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I am trying to understand a piece of the proof that the exponential map is locally one-to-one and onto in Brian Hall's Lie Groups, Lie Algebras, and Representations.

Theorem 3.42. For $\epsilon\in(0,\log 2)$, let $U_\epsilon=\{X\in M_n(\mathbb{C})|\|X\|<\epsilon\}$ and let $V_\epsilon=\exp(U_\epsilon)$. Suppose $G\subset GL(n;\mathbb{C})$ is a matrix Lie group with Lie algebra $\mathfrak{g}$. Then there exists $\epsilon\in(0,\log 2)$ such that for all $A\in V_\epsilon$, $A\in G$ if and only if $\log A\in\mathfrak{g}$.

The first step of the proof in the book is defining the map $$ \Phi(X+Y):=e^Xe^Y,\quad X\in \mathfrak{g},\ Y\in \mathfrak{g}^\perp $$ where $\mathfrak{g}^\perp$ is the orthogonal complement of $\mathfrak{g}$ in $M_n(\mathbb{C})\cong R^{2n^2}$.

The book claims that the following calculation shows the derivative of $\Phi$ at the point $0\in\mathbb{R}^{2n^2}$ is the identity:

$$\frac{d}{dt}\Phi(tX,0)\mid_{t=0}=X,\quad \frac{d}{dt}\Phi(0,tY)\mid_{t=0}=Y. \tag{0} $$

I do not understand why:

  • By definition of $\Phi$, it has only one variable in $M_n(\mathbb{C})$. What does the expression $\Phi(tX,0)$ mean?
  • Can one show by definition that the derivative of $\Phi$ at the point $0\in\mathbb{R}^{2n^2}$ is indeed the identity: $$ \lim_{t\to 0}\frac{1}{t}\bigg(\Phi(0+tZ)-\Phi(0)\bigg)=Z\,?\tag{1} $$

  • How does (0) imply (1)?

Following the definition of $\Phi$, (1) is equivalent to $$ \lim_{t\to 0}\frac{1}{t}\bigg(e^{tX}e^{tY}-I\bigg)=X+Y,\quad X\in\mathfrak{g},\ Y\in\mathfrak{g}^\perp.\tag{2} $$ But I fail to see how this is true.

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  • $\begingroup$ By definition of the exponential functions, $$ e^{tX}e^{tY}= \big(I+tX+\frac{1}{2}t^2X^2+\frac{1}{3!}t^3X^3+\cdots\big) \big(I+tY+\frac{1}{2}t^2Y^2+\frac{1}{3!}t^3Y^3+\cdots\big) =I+t(X+Y)+O(t^2)\,, $$ which implies (2). But still I can't tell why (0) implies (1). $\endgroup$ – user6 Mar 2 at 23:33
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    $\begingroup$ It looks like they meant to write $\Phi(X,Y)$, not $\Phi(X+Y)$. Since one has a direct-sum decomposition $M_n(\Bbb C) = \mathfrak g\oplus \mathfrak g^\perp$, there's not much difference, as $(X,Y)$ uniquely determines $X+Y$ and vice versa. $\endgroup$ – Ted Shifrin Mar 3 at 0:31
  • $\begingroup$ @TedShifrin: That makes sense. Thanks! Do you have an idea how (0) implies that the derivative of $\Phi$ at $0$ is the identity map? It seems that (0) only proves two "directions". $\endgroup$ – user6 Mar 3 at 0:44
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    $\begingroup$ The matrix of the derivative (using the map as I said it) will be the identity if you choose any basis for $\mathfrak g$ and any basis for $\mathfrak g^\perp$. Or just use the direct sum decomposition ... $\endgroup$ – Ted Shifrin Mar 3 at 0:46
  • $\begingroup$ The exponential is not locally injective in general, by an argument of Dixmier, see math.stackexchange.com/a/1592257/35400. It's only locally injective around 0. $\endgroup$ – YCor Mar 3 at 20:51
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Thanks to Ted's comment, I think I have an answer to my questions.

  • By context, $(tX,0)$ should be understood as an element in $\mathfrak{g}$ while $M_n(\mathbb{C})=\mathfrak{g}\oplus \mathfrak{g}^\perp$.
  • The function $\Phi$ is smooth by its definition and thus $$ \Phi'(0)(Z)=\lim_{t\to 0}\frac{\Phi(0+tZ)-\Phi(0)}{t}\,, $$ where the right hand size is the directional derivative of $\Phi$ at $0$ in the direction $Z$. On the other hand, if one writes $Z$ as $X+Y$ where $X\in\mathfrak{g}$ and $Y\in\mathfrak{g}^\perp$, then (0) implies that $$ \Phi'(0)(Z)=\Phi'(0)(X)+\Phi'(0)(Y)=X+Y=Z. $$
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