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Let $V$ be an $n$-dimensional vector space over a field $F$. Let $\mathrm{Gr}(V,d)$ be the set (Grassmann variety) of all $d$-dimensional subspaces of $V$. We can regard $\mathrm{Gr}(V,d)$ as a subset of the vector space $\Lambda^d(V)$ (of dimension ${n\choose d}$) via the Plücker embedding. Thus $\mathrm{Gr}(V,d)$ defines a matroid $\mathrm{Pl}(V,d)$ (where independence corresponds to linear independence in $\Lambda^d(V)$). Can anything interesting be said about this matroid?

In particular, there is a "natural" matroid structure on $\mathrm{Gr}(V,d)$, namely, the matroid corresponding to the $d$th Dilworth truncation $D_d \mathbb{B}(V)$ of the lattice $\mathbb{B}(V)$ of all subspaces of $V$. Is $D_d \mathbb{B}(V)$ isomorphic to $\mathrm{Pl}(V,d)$?

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  • $\begingroup$ Can you give a quick self-contained definition of a Dilworth truncation? I must have forgotten it, and the reference isn't all that useful. $\endgroup$
    – darij grinberg
    Commented Oct 16, 2022 at 2:29
  • $\begingroup$ Is the Dilworth truncation of a matroid the same as defined in Definition 12.2.1 of H. Narayanan, Submodular functions and electrical networks, 2nd edition 2019, applied to the rank function of the matroid? $\endgroup$
    – darij grinberg
    Commented Oct 16, 2022 at 2:33
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    $\begingroup$ @darijgrinberg: a quick (but not so easy to work with) definition is the following. Let $L$ be a geometric lattice of rank $n$. Let $1\leq d\leq n-1$. Then the matroid $M$ corresponding to $D_dL$ has as points the elements of $L$ of rank $d$. A subset $S$ of points of $M$ is independent if the join of any $k$ elements of $S$ has rank at least $d+k-1$ in $L$. $\endgroup$
    – Richard Stanley
    Commented Oct 16, 2022 at 2:52
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    $\begingroup$ is the following correct reformulation? "if $A$ is a collection of $d$-dimensional subspaces of $V$, then the corresponding Plücker elements of $\Lambda^d(V)$ are linearly independent if and only if the span of any $k$ subspaces from $A$ have dimension at least $d+k-1$"? $\endgroup$
    – Fedor Petrov
    Commented Oct 16, 2022 at 9:03
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    $\begingroup$ (if so, then this seems to be not true for three 2-dimensional coordinate subspaces in $\mathbb{R}^3$) $\endgroup$
    – Fedor Petrov
    Commented Oct 16, 2022 at 10:41

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