How does the triangulation of $S^k$ and $B^{k}$ look like?

Question

A $k$-simplex is a convex hull of $k+1$ points (which are called vertices) in general position in $\mathbb{R}^n$ (for $k\le n$). A face of a simplex is a simplex spanned by a subset of its vertex set. And a collection $C$ of simplices in $\mathbb{R}^n$ is called a simplicial complex if every two simplices in $C$ meet, if at all, in a face common to both. The union of all simplices in $C$ is denoted by $||C||$. A triangulation of a topological space $X$ is a simplicial complex $C$ such that $||C||$ is homeomorphic to $X$.

I am wondering how does the triangulations of a $k$-sphere, i.e., a topological space homeomorphic to $S^k:=\{\mathbf{x}\in \mathbb{R}^{k+1}:||x||_2=1\}$, and of a $k$-disk, i.e., a topological space homeomorphic to $B^k:=\{\mathbf{x}\in \mathbb{R}^{k}:||x||_2\le 1 \}$, look like?

The only way I can imagine for this kind of triangulation looks like the following: you can construct it iteratively via adding $k$-simplices one by one, sharing some of its proper faces with previous added $k$-simplices. (So the triangulation consists of all these $k$-simplices and their faces.)

Question: Is there any other kind of triangulation not generated in this way? For example, it is possible to use some simplices of higher or lower dimension essentially? ("Essentially" means that we don't count the case that adding some lower dimension simplices first as faces of some later added $k$-simplex.)

Answer 1

Any triangulation of $B^k$ or $S^k$ can only involve simplices of dimension $k$ or less. This is easiest to show using the concept of topological dimension. It can be shown that the dimension of a simplicial complex is equal to the highest dimension simplex it contains, while the topological dimension of $B_k$ and $S_k$ are both $k$. See https://en.wikipedia.org/wiki/Lebesgue_covering_dimension for some more info.

It can also be shown that in any triangulation of $B^k$ or $S^k$, that any $\ell$-simplex with $\ell<k$ appears as a face of some other $k$-simplex in the triangulation. One way to see this is to note that $B^k$ and $S^k$ are manifolds (the former being a manifold with boundary); every interior point of $B^k$ or $S^k$ has a neighborhood which is homeomorphic to a connected open subset of $\mathbb R^k$, and every border point is homemorphic to a connected open set in the half-space $H^k=\{(x_1,\dots,x_k)\in \mathbb R^k\mid x_1 \ge 0\}$. However, the neighborhood of a point on an $\ell$-simplex with $\ell<k$ does not look like either of these, again since the topological dimension is too small.