Is the simplicial join of two spherical simplicial complexes itself spherical?

Question

I think this ought to be true, but I am struggling to see why.

Of course if one of the spheres is $S^0$ then this is trivially true, as we are just glueing two cones along their boundary. I'm not seeing a general argument though.

Answer 1

This is true for the topological join, and the simplicial join of two finite complexes is homeomorphic to their topological join. So, given integers $n,i,j$ with $n=i+j+1$, I just have to describe how $S^n$ is the topological join of $S^i$ and $S^j$. If you work through this description in the special case described in your question, and then perhaps in a couple of other familiar cases such as $i=0$, $j=1$ and $i=j=1$, perhaps you'll be able to understand the general case.

Let $S^n$ be the unit sphere in $n+1$ dimensional space with coordinates $x_1,...,x_{i+j+2}=x_{n+1}$. Then think of $S^i \subset S^n$ as the unit $i$ dimensional subsphere in the coordinate subspace $\{(x_1,...,x_{i+1},0,...,0)\}$, and $S^j \subset S^n$ as the unit $j$ dimensional subsphere in the coordinate subspace $\{(0,....,0,x_{i+2},...,x_{i+j+2})\}$.

Next, given $p \in S^i$ and $q \in S^j$, I'll describe the join segment $[p,q] \in S^n$. Thinking of $p,q$ as unit vectors in $\mathbb{R}^{n+1}$, they are evidently orthogonal to each other, and hence they span a 2-dimensional subspace. Let $C(p,q)$ be the convex cone generated by $p,q$, specifically $$C(p,q) = \{sp+tq \,\bigm|\, s,t \ge 0\} $$ Then the join segment is defined to be $[p,q] = C(p,q) \cap S^n$. It's not hard to see that if the ordered pairs $(p,q)$, $(p',q')$ are unequal then the two join segments $[p,q]$, $[p',q']$ have disjoint interiors.

One can parameterize the join segment $[p,q]$ by the interval $[0,1]$ using the function $$f_{p,q}(t) = \frac{(1-t)p + tq}{\bigl|(1-t)p+tq\bigr|} $$

The topological join itself is defined as the quotient space of $S^i \times S^j \times [0,1]$ where $S^i \times 0$ is collapsed to a single point and $S^j \times 1$ is collapsed to a single point. Define a function $$F : S^i \times S^j \times [0,1] \to S^n, \qquad F(p,q,t) \mapsto f_{pq}(t) $$ This is pretty obviously continuous. Also, it has the same point pre-images as the decomposition of $S^i \times S^j \times [0,1]$ that defines the topological join (because of interior disjointness of join segments, as described earlier). So by general theorems of quotient maps, the function $F$ descends to a homeomorphism from the topological join of the spheres $S^i$ and $S^j$ to the sphere $S^n$.