Realizing spherical complexes as convex polytope

Question

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. Equivalently, it is the complex obtained by intersecting a complete polyhedral fan with the a (d−1)-dimensional sphere centered at the vertex of the fan.

We know that every combinatorial information of a spherical complex can be realized as a convex polytope in dimension 3 by considering the so-called canonical polytope. The proof from Wikipedia seems to depend on the circle packing theorem and the midsphere theorem.

So my question is:

Is any spherical polytope of dimension $n$ isomorphic to a convex polytope as abstract polytope for $n > 3$?

According to This question the midsphere theorem is still unknown for higher dimension, so I suspect there is a counterexample. Any idea or reference is appreciated.

EDIT:

(1) The definition of spherical polytope is edited several times (see the discussion on comment)

(2) I just saw this wiki where it is stated that "convex polytopes in p-space are equivalent to tilings of the (p−1)-sphere." which seems to give a positive answer of my question, but no reference is provided there.

Answer 1

Another way to phrase your question is "whether every complete fan is (combinatorially equivalent to) the face fan (or the normal fan) of a convex polytopes". The answer is No in dimension $n\ge 4$ and I will provide an example derived from the following non-polytopal 4-diagram:

This picture is taken from Example 5.10 in Günter Ziegler's book "Lectures on Polytopes". The book explains why this diagram is not the Schlegel diagram of any 4-dimensional polytope. It is important to note that the "outer cell" of the diagram is a simplex.

Now, consider this diagram $D$ embedded into a 3-dimensional affine subspace of $\Bbb R^4$ that does not pass though the origin. Let $|D|$ be the simplex that forms the "outer cell" of $D$. Let $x\in\Bbb R^4$ be a point for which the convex hull $\Delta:=\mathrm{conv}(|D|\cup\{x\})$ contains the origin in its interior. This convex-hull is a 4-simplex and $|D|$ is one of its facets. Take the face fan of the simplex $\Delta$ and subdivide the cone over the facet $|D|$ into the cones over the cells of $D$. This is a complete fan in which no cone contains antipodal points. You can intersect it with a sphere to obtain a spherical complex.

Suppose now that $P$ is a polytope combinatorially equivalent to this complex. Let $y\in P$ be the vertex that corresponds to the vertex $x$ of the complex. Its four neighbors in $P$ span a simplex $\Delta'$ (that corresponds to $|D|$ in the fan above). But deleting $y$ from $P$ (i.e. taking the convex hull of all vertices of $P$ except for $y$) leaves a polytope for which $\Delta'$ is a facet (here we use that $\Delta'$ resp. $|D|$ is a simplex) and whose Schlegel diagram based at $\Delta'$ is exactly $D$. This is a contradiction.