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This week i tried and practice the way to calculate double schubert polynomials with a square tiling. even though i couldnt connect the tilings i counted some of the blank squares in even and odd matrices with 2 particular tiles in the bottom right corner, the cross tile and one of the elbow tiles. I think they are the only ones to do it with. Anyways I had another condition to add only cross tiles in the diagonal aside from the 2 corner pieces i mentioned. so i tried and count how many rows and columns were necessary to complete the tiling if the farthest out cross tile from the corner exits through the closest edges and the closer tilings also exit through the next closest ones- the closer to the corner the further out they exit through the top part of the crosses. in the graphic i made its easier to see. the cross has two parts that exit: two edges that leave through the bottom and side and the side of the cross pointed toward the opposite corner exits a few rows and columns in the other direction. it took a lot of attempts. after i tried and alter the cross tiles in the diagonal to an elbow tile and it doesnt leave all the edges with a line exiting them so i forgot about it for now.

in the odd nxn squares and even n-1 x n-1 squares there is the same number of blank squares. the odd squares blank can be moved to the last indices possible of the even squares- aside from that i can not really tell how a square nxn tiling and square n-1 x n-1 tiling are supposed to calculate two different schubert polynomials.

here is another graphic with a 3 by 3 that has the outer blank diagonal shifted in to indices 2,2 beside a 2 by 2 without the possibility of shifting the blank square in at all. That is my best guess why the different sizes are different polynomials. can anybody explain more precisely?

even and odd square matrices

shifted blank square

Priocapybara50
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1 Answers1

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Technically there's an object that finds the coefficient of a term in power series', there should be one that finds the available indices of blank squares.

In another question about Stanley introduction to hyperplane arrangements multiple types of coefficients of the variable term $t^{k}$ are able to be expressed by $[t^{k}] \chi (t)$.

Ohio skateboarder . 7
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  • that is a good explanation... & i noticed that you originally decided that square into two topological spaces X and Y where Y is the rest of the square aside from the square with the cross diagonal parts and the first row and column that has the top part of the first or last cross tiles top part to exit from, an elbow shape of tiles that is where the tiles in the diagonal except for one tile can exit out of the y index from the x index. It's interesting I will do some now, I reviewed some real analysis after I quit finding the right Schubert polynomials. does that elbow space count as a elbow? – Priocapybara50 May 30 '23 at 21:41
  • In this situation the double elbow tile, like seen in isosurfaces textbooks, is not a tile. Yet the square tilings can necessarily have the elbow space except if the corner tile in the final indices is the only tile an elbow tile because further diagonal pieces are one of six tiles in the tiling with six squares, that are a line beginning at every first x index at the bottom of the square that are lines exiting through every different y index at the side of the square. It's a type of bijective set of lines between two sides. – Ohio skateboarder . 7 May 30 '23 at 21:49