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There are well known identities relating elementary symmetric polynomials in $k$ variables and complete homogeneous symmetric polynomials in the same set of variables. For example,

$$e_{2,k}(x) = h_{1,k}(x)^2 - h_{2,k}(x).$$

I'd like to do the same but for the factorial versions that also have $y$ variables. There is some subtlety that needs to be worked out, in particular the $y$ variables need to be altered to make the same identities work. For example, $$E_{2,4}(x_1,x_2,x_3,x_4;y_1,y_2,y_3)=H_{1, 4}(x_1, x_2, x_3, x_4; y_1, y_2, y_3, y_5)H_{1, 4}( x_1, x_2, x_3, x_4; y_1, y_2, y_3, y_4) - H_{2,4}(x_1,x_2,x_3,x_4;y_1,y_2,y_3,y_4,y_5).$$

I have a systematic way of computing this programmatically, I'm just looking for a formulaic way rather than algorithmic, which would very likely be much faster.

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In case anyone ever has this problem again, I found this in formula (5.30) in the paper Bertiger, Anna, Elizabeth Milićević, and Kaisa Taipale. "Equivariant quantum cohomology of the Grassmannian via the rim hook rule." Algebraic Combinatorics 1.3 (2018): 327-352.

The relevant work is that for any $s$, there is the shifted identity

$$\sum_{r=0}^k (-1)^rE_{r,k}(x;y)H_{s-r,k}(x;\tau^{1-s}y)=0$$ The trouble I had is there are no variables with negative indices, which would be implied by the shift operator. Turns out if you introduce them, they cancel anyway. For my purposes it's most advantageous to stick particular values in there, but of course 0 would also work.

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  • $\begingroup$ Don't forget to accept your own answer, if it resolves the question! $\endgroup$ Commented May 13 at 0:37
  • $\begingroup$ @LSpice I'm almost certain it does, but my initial attempts to code it up didn't work. The problem is probably on my end. I will wait until I have a working implementation. $\endgroup$ Commented May 13 at 1:50

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