[proofplan] We identify the centre inside the maximal torus by showing that it is precisely the subgroup on which every root is trivial. Since $K$ is simply connected, the character lattice of $T$ is the full weight lattice $P$, and the subgroup generated by the roots is $Q$; the standard evaluation pairing between a compact torus and its character lattice then identifies this root-annihilator with $\operatorname{Hom}(P/Q,S^1)$. For the descent statement, a representation factors through the quotient $K/F$ exactly when $F$ acts trivially, and [Schur's lemma](/theorems/2414) reduces the central action on an irreducible representation to a scalar read off from the highest weight. [/proofplan] [step:Identify central elements as elements of $T$ killed by every root] Let $\mathfrak{k}$ denote the [Lie algebra](/page/Lie%20Algebra) of $K$, let $\mathfrak{t}$ denote the Lie algebra of $T$, and let $\mathfrak{k}_{\mathbb C}:=\mathfrak{k}\otimes_{\mathbb R}\mathbb C$ be the complexified Lie algebra. For each $\alpha\in R$, let $\mathfrak{k}_{\mathbb C,\alpha}:=\{X\in\mathfrak{k}_{\mathbb C}:\operatorname{Ad}_s(X)=\alpha(s)X\text{ for every }s\in T\}$ be the corresponding root space. Define $Z_T:=\{t\in T:\alpha(t)=1\text{ for every }\alpha\in R\}$. We first prove that \begin{align*} Z(K)=Z_T. \end{align*} Let $z\in Z(K)$. Since $K$ is compact and connected and $T$ is a maximal torus, [citetheorem:9713] implies that $z$ is conjugate to an element of $T$. Since $z$ is central, its [conjugacy class](/page/Conjugacy%20Class) is the singleton $\{z\}$, so $z\in T$. For $z\in T$, the adjoint action $\operatorname{Ad}_z$ acts on the root space $\mathfrak{k}_{\mathbb C,\alpha}$ by the scalar $\alpha(z)$. If $z\in Z(K)$, then conjugation by $z$ is the identity automorphism of $K$, so its differential $\operatorname{Ad}_z:\mathfrak{k}_{\mathbb C}\to\mathfrak{k}_{\mathbb C}$ is the identity [linear map](/page/Linear%20Map). Hence $\alpha(z)=1$ for every $\alpha\in R$, and therefore $z\in Z_T$. Conversely, let $t\in Z_T$. The element $t$ acts as the identity by the adjoint representation on $\mathfrak{t}_{\mathbb C}$ because $T$ is abelian. It also acts as the identity on every root space $\mathfrak{k}_{\mathbb C,\alpha}$ because $\alpha(t)=1$. Since the complexified Lie algebra decomposes as \begin{align*} \mathfrak{k}_{\mathbb C}=\mathfrak{t}_{\mathbb C}\oplus \bigoplus_{\alpha\in R}\mathfrak{k}_{\mathbb C,\alpha}, \end{align*} the operator $\operatorname{Ad}_t$ is the identity on $\mathfrak{k}_{\mathbb C}$, hence also on $\mathfrak{k}$. Thus $t$ lies in the kernel of the adjoint homomorphism \begin{align*} \operatorname{Ad}:K\to GL(\mathfrak{k}). \end{align*} For a connected Lie group, the kernel of $\operatorname{Ad}$ is $Z(K)$. Indeed, if $\operatorname{Ad}_t$ is the identity, then the conjugation map $c_t:K\to K$, $x\mapsto txt^{-1}$, satisfies \begin{align*} c_t(\exp X)=\exp(\operatorname{Ad}_t X)=\exp X \end{align*} for every $X\in\mathfrak{k}$. The subgroup generated by $\exp(\mathfrak{k})$ is open because $\exp$ maps a neighbourhood of $0\in\mathfrak{k}$ onto a neighbourhood of the identity element $e\in K$; it is also closed because its cosets are open. Since $K$ is connected, this subgroup is all of $K$. Hence $c_t=\operatorname{id}_K$, so $t\in Z(K)$. [guided] The goal is to