Do representations of same dimension implies isomorphic closed orbits?

Question

Let us recall this fact. Let $G$ be a semisimple algebraic group over $\mathbb C$ and let $V,V'$ be two irreducible $G$-representations. We denote by $X,X'$ the unique closed $G$-orbits contained in $\mathbb P V, \mathbb P V'$ respectively. We know that if $$ \mathbb P V \supset X \cong X' \subset \mathbb P V' $$ as projective $G$-varieties, then $\mathbb PV \cong \mathbb PV'$ as projective spaces. In particular, $\dim V=\dim V'$.

I want to understand the inverse direction: if I have two irreducible $G$-representations $W,W'$ of the same dimension, should I conclude that the closed $G$-orbits $Y \subset \mathbb P W, Y' \subset \mathbb P W'$ are isomorphic as projective $G$-varieties?

Answer 1

No, this is not true. For instance the symplectic group $$ G = \mathrm{Sp}(V), $$ where $V$ is a symplectic vector space of dimension 6 has two irreducible representations of dimension 14: $$ V_1 = \wedge^2V^\vee / \langle \omega \rangle, \qquad\text{and}\qquad V_2 = \wedge^3V^\vee / (V^\vee \wedge \omega), $$ where $\omega$ is the symplectic form. These representations are not isomorphic, as well as the corresponding closed orbits, which are the isotropic Grassmannians $$ X_1 = \mathrm{IGr}(2,V) \qquad\text{and}\qquad X_2 = \mathrm{IGr}(3,V). $$

By the way, the inverse is also not true unless the line bundles associated with the embeddings of $X \to \mathbb{P}(V)$ and $X' \to \mathbb{P}(V')$ are identified by the isomorphism $X \cong X'$.