This is the spin-off of the question I previously asked.

First, let me remind you some notation from that question:

$G_0$ - compact, simply connected Lie group giving rise (by complexification) to a semi-simple Lie Group $G$

$\mathfrak{g}$ - Lie algebra of $G$

$\delta=\frac{1}{2}\sum_{\alpha>0}\alpha$ (summation over all possitive roots of $\mathfrak{g}$)

$V_{\lambda}$ - finite dimensional complex vector space on which $\mathfrak{g}$ is irreducibly represented (with the highest weight $\lambda$ and highest weight vector $v_\lambda$).

$\mathbb{P}V_\lambda$ - complex projective space of $V_\lambda$

$\pi: V_\lambda\rightarrow \mathbb{P}V_\lambda$ - mapping onto the projective space

$O_{v_{\lambda}}^0 $ - orbit of $G_0$ through $v_\lambda$

$\Omega:V_\lambda\otimes V_\lambda \rightarrow V_\lambda \otimes V_\lambda$ - second order Casimir represented on $V_\lambda\otimes V_\lambda$

Now, my the true question starts:

In the article by Lichteinstein ( check here )it is proven that the necessary and sufficient condition for $\pi(v)$ to be in $\pi(O_{v_{\lambda}}^0)$ is the following:

$\Omega (v\otimes v) =\langle 2\lambda+2\delta,2\lambda\rangle (v\otimes v)$

In the above $\langle\cdot ,\cdot \rangle $ is a standard inner product on Cartan algebra dual $\mathfrak{h}^*$ .

I came up with the way of reasoning that might simplify arguments for this statement - they rely on a rather heavy (from my perspective) machinery. $\Omega$ decomposes into a direct sum of multiples of identity acting on irreducible representations $V_\alpha$ of $\mathfrak{g}$ on which $V_\lambda\otimes V_\lambda$ decomposes:

$\Omega=\sum_{\alpha\in\mathcal{A}}\beta(\alpha)Id_{V_\alpha}$

$\beta(\alpha)$ is some function of the weight $\alpha$- see above, $\mathcal{A}$ denotes some collection of weights - there may be some overlappings (if that is the case corresponding spaces $V_\alpha$ and $V'_\alpha$ are different). .

Now, it is easy to see that in the above sum there is only one $\alpha_0=2\lambda$ (it is the weight of the irreducible representation generated on $V_\lambda\otimes V_\lambda$ by $v_\lambda\otimes v_\lambda$ ). As $\Omega$ commutes with all unitary operators in the representation of $G_0$ (see notation above) in $V_\lambda \otimes V_\lambda$, clearly: $\Omega(v\otimes v)=\beta(2 \lambda) v\otimes v$ for $v\in O_{v_\lambda}^0$.

In order to have the complete characterization of $\pi(O_{v_\lambda}^0)$ I only have to know that $v \otimes v \in V_{2 \lambda}$ implies that $v\otimes v$ is proportional to some $\Phi(g) v_\lambda \otimes \Phi(g) v_\lambda$, where $\Phi$ - representation of $G_0$ in $V_\lambda$, $g\in G_0$.

One possibility would be to allow $G_0$ to act transitively on $\pi(V_{2\lambda})$ where $V_{2 \lambda}$ treated as a subspace of $V_\lambda\otimes V_\lambda$. Yet, this condition is not fulfilled for large dimensions of $V_\lambda$. Let me state the final question.

When $v \otimes v \in V_{2 \lambda}$ ( $V_{2 \lambda}$ is treated as as subspace of $V_{\lambda}\otimes V_\lambda$) implies that $v\otimes v$ is proportional to some $\Phi(g) v_\lambda \otimes \Phi(g) v_\lambda$?

This is the spin-off of the question I previously asked.

First, let me remind you some notation from that question:

$G_0$ - compact, simply connected Lie group giving rise (by complexification) to a semi-simple Lie Group $G$

$\mathfrak{g}$ - Lie algebra of $G$

$\delta=\frac{1}{2}\sum_{\alpha>0}\alpha$ (summation over all possitive roots of $\mathfrak{g}$)

$V_{\lambda}$ - finite dimensional complex vector space on which $\mathfrak{g}$ is irreducibly represented (with the highest weight $\lambda$ and highest weight vector $v_\lambda$).

$\mathbb{P}V_\lambda$ - complex projective space of $V_\lambda$

$\pi: V_\lambda\rightarrow \mathbb{P}V_\lambda$ - mapping onto the projective space

$O_{v_{\lambda}}^0 $ - orbit of $G_0$ through $v_\lambda$

$\Omega:V_\lambda\otimes V_\lambda \rightarrow V_\lambda \otimes V_\lambda$ - second order Casimir represented on $V_\lambda\otimes V_\lambda$

Now, my the true question starts:

In the article by Lichteinstein ( check here )it is proven that the necessary and sufficient condition for $\pi(v)$ to be in $\pi(O_{v_{\lambda}}^0)$ is the following:

$\Omega (v\otimes v) =\langle 2\lambda+2\delta,2\lambda\rangle (v\otimes v)$

In the above $\langle\cdot ,\cdot \rangle $ is a standard inner product on Cartan algebra dual $\mathfrak{h}^*$ .

I came up with the way of reasoning that might simplify arguments for this statement - they rely on a rather heavy (from my perspective) machinery. $\Omega$ decomposes into a direct sum of multiples of identity acting on irreducible representations $V_\alpha$ of $\mathfrak{g}$ on which $V_\lambda\otimes V_\lambda$ decomposes:

$\Omega=\sum_{\alpha\in\mathcal{A}}\beta(\alpha)Id_{V_\alpha}$

$\beta(\alpha)$ is some function of the weight $\alpha$- see above, $\mathcal{A}$ denotes some collection of weights - there may be some overlappings (if that is the case corresponding spaces $V_\alpha$ and $V'_\alpha$ are different). .

Now, it is easy to see that in the above sum there is only one $\alpha_0=2\lambda$ (it is the weight of the irreducible representation generated on $V_\lambda\otimes V_\lambda$ by $v_\lambda\otimes v_\lambda$ ). As $\Omega$ commutes with all unitary operators in the representation of $G_0$ (see notation above) in $V_\lambda \otimes V_\lambda$, clearly: $\Omega(v\otimes v)=\beta(2 \lambda) v\otimes v$ for $v\in O_{v_\lambda}^0$.

In order to have the complete characterization of $\pi(O_{v_\lambda}^0)$ I only have to know that $v \otimes v \in V_{2 \lambda}$ implies that $v\otimes v$ is proportional to some $\Phi(g) v_\lambda \otimes \Phi(g) v_\lambda$, where $\Phi$ - representation of $G_0$ in $V_\lambda$, $g\in G_0$.

One possibility would be to allow $G_0$ to act transitively on $\pi(V_{2\lambda})$ where $V_{2 \lambda}$ treated as a subspace of $V_\lambda\otimes V_\lambda$. Yet, this condition is not fulfilled for large dimensions of $V_\lambda$. Let me state the final question.

When $v \otimes v \in V_{2 \lambda}$ ( $V_{2 \lambda}$ is treated as as subspace of $V_{\lambda}\otimes V_\lambda$) implies that $v\otimes v$ is proportional to some $\Phi(g) v_\lambda \otimes \Phi(g) v_\lambda$?