[proofplan] Restrict the representation to the maximal torus $T$ and decompose $V$ into finitely many weight spaces. The connectedness assumption belongs to the compact connected Lie group setting in which the maximal torus and root data are fixed; the argument below only uses the resulting torus $T$, its root spaces, and the chosen positive system. The finite set of all nonzero infinitesimal weights admits a maximal element for the partial order generated by adding positive roots. For a vector in a maximal weight space, the [Lie algebra](/page/Lie%20Algebra) relation between $\mathfrak t_{\mathbb C}$ and the root space $\mathfrak g_\alpha$ shows that positive root vectors shift weight $\lambda$ to weight $\lambda+\alpha$. Maximality forces the shifted weight space to vanish, so every positive root space annihilates the chosen vector. [/proofplan]

[step:Decompose the representation into finitely many torus weight spaces]Let \begin{align*} \rho_T:T\to GL(V) \end{align*} denote the restriction of $\rho$ to $T$, and let $\exp_T:\mathfrak t\to T$ denote the Lie exponential map of the torus $T$. Since $T$ is compact abelian, averaging any Hermitian [inner product](/page/Inner%20Product) over $T$ makes every $\rho_T(t)$ unitary, and the commuting normal operators $\{\rho_T(t):t\in T\}$ are simultaneously diagonalizable on the finite-dimensional space $V$. Thus there is a finite set $\Gamma\subset X^*(T)$ of torus characters such that, with \begin{align*} E_\chi:=\{u\in V:\rho_T(t)u=\chi(t)u\text{ for every }t\in T\}, \end{align*} one has \begin{align*} V=\bigoplus_{\chi\in\Gamma}E_\chi, \end{align*} where $E_\chi\ne\{0\}$ for every $\chi\in\Gamma$, and the torus-character eigenspace for any $\chi\notin\Gamma$ is zero. For each $\chi\in\Gamma$, let $\nu_\chi\in\mathfrak t_{\mathbb C}^*$ denote the complex-linear extension of $d\chi_e:\mathfrak t\to\mathbb C$. Differentiating the identity $\rho_T(\exp_T(sY))u=\chi(\exp_T(sY))u$ at $s=0$ shows that $E_\chi\subset V_{\nu_\chi}$. Conversely, if $u\in V_{\nu_\chi}$ and $Y\in\mathfrak t$, then the map $s\mapsto \rho_T(\exp_T(sY))u$ solves the linear [ordinary differential equation](/page/Ordinary%20Differential%20Equation) $f'(s)=\nu_\chi(Y)f(s)$ with initial value $u$, so \begin{align*} \rho_T(\exp_T(sY))u=\exp(s\nu_\chi(Y))u=\chi(\exp_T(sY))u. \end{align*} Since the exponential map of a compact torus is surjective, this gives $u\in E_\chi$. Hence $E_\chi=V_{\nu_\chi}$. If $\chi,\psi\in\Gamma$ and $\nu_\chi=\nu_\psi$, then $E_\chi=V_{\nu_\chi}=V_{\nu_\psi}=E_\psi$, so the directness of the torus-character decomposition and the nonzero conditions $E_\chi\ne\{0\}$ and $E_\psi\ne\{0\}$ force $\chi=\psi$. Define \begin{align*} \Omega:=\{\nu_\chi:\chi\in\Gamma\}\subset\mathfrak t_{\mathbb C}^*. \end{align*} Then \begin{align*} V=\bigoplus_{\nu\in\Omega}V_\nu, \end{align*} with $V_\nu\ne\{0\}$ for every $\nu\in\Omega$. Moreover, $\Omega$ is the full set of nonzero infinitesimal weights: if $u\in V_\eta$ and $u=\sum_{\chi\in\Gamma}u_\chi$ is its decomposition with $u_\chi\in E_\chi=V_{\nu_\chi}$, then applying $d\rho_{\mathbb C}(H)$ for arbitrary $H\in\mathfrak t_{\mathbb C}$ gives \begin{align*} \sum_{\chi\in\Gamma}(\nu_\chi(H)-\eta(H))u_\chi=0. \end{align*} The direct-sum decomposition forces $u_\chi=0$ unless $\nu_\chi=\eta$, so $V_\eta=\{0\}$ whenever $\eta\notin\Omega$. Since $V\ne\{0\}$, the set $\Omega$ is nonempty.[/step]

