The short version:
Can the theory of weights for SL(n,C) be explained concretely in terms of raising and lowering operators on spaces of polynomials?
A deleted question asked how to prove SL(3,C) acts irreducibly on the space of homogenous polynomials of fixed degree. I think it was deleted as a homework question, but it has never been a homework question at my university (at least not in the past decade), and so I thought it might be reasonable to try and work out these ideas in a setting that makes sense to me:
Let K be a field of characteristic 0 and let SL(n,K) be the group of n×n matrices with entries from K and determinant 1. Let V be the K-vector space with basis the monomials xi of total degree w, where i = (i1,i2,...,in) is a multi-index with n entries, all non-negative integers whose sum is w. V has dimension w+1. SL(n,K) operates on V in the one true manner, by substitution of variables. In particular, for j = 2,3,...,n, the group SL(n,K) contains the elementary substitutions Ej which takes xj to x1+xj and takes xk to xk when j≠k.
Define the lowering operator Lj as Ej−1, so it takes a polynomial f to the difference between Ej(f) and f itself. The L are called lowering operators because they lower the degree of f when considered as a polynomial in xj, though they hold the total degree constant. This can be checked by applying it to monomials. For any individual monomial, factor it as xja⋅xi where ij=0, then Lj(xja⋅xi) = (x1+xj)a.xi − xja⋅xi = ( (x1+xj)a−xja )⋅xi. The leading term, when considered as a polynomial in xj is then ax1xja−1xi, so it has degree a−1 in xj.
If W is a submodule of V containing some nonzero polynomial f, then it must have nonzero degree in one of the variables xj. If any of these j are greater than 1, then apply Lj to transfer the degree from j to 1. Repeat until one is left with a⋅x1w. Since each Lj takes W to W and W is closed under division by nonzero scalars, we must have that any non-empty SL(n,K) invariant subspace of V contains x1w. In fact, we only use that it is closed under the maximal unipotent subgroup generated by the Ej.
For instance, if n=3, then we can more simply write x1=x, x2=y, and x3=z. If w=4, and W contains xyyz+2xyzz, then apply L3 = Lz to get xyy(x+z)+2xy(x+z)(x+z) − xyyz+2xyzz = xyyx + 2xyxx + 4xyxz = 2xxxy + xxyy + 4xxyz. The total degree remained the same, but shifted from z to x, at the cost of increasing the leading coefficient. Applying Lz again, one gets the first two terms vanish, and one is left with 4xxy(x+z) − 4xxyz = 4xxxy. Applying Ly one gets the pure 4xxx(x+y)−4xxxy = 4xxxx.
Notice that there were multiple paths we could have taken to move from f to xw. I think these paths are basically what are called "weights", and the lowering operators are called (simple, positive) "roots".
Now we must recover all other monomials from this monomial, and so we define the raising operators. We begin with the elementary Fj in SL(n,K) defined by taking x1 to x1+xj and xk to xk for k≠j. The raising operator Rj is then Fj−1 which takes a polynomial f to the difference between Fj(f) and f itself. Rj sacrifices a degree in x1 to provide a degree in xj. Applying each Rj operator ij times to x1w we obtain a polynomial with nonzero xi term, but with a great many other terms that seem difficult to control.
How does one use the raising operator to recover the other monomials?
For instance, to show xxyz is in W, I imagine that we should apply Rz to xxxx to get (x+z)(x+z)(x+z)(x+z) − xxxx = 4xxxz + 6xxzz + 4xzzz + zzzz. But then, how does one isolate 4xxxz? Here is my current method, which seems overly complex compared to lowering: Applying Lz 3 times and subtract 60xxxx (already known to be in W) takes zzzz to a multiple of xxxz, but it does so by zero-ing out the other terms. Applying Ry to xxxz gives (x+y)(x+y)(x+y)z−xxxz = 3xxyz + 3xyyz + yyyz. Now apply Ly twice to zero out the first two terms and take yyyz to 12xxxz + 6xxyz. Since xxxz is known to be in W, we can remove it and rescale to get xxyz in W.
This random sort of zig-zag makes it hard to keep track of which "roots" we've been applying (that is, which raising and lowering operators). In the Lie algebra case, I thought things were a bit cleaner. At any rate, successfully applying these operators should have the side-effect of detecting in a very concrete way the "highest weight", if I understand correctly. However, I cannot yet check, since I cannot yet successfully keep track of how I am applying these operators.
Is there a clear relationship between the paths one takes using these operators and the theory of weights? For instance, is x1w a highest weight vector?
It seems to me that SL(n,K) should have other irreducible representations other than just these polynomial representations, just because the paths defined by the operators are so symmetric. I think for SL(2,K) this is basically all there is, perhaps allowing for field automorphisms to be applied first.
Can all of the finite dimensional irreducible representations of SL(n,K) be explained in terms of actions on polynomials by substitutions and field automorphisms?
Assuming that this stuff makes sense, it would be handy to try it out on a different weight lattice.
Is there a version of this sort of description for any of the other classical groups or for G2?
It would be nice if say the symplectic group operated by substitutions on some subspace of homogenous polynomials, possibly restricted in some way, and that one could then find the raising and lowering operators. Any group where the action is natural and is likely to work well in the setup of Dickson's Linear Groups is fine, but I suspect the symplectic group might be the one with the least complications.