[proofplan]
We decompose $V$ into its $T$-weight spaces and choose a maximal weight for the positive-root order. Root operators shift weights by roots, so maximality forces every vector of the chosen weight to be killed by all positive root spaces. Irreducibility and the PBW triangular decomposition then show that every weight generated from such a vector is obtained by subtracting a nonnegative integral combination of positive roots, which proves uniqueness of the highest weight. Finally, after averaging to obtain a $G$-invariant Hermitian [inner product](/page/Inner%20Product), we use orthogonality of distinct torus weight spaces and irreducibility to prove that the highest weight space is one-dimensional.
[/proofplan]
[step:Reduce the torus case to Schur irreducibility]
Suppose first that $\Phi=\varnothing$. Then the complexified [Lie algebra](/page/Lie%20Algebra) $\mathfrak g_{\mathbb C}$ equals $\mathfrak t_{\mathbb C}$, because the root space decomposition has no nonzero root spaces. Since $G$ is compact and connected and $T$ is a maximal torus, this means $G=T$ is abelian.
By [citetheorem:9726], applied to the restriction $\pi|_T:T\to GL(V)$, there is a finite subset $\Omega(V)\subset \mathfrak t_{\mathbb C}^*$ such that
\begin{align*}
V=\bigoplus_{\mu\in\Omega(V)}V_\mu.
\end{align*}
Each $V_\mu$ is $G$-stable because $G=T$ and $T$ acts on $V_\mu$ by the character with differential $\mu$. Since $\pi$ is irreducible and $V\ne 0$, exactly one weight space occurs and it equals $V$. If $\dim_{\mathbb C}V>1$, every one-dimensional subspace of $V$ would be $G$-stable, because $G$ acts by scalars on $V$. Hence $\dim_{\mathbb C}V=1$.
Since $\Phi^+=\varnothing$, the condition that positive root spaces annihilate a vector is vacuous. Thus the unique weight of $V$ is the unique highest weight, and its weight space is one-dimensional. This proves the theorem in the torus case.
[/step]
[step:Choose a maximal weight and prove it is killed by all positive root spaces]
Assume now that $\Phi\ne\varnothing$. By [citetheorem:9726] applied to the restricted representation $\pi|_T:T\to GL(V)$, the finite-dimensional $T$-module $V$ decomposes as the [direct sum](/page/Direct%20Sum) of its infinitesimal weight spaces $V_\mu$. Let $\Omega(V)\subset\mathfrak t_{\mathbb C}^*$ denote the finite set of weights $\mu$ for which $V_\mu\ne\{0\}$. Define a partial order $\preceq$ on $\mathfrak t_{\mathbb C}^*$ by declaring
\begin{align*}
\mu\preceq\nu
\end{align*}
when
\begin{align*}
\nu-\mu=\sum_{\alpha\in\Phi^+}n_\alpha\alpha
\end{align*}
for integers $n_\alpha\ge 0$, with all but finitely many $n_\alpha$ equal to $0$. Since $\Omega(V)$ is finite, choose a maximal element $\lambda\in\Omega(V)$ for this order, and choose $v\in V_\lambda$ with $v\ne 0$.
Let $\alpha\in\Phi^+$ and let $X\in(\mathfrak g_{\mathbb C})_\alpha$. For every $H\in\mathfrak t_{\mathbb C}$, the root-space relation gives
\begin{align*}
[H,X]=\alpha(H)X.
\end{align*}
Using that $d\pi$ is a Lie algebra representation, we compute
\begin{align*}
d\pi(H)d\pi(X)v=d\pi(X)d\pi(H)v+d\pi([H,X])v.
\end{align*}
Since $v\in V_\lambda$, this becomes
\begin{align*}
d\pi(H)d\pi(X)v=(\lambda(H)+\alpha(H))d\pi(X)v.
\end{align*}
Thus $d\pi(X)v\in V_{\lambda+\alpha}$. If $d\pi(X)v\ne 0$, then $\lambda+\alpha\in\Omega(V)$ and $\lambda\prec\lambda+\alpha$, contradicting the maximality of $\lambda$. Therefore
\begin{align*}
d\pi(X)v=0.
\end{align*}
So $\lambda$ supports a nonzero vector killed by every positive root space.
[guided]
The first task is to locate a weight that cannot be raised by any positive root. Let $\Omega(V)$ be the finite set of weights of the $T$-module $V$. The finiteness comes from the finite-dimensional weight-space decomposition for compact tori, applied to the restricted representation $\pi|_T:T\to GL(V)$. We order weights by positive roots: for $\mu,\nu\in\mathfrak t_{\mathbb C}^*$, write $\mu\preceq\nu$ if
\begin{align*}
\nu-\mu=\sum_{\alpha\in\Phi^+}n_\alpha\alpha
\end{align*}
with integers $n_\alpha\ge 0$. Because $\Omega(V)$ is finite, it has at least one maximal element for this order. Choose such a maximal weight $\lambda\in\Omega(V)$ and choose $v\in V_\lambda$ with $v\ne 0$.
