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A simplicial complex $K$ on a vertex set $[m] = \{1,...,m \}$ - where here we say vertex set to mean an ambient set of vertices on which $K$ is defined - is self-dual if it is equal to its Alexander dual $\widehat{K}$, where $\widehat{K}$ is the simplicial complex on $[m]$ whose simplices are the complements of the missing faces of $K$. That is, simplices of $\widehat{K}$ are given by $$\widehat{K} = \{J \subseteq [m] \; | \; [m] \backslash J \notin K \}.$$

An example of a self dual complex is $\partial \Delta^2$ disjoint unioned with one vertex: $K = \{\emptyset, \{ 1\} \{2 \} \{ 3\} \{ 4\} \{ 1,2\} \{2,3 \} \{ 1,3\} \}$.

Timotejević's 2019 note explores the combinatorial structure of such complexes. For example, theorem 2.4 therein says that a complex $K$ is self dual if and only if for arbitrary $A \subseteq [m]$, $A \in K \iff [m] \backslash A \notin K$.

One may also define a simplicial complex to be self-dual if it is isomorphic, as a simplicial complex, to its Alexander dual. With this definition, the above theorem does not hold. For example, consider the simplicial complex on four vertices $$K = \{\emptyset, \{ 1\} \{2 \} \{ 3\} \{ 4\} \{ 1,2\} \{2,3 \} \{ 3,4\} \}.$$ This is isomorphic to its Alexander dual $\widehat{K} = \{\emptyset, \{ 1\} \{2 \} \{ 3\} \{ 4\} \{ 1,2\} \{1,3 \} \{ 2,4\} \}$, and is therefore self-dual in this sense.

What is known about self-dual complexes in the second sense? What is known about their combinatorial structure?

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    $\begingroup$ Note that one thing that makes this a little unnatural is that it depends on $m$. $\endgroup$ Commented May 6, 2021 at 14:26
  • $\begingroup$ I understand. Thank you for your remarks. When I have been saying "vertex set", I have always meant an ambient set on which my complex is defined, rather than the singleton simplices in my complex. I'll edit my question to clarify my meaning. $\endgroup$
    – Matt
    Commented May 7, 2021 at 19:08
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    $\begingroup$ I'm going to delete my other comments because they were getting a bit repetitive, and my first comment is really the point. $\endgroup$ Commented May 7, 2021 at 19:18

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