Gap in an argument in Fulton & Harris?

Question

I'm reading through the two chapters in Fulton and Harris on the representation theory of $\mathfrak{sl}(3,\mathbb{C})$, in preparation for lecturing on them this week. I'll use F&H's notation, so that $\Gamma_{a,b}$ is the irreducible representation of highest weight $a e_1 - b e_3$ (with $e_i$ the character taking a diagonal matrix to its $(i,i)$ entry).

In the discussion of the multiplicities of the weights occurring in $\Gamma_{a,b}$, it seems to me that there's a gap -- so I am wondering what I'm missing here (or if perhaps there really is a gap). The issue arises at the sentence "To begin with..." on page 184, so let me give a quick summary of the argument to that point. Let $V$ be the standard representation of $\mathfrak{sl}(3,\mathbb{C})$ on $\mathbb{C}^3$, and $V^*$ its dual. The weights of ` $W :=\mathrm{Sym}^a V \otimes \mathrm{Sym}^b V^*$ lie on shrinking concentric hexagons $H_0,H_1,\ldots$ (with $H_0$ the outermost hexagon, and with $H_i$ degenerating to a triangle for $i$ sufficiently large). The multiplicities of $W$ are constant along each hexagon. For the argument to go through it suffices to show that the only highest vectors of $W$ are the unique ones (up to scaling) that occur at the unique dominant vertices of each $H_i$ up to $i = \textrm{min}(a,b)$, each of which contributes a $\Gamma_{a-i,b-i}$ to $W$.

The claim in the sentence "To begin with..." is that this follows for essentially combinatorial reasons: if there were a highest weight vector of weight $\alpha$, where $\alpha$ lies on $H_i$ but is not the dominant vertex, then "the multiplicity of $\alpha$ in $W$ would be strictly greater than [the multiplicity of the dominant vertex of $H_i$]." But why? Obviously this is correct if one considers only the contributions from highest weight vectors lying on $H_i$ or in its interior. But it seems a priori possible that some $\Gamma_{a-j,b-j}$ with $j < i$ (i.e., the irreducible constituent coming from one of the higher weight vectors on a hexagon lying outside $i$) could contribute a larger multiplicity to the dominant vertex of $H_i$ than to $\alpha$. If you haven't yet proved that the multiplicities of $\Gamma_{a,b}$ are also constant along the hexagons, I don't immediately see how this argument is going to give an inductive proof of that claim. Am I missing something elementary here, or does one have to work harder than F&H claim in order to finish the argument?

Answer 1

I'm still on the fence as to the pedagogical value of the "lecture" approach taken by F&H, but anyway it seems essential in a formal classroom presentation to articulate clearly which features of irreducible representations are or aren't known in general at each step of the exploration of examples in low ranks. F&H defer to later chapters the rigorous general treatment including the Weyl group and its action on weights. Initially they work out some details of examples starting with type A. These are illuminating and especially useful in concrete applications to algebraic geometry. But eventually the student has to confront more intricate behavior, especially in the case of tensor product decompositions.

Having said this, I suspect that the informal discussion in F&H may be somewhat oversimplified on page 184. Though I'd have to review the previous discussion more thoroughly to be sure about that. The structure of the weight diagram for an arbitrary irreducible representation of $\mathfrak{sl}_3(\mathbb{C})$ does depend ultimately on the Weyl group symmetry involved, without which it's hard to develop from first principles. Along the way you have to get control over weight strings in terms of the restriction to rank one, etc. The particular type of tensor product studied here in rank two is very nice but also very special in having summands with multiplicity one. But how to make it all rigorous without relying on too much general theory is a balancing act.

My own solution, for what it's worth, is to have students take turns talking about this material and arguing about the details. They can learn a lot that way.