We now prove that no positive root operator can move $v$ upward. Fix $\alpha\in\Phi^+$ and $X\in(\mathfrak g_{\mathbb C})_\alpha$. The defining property of the root space is
\begin{align*}
[H,X]=\alpha(H)X
\end{align*}
for every $H\in\mathfrak t_{\mathbb C}$. Since $d\pi:\mathfrak g_{\mathbb C}\to\mathfrak{gl}(V)$ is a Lie algebra representation, it preserves brackets. Therefore
\begin{align*}
d\pi(H)d\pi(X)v=d\pi(X)d\pi(H)v+d\pi([H,X])v.
\end{align*}
Because $v$ has weight $\lambda$, we have $d\pi(H)v=\lambda(H)v$. Substituting this and the root-space identity gives
\begin{align*}
d\pi(H)d\pi(X)v=\lambda(H)d\pi(X)v+\alpha(H)d\pi(X)v.
\end{align*}
Equivalently,
\begin{align*}
d\pi(H)d\pi(X)v=(\lambda+\alpha)(H)d\pi(X)v
\end{align*}
for every $H\in\mathfrak t_{\mathbb C}$. This says precisely that $d\pi(X)v$ is either zero or a vector of weight $\lambda+\alpha$.
If $d\pi(X)v$ were nonzero, then $\lambda+\alpha$ would be a weight of $V$. But $\lambda\prec\lambda+\alpha$ because $\alpha$ is a positive root, contradicting the chosen maximality of $\lambda$. Hence
\begin{align*}
d\pi(X)v=0.
\end{align*}
Thus $v$ is a highest weight vector of weight $\lambda$.
[/guided]
[/step]
[step:Use connectedness to identify the generated Lie algebra module with $V$]
Let $U(\mathfrak g_{\mathbb C})$ denote the universal enveloping algebra of the complex Lie algebra $\mathfrak g_{\mathbb C}$. Define
\begin{align*}
M:=U(\mathfrak g_{\mathbb C})v\subset V
\end{align*}
to be the complex vector subspace generated by the action of $U(\mathfrak g_{\mathbb C})$ on $v$. Let
\begin{align*}
\exp_G:\mathfrak g\to G
\end{align*}
denote the Lie exponential map of $G$. The subspace $M$ is stable under $\mathfrak g_{\mathbb C}$ by construction.
We claim that $M$ is $G$-stable. Since $M$ is finite-dimensional and stable under $\mathfrak g$, the solution of the linear [ordinary differential equation](/page/Ordinary%20Differential%20Equation)
\begin{align*}
\frac{d}{ds}m(s)=d\pi(X)m(s)
\end{align*}
with $m(0)\in M$ remains in $M$ for every $X\in\mathfrak g$ and every $s\in\mathbb R$. Hence $\pi(\exp_G(sX))M\subset M$. The connected Lie group $G$ is generated by exponentials from $\mathfrak g$: the subgroup generated by $\exp_G(\mathfrak g)$ contains the image under multiplication of a neighbourhood of $0\in\mathfrak g$, hence contains an open neighbourhood of the identity in $G$, and an open subgroup of a connected topological group is the whole group. Therefore $\pi(g)M\subset M$ for every $g\in G$.
Since $v\ne 0$, the subspace $M$ is nonzero. By irreducibility of the $G$-representation $V$, the nonzero $G$-stable subspace $M$ must equal $V$:
\begin{align*}
U(\mathfrak g_{\mathbb C})v=V.
\end{align*}
[/step]
[step:Apply triangular decomposition to locate all weights below $\lambda$]
Let
\begin{align*}
\mathfrak n^+:=\bigoplus_{\alpha\in\Phi^+}(\mathfrak g_{\mathbb C})_\alpha
\end{align*}
and
\begin{align*}
\mathfrak n^-:=\bigoplus_{\alpha\in\Phi^+}(\mathfrak g_{\mathbb C})_{-\alpha}.
\end{align*}
By the compact Lie algebra root-space structure theorem, as encoded in [citetheorem:9728], the complexified Lie algebra decomposes with respect to $\mathfrak t_{\mathbb C}$ as
\begin{align*}
\mathfrak g_{\mathbb C}=\mathfrak t_{\mathbb C}\oplus\bigoplus_{\alpha\in\Phi}(\mathfrak g_{\mathbb C})_\alpha.
