I can suggest some names, which I have learned from others who research these things.
- A 4-simplex, or 4D analogue of a triangle, is called a Pentachoron , describing a regular, 5-sided 4D polytope. Also called a 5-cell. These are the n-simplex.
There are 4 types of ring torus objects in 4D, which can be seen visualized here. The general name of hypertorus works well:
Spheritorus : sphere-bundle over the circle :$S^2$ x $S^1$ $$\left(\sqrt{x^2+y^2} -a\right)^2 +z^2+w^2 = b^2$$
Torisphere : circle-bundle over the sphere : $S^1$ x $S^2$
$$\left(\sqrt{x^2+y^2+z^2} -a\right)^2 +w^2 = b^2$$
3-torus : circle over circle over circle : $T^3$
$$\left(\sqrt{\left(\sqrt{x^2+y^2}-a\right)^2+z^2}-b\right)^2+w^2 = c^2$$
Tiger : circle-bundle over the flat 2-torus (Clifford torus)
$$\left(\sqrt{x^2+y^2} -a\right)^2 +\left(\sqrt{z^2+w^2} -b\right)^2 = c^2$$
As for the bar -> cylinder -> duocylinder, I'm not sure exactly what sequence you are using here. The best fit I can see is describing a specific bisecting rotation around an n-1 plane into n+1 dimensions. In this case, the next 5D shape is called a Cylspherinder , a cartesian product of a $D^2$ and $D^3$ (solid disk times solid sphere). But, you can also make a Spherinder (sphere prism, another type of 4D cylinder) from a rotation of a cylinder into 4D.
Cylinder: $\left|\sqrt{x^2+y^2} -z\right|+\left|\sqrt{x^2+y^2} +z\right| = a$
Duocylinder: $\left|\sqrt{x^2+y^2} -\sqrt{z^2+w^2}\right|+\left|\sqrt{x^2+y^2} +\sqrt{z^2+w^2}\right| = a$
Spherinder : $\left|\sqrt{x^2+y^2+z^2} -w\right|+\left|\sqrt{x^2+y^2+z^2} +w\right| = a$
Cylspherinder : $\left|\sqrt{x^2+y^2+z^2} -\sqrt{w^2+v^2}\right|+\left|\sqrt{x^2+y^2+z^2} +\sqrt{w^2+v^2}\right| = a$
I guess you can call these n-cylinders, but there are even more types of these than just product of n-balls and n-cubes. You can also include the product of n-ball and n-simplex as well. In fact, any shape with both flat and curved cells can fit into this group (product of 2-ball (and higher) with any genus-0 object)
Cyltrianglinder : $\left|\big||x|+2y\big|+|x| -2\sqrt{z^2+w^2}\right|+\left|\big||x|+2y\big|+|x| +2\sqrt{z^2+w^2}\right| = a$
Given any tetrahedron $T$ with vertices $p_1, p_2, p_3, p_4$. Let
- $r$ and $u$ be the in-radius and in-center.
- $(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$ be the barycentric coordinates of $u$ with respect to $T$. i.e. a list of $4$ numbers satisfy:
$$u = \alpha_1 p_1 + \alpha_2 p_2 + \alpha_3 p_3 + \alpha_4 p_4
\quad\text{ and }\quad
\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 1$$
- For each $i$, let $A_i$ be the area of the face of $T$ opposite to $p_i$ and $h_i$ be the height of $p_i$ to that face.
- $V$ be the volume of $T$.
We know that for each $i$, $\displaystyle\;\alpha_i = \frac{r}{h_i}$, together with the fact:
$$
h_1A_1 = h_2A_2 = h_3 A_3 = h_4 A_4 = 3V
$$
We have $\displaystyle\;\alpha_i = \frac{r}{h_i} = \frac{rA_i}{3V}$ and
$\displaystyle\;\sum_{i=1}^4 \alpha_i = 1$
reduces to $\displaystyle\;\frac{r}{3V}\sum_{i=1}^4 A_i = 1$.
As a result,
$$u = \sum_{i=1}^4 \alpha_i p_i = \frac{r}{3V}\sum_{i=1}^4 A_i p_i
= \frac{\sum\limits_{i=1}^4 A_i p_i}{\sum\limits_{i=1}^4 A_i}
$$
i.e the in-center is the area weighted average of the vertices.
The actual computation of the coordinates of in-center for this problem is left as an exercise.
Best Answer
Let $A=\frac{n+1}{n}I_{n+1}-\frac{1}{n}\mathbf1\mathbf1^T$. That is, $$ a_{ij}=\begin{cases}1&\text{ if } i=j,\\ \frac{-1}{n}&\text{ if } i\ne j.\end{cases} $$ $A$ has a simple zero eigenvalue and also an eigenvalue $\frac{n+1}{n}$ of multiplicity $n$. Hence it can be orthogonally diagonalised as $Q^T\operatorname{diag}\left(\frac{n+1}{n},\ldots,\frac{n+1}{n},0\right)\,Q$. If we multiply $Q$ by $\sqrt{\frac{n+1}{n}}$ and drop its last row to form an $n\times(n+1)$ matrix $V$, we get $V^TV=A$, i.e. the columns of $V$ are equidistant vectors on the unit sphere.
Edit. $V$ can be constructed recursively as follows (see also the answer by Martin Argerami to the question cited by the OP). Let \begin{align} V_1&=\pmatrix{1&-1}\in\mathbb R^{1\times2},\\ V_k&=\pmatrix{1&-\frac1k\mathbf1^T\\ \mathbf0&\sqrt{1-\frac1{k^2}}\,V_{k-1}}\in\mathbb R^{k\times(k+1)}. \end{align} Then the columns of each $V_k$ are $k+1$ equidistant vectors on the unit sphere in $\mathbb R^k$, and such a set is unique up to orthogonal transformation. It is easy to prove by mathematical induction that $V_n^TV_n=A$.