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Let $P$ be a spherical complex, which essentially means a tiling of a sphere, let us say the $(d-1)$-dimensional sphere $\mathbb{S}^{d-1}$ in $\mathbb{R}^d$ to fix notation, where each cell is a spherical polytope. These are higher-dimensional analogues of spherical polyhedra.

My question is: is there a known construction of a combinatorially dual spherical complex, say $Q$ (assuming such a $Q$ always exists)?

So I am essentially asking for a spherical analogue of polar duality (polar duality assigns a dual convex polytope to a given convex polytope).

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    $\begingroup$ @SamHopkins, yes I am not sure if this is standard terminology or not. I have read in a book that a spherical polytope is the intersection of some closed hemispheres which is non-empty and does not contain a pair of antipodal points. A spherical complex is a tiling of the whole $(d-1)$-dimensional sphere by spherical polytopes. $\endgroup$ – Malkoun May 10 '20 at 17:10
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    $\begingroup$ If so you might be able to use duality of fans to get what you want. $\endgroup$ – Sam Hopkins May 10 '20 at 17:18
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    $\begingroup$ @SamHopkins, I am new to this area. I don't know what is a "complete polyhedral fan" for instance. $\endgroup$ – Malkoun May 10 '20 at 17:18
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    $\begingroup$ A (convex) polyhedral cone is this: en.wikipedia.org/wiki/…. A fan is a collection of polyhedral cones which intersect properly (the intersection of any two cones in the fan is a cone in the fan which is a common face of both). "Complete" just means it fills up all of space. Intuitively if you intersect a polyhedral fan with a sphere centered at the origin you should get a spherical complex in your sense; and it would seem the reverse procedure is possible too. $\endgroup$ – Sam Hopkins May 10 '20 at 17:22
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    $\begingroup$ see my edit. Now I'm less certain that this duality makes sense. $\endgroup$ – Sam Hopkins May 10 '20 at 19:22
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As discussed in the comments, this stuff is more commonly described using the language of fans. A (convex, polyhedral) cone in $\mathbb{R}^{d}$ is the intersection of finitely-many half-spaces through the origin. A (polyhedral) fan is a collection of cones that intersect properly: the intersection of any two cones in the fan is again a cone in the fan which is a common face of both cones. A fan is complete if the union of all the cones in the fan is all of $\mathbb{R}^d$. By intersecting a complete fan with a sphere centered at the origin we obtain a spherical complex in the sense of the question-asker. This procedure should also be reversible by "coning over" a spherical complex. Cones have dual cones, and in this way we get dual fans. This gives the desired combinatorial duality for spherical complexes.

EDIT: Whoops, now I am actually less sure of exactly how duality of cones leads to duality of fans. The dual cones of a fan will not fit together into a fan. So this does not answer the question. (Of course, if our fan happens to be polytopal, then we can use polar duality of polytopes.)

EDIT 2: I asked Vic Reiner about this question, and he gave me a lot of good information. He pointed out that the question of the existence of "dual" CW complexes is a difficult and subtle point, as discussed for instance in this other MO question. However, for a PL cell decomposition of $\mathbb{S}^d$ there exists a dual PL cell decomposition of $\mathbb{S}^d$ with a dual face lattice, as proved in Proposition 4.7.26(iv) on pg. 214 of the "Oriented Matroids" book by Björner et al. The spherical complexes you describe (a.k.a. polyhedral fans) will certain be PL, but this result still does not quite answer your question because it is not clear that the dual PL cell decomposition will correspond to a polyhedral fan.

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  • $\begingroup$ No, what's troubling me is that if I take all the cones in a fan and take their duals, they will not fit together into a fan. $\endgroup$ – Sam Hopkins May 10 '20 at 19:25
  • $\begingroup$ Could you give an example where the dual cones of a fan do not fit together into a fan please? $\endgroup$ – Malkoun May 10 '20 at 19:26
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    $\begingroup$ Even the most basic example of say 6 equiangular cones in $\mathbb{R}^2$. These cones are acute so their duals will be too big. $\endgroup$ – Sam Hopkins May 10 '20 at 19:27
  • $\begingroup$ What you kinda want to do is take the normal fan of a fan. But I'm not sure that makes sense. $\endgroup$ – Sam Hopkins May 10 '20 at 19:30
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    $\begingroup$ At any rate you should unaccept this answer, because while it might be useful for you as a starting point, it doesn't answer your question. I'm now convinced your question is very interesting, though, and I'm surprised I've never heard discussion of "dual polyhedral fans" before. One possible answer is that only genuinely polytopal fans have duals: an analogous thing is that only planar graphs have duals. $\endgroup$ – Sam Hopkins May 10 '20 at 19:49

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