replace the abstract centre $Z(K)$ by an explicit subgroup of the torus $T$. For each $\alpha\in R$, the root space $\mathfrak{k}_{\mathbb C,\alpha}$ means $\{X\in\mathfrak{k}_{\mathbb C}:\operatorname{Ad}_s(X)=\alpha(s)X\text{ for every }s\in T\}$. Define $Z_T:=\{t\in T:\alpha(t)=1\text{ for every }\alpha\in R\}$. We prove that $Z(K)=Z_T$. First take $z\in Z(K)$. The [maximal torus theorem](/theorems/9713) [citetheorem:9713] says that every element of the compact connected Lie group $K$ is conjugate to an element of the fixed maximal torus $T$. Thus there are $h\in K$ and $t\in T$ such that \begin{align*} hzh^{-1}=t. \end{align*} Because $z$ is central, $hzh^{-1}=z$, and hence $z=t\in T$. Now we check that every root is trivial on $z$. For each root $\alpha\in R$, the corresponding root space $\mathfrak{k}_{\mathbb C,\alpha}$ is an eigenspace for the adjoint action of $T$: for every $v\in \mathfrak{k}_{\mathbb C,\alpha}$, \begin{align*} \operatorname{Ad}_z(v)=\alpha(z)v. \end{align*} Since $z$ is central, conjugation by $z$ is the identity map on $K$, and therefore its differential is the identity linear map on $\mathfrak{k}_{\mathbb C}$. Applying this identity operator to any non-zero $v\in\mathfrak{k}_{\mathbb C,\alpha}$ gives \begin{align*} v=\operatorname{Ad}_z(v)=\alpha(z)v. \end{align*} Thus $\alpha(z)=1$. Since this holds for every root $\alpha\in R$, we have $z\in Z_T$. Conversely, suppose $t\in Z_T$. We prove that $t$ is central. The torus $T$ is abelian, so $\operatorname{Ad}_t$ is the identity on its Lie algebra $\mathfrak{t}$, and therefore also on $\mathfrak{t}_{\mathbb C}$. On a root space $\mathfrak{k}_{\mathbb C,\alpha}$, the operator $\operatorname{Ad}_t$ acts by multiplication by $\alpha(t)$, and $\alpha(t)=1$ by the definition of $Z_T$. Hence $\operatorname{Ad}_t$ is the identity on every summand in the root decomposition \begin{align*} \mathfrak{k}_{\mathbb C}=\mathfrak{t}_{\mathbb C}\oplus \bigoplus_{\alpha\in R}\mathfrak{k}_{\mathbb C,\alpha}. \end{align*} Therefore $\operatorname{Ad}_t$ is the identity on $\mathfrak{k}_{\mathbb C}$ and hence on the real Lie algebra $\mathfrak{k}$. It remains to translate this infinitesimal statement back to the group. The [kernel of the adjoint representation](/theorems/3752) of a connected Lie group is its centre. Indeed, if $\operatorname{Ad}_t=\operatorname{id}_{\mathfrak{k}}$, then the conjugation automorphism \begin{align*} c_t:K&\to K \end{align*} \begin{align*} x&\mapsto txt^{-1} \end{align*} satisfies \begin{align*} c_t(\exp X)=\exp(\operatorname{Ad}_t X)=\exp X \end{align*} for each $X\in\mathfrak{k}$. Let $H\le K$ be the subgroup generated by $\exp(\mathfrak{k})$. Since the exponential map sends a neighbourhood of $0\in\mathfrak{k}$ onto a neighbourhood of the identity element $e\in K$, the subgroup $H$ contains an open neighbourhood of $e$ and is therefore open. Its complement is a union of open cosets of $H$, so $H$ is also closed. Because $K$ is connected and $H$ is non-empty, open, and closed, we have $H=K$. Thus $c_t$ is the identity map on all of $K$. Hence $txt^{-1}=x$ for every $x\in K$, which means $t\in Z(K)$. [/guided] [/step] [step:Use the torus character pairing to dualise the quotient $P/Q$] The evaluation pairing of the compact torus $T$ with its character lattice is the [group