# A specific question regarding a proof in Knapp's book

I got stuck in an apparently trivial point within the proof of Lemma 3.13 on p. 55 of Knapp's Representation Theory of Semisimple Groups. The author concludes in the first paragraph that $f_v$ must be of class $\mathcal{C}^1$. In local coordinates around the identity element, this amounts to the existence and continuity of all directional derivatives. Explicitly, one has to show for each direction $X\in\mathfrak{g}$ that the map

$$Y\mapsto\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\Phi(e^{Y+tX})v$$

makes sense and is continuous in a neighborhood of $0$ in $\mathfrak{g}$. This would follow at once from the hypothesis if we had $e^{Y+tX}=e^Y e^{tX}$ which is of course false in general. Actually, I cannot even see at the moment why the maps above are meaningful.

Recently, I asked essentially the same question to some knowledgeable people around me and also here in MathOverflow, and got quite different answers. Some referred to the Campbell-Baker-Hausdorff formula while others made no mention of it. I pondered upon each answer, but unfortunately none of them led me to an understanding of the issue.

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You write 'This would follow at once from the hypothesis' Which hypothesis? –  anton Jun 10 '13 at 19:58
If I understand correctly, the expression on the right-hand side is defined to be the solution to a certain ordinary differential equation whose initial data is determined by $Y$, so your result follows from the fact that solutions to ODE's depend smoothly on the initial conditions. The CBH formula exhibits the solution explicitly (i.e. it gives a formula for $e^{Y + tX}$), allowing you to deduce smoothness without resorting to general theorems in analysis. –  Paul Siegel Jun 10 '13 at 20:03
@Murat: In a case like this you'd probably do best to try first contacting the author himself by email. Though Knapp is technically retired at Stony Brook, he still has an email address there (and has posted various errata for his books, though this isn't one of them). Naturally you might not get a direct answer from him, but in the absence of other advice it's worth trying. (Check the faculty list on the SUNY Stony Brook math department homepage.) –  Jim Humphreys Jun 10 '13 at 20:04
Surely there are other possibilities, too, but, as Paul Siegel says, something like general, qualitative (not formulaic) change-of-variables (a.k.a., inverse function theorem...) should suffice. It's not about Lie groups, etc. Asking or wanting a formula is a bit misguided, even if natural. –  paul garrett Jun 10 '13 at 23:56
@anton: The hypothesis is that $\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\Phi(e^{tX})v$ exists for every $X\in\mathfrak{g}$. If we had $e^{Y+tX}=e^Y e^{tX}$ then $\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\Phi(e^{Y+tX})v = \Phi(e^Y)\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\Phi(e^{tX})v$ whence the result. –  Murat Güngör Jun 11 '13 at 6:21
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## 1 Answer

Explicitly, what Knapp writes, amounts to in detail: $\Phi$ is a representation of $G$ on a Hilbert space, So $\Phi(g)$ is bounded linear operator. Then: $X\mapsto\frac{d}{dt}|_0 \Phi(g.e^{tX})v = \frac{d}{dt}|_0 \Phi(g).\Phi(e^{tX})v = \Phi(g)\frac{d}{dt}|_0 \Phi(e^{tX})v$.
The last equality holds by the chain rule, since we are on a Hilbert space.

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Yes, I am aware of these circumstances. (By the way, I would think that the last equality is a consequence of the continuity of $\Phi(g)$ rather than chain rule.) However, I do not think—or at least I cannot see how—this resolves my problem. –  Murat Güngör Jun 11 '13 at 6:45
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