# Can adding a single element to a Lie group make it infinite-dimensional?

Suppose that I have a finite-dimensional Lie group, $$F\subset H=\text{Diff}(\mathcal{M})$$, of diffeomorphisms on some smooth finite-dimensional manifold $$\mathcal{M}$$. Suppose that there is some element in the larger group, $$g\in H$$, which is not in the smaller group $$g\notin F$$. Let $$G\subset H$$ be the smallest Lie subgroup of $$H$$ which contains both $$F$$ and $$g$$.

$$G:= \text{Closure}(\{g\}\cup F)\subset H$$

I am interested in bounding the size of $$G$$ (i.e., the dimensionality of $$G$$) in terms of the dimensionalities of $$F$$ and of $$\mathcal{M}$$. For instance, one might conjecture that adding a single element $$g$$ to a Lie group $$F$$ will never increase its dimension by more than a factor of $$\text{dim}(\mathcal{M})$$. Namely, one might conjecture $$\text{dim}(G)\leq\text{dim}(\mathcal{M}) * \text{dim}(F)$$. At the end of the day, all I need is proof that $$G$$ is finite-dimensional (even if $$\text{dim}(G) < \infty$$ is arbitrarily large). Is it possible that adding a single element $$g$$ to a finite-dimensional Lie group $$F$$ would make it infinite dimensional?

(* Edited to emphasize the problem of interest with $$F\subset H=\text{Diff}(\mathcal{M})$$ being diffeomorphisms).

• By taking $F = \{1\}$ we can ask a simpler question "can a single element generate an infinite-dimensional Lie group" Jun 5 at 21:41
• @lisyarus: is your simpler question equivalent to the original? Jun 5 at 22:00
• @lisyarus I think the answer to your simpler question is no, right? Let $g$ be the single element. Can't we associate to $g$ an element of the Lie algebra, $a$? Isn't the Lie algebra of the closure of $g$ just the closure of the Lie bracket on $a$? But $a$ commutes with itself, so nothing else is generated. What makes the problem with $F\neq\{1\}$ interesting is that this new element of the Lie algebra might fail to commute with things in $F$ and lead to $G$ being a much larger group than $F$. Jun 5 at 22:21
• @DanielGrimmer The problem with this is that the Lie group can be generated by a single element while the Lie algebra is not. See my answer below. Jun 5 at 23:51

There is no bound on the dimensionality of $$G$$. In fact, a single element can generate a Lie group of arbitrarily large dimension. Thus the $$n$$-dimensional torus $$\mathbb T^n$$ is the closure of the group generated by a single element $$(e^{2\pi i \omega_1}. \ldots, e^{2\pi i\omega_n})$$ if $$1$$ and the $$\omega_j$$ are linearly independent over the rationals.
• Thank you for the insightful reply. For my purposes, it is relevant whether the dimensionality of G is literally infinite or merely arbitrarily large but finite. Could there be an element $g\in H$ with $dim(H)=\infty$ such that $g$ is a member only of infinite-dimensional Lie subgroups of $H$? Said differently, could there be an element $g\in H$ which falls outside of the union of all of the finite-dimensional Lie subgroups of $H$? Does the answer to this question change if we demand that $H=\text{Diff}(\mathcal{M})$ are the diffeomorphisms of some finite-dimensional smooth manifold? Jun 6 at 5:01
• Your example increases the dimension exactly by a factor of $dim(M)$ which is the bound OP conjectured. The actual question is whether anything bigger than that can happen. Jun 6 at 8:08
It's probably easier to work with Lie algebras rather than Lie groups here. If we take $$M = \mathbb{R}^2$$ then $$Lie(Diff(\mathbb{R}^2)$$ is the Lie algebra of smooth vector fields on $$\mathbb{R}^2$$.
Let's take say $$y^2 \frac{d}{dx}$$ and $$x^2 \frac{d}{dy}$$. Either one of these generates a finite (one) dimensional Lie algebra, but together they generate something infinite dimensional.