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In the paper "Quantum state transformations and the Schubert calculus" by Sumit Daftuar and Patrick Hayden (Annals of Physics 315 (2005) 80-122) on page 91, we have following notations:

$A_r$ denotes the $r$-dimensional vector space spanned by eigenvectors corresponding to the $r$ largest eigenvalues $\lambda_1 < \ldots < \lambda_r$ of $A$ (I am assuming non degenerate ones for simplicity), where $A$ is some given, Hermitian matrix in $\mathbb{C}^{n \times n}$. $\pi$ is a given binary sequence of length $n$ and weight $r$.

Now, certain Schubert cells are defined as: $$ S_\pi = \{ V \leq \mathbb{C}^n \mid \mathrm{dim}(V \cap A_i)/(V \cap A_{i-1}) = \pi(i), \ 1 \leq i \leq n\}$$ The relation $V \leq \mathbb{C}^n$ here means $V$ is a subspace of $\mathbb{C}^n$ and $\pi(i)$ is the $i$-th term of the sequence $\pi$.

To ease notation, the sequence $\ell_1 < \ldots <\ell_r$ reflects the indices at which $\pi$ equals $1$.

I am baffled by the definition of $S_\pi$ since to me it seems as if there is always only one element contained in $S_\pi$, namely the subspace spanned by the eigenvectors corresponding to the eigenvalues $\lambda_{\ell_j}$, $j = 1,\ldots,r$, of $A$.

From my yet very limited knowledge about Schubert calculus and from the subsequent analysis in the paper, I assume that the following is meant: $$ S_\pi = \{ V \leq \mathbb{C}^n \mid \mathrm{dim}(V \cap A_{\ell_j}) \geq j, \ j = 1,\ldots,r \} $$ where $\ell_{j}$ is defined by $\pi$ as above. In that sense $A_{\ell_1} \subset \ldots \subset A_{\ell_r}$ is the relevant flag. Can someone confirm this?

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I am baffled by the definition of $S_\pi$ since to me it seems as if there is always only one element contained in $S_\pi$, namely the subspace spanned by the eigenvectors corresponding to the eigenvalues $\lambda_{\ell_j}$, $j = 1,\ldots,r$, of $A$.

This logic is wrong. Try the simplest nontrivial case -- the case when $n=2$ and $r=1$ and $\pi=\left(0,1\right)$. Then, $S_\pi$ contains all $1$-dimensional subspaces $V$ of the $2$-dimensional vector space $\mathbb{C}^n$ satisfying $V \cap A_1 = 0$ (and $\dim\left(V \cap A_2\right) = 1$, but this is automatically satisfied because $A_2 = \mathbb{C}^n$ and $V$ is $1$-dimensional). These are all $1$-dimensional subspaces of $\mathbb{C}^n$ except for $A_1$ itself (which lies in $S_{\left(1,0\right)}$ instead).

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  • $\begingroup$ I feel so embarrassed... thank you. $\endgroup$
    – Sebastian K.
    Commented Jan 18, 2019 at 11:31
  • $\begingroup$ At least, if I am not mistaken, the set $S_\pi$ (my lower definition) is the closure of the correct $S_\pi$ (as defined by the authors). So their definition should in fact be a Schubert cell, while mine is the Schubert variety. It seems like (English) Wikipedia is again not the best source, since their definition of Schubert cells seems to be the one of Schubert variety instead. I am not yet sure how distinguished those two definitions are in this field of mathematics... $\endgroup$
    – Sebastian K.
    Commented Jan 18, 2019 at 14:21
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    $\begingroup$ @SebastianK.: Yes, their definition is a Schubert cell, while yours is the corresponding Schubert variety (i.e., the Zariski closure of the Schubert cell). I think the definitions are fairly standard, except that most authors use the standard basis instead of the eigenbasis of $A$ (of course, the difference is insubstantial, since any basis can be transformed into any other by an automorphism of the vector space), and that some authors use row-equivalence classes of $r\times n $-matrices instead of vector subspaces (but again, this is in bijection). $\endgroup$
    – darij grinberg
    Commented Jan 18, 2019 at 14:30

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