Characterization of differentiability via Lie derivatives

Question

Yesterday I asked this question in MathOverflow but did not receive an answer yet. I want to try my chance here too, since I am in kind of a hurry. Answers will be much appreciated.

I intend to propose as a project the proof of the statement below, but I want to make sure that it is not already proved somewhere else before.

Let $G$ be a Lie group, and $f$ a real-valued function on $G$. The expression $f \in \mathcal{C}^k(G)$ makes sense, and this would be the case even if $G$ were merely a smooth manifold. On the other hand, the Lie group structure on $G$ enables one to speak of directional derivatives of $f$. Indeed, the Lie algebra $\mathfrak{g}$ of $G$ is canonically isomorphic to the space of left-invariant derivations of $\mathcal{C}^\infty(G)$; under this isomorphism, each $X \in \mathfrak{g}$ is associated with the (left) Lie derivative operator $\mathcal{L}_X$ given by $$(\mathcal{L}_X f)(y) := \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} f(y e^{tX})$$ where $f \in \mathcal{C}^\infty(G)$ and $y \in G$. It is natural to call $\mathcal{L}_X f$ as the (left) directional derivative of $f$ along $X$. Taking the above equality as a definition, one may expect as in elementary analysis that being in $\mathcal{C}^k$ is equivalent to having continuous directional derivatives of order $k$. This is what our statement says:

Statement. Let $G$ be a Lie group, and $f$ a real-valued function on $G$. For each $k \in \mathbb{N}$, $f \in \mathcal{C}^k(G)$ if and only if $(\mathcal{L}_{X_1} \cdots \mathcal{L}_{X_k})f$ exists and is continuous for all $X_1,\ldots,X_k \in \mathfrak{g}$.

I searched quite a while for this statement in the literature but could not find anything. (The proof is not so trivial as you might think at a first glance. Please have a look at this question and p. 15 of this essay also.) Have you ever seen it somewhere? If so, could you please give a reference?

Answer 1

Let us consider what happens on $\mathbb{R}^n$. We denote by $X_1,\dots,X_n$ the standard vector fields. Then it is well known that $f$ is $C^k$ iff for any $i_1,\dots,i_k$, the directional derivatives $X_{i_1}.(X_{i_2}(\dots.(X_{i_k}.f)\dots)$ exist and are continuous.

Now take any other vector fields $Y_1,\dots,Y_n$ giving a basis of $T_x \mathbb{R}^n$ for $x$ in a neighborhood of $x_0$. Then you can write $Y_i = \lambda_i^j X_j$, with smooth $\lambda_i^j$. This shows that if $f$ is $C^k$ in a neighborhood of $x_0$ then the directional derivatives $Y_{i_1}.(Y_{i_2}(\dots.(Y_{i_k}.f)\dots)$ exist and are continuous. Reversing the argument, this shows that in order to show that a map is $C^k$, you can compute directional derivatives for any vector fields giving a basis of the tangent space.

Now, for your Lie group, $L_X f$ is just the directional derivative of $f$ in the direction of the left-invariant vector field associated to $X$. I think this proves what you want. Or am I missing something ?

Answer 2

If you could show that $(\mathcal{L}_{X_1} \cdots \mathcal{L}_{X_m})f$ exists and is continuous for all $X_1,\ldots,X_m \in \mathfrak{g}$, for all $m \le k$, then perhaps you could show by induction that $f \in C^m$ for all $m \le k$. Start with the case $m=0$ (which is trivial), and then notice that on any coordinate patch containing $g \in G$, that the highest order term in $(\mathcal{L}_{X_1} \cdots \mathcal{L}_{X_m})f$ is simply $\partial^m f/(\partial x_1 \cdots \partial x_m)$, where $x_i$ is a coordinate in the direction of $X_i$ at the point $p$ mapped to the coordinate patch. And the lower order terms are smooth functions times lower order derivatives, so they are automatically continuous (by the inductive hypothesis).

So how to get the smoothness $(\mathcal{L}_{X_1} \cdots \mathcal{L}_{X_m})f$ from the smoothness of $(\mathcal{L}_{X_1} \cdots \mathcal{L}_{X_m})f$? Well you can create the operation "integration along $X_i$", and this is a bounded map from $C^n$ to $C^n$ for any $n \ge 0$, at least of you stay within a compact subset of $G$. So apply these operations, and it should work out.

Answer 3

Around every point $p \in G$ here exists a local chart $(U,h)$ such that $Dh(\{X_1, \ldots, X_n\}) = \{e_1, \ldots, e_n\}$ the standard basis of $\mathbb{R}^n$. It follows that if $\gamma_i$ is an integral curve of $X_i$ in $U$, then $h \circ \gamma_i$ is just a straight line in the $x_i$ direction. Hence if $f \colon G \to \mathbb{R}$ has a continuous Lie derivative with respect to $X_i$, then $f \circ h^{-1} \colon h(U) \to \mathbb{R}$ has a continuous partial derivative with respect to $x_i$.

We therefore conclude that if $f$ has continuous Lie derivatives with respect to each $X_i$, then $f \circ h^{-1}$ has continuous partial derivatives, and hence is of class $C^1$. But if $f \circ h^{-1}$ is of class $C^1$, then so is $f$. It is now clear how to generalise to $C^k$ for any $k$.