How to understand that minors are matrix elements in fundamental representations of $SL_n$?

Question

In the video, Lecture 3 of June 14, 49:00-53:00, it is said that "minors are matrix elements in fundamental representations of $SL_n$". What are fundamental representations of $SL_n$? How to understand that minors are matrix elements in fundamental representations of $SL_n$? Thank you very much.

Answer 1

The notion of fundamental representations is defined for any finite-dimensional semisimple Lie algebra ${\mathfrak g}$ with a fixed choice of a Cartan subalgebra ${\mathfrak h}$ and a set of positive roots $\Phi^+\subset\Phi({\mathfrak g},{\mathfrak h})\subset{\mathfrak h}^{\ast}$ of ${\mathfrak g}$ with respect to ${\mathfrak h}$: Namely, if $\{\alpha_1,...,\alpha_n\}$ is an enumeration of the simple roots corresponding to $\Phi^+$, and if $\alpha_1^{\vee},\ldots,\alpha_n^{\vee}$ are the associated coroots in ${\mathfrak h}$, then the $i$-th fundamental weight ${\mathfrak g}$ is the unique $\omega_i\in{\mathfrak h}^{\ast}$ such that $\omega_i(\alpha_j^{\vee})=\delta_{ij}$. The $i$-th fundamental representation is the unique irreducible representation with highest weight $\omega_i$.

Consider now the case of ${\mathfrak g} = {\mathfrak s}{\mathfrak l}_n({\mathbb C})$ with ${\mathfrak h} := \text{diag}$, $\Phi^+ := \{\varepsilon_i - \varepsilon_j\ |\ i<j\}$ (where $\varepsilon_i:{\mathfrak h}=\text{diag}\to{\mathbb C}$ is the projection onto the $i$-th diagonal entry) and $\alpha_i^{\vee} = \text{diag}(0,\ldots,0,1,-1,0,\ldots,0)$. Then the $i$-th fundamental weight $\omega_i$ is given by $\varepsilon_1+\ldots+\varepsilon_i$, and the $i$-th fundamental representation is the natural representation of ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$ on $\bigwedge^i {\mathbb C}^n$: If $e_1,\ldots,e_n$ is the standard basis of ${\mathbb C}^n$, then the ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$-module $\bigwedge^i {\mathbb C}^n$ is spanned by the highest weight vector $e_1\wedge\ldots\wedge e_i$, which has weight $\varepsilon_1+\ldots+\varepsilon_i=\omega_i$ since each $e_j$ has weight $\varepsilon_j$ in the natural representation of ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$ on ${\mathbb C}^n$.

Finally, the matrix coefficients of ${\mathfrak s}{\mathfrak l}_n({\mathfrak C})$ with respect to the basis $\{e_{j_1}\wedge\ldots\wedge e_{j_i}\ |\ j_1<\ldots<j_i\}$ of $\bigwedge^i{\mathbb C}^n$ are precisely the $i$-minors.

Answer 2

For example, in the case of $SL_3$, we have $$ V_{\omega_1} = \langle b_{11}, b_{12}, b_{13} \rangle. $$ is the fundamental module of $SL_3$ with highest weight $\omega_1$. The weight of $b_{11}$ is $\omega_1$. The weight of $b_{12}$ is $-\omega_1 + \omega_2$. The weight of $b_{13}$ is $-\omega_2$.

We have $$ V_{\omega_2} = \langle b_{11}b_{22} - b_{12}b_{21}, b_{12}b_{23}-b_{13}b_{22}, b_{11}b_{23}-b_{13}b_{21} \rangle. $$ The weight of $b_{11}b_{22} - b_{12}b_{21}$ is $\omega_2$.