We have $gl(3)\subset gl(4)$. I have a $gl(n)$ $V(3,2,1,0)$-module.
I want to know two things:
1) What is the $gl(3,\Bbb C)\subset gl(4,\Bbb C)$ branching rule for the $gl(n)$ $V(3,2,1,0)$-module.
2) What is the highest weight vector of $gl(n-1)$ as a subalgebra of $gl(n)$ with highest weight vector $(3,2,1,0)$?
It seems the Branching rule requires $(3,2,1,0)=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$ which is our weight of our highest weight vector, to give us $(\mu_1,\mu_2,\mu_3)$ for the highest weight vector of the $gl(3)$-submodule, via the rule $\lambda_i - \mu_i \geq 0$ and $\mu_i - \lambda_{i+1}\geq 0$(integers).
So we have:
\begin{align} 3-\mu_1 \geq 0\quad\quad\mu_1 -2\geq 0\\ 2-\mu_2 \geq 0\quad\quad \mu_2 -1\geq 0\\ 1-\mu_3 \geq 0\quad\quad \mu_3 - 0 \geq0 \end{align} Giving us: \begin{align} 3\geq \mu_1 \geq 2\\ 2\geq \mu_2 \geq 1\\ 1\geq \mu_3 \geq 0\\ \end{align}
Which gives us $(\mu_1,\mu_2,\mu_3)=(3,2,1),(3,2,0),(3,1,1),(3,1,0),(2,2,1),(2,2,0),(2,1,1),(2,1,0)$
Then (for 2)) all of these are highest weight vectors of the $gl(3)$-submodule?
And the branching rule (for 1)) I 'think' says that :
$$V(3,2,1,0)|_{gl(3)} = V'(3,2,1) \oplus V'(3,2,0) \oplus V'(3,1,1) \oplus V'(3,1,0) \oplus V'(2,2,1) \oplus V'(2,2,0 )\oplus V'(2,1,1) \oplus V'(2,1,0)$$
Maybe I could verify if I knew the dimension of the LHS&RHS, which I do know how to determine using Weyl's dimension formula, but it would be extremely long winded, and I suspect hopefully there is a better way.