The dimension of any irreducible $\frak{sl}\_n$-representation $V$ is clearly equal to the dimension of its dual representation $V^\*$. Does the dimension of an irreducible $\frak{sl}\_n$-representation determine it uniquely up to its dual? What happens for the $B,C$, and $D$ series?



Dimensions of $\frak{sl}_n$-representations
Recall that the dimension of the $\mathfrak{sl}\_n$ representation indexed by partition $\lambda$ is the number of semistandard Young tableaux of shape $\lambda$ with entries in $\{1,2,\ldots,n\}$.


You can check that there are 15 SSYTs of shape $(3,1)$ with entries in $\{1,2,3\}$, and similarly 15 SSYTs of shape $(4)$ with entries in $\{1,2,3\}$.


Furthermore these representations are not dual: $(3,1)=\omega\_1+2\omega\_2$ while $(4)=4\omega\_1$.


So the answer is **no**: not even irreducible $\mathfrak{sl}\_3$ representations are determined by their dimensions, up to duality.


