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?



Typo in a paper definition of Schubert cells?

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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 $A$.
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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).