\end{align*}
Separating positive and negative roots gives the triangular decomposition
\begin{align*}
\mathfrak g_{\mathbb C}=\mathfrak n^-\oplus\mathfrak t_{\mathbb C}\oplus\mathfrak n^+.
\end{align*}
By the Poincare-Birkhoff-Witt theorem applied to an ordered basis obtained by first listing a basis of $\mathfrak n^-$, then a basis of $\mathfrak t_{\mathbb C}$, and then a basis of $\mathfrak n^+$, $U(\mathfrak g_{\mathbb C})$ is spanned by products of elements from $U(\mathfrak n^-)$, then $U(\mathfrak t_{\mathbb C})$, then $U(\mathfrak n^+)$. Since $\mathfrak n^+v=\{0\}$ and $U(\mathfrak t_{\mathbb C})$ acts on $v$ by scalars determined by $\lambda$, the equality $V=U(\mathfrak g_{\mathbb C})v$ implies that $V$ is spanned by vectors obtained by applying monomials in negative root vectors to $v$.
More explicitly, every weight appearing in $V$ has the form
\begin{align*}
\lambda-\sum_{\alpha\in\Phi^+}n_\alpha\alpha
\end{align*}
with integers $n_\alpha\ge 0$. This follows by repeated use of the root-shifting calculation: if $Y\in(\mathfrak g_{\mathbb C})_{-\alpha}$ and $z\in V_\mu$, then $d\pi(Y)z\in V_{\mu-\alpha}$.
Consequently, if $\eta$ is a weight of $V$ and $V_\eta$ contains a nonzero vector killed by all positive root spaces, then the same connectedness and irreducibility argument gives $V=U(\mathfrak g_{\mathbb C})z$ for any such nonzero vector $z\in V_\eta$. Applying the preceding weight-lowering formula with $\eta$ in place of the original highest weight gives
\begin{align*}
\lambda=\eta-\sum_{\alpha\in\Phi^+}m_\alpha\alpha
\end{align*}
for some integers $m_\alpha\ge 0$. On the other hand, the formula obtained from $v$ gives
\begin{align*}
\eta=\lambda-\sum_{\alpha\in\Phi^+}n_\alpha\alpha
\end{align*}
for some integers $n_\alpha\ge 0$. Adding these two identities yields
\begin{align*}
\sum_{\alpha\in\Phi^+}(m_\alpha+n_\alpha)\alpha=0.
\end{align*}
Since all coefficients are nonnegative and $\Phi^+$ lies in an open half-space determined by the chosen positive system, this forces $m_\alpha+n_\alpha=0$ for every $\alpha\in\Phi^+$. Hence $\eta=\lambda$. Thus the highest weight is unique.
[/step]
[step:Average the inner product and prove orthogonality of distinct torus weight spaces]
Choose any Hermitian inner product $(\cdot,\cdot)_0$ on $V$, linear in the first variable. By [citetheorem:9712], there exists a $G$-invariant Hermitian inner product $(\cdot,\cdot)_V$ on $V$. Its restriction to $T$ is $T$-invariant.
Let $a\in V_\mu$ and $b\in V_\nu$ with $\mu\ne\nu$. By [citetheorem:9726], the infinitesimal weights $\mu$ and $\nu$ are the differentials of distinct torus characters. Let
\begin{align*}
\chi_\mu:T\to S^1
\end{align*}
and
\begin{align*}
\chi_\nu:T\to S^1
\end{align*}
denote the corresponding continuous characters, so $\pi(t)a=\chi_\mu(t)a$ and $\pi(t)b=\chi_\nu(t)b$ for every $t\in T$. Since the characters are distinct, choose $t\in T$ such that $\chi_\mu(t)\ne\chi_\nu(t)$. The $T$-invariance of $(\cdot,\cdot)_V$ gives
\begin{align*}
(a,b)_V=(\pi(t)a,\pi(t)b)_V=\chi_\mu(t)\overline{\chi_\nu(t)}(a,b)_V.
\end{align*}
Because $\chi_\mu(t)$ and $\chi_\nu(t)$ lie in $S^1$ and are unequal, $\chi_\mu(t)\overline{\chi_\nu(t)}\ne 1$. Hence $(a,b)_V=0$. Thus distinct $T$-weight spaces are orthogonal.
[guided]
The multiplicity-one argument needs two facts from compactness: an invariant inner product and orthogonality of different torus weights. Start with an arbitrary Hermitian inner product $(\cdot,\cdot)_0$ on $V$, linear in the first variable. Since $G$ is compact and $\pi$ is finite-dimensional, [citetheorem:9712] gives a $G$-invariant Hermitian inner product $(\cdot,\cdot)_V$. In particular,
\begin{align*}
(\pi(t)a,\pi(t)b)_V=(a,b)_V
\end{align*}
for every $t\in T$ and all $a,b\in V$.
