# A movement requires 1 dimension, a rotation requires 2 dimensions, a what requires three dimensions?

Movement A zero-dimensional object cannot move. A one-dimensional object can move in one dimension (the x-axis). A two-dimensional object can move in two dimensions (the x-axis and y-axis). A three dimensional object can move in three dimensions (the x-, y-, and z-axis).

In other words: There are n different ways for an n-dimensional object to move.

Rotation A zero- and a one-dimensional object cannot rotate. A two-dimensional object can rotate in one way (in the xy-plane). A three dimensional object can rotate in three ways (in the xy-, xz-, and yz-plane). And a four dimensional object can rotate in 6 ways (xy-, xz-, yz-, xw-, yw-, and zw-plane).

In other words: There are n(n-1)/2 different ways for an n-dimensional object to rotate.

Now for the question, what do you call the next thing?

I do know that the next thing has some properties if you follow the same pattern. But I have no idea what it is and I was hoping you guys could help me out.

Here are some of the properties that I have extrapolated:

The next thing

A zero-, one- and two-dimensional object cannot do it. A three dimensional object can do it in one way (along xyz). A four dimensional object can do it in four ways (along xyz, xyw, xzw and yzw. And if you want to extrapolate further... an n-dimensional object can do it in n(n-1)(n-2)/6 different ways.

So yeah, what do you call this thing?

• Are you referring to something like hyperspherical propagation ? e.g., a hypersphere that is getting big and biger along the time axes ! ? :) – Cardinal Aug 1 '15 at 12:43
• The generator of translations in $1$ dimension is $\epsilon_{i2}\partial/\partial x_i$. The generator of rotations in $2$ dimensions is $\epsilon_{ij3}x_i\partial/\partial x_j$. The next thing might have the generator $\epsilon_{ijk4}x_ix_j\partial/\partial x_k$ in three dimensions, but unfortunately that vanishes by antisymmetry. (Here $\epsilon$ is the Levi-Civita symbol.) – joriki Aug 1 '15 at 12:45
• @joriki I do not know. I read the wikipedia article of that, but unfortunately, I do not understand it. – Paul Aug 1 '15 at 12:57
• @Cardinal I don't think i'm referring to hyperspherical propagation, since that does have an application in one dimension. (as far as I know anyway). EDIT: Also, I think I am referring to only objects with spacial dimensions. I.e. where time isn't a dimension. – Paul Aug 1 '15 at 13:04
• One way of interpreting your question is as follows. The Lie group $\mathbb R^n$ acts on $\mathbb R^n$ effectively and transitively (by translation); each has dimension $n$. The Lie group $O(n+1)$ acts on $S^n$ effectively and transitively by rotation; $O(n+1)$ has dimension $n(n+1)/2$. Is there an $n(n-1)(n-2)/6$-dimensional Lie group that acts effectively and transitively on some $n$-manifold? But it is a theorem of Montgomery and Zippin that a compact group $G$ acting transitively and effectively on a connected $n$-manifold can have dimension at most $n(n+1)/2$. – user98602 Aug 1 '15 at 18:27

The Lie group $\mathbb R^n$ acts effectively and transitively on the smooth manifold $\mathbb R^n$; each of these is $n$-dimensional. The special orthogonal group $SO(n+1)$ acts effectively and transitively on $S^n$; the former has dimension $n(n+1)/2$, and $S^n$ has dimension $n$.
Can we improve this? Are there, for instance, $n(n-1)(n-2)/6$-dimensional groups that act on $n$-dimensional manifolds?
If the group $G$ is compact, the answer is a firm no. It is a result of Montgomery and Zippin that a compact topological group acting effectively and transitively on a manifold of dimension $n$ must have dimension at most $n(n+1)/2$. If you want $G$ to be a Lie group acting smoothly, this is easier: first prove it if $G$ acts by isometries on a Riemannian manifold by induction; then if you have a transitive effective action on $M$, average the metric over $G$ to get a $G$-invariant metric on $M$, hence an action by isometries. That $G$ is compact is important here!
I don't know what bounds one has in the case that $G$ is non-compact - I'd like to hear from somebody if they do.