Understanding the proof of the theorem that the exponential map is locally one-to-one at 1 and onto

Question

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.

Answer 1

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 $$ \Phi'(0)(Z)=\Phi'(0)(X)+\Phi'(0)(Y)=X+Y=Z. $$