isomorphism](/page/Group%20Isomorphism) $\operatorname{ev}:T\to \operatorname{Hom}(X^*(T),S^1)$ defined by $\operatorname{ev}(t)(\lambda)=\lambda(t)$. This is the standard Pontryagin duality identification for compact tori: $X^*(T)$ is a free abelian lattice, and its characters separate points of $T$. Since $K$ is simply connected and semisimple, the character lattice $X^*(T)$ is the full weight lattice $P$. Under this identification, the root lattice is \begin{align*} Q=\mathbb ZR\subset P. \end{align*} By the previous step, \begin{align*} Z(K)=\{t\in T:\alpha(t)=1\text{ for every }\alpha\in R\}. \end{align*} Equivalently, \begin{align*} Z(K)=\{t\in T:\lambda(t)=1\text{ for every }\lambda\in Q\}. \end{align*} Therefore $\operatorname{ev}$ restricts to a homomorphism $\Phi:Z(K)\to \operatorname{Hom}(P/Q,S^1)$ defined by $\Phi(z)(\lambda+Q)=\lambda(z)$. This map is well-defined because every element of $Q$ is trivial on $Z(K)$. The map $\Phi$ is injective because characters in $X^*(T)=P$ separate points of $T$. To prove surjectivity, let $\psi:P/Q\to S^1$ be a [group homomorphism](/page/Group%20Homomorphism). Define $\tilde{\psi}:P\to S^1$ by $\tilde{\psi}(\lambda)=\psi(\lambda+Q)$. By the torus character pairing, there exists $t\in T$ such that $\lambda(t)=\tilde{\psi}(\lambda)$ for every $\lambda\in P$. If $\alpha\in R$, then $\alpha\in Q$, so \begin{align*} \alpha(t)=\tilde{\psi}(\alpha)=\psi(Q)=1. \end{align*} Thus $t\in Z(K)$, and $\Phi(t)=\psi$. Hence $\Phi$ is surjective, and therefore \begin{align*} Z(K)\cong \operatorname{Hom}(P/Q,S^1). \end{align*} [/step] [step:Characterize when a representation descends to the central quotient] Let $F\le Z(K)$ be a finite central subgroup, and let $q:K\to G=K/F$ be the quotient homomorphism. Let $\rho:K\to GL(V)$ be an irreducible finite-dimensional complex representation with highest weight $\lambda\in P_+$, and let $L_\lambda\subset V$ denote its highest-weight line. The representation $\rho$ descends to a representation of $G$ if and only if there exists a representation $\bar{\rho}:G\to GL(V)$ such that \begin{align*} \rho=\bar{\rho}\circ q. \end{align*} This is equivalent to $F\subset \ker\rho$, because $\ker q=F$ and homomorphisms out of a quotient are exactly homomorphisms that are trivial on the quotient kernel. It remains to relate $F\subset\ker\rho$ to the highest-weight line. Let $f\in F$. Since $f\in Z(K)$, the operator $\rho(f):V\to V$ commutes with $\rho(k)$ for every $k\in K$: \begin{align*} \rho(f)\rho(k)=\rho(fk)=\rho(kf)=\rho(k)\rho(f). \end{align*} By Schur's lemma, because $\rho(f)$ is a $K$-intertwiner of the irreducible complex representation $V$, there exists a scalar $c_f\in\mathbb C$ such that \begin{align*} \rho(f)=c_f\,\operatorname{id}_V. \end{align*} Since $f$ has finite order and $\rho(f)$ is invertible, $c_f\in S^1$. Therefore $\rho(f)$ is the identity on all of $V$ if and only if it is the identity on the non-zero line $L_\lambda$. Hence $F\subset\ker\rho$ if and only if every element of $F$ acts as the identity on $L_\lambda$. This proves the descent criterion. Equivalently, under the central-character isomorphism above, $f\in F$ acts on $L_\lambda$ by the scalar $\lambda(f)$; hence the same condition is \begin{align*} \lambda(f)=1\quad\text{for every }f\in F. \end{align*} This completes the proof. [/step]