Now let $a\in V_\mu$ and $b\in V_\nu$ with $\mu\ne\nu$. The weight-space decomposition theorem for compact tori, [citetheorem:9726], says that these infinitesimal weights come from torus characters. Define
\begin{align*}
\chi_\mu:T\to S^1
\end{align*}
and
\begin{align*}
\chi_\nu:T\to S^1
\end{align*}
to be the corresponding continuous characters. Thus $\pi(t)a=\chi_\mu(t)a$ and $\pi(t)b=\chi_\nu(t)b$ for every $t\in T$. Since $\mu\ne\nu$, the characters are distinct, so there is $t\in T$ with $\chi_\mu(t)\ne\chi_\nu(t)$.
Apply $T$-invariance at this element $t$. We get
\begin{align*}
(a,b)_V=(\pi(t)a,\pi(t)b)_V=\chi_\mu(t)\overline{\chi_\nu(t)}(a,b)_V.
\end{align*}
Because both character values lie on the unit circle and are unequal, $\chi_\mu(t)\overline{\chi_\nu(t)}\ne 1$. Therefore the last displayed identity forces $(a,b)_V=0$. This proves that distinct $T$-weight spaces are orthogonal.
[/guided]
[/step]
[step:Prove that the highest weight space is one-dimensional]
Let $v_1\in V_\lambda$ be a nonzero highest weight vector. By the preceding generation argument,
\begin{align*}
U(\mathfrak g_{\mathbb C})v_1=V.
\end{align*}
Let $w\in V_\lambda$ be arbitrary. We first check that $w$ is automatically killed by all positive root spaces. If $\alpha\in\Phi^+$ and $X\in(\mathfrak g_{\mathbb C})_\alpha$, the root-shifting calculation gives $d\pi(X)w\in V_{\lambda+\alpha}$. Since the preceding uniqueness argument shows that $\lambda$ is the unique highest weight, no weight strictly above $\lambda$ occurs; hence $V_{\lambda+\alpha}=\{0\}$ and $d\pi(X)w=0$. Thus every vector in $V_\lambda$ is a highest weight vector.
Define
\begin{align*}
c:=\frac{(w,v_1)_V}{(v_1,v_1)_V}\in\mathbb C
\end{align*}
and
\begin{align*}
u:=w-cv_1\in V_\lambda.
\end{align*}
Then $u$ is also killed by all positive root spaces and satisfies
\begin{align*}
(u,v_1)_V=0.
\end{align*}
We prove that $u=0$. Let $A\in U(\mathfrak g_{\mathbb C})$. By the PBW triangular decomposition and the fact that positive root operators kill $u$, the vector $Au$ is a linear combination of vectors of weights of the form
\begin{align*}
\lambda-\sum_{\alpha\in\Phi^+}n_\alpha\alpha.
\end{align*}
If some coefficient $n_\alpha$ is nonzero, the corresponding weight is not $\lambda$. Distinct weight spaces are orthogonal for the $T$-invariant Hermitian inner product by the preceding step. Therefore the corresponding vector is orthogonal to $v_1$. If all coefficients are zero, the corresponding term is a scalar multiple of $u$, which is orthogonal to $v_1$ by construction. Hence
\begin{align*}
(Au,v_1)_V=0
\end{align*}
for every $A\in U(\mathfrak g_{\mathbb C})$.
Thus the nonzero submodule $U(\mathfrak g_{\mathbb C})u$ would be orthogonal to $v_1$. If $u\ne 0$, irreducibility and connectedness give
\begin{align*}
U(\mathfrak g_{\mathbb C})u=V.
\end{align*}
This would imply that $v_1$ is orthogonal to all of $V$, including itself, contradicting
\begin{align*}
(v_1,v_1)_V>0.
\end{align*}
Therefore $u=0$, so $w=cv_1$. Hence $V_\lambda$ is spanned by $v_1$, and
\begin{align*}
\dim_{\mathbb C}V_\lambda=1.
\end{align*}
[/step]
[step:Conclude uniqueness and multiplicity one]
We have constructed a weight $\lambda$ for which $V_\lambda$ contains a nonzero vector annihilated by every positive root space. The triangular-decomposition argument proves that no other weight can have this property. The invariant-inner-product argument proves that all highest weight vectors of weight $\lambda$ are scalar multiples of one another. Therefore $\lambda$ is the unique highest weight of $V$ with respect to $(T,\Phi^+)$ and its weight space has complex dimension one:
\begin{align*}
\dim_{\mathbb C}V_\lambda=1.
\end{align*}
This proves the theorem.
[/step]
