I have one question which might be stupid. When we want to find which Schubert cell a point of the Grassmannian belongs to, we cut off the upside-down staircase from the matrix, all entries in the staircase should be 0, right? Why must the first column of the matrix be a zero column? ]]>

Also, I think the sentence “…since any transition matrix to another basis must be integral and orthogonal with respect to the Hall inner product…” is a little bit confusing, because the transition matrix is orthogonal as a matrix, and it does not need any reference to inner product. ]]>

Oh a pedagogical tool! Now I understand completely.

]]>Hi David,

I think I fixed your formatting… not sure why it was giving you trouble before! That is super interesting, I wasn’t aware of these generalized “cyclic Knuth” moves, thanks for sharing!

Maria

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