On a technical fact used in the proof of density of smooth vectors in a representation

Question

My question concerns a fact used in Knapp's Representation Theory of Semisimple Groups implicitly in the proof of Lemma 3.13, which is crucial for the proof of density of smooth vectors in a representation.

Let $G$ be a linear Lie group, say $\mathrm{SL}(2,\mathbb{R})$, and $f:G\to\mathbb{R}$ any function. If the tangent space $\mathrm{T}_g G$ is identified with $\mathfrak{g}$ via left translation by $g$ as usual, then for each $X\in\mathfrak{g}$ one can speak of the directional derivative of $f$ with respect to $X$ as a function of $G$:

\begin{align*} \mathrm{d}_X f: G & \to\mathbb{R}.\\ g & \mapsto \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}f\left(ge^{tX}\right) \end{align*}

I want to show the following:

$\qquad\qquad$ $f$ is of class $\mathcal{C}^1$ $\iff$ for each $X\in\mathfrak{g}$ the mapping $\mathrm{d}_X f$ exists and is continuous.

The forward implication is easy. Conversely, let $\widetilde{f}:\mathfrak{g}\to\mathbb{R}$ be the pullback of $f$ around $1$ via the exponential map. We are to show that $\widetilde{f}$ is $\mathcal{C}^1$ in a neighborhood of $0,$ or equivalently, that $\widetilde{f}$ has continuous directional derivatives in a neighborhood of $0$. Choose a direction $X\in\mathfrak{g}$ arbitrarily. We want to show that the directional derivative of $\widetilde{f}$ with respect to $X$, namely the map

\begin{align*} \mathfrak{g} & \to\mathbb{R},\\ Y & \mapsto \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}f\left(e^{Y+tX}\right) \end{align*}

exists and is continuous in a neighborhood of $0$. If we had $e^{Y+tX}=e^Y e^{tX}$ then our hypothesis on $\mathrm{d}_X f$ would immediately imply the result, but of course one cannot expect this equality to hold true since $X$ and $Y$ do not commute in general. How can I complete the proof?

Answer 1

Let me try to answer your question. Our problem is to determine what is the vector $\frac{d}{dt}|_{t=0}\exp(Y+tX)$.

Let $L_{\exp Y}$ be the left translation by $\exp Y$. As you mentioned before, our first expectation is that $\frac{d}{dt}|_{t=0}\exp(Y+tX)=L _{ \exp Y} X$. However, we know that this is not true since the Lie algebra is not abelian.

In fact the correct form, which is standard in Lie theory, is \begin{equation} \frac{d}{dt}|_{t=0}\exp(Y+tX)=L _{ \exp Y}(\frac{1-e^{-ad Y}}{ad Y}X), \end{equation} where $\frac{1-e^{-ad Y}}{ad Y}$ is a power series of $ad Y$ with radius of convergence $>0$.

To prove this, it is sufficient to calculate $\frac{d}{dt}|_{t=0}\exp(-Y)\exp(Y+tX)$, and for this we can use the Baker–Campbell–Hausdorff formula (I think there are other simpler proofs of the above result but I cannot remember).

Then we can show that $F$ is a $C^1$ function, since the right hand side of the above formula is continuous with respect to $X$ and $Y$ in a small neighborhood of $0$.

Answer 2

Look here for a complete proof shorter than the one in Knapp.