Do we have $\mathbb{C}[\text{SL}_n] = \bigoplus_{\lambda, \,\text{ht}(\lambda)\leq n} V_{\lambda} $?

Question

The coordinate algebra $$\mathbb{C}[\text{SL}_n]=\mathbb{C}\big[x_{ij}: i, j \in \{1, \ldots, n\}\big]/\big(\det(x_{ij}) - 1\big)$$ is a representation of $\text{SL}_n$: $$(g'.f)(g)=f(g'^T g)\,.$$ Let $$V_{\lambda} = \langle e_T : T \text{ is a semistandard Young tableau of shape $\lambda$} \rangle\,,$$ where $\lambda$ is a Young diagram, $$e_T = \prod_c e_c\,,$$ where $c$ ranges over the columns of $T$, and for $c = (c_1, \ldots, c_l)^T$, $e_c$ is the minor of $x=(x_{ij})$ consisting of the first $l$ columns and rows $c_1, \ldots, c_l$. Then $V_{\lambda}$ is an irreducible representation of $\text{SL}_n$.

Do we have $$\mathbb{C}[\text{SL}_n] = \bigoplus_{\lambda,\, \text{ht}(\lambda)\leq n} V_{\lambda}\,?$$

Let $U$ acts on $\text{SL}_n$ by right multiplication. Then $\mathbb{C}[\text{SL}_n/U]$ is generated by all minors of $x=(x_{ij})$ which uses the first $l$ columns.

Do we have $$\mathbb{C}[\text{SL}_n/U] = \bigoplus_{\lambda,\, \text{ht}(\lambda)\leq n} V_{\lambda}\,?$$

Thank you very much.

Edit: Maybe the correct equalities are

(1) $\mathbb{C}[\text{SL}_n] = \bigoplus\limits_{\lambda,\, \text{ht}(\lambda)\leq n} V_{\lambda} \otimes V_{\lambda}^*$ and

(2) $\mathbb{C}[\text{SL}_n/U] = \bigoplus\limits_{\lambda,\, \text{ht}(\lambda)\leq n} V_{\lambda}$?

Maybe (1) follows from the algebraic Peter-Weyl theorem?

Answer 1

As Nate mentioned in his comment, (1) is just Peter-Weyl. See this post for details and references.

For (2) I assume you mean $U$ to be the unipotent radical of a Borel. You're almost correct, it should be a direct sum over the duals $$k[G/U] = \bigoplus V_\lambda^*$$ where $\lambda$ ranges over all irreducibles. The way to see this is a function on $G/U$ should be a function on $G$ which is invariant under multiplication by $U$ on the left. To see what this is, note that the Peter-Weyl decomposition is really a decomposition by $G \times G$-representations $$k[G] = \bigoplus V_\lambda \boxtimes V_\lambda^*$$ where one $G$ acts on the left factor, the other on the right factor. Insisting that right multiplication is $U$-invariant means taking $U$-invariant vectors of the left factor. The only $U$-invariant of $V_\lambda$ is the highest weight vector $v_\lambda$, so we have $$k[G/U] = \bigoplus \langle v_\lambda \rangle \otimes V_\lambda^* \simeq \bigoplus V_\lambda^*$$

Note: I might have messed up left/right and duals. Of course duality is an involution on the category, so it's the same expression