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

]]>maybe (EMA)

although i prefer brackets since you don’t need to use the shift key or reach as far, plus it won’t be confused with a government agency: BEMA or [EMA]

]]>Your blog ate the formatting of my reply. Let’s see if the usual WordPress syntax works better: Consider circular orderings of $[n]$, where we can change $ki\cdots j \cdots$ to $ik \cdots j \cdots$ for $i \lt j \lt k$. In other words, we can move $k$ past $i$ as long as $j$ occurs somewhere in the word. That is clearly more general than your cyclic Knuth, as you only allow $kij \to ikj$ and $jki \to jik$.

It turns out that the set of cyclic orders of [n] becomes a graded poset of length $\binom{n}{3}$ under this operation, with unique minimal element $n(n-1) \cdots 21$ and unique maximal element $12\cdots (n-1)n$. See https://arxiv.org/abs/0909.5324 for a more general statement related to an arbitrary affine Coxeter group, and see 2010 USAMO Problem 2 for a fun restatement.

Let me know if this seems inteeresting/useful to you!

]]>Anyway, let me know if it seems interesting!

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