Exercise 11, chapter 2 in Lie Groups, Lie Algebras, and Representations: An Elementary Introduction

Question

I am reading the book: Lie Groups, Lie Algebras, and Representations: An Elementary Introduction by Brian C. Hall. I am stuck at the following exercise: exercise 11, chapter 2 . Can you help me?

Suppose $G$ is a matrix lie group in $GL(n, \mathbb{C})$ with the lie algebra $\mathfrak{g}$. suppose that $A$ is in G and that $||A-I||<1$,so the power series for $\log A$ is convergent. Is it necessarily the case that $\log A$ is in $\mathfrak{g}$？ prove or give a counterexample.

In this book, $\log A$ is defined as a power series: $$ log A = \sum _{n=1}^{\infty} (-1)^{n+1}(A-I)^n/n$$ whenever the series converges.

I know that if $A$ is small enough, then we can prove that it is true.

Answer 1

I reposted a closely related question here and got a good answer.

To summarize: No, $\log(A)$ may not be in $\mathfrak{g}$. Counterexamples can be found in the group of $n$-th roots of unity for large $n$, or in the line on the torus $$G_a = \left\{\begin{pmatrix} e^{2\pi it} & 0 \\ 0 & e^{2\pi ita} \end{pmatrix} : t \in \mathbb{R} \right\}$$ for $a \in \mathbb{Q}$ with large denominator.

Answer 2

In the cases of ordinary numbers, under the definition $$ \log x = \sum _{n=1}^{\infty} (-1)^{n+1}(x-1)^n/n, $$ we can expand $$ e^{\log x} = C_0 + C_1 (x-1)+ C_2 (x-1)^2 + \cdots .$$ Because $ e^{\log x} =x $, we should get $$ C_0 =1, C_1 =1 ,C_i=0 \, (i\geq 2) . $$ In the cases of matricies, under the definition $$ \log A = \sum _{n=1}^{\infty} (-1)^{n+1}(A-I)^n/n, $$ we can also expand $ e^{\log x} . $ Because $ \left[ (A-I)^k,(A-I)^l \right]=0, $ the coefficients of this series are the same as in the cases of ordinary numbers. So $$ e^{\log A} = I + (A-I) = A $$ Thus it is necessarily the case that $\log A$ is in $\mathfrak{g}$.