Are any higher-dimensional analogues of triangles flexible? I don't know if my terminology is correct, so bear with.
Triangles are 'inflexible'. By that, I mean that if you consider a triangle to be a graph of 3 edges and 3 vertices in a closed loop then you cannot change the angle between 2 edges at any vertex without changing the length of at least one of the other edges in the graph.
This is in contrast to (say) a square, where you can change the angle between two edges at a vertex without having to change the length of any of the edges.
The same effect persists with a pyramid with a triangular base (which I guess is considered to be a 3-dimensional triangle), but does that same effect persist with higher dimensional triangles? My gut feeling says that it should, but I have found my gut to be pretty useless at dealing with objects of dimensions > 3.
 A: Choose $n+1$ points in general position. That is, no $3$ of the points will be on the same line, no $4$ of the points will be in the same plane, no $5$ of the points will be in the same 3-hyperplane, etc. This requires an $n$-dimensional space to accomplish.
Now $n$ of those points, $P_1, \dots, P_n$, will determine an $(n-1)$-hyperplane. Establish a coordinate system $(x_1,\dots,x_n)$ with that hyperplane being $x_n = 0$. Each of the $P_i$ will have coordinates $(p_{i1},\dots,p_{in})$ under this system. But the final point, $P_0$, being in general position, will not be on that hyperplane. So $p_{0n} \ne 0$. Let $r_i$ be the distance from $P_0$ to $P_i$. Suppose some other point $Q$ in that $n$-dimensional space is the same distance from each of $P_1, \dots,P_n$ as $P_0$. Then we have $n$ equations in the coordinates of $Q$:
$$\forall i, 1\le i \le n, \sum_{k=1}^n (q_k - p_{ik})^2 = r_i^2$$
These equations are unfortunately quadratic. but if we subtract off the equation for $i = n$ from the others, we get $n-1$ linear equations in the $q_k$:
$$2\sum_{k=1}^n (p_{nk}-p_{ik})q_k = r_i^2 - r_n^2 + \sum_{k=1}^n p_{nk}^2 -p_{ik}^2$$
Because the points are in general position, this system will be non-singular. But because of the way we chose the coordinate system, $p_{in} = 0$ for all $i > 0$, so none of those equations includes a term in $q_n$. I.e., it is a non-singular system of $n-1$ linear equations in $n-1$ unknowns, and therefore has a unique solution, for $q_1,\dots,q_{n-1}$. This leaves only $q_n$ as unknown. Further, since $p_{01},\dots, p_{0(n-1)}$ are already known to be a solution, we must have $q_k = p_{0k}$ for all $k < n$. That is, $P_0$ and $Q$ can only differ in their last coordinate. All possible points $Q$ must lie on the line connecting $P_0$ to its image in the hyperplane $x_n = 0$. If we take any one of the original quadratic equations, it is the equation $|Q - P_i|^2 = r_i^2$ of a hypersphere about $P_i$. This hypersphere intersects the line on which $Q$ must reside at $P_0$. Since a line intersects a hypersphere in at most two points, there is exactly one other point that $Q$ can be, which will be the reflection of $P_0$ through the hyperplane $x_n = 0$.
The point of all of this is that you cannot move $P_0$ around without changing its distances to the other points. That would require other points near $P_0$ that share the same distances. But there is only one other such point, and it is on the other side of the hyperplane, so no continuous movement is possible. Provided all distances between the $n+1$ points are fixed, the system is rigid in that $n$-hyperspace.
If we go up a dimension or more, there are infinitely many points at the same distance from the $P_i$ as $P_0$. But these other points just represent rotations of the system, not flexing.
The vertices and edges of an $n$-simplex are exactly such a system. There is an edge connecting every two vertices, so the system is rigid.
