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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?

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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 $0$.

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    $\begingroup$ One does not need the BCH formula for such sort of computation, usually by expanding out the exponential function to Taylor series, one can show that the linear term is commutative directly by hand, and the non-commutativity only encountered in higher order terms, which vanish anyway as $t \to 0$. $\endgroup$
    – Asaf
    May 26, 2013 at 16:05
  • $\begingroup$ @ Asaf I also think we do not need to use BCH formula. But I'm not sure what do you mean by "linear term is commutative". $\endgroup$ May 26, 2013 at 16:25
  • $\begingroup$ @Zhaoting More explicitly, the formula you mention is $\left.\frac{d}{dt}\right|_{t=0} e^{Y+tX}=X-\frac{1}{2!}[Y,X]+\frac{1}{3!}[Y,[Y,X]]-+\cdots$. However, we do not need this to deduce continuity of the left-hand side with respect to $X$ and $Y$: the mapping $(X,Y,t)\mapsto e^{Y+tX}$ is smooth as well as its partial derivative with respect to $t$. To put it differently, I do not see where in your argument we make use of the hypothesis that the $\mathrm{d}_X f$ exist and be continuous. I cannot even see at the moment why the map we ought to show to be continuous actually exists. $\endgroup$ May 27, 2013 at 9:07
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Look here for a complete proof shorter than the one in Knapp.

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  • $\begingroup$ Thank you, Professor Michor. Unfortunately, I am not so familiar with differential geometry; nevertheless, could you please cite the exact statement in the paper that corresponds to Knapp's Lemma 3.13 or Theorem 3.15? $\endgroup$ Jun 11, 2013 at 6:37

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