Matrix logarithm not in Lie algebra

Question

In Hall's Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, he defines the Lie algebra of a matrix Lie group $G$ as the set $\mathfrak{g}$ of all matrices $X$ such that $e^{tX}\in G$ for all $t\in\mathbb{R}$. Here the exponential of matrices is defined using the power series. Similarly, he defines the logarithm of a matrix $A$ using power series: \begin{equation} \log A=\sum_{m=1}^{\infty}(-1)^{m+1}\frac{(A-I)^m}{m}. \end{equation} It is known that this series converges when $\|A-I\|<1$, where $\|\cdot\|$ is the Hilbert-Schmidt norm.

Now, in Exercise 3.7, we are asked the following question:

Given an $A$ in a matrix Lie group such that $\|A-I\|<1$ (so that the series above converges), is it always true that $\log A$ is in $\mathfrak{g}$? Prove or give a counterexample.

My idea is the following: We know that the exponential map $exp:\mathfrak{g}\to G$ is a local diffeomorphism between a small neighbourhood $U$ of $0$ in $\mathfrak{g}$ and a small neighbourhood $V$ of $I$ in $G$. However, $V$ may be very small such that it may not contain some $A$ that satisfies $\|A-I\|<1$ (that is, although $A$ is already closed to $I$, it may still not in $V$). In this case, $\log A$ may not necessarily inside $\mathfrak{g}$. But then when I tried to find some counterexamples, they are all outside the radius $1$ ball of $I$ (i.e., these examples $A$ are such that $\|A-I\|>1$). Thus, I am lost again.

Any hint, suggestion, comment and answer are much appreciated.

Answer 1

Let $G$ be a finite cyclic subgroup of $S^1 \subset GL(1,\mathbb{C})$ of 12 elements, for example. Clearly its Lie algebra $\mathfrak{g}$ is zero, but there is an element $A\in G$ with $0\neq \Vert A-I \Vert = |A-1| < 1$. This $A$ satisfies $\log A \neq 0$ so we are done.

Answer 2

In addition to the example in a discrete group given by Junyoung Park, we can construct an example in a continuous group with non-trivial Lie algebra.

Let $n$ be a positive integer, $\epsilon > 0$ a real number and $\omega = e^{\frac{2\pi i}{n}}$. Consider the $n \times n$ complex matrix

$$ A_{n,\epsilon} = \begin{bmatrix} \omega & \epsilon & & & \\ & \omega & \epsilon & & \\ & & \omega & \epsilon & \\ & & & & \ddots \\ & & & & & \omega & \epsilon \\ & & & & & & \omega \\ \end{bmatrix} $$

which is easily seen to be in $SL(n, \mathbb{C})$. Compute the Hilbert-Schmidt norm

$$ \|A_{n,\epsilon} - I\|^2 = {\rm Tr}\,{(A_{n,\epsilon} - I)(A_{n,\epsilon} - I)^*} = n |\omega - 1|^2 + (n - 1) \epsilon^2 = 4n \sin^2\frac{\pi}{n} + (n - 1) \epsilon^2 $$

to see that $\|A_{n,\epsilon} - I\| < 1$ if and only if

$$ \epsilon^2 < \frac{1}{n - 1}\left(1 - 4n \sin^2\frac{\pi}{n}\right). $$

The right hand side is positive when $n \ge 40 > 4\pi^2$. Therefore, there exist $n$ and $\epsilon$ such that $\|A_{n,\epsilon} - I\| < 1$.

Now, if $\|A_{n,\epsilon} - I\| < 1$ then $\log A_{n,\epsilon}$ converges and $e^{\log A_{n,\epsilon}} = A_{n,\epsilon}$. Let $X_{n,\epsilon} = \log A_{n,\epsilon}$. We will prove that $X_{n,\epsilon} \notin \mathfrak{sl}(n, \mathbb{C})$.

We begin by showing that $X_{n,\epsilon}$ has only one eigenvalue. Suppose that $\lambda_1 \ne \lambda_2$ are eigenvalues of $X_{n,\epsilon}$ and denote by $v_1$ and $v_2$ respective eigenvectors. Then $v_1$ and $v_2$ are linearly independent and are eigenvectors of $A_{n,\epsilon}$. However, this is a contradiction because we know that $A_{n,\epsilon}$ has exactly one one-dimensional invariant subspace as a consequence of its Jordan normal form. Therefore, $X_{n,\epsilon}$ has only one eigenvalue, $\lambda_1 = \lambda_2 = \lambda$.

Finally, suppose that $X_{n,\epsilon} \in \mathfrak{sl}(n, \mathbb{C})$. Then ${\rm Tr}\,X_{n,\epsilon} = 0$ and so $\lambda = 0$. However, this implies that $1$ is an eigenvalue of $A_{n,\epsilon}$ which is not true. Therefore, $X_{n,\epsilon} \notin \mathfrak{sl}(n, \mathbb{C})$.