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

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.

• 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). – user6 Mar 2 at 23:33
• 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. – Ted Shifrin Mar 3 at 0:31
• @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". – user6 Mar 3 at 0:44
• 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 ... – Ted Shifrin Mar 3 at 0:46
• 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. – YCor Mar 3 at 20:51

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