[guided]We first use the torus part of the representation, because compact tori have completely reducible one-dimensional character behavior. Define the restricted representation \begin{align*} \rho_T:T\to GL(V) \end{align*} by $\rho_T(t):=\rho(t)$ for $t\in T$. The compact-torus decomposition used here follows directly from unitary diagonalization: average a Hermitian inner product over $T$, observe that the operators $\rho_T(t)$ are commuting normal operators, and diagonalize them simultaneously. This decomposes $V$ by group characters of $T$. Thus there is a finite set $\Gamma\subset X^*(T)$ such that, for \begin{align*} E_\chi:=\{u\in V:\rho_T(t)u=\chi(t)u\text{ for every }t\in T\}, \end{align*} one has \begin{align*} V=\bigoplus_{\chi\in\Gamma}E_\chi. \end{align*} The construction also gives $E_\chi\ne\{0\}$ for $\chi\in\Gamma$ and no nonzero eigenspace for torus characters outside $\Gamma$. The statement of the present theorem uses infinitesimal weights, so we must connect the two conventions. For a character $\chi\in\Gamma$, define $\nu_\chi\in\mathfrak t_{\mathbb C}^*$ to be the complex-linear extension of the differential $d\chi_e:\mathfrak t\to\mathbb C$. If $u\in E_\chi$, then for every $Y\in\mathfrak t$ and every real $s$, the character equation gives \begin{align*} \rho_T(\exp_T(sY))u=\chi(\exp_T(sY))u. \end{align*} Differentiating this identity at $s=0$ gives \begin{align*} d\rho_{\mathbb C}(Y)u=d\chi_e(Y)u. \end{align*} After complex-linear extension from $\mathfrak t$ to $\mathfrak t_{\mathbb C}$, this says $u\in V_{\nu_\chi}$. Hence $E_\chi\subset V_{\nu_\chi}$. Conversely, suppose $u\in V_{\nu_\chi}$. Fix $Y\in\mathfrak t$ and define \begin{align*} f_Y:\mathbb R\to V \end{align*} by \begin{align*} f_Y(s):=\rho_T(\exp_T(sY))u. \end{align*} Because $u$ has infinitesimal weight $\nu_\chi$, differentiating the orbit curve gives \begin{align*} f_Y'(s)=\nu_\chi(Y)f_Y(s). \end{align*} The scalar multiple curve $s\mapsto \exp(s\nu_\chi(Y))u$ satisfies the same linear ordinary differential equation and the same initial condition at $s=0$. Uniqueness for linear ordinary differential equations therefore gives \begin{align*} \rho_T(\exp_T(sY))u=\exp(s\nu_\chi(Y))u=\chi(\exp_T(sY))u. \end{align*} The exponential map of a compact torus is surjective, so every $t\in T$ has the form $t=\exp_T(Y)$ for some $Y\in\mathfrak t$. Hence $\rho_T(t)u=\chi(t)u$ for every $t\in T$, and $u\in E_\chi$. We have proved $E_\chi=V_{\nu_\chi}$. If two characters $\chi,\psi\in\Gamma$ had the same infinitesimal weight, then the preceding equality would give $E_\chi=E_\psi$. Since the torus-character decomposition is direct and both spaces are nonzero, this forces $\chi=\psi$. Thus the passage from torus characters in $\Gamma$ to infinitesimal weights in $\Omega$ does not identify two distinct nonzero summands. Now define \begin{align*} \Omega:=\{\nu_\chi:\chi\in\Gamma\}\subset\mathfrak t_{\mathbb C}^*. \end{align*} The preceding identification turns the torus-character decomposition into the infinitesimal-weight decomposition \begin{align*} V=\bigoplus_{\nu\in\Omega}V_\nu. \end{align*} Finally, we record the point needed later: $\Omega$ is the full set of nonzero infinitesimal weights. Indeed, if $u\in V_\eta$ and $u=\sum_{\chi\in\Gamma}u_\chi$ with $u_\chi\in E_\chi=V_{\nu_\chi}$, then for every $H\in\mathfrak t_{\mathbb C}$ we get \begin{align*} \sum_{\chi\in\Gamma}(\nu_\chi(H)-\eta(H))u_\chi=0. \end{align*} Since the sum is direct, each nonzero $u_\chi$ must satisfy $\nu_\chi=\eta$. Thus if $\eta\notin\Omega$, every component $u_\chi$ is zero and $u=0$. Therefore $V_\eta=\{0\}$ for all $\eta\notin\Omega$. The nonzero hypothesis on $V$ then ensures that $\Omega$ is nonempty.[/guided]

[step:Choose a maximal weight for the positive-root order] Define a relation $\preceq$ on $\mathfrak t_{\mathbb C}^*$ by declaring $\nu\preceq\eta$ when there exist integers $m_\alpha\ge 0$, all but finitely many equal to $0$, such that \begin{align*} \eta-\nu=\sum_{\alpha\in R^+}m_\alpha\alpha. \end{align*} This relation is antisymmetric on weights: after choosing the simple roots contained in $R^+$, every element of the positive-root cone has nonnegative simple-root coordinates, so a vector and its negative can both lie in this cone only when both are $0$. Hence the strict relation induced by $\preceq$ has no cycles. Since $\Omega$ is finite and nonempty, there exists $\lambda\in\Omega$ such that no element of $\Omega$ is strictly larger than $\lambda$ for this relation. Choose a vector $v\in V_\lambda\setminus\{0\}$. [/step]

