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).

  • $\begingroup$ By taking $F = \{1\}$ we can ask a simpler question "can a single element generate an infinite-dimensional Lie group" $\endgroup$
    – lisyarus
    Jun 5 at 21:41
  • $\begingroup$ @lisyarus: is your simpler question equivalent to the original? $\endgroup$
    – Rob Arthan
    Jun 5 at 22:00
  • 1
    $\begingroup$ @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 $F$. $\endgroup$ Jun 5 at 22:21
  • $\begingroup$ @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. $\endgroup$ Jun 5 at 23:51

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Jun 6 at 5:01
  • $\begingroup$ 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. $\endgroup$
    – quarague
    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.

  • $\begingroup$ That is a very clean example, thank you so much! $\endgroup$ Jun 7 at 8:51

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