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For the chosen simple roots $\Delta$, the open fundamental chamber of a root system isEquivalently $\lambda (h_{\alpha })>0$ for every simple coroot. Since all positive roots are nonnegative combinations of simple roots, requiring the inequality for all positive roots gives the same chamber. It is a connected component of the complement of the root reflection hyperplanes; the Weyl group permutes those chambers simply transitively.

The Closed dominant Weyl chamber is obtained by replacing the inequalities by $\ge 0$. It is the cone generated by the fundamental weights. When the fundamental Weyl chamber is taken to include its walls, this closed convention is intended; both conventions determine the same positive system and boundary rays.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

Take the simple roots $L_1-L_2$ and $L_2-L_3$. A weight represented by $(k_1,k_2,k_3)$ is in the Closed dominant Weyl chamber exactly when $k_1\ge k_2\ge k_3$. Among nonnegative triples of total four, the possibilities are $(4,0,0),(3,1,0),(2,2,0),(2,1,1)$. Using $L_1+L_2+L_3=0$, their explicit weight space bases areEach listed tensor has the required summed weight, and the monomial basis from part (i) shows linear independence and exhausts all possible tensors of that weight. Thus these are all the requested bases.

Let $x_1,x_2,x_3$ be the coordinate basis of the defining representation $W$ of ${\mathfrak{sl}}_3$. Write $s_{ij}=x_i{x}_j$ for $1\le i\le j\le 3$. The six $s_{ij}$ form a basis of the symmetric power ${\mathrm{Sym}}^{2}W$, with weight $L_i+L_j$. A basis for the tensor product of Lie algebra representations $T$ consists of all 36 ordered tensorsThe order of the two factors matters; there is no quotient identifying their interchange.

Each weight of a representation has the form $k_1{L}_1+k_2{L}_2+k_3{L}_3$, where the nonnegative integers satisfy $k_1+k_2+k_3=4$. Its weight multiplicity is the number of ways to split this triple into two nonnegative triples, each summing to two. The possible patterns, including every permutation, areFor example, $(2,2,0)$ splits as $(2,0,0)+(0,2,0)$, the reverse split, or $(1,1,0)+(1,1,0)$, giving multiplicity three. For $(2,1,1)$, the four splits pair $(2,0,0)$ with $(0,1,1)$ and $(1,1,0)$ with $(1,0,1)$, each in both orders. The multiplicities sum to $3+12+9+12=36$, verifying the dimension.

The following weight diagram uses Dynkin labels $(p,q)=(k_1-k_2,k_2-k_3)$; the circle numbers give multiplicities and orange points mark the Closed dominant Weyl chamber.

Choose a positive system $R^{+}$ and its simple roots $\Delta =\{{\alpha }_1,\dots ,{\alpha }_{\ell }\}$. The corresponding fundamental Weyl chamber is the open coneIts Closed dominant Weyl chamber replaces all strict inequalities by weak inequalities. The facts used here are that every positive root is a nonnegative integral combination of the simple roots, and that the Weyl chambers are the connected components left after removing the root-orthogonal hyperplanes. Hence the displayed inequalities describe one such component. The Weyl group permutes these components simply transitively, and every orbit in $E$ meets the Closed dominant Weyl chamber in exactly one point. A point on a wall may have a nontrivial stabilizer, which is why uniqueness of the chamber point does not always mean uniqueness of the group element carrying it there.

Let $v$ be a highest-weight vector of weight $\lambda$. For a simple root ${\alpha }_i$, choose its sl2 subalgebra associated with a root, with operators $E_i,F_i,H_i$ satisfying $[E_i,F_i]=H_i$, $[H_i,F_i]=-2F_i$, and $H_i={\alpha }_i^{\vee }$. We have $E_{i}v=0$ and $H_{i}v=\lambda (H_i)v$.

Since $F_i$ decreases the weight by ${\alpha }_i$ and there are only finitely many weights of a representation, let $n\ge 0$ be maximal with $F_i^{n}v\ne 0$. Induction using the two bracket relations gives the sl2 highest-weight lowering formulaApplying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^{n}v$ on the right forces $\lambda (H_i)=n$. This holds for every simple root, soThus the highest weight is a dominant integral weight, in particular an element of the Closed dominant Weyl chamber.

We induct on the Coxeter length $\ell (w)$. There is nothing to prove when $\ell (w)=0$. Otherwise choose a simple root $\alpha$ such thatequivalently, $w\alpha$ is negative. Since $\lambda$ and $\mu =w\lambda$ lie in the Closed dominant Weyl chamber,Therefore $⟨\lambda ,{\alpha }^{\vee }⟩=0$, and the simple reflection $s_{\alpha }$ fixes $\lambda$. Moreover,The induction hypothesis writes $w{s}_{\alpha }$ as a product of simple reflections that fix $\lambda$. Multiplying on the right by $s_{\alpha }$ gives the required expression for $w$. This is the Weyl stabilizer of a dominant point lemma.