[step:Show that positive root spaces raise weights]Fix $\alpha\in R^+$ and $X\in\mathfrak g_\alpha$. Let \begin{align*} w:=d\rho_{\mathbb C}(X)v\in V. \end{align*} For every $H\in\mathfrak t_{\mathbb C}$, the representation property of the differential gives \begin{align*} d\rho_{\mathbb C}(H)w=d\rho_{\mathbb C}(H)d\rho_{\mathbb C}(X)v. \end{align*} Using the Lie algebra homomorphism identity for $d\rho_{\mathbb C}$, we obtain \begin{align*} d\rho_{\mathbb C}(H)d\rho_{\mathbb C}(X)v=d\rho_{\mathbb C}([H,X])v+d\rho_{\mathbb C}(X)d\rho_{\mathbb C}(H)v. \end{align*} Because $X\in\mathfrak g_\alpha$, one has $[H,X]=\alpha(H)X$. Because $v\in V_\lambda$, one has $d\rho_{\mathbb C}(H)v=\lambda(H)v$. Therefore \begin{align*} d\rho_{\mathbb C}(H)w=\alpha(H)d\rho_{\mathbb C}(X)v+\lambda(H)d\rho_{\mathbb C}(X)v. \end{align*} Hence \begin{align*} d\rho_{\mathbb C}(H)w=(\lambda+\alpha)(H)w. \end{align*} Since this holds for every $H\in\mathfrak t_{\mathbb C}$, we have \begin{align*} w\in V_{\lambda+\alpha}. \end{align*}[/step]

[guided]The key point is that root vectors shift weights by their roots. We prove this directly from the Lie bracket relation, rather than treating it as a black box. Fix a positive root $\alpha\in R^+$ and a root vector $X\in\mathfrak g_\alpha$. Define \begin{align*} w:=d\rho_{\mathbb C}(X)v. \end{align*} To prove that $w$ has weight $\lambda+\alpha$, we must check the defining eigenvalue equation against every $H\in\mathfrak t_{\mathbb C}$. Let $H\in\mathfrak t_{\mathbb C}$. Since $d\rho_{\mathbb C}$ is a Lie algebra representation, it satisfies \begin{align*} d\rho_{\mathbb C}([H,X])=d\rho_{\mathbb C}(H)d\rho_{\mathbb C}(X)-d\rho_{\mathbb C}(X)d\rho_{\mathbb C}(H). \end{align*} Rearranging this identity and applying both sides to $v$ gives \begin{align*} d\rho_{\mathbb C}(H)d\rho_{\mathbb C}(X)v=d\rho_{\mathbb C}([H,X])v+d\rho_{\mathbb C}(X)d\rho_{\mathbb C}(H)v. \end{align*} Now we use the two eigenvalue relations available to us. The root-space condition $X\in\mathfrak g_\alpha$ means \begin{align*} [H,X]=\alpha(H)X. \end{align*} The weight-space condition $v\in V_\lambda$ means \begin{align*} d\rho_{\mathbb C}(H)v=\lambda(H)v. \end{align*} Substituting these into the previous display yields \begin{align*} d\rho_{\mathbb C}(H)w=\alpha(H)d\rho_{\mathbb C}(X)v+\lambda(H)d\rho_{\mathbb C}(X)v. \end{align*} Since $w=d\rho_{\mathbb C}(X)v$, this becomes \begin{align*} d\rho_{\mathbb C}(H)w=(\lambda+\alpha)(H)w. \end{align*} This is exactly the defining condition for $w$ to lie in $V_{\lambda+\alpha}$. Thus applying a positive root vector to a vector of weight $\lambda$ produces either $0$ or a vector of weight $\lambda+\alpha$.[/guided]

[step:Use maximality to force all positive root vectors to annihilate the chosen vector] For the fixed $\alpha\in R^+$, the relation defining $\preceq$ gives \begin{align*} \lambda\preceq\lambda+\alpha. \end{align*} The maximality of $\lambda$ among weights in $\Omega$ implies that $\lambda+\alpha\notin\Omega$. Since $\Omega$ was established as the full set of nonzero infinitesimal weights, the weight space $V_{\lambda+\alpha}$ is zero. From the previous step, \begin{align*} d\rho_{\mathbb C}(X)v\in V_{\lambda+\alpha}=\{0\}. \end{align*} Thus \begin{align*} d\rho_{\mathbb C}(X)v=0. \end{align*} Since $\alpha\in R^+$ and $X\in\mathfrak g_\alpha$ were arbitrary, this proves that $v$ is annihilated by every positive root space. If $R^+=\varnothing$, this conclusion is vacuous, and the vector $v\in V_\lambda\setminus\{0\}$ chosen above already satisfies the required condition. Hence $v$ is a highest weight vector for $V$ relative to $(T,R^+)$. [/step]