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 $\pi(i)$.

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?