Let $V$ be a finite dimensional vector space over a finite field $F$. (The case $F = \mathbb{Z}/2\mathbb{Z}$ is the case I most care about.) Consider the poset of linearly independent subsets of $V$ ordered by inclusion. Is this poset shellable? There is an obvious lexicographic ordering that might be of use.
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$\begingroup$ shellable? lol why is that a thing. you may want to define that for myself and others. $\endgroup$– Douglas SirkCommented Aug 28, 2019 at 21:13
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$\begingroup$ shellable: en.wikipedia.org/wiki/Shelling_(topology) $\endgroup$– Carlo BeenakkerCommented Aug 28, 2019 at 21:15
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$\begingroup$ @DouglasSirk Add to Carlo Bennaker's wikipedia link that a poset is called shellable iff its order complex is shellable. $\endgroup$– Andreas BlassCommented Aug 28, 2019 at 21:16
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$\begingroup$ The collection of linearly independent subsets of $V$ ordered by inclusion is naturally a simplicial complex (indeed, it is evidently the independence complex of a matroid realizable over your field $F$). Do you want perhaps to know if this independence complex is shellable? Asking if the poset of faces is shellable is the same as asking if the barycentric subdivision of this independence complex is shellable? $\endgroup$– Sam HopkinsCommented Aug 28, 2019 at 21:41
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$\begingroup$ But to answer your question: the independence complex of any matroid is shellable- see Björner "The homology and shellability of matroids and geometric lattices"; and barycentric subdivision preserves shellability- see Björner "Shellable and Cohen-Macaulay Partially Ordered Sets". $\endgroup$– Sam HopkinsCommented Aug 28, 2019 at 21:52
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(Upgrading comments to an answer.)
The collection of linearly independent subsets of a finite-dimensional vector space $V$ over a finite field $F$ ordered by inclusion is naturally a finite simplicial complex: it is evidently the independence complex of a matroid realizable over $F$.
You might actually want to know if this independence complex is shellable. Asking if the poset of faces is shellable is the same as asking if the barycentric subdivision of this independence complex is shellable.
At any rate: the independence complex of any (finite) matroid is shellable: see Björner, "The homology and shellability of matroids and geometric lattices" (citation below). And moreover, barycentric subdivision preserves shellability: see Björner, "Shellable and Cohen-Macaulay partially ordered sets".
Hence, whatever your intended question, the answer is: yes, this complex is shellable.
Björner, Anders, Shellable and Cohen-Macaulay partially ordered sets, Trans. Am. Math. Soc. 260, 159-183 (1980). ZBL0441.06002.
Björner, Anders, The homology and shellability of matroids and geometric lattices, Matroid applications, Encycl. Math. Appl. 40, 226-283 (1992). ZBL0772.05027.
answered Aug 29, 2019 at 2:32
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$\begingroup$ (You can accept the answer if it’s what you want.) $\endgroup$ Commented Aug 29, 2019 at 15:33
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$\begingroup$ (By the way, an issue I’m slightly eliding is that when discussing the topology of a poset with a minimal element- in this case, the empty set, or empty face- it is conventional to omit this minimal element when taking the order complex: otherwise the complex you get is always contractile, as it is a cone. In order for “order complex of face poset=barcentryic subdivision of original complex” to be true you have to do this removal of the empty face.) $\endgroup$ Commented Aug 29, 2019 at 18:41
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$\begingroup$ I think the empty set is considered linearly independent; it amounts to augmenting the simplicial complex. A maximum element would be adding a cone point. $\endgroup$– rickCommented Aug 30, 2019 at 1:52
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$\begingroup$ I’m saying: if you naively take the order complex of a poset with a minimum (or with a maximum) you will get a complex that’s a cone (hence in particular necessarily contractible). $\endgroup$ Commented Aug 30, 2019 at 3:23