda$ on that circle gives \begin{align*} \langle \lambda,\alpha^\vee\rangle\in\mathbb Z. \end{align*} By definition, the full weight lattice is \begin{align*} P:=\{\nu:\langle \nu,\alpha^\vee\rangle\in\mathbb Z\text{ for every }\alpha\in R\}. \end{align*} Thus $\lambda\in P$. Since $\lambda\in X^*(T)$ was arbitrary, \begin{align*} X^*(T)\subset P. \end{align*} Together with the previous step, this proves \begin{align*} Q\subset X^*(T)\subset P. \end{align*} [/step] [step:Identify the character lattice of a central quotient] Let $C\le Z(\widetilde G)$ be a finite central subgroup, and define \begin{align*} G_C:=\widetilde G/C. \end{align*} Let \begin{align*} q_C:\widetilde G\to G_C \end{align*} be the quotient homomorphism, and let \begin{align*} T_C:=q_C(\widetilde T) \end{align*} be the image torus in $G_C$. The restriction \begin{align*} q_{C,T}:\widetilde T\to T_C \end{align*} is surjective and has kernel $C\cap \widetilde T$. Since $C\le Z(\widetilde G)$ and every central element of a compact connected Lie group lies in every maximal torus, we have $C\subset \widetilde T$. Hence \begin{align*} \ker q_{C,T}=C. \end{align*} A character $\lambda\in X^*(\widetilde T)=P$ descends to a character of $T_C$ precisely when it is constant on the fibres of $q_{C,T}$. Since the fibres are cosets of $C$, this is equivalent to \begin{align*} \lambda(c)=1\text{ for every }c\in C. \end{align*} Therefore \begin{align*} X^*(T_C)=\{\lambda\in P:\lambda(c)=1\text{ for every }c\in C\}=L_C. \end{align*} [/step] [step:Recover every intermediate lattice from a finite central subgroup] By [citetheorem:9752], there is a natural perfect pairing between the finite abelian groups $Z(\widetilde G)$ and $P/Q$: \begin{align*} Z(\widetilde G)\times P/Q&\to S^1 \end{align*} \begin{align*} (z,\lambda+Q)&\mapsto \lambda(z). \end{align*} Let $L$ be an intermediate lattice with \begin{align*} Q\subset L\subset P. \end{align*} Define the finite subgroup \begin{align*} C_L:=\{z\in Z(\widetilde G):\lambda(z)=1\text{ for every }\lambda\in L\}. \end{align*} This is the annihilator of the subgroup $L/Q\le P/Q$ under the displayed perfect pairing. Finite Pontryagin duality gives that the double annihilator of $L/Q$ is $L/Q$. Hence \begin{align*} \{\lambda\in P:\lambda(c)=1\text{ for every }c\in C_L\}=L. \end{align*} Thus $L=L_{C_L}$, so every intermediate lattice occurs as the character lattice of the torus in the quotient $\widetilde G/C_L$. Conversely, start with a finite central subgroup $C\le Z(\widetilde G)$. The image $L_C/Q$ is the annihilator of $C$ inside $P/Q$. If we then form the subgroup attached to $L_C$, we get \begin{align*} C_{L_C}=\{z\in Z(\widetilde G):\lambda(z)=1\text{ for every }\lambda\in L_C\}. \end{align*} This is the double annihilator of $C$ under the same perfect pairing. Finite Pontryagin duality gives that the double annihilator of a subgroup of a finite abelian group is the subgroup itself, so \begin{align*} C_{L_C}=C. \end{align*} Thus the assignment $C\mapsto L_C$ is injective as well as surjective onto the intermediate lattices. If $C_1\subset C_2$, then every character trivial on $C_2$ is trivial on $C_1$, so \begin{align*} L_{C_2}\subset L_{C_1}. \end{align*} Thus the correspondence is inclusion-reversing. Taking $C=Z(\widetilde G)$ gives the annihilator of all of $Z(\widetilde G)$, which is exactly $Q$ under the perfect pairing with $P/Q$. Therefore the adjoint quotient $\widetilde G/Z(\widetilde G)$ corresponds to the root lattice $Q$. This proves both the lattice bounds and the stated classification of finite central quotients by intermediate lattices. [/step]