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Consider the following theorem from Hall's Lie Groups, Lie Algebras and Representations:

Theorem 2.27: For $0 < \varepsilon < \textrm{ln} 2$, let $U_\varepsilon = \{X \in M_n(\Bbb{C}) | ||x|| < \varepsilon\}$ and let $V_\varepsilon = \exp(U_\varepsilon)$. Suppose $G \subseteq \textrm{GL}_n(\Bbb{C})$ is a matrix Lie group with Lie algebra $\mathfrak{g}$. Then there exists $\varepsilon \in (0, \textrm{ln}(2))$ such that for all $A \in V_{\varepsilon}$, $A$ is in $G$ if and only if $\log A$ is in $\mathfrak{g}$.

The norm here is defined to be: If $A = (a_{ij})$ then $||A|| = \sqrt{\sum_{i,j} |a_{i,j}|^2}$.

Now I am wondering if it is possible to have a matrix $A \in G$ with $|| A - I|| < 1$, but that $\log A \notin \mathfrak{g}$. Now to find such an $A$, I must necessarily have that $A$ not be the exponential of any matrix $X \in \mathfrak{g}$, for otherwise the hypotheses of the theorem will be satisfied. Now I tried doing this with $\textrm{SU}(2)$ by finding a matrix that has only 1 eigenvalue of multiplicity 2. However this imposes the condition that the $a_{11}$ entry of such an $A$ have real part 1, which would mean that $||A- I||<1$ is not satisfied.

I just realised $\textrm{SL}_2(\Bbb{R})$ will not work as well because to have eigenvalue of multiplicity $1$, I would need that the trace of my $A$ be $\pm 2$. I don't think for such an $A$ the condition $||A - I|| < 1$ would be satisfied. $\textrm{U}_n$ would also not work to give an example because it is known the exponential map $\exp : \mathfrak{u}(n) \longrightarrow \textrm{U}(n)$ is surjective.

My question is: For me to find an example of $A \in G$ with $||A - I||<1$, but $\log A \notin \mathfrak{g}$ it must be the case that $\exp$ does not map open balls to open balls. Does anyone have an example to show this? Also how do I go about finding such a matrix $A$?

Thanks.

Edit: Proof that $\exp : \mathfrak{u}(n) \to \textrm{U}(n)$ is surjective. Take any unitary matrix $U$. Now a more advanced version of the Complex Spectral Theorem tells us that there is a unitary matrix $P$ consisting of eigenvectors of $U$ such that

$$PUP^{-1} = D$$

with $D$ diagonal. Now since all the eigenvalues of $U$ lie on the unit circle, we can write $D = e^{C}$ for some $C$ of the form

$$\left(\begin{array}{cccc} i\theta_1 & \\ & i\theta_2 \\ & & \ddots \\ &&&i\theta_n\end{array}\right)$$

for $0 \leq \theta_k < 2\pi$ for all $1 \leq k \leq n$. Now it is clear that $C \in \mathfrak{u}(n)$ and since $P \in G$, we have that $P^{-1}CP \in \mathfrak{u}(n)$. Since $U = e^{P^{-1}CP}$, it follows that $\exp$ is surjective.

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Hall must have a standing hypothesis that $G$ is connected. Otherwise, let $n=1$ and let $G$ be the group of $N$-th roots of unity. Then $2 \pi i/N$ is within a ball of radius $2 \pi/N$ over zero, and its exponential is $G$, but $2 \pi i/N$ is not in $\mathfrak{g}=(0)$. –  David Speyer Sep 2 '12 at 15:21
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1 Answer

up vote 5 down vote accepted

It is true that matrix exponential is not an open map. For example, look at $\exp$ in a neighborhood of $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & - 2 \pi i \end{smallmatrix} \right)$. All matrices near $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right)$ are diagonalizable (over $\mathbb{C}$) so their exponentials are diagonalizable. However, $\exp \left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ and an arbitrarily small neighborhood of $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ contains matrices of the form $\left( \begin{smallmatrix} 1 & \epsilon \\ 0 & 1 \end{smallmatrix} \right)$ which are not diagonalizable.

Note that the matrix $\left( \begin{smallmatrix} 1 & \epsilon \\ 0 & 1 \end{smallmatrix} \right)$ does have logarithms, for example $\left( \begin{smallmatrix} 0 & \epsilon \\ 0 & 0 \end{smallmatrix} \right)$. But these logarithms have Jordan blocks of size $2$, and thus can't get near $\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right)$.

Similar examples can be made with $\left( \begin{smallmatrix} 0 & 2\pi \\ - 2\pi & 0 \end{smallmatrix} \right)$, if you'd like real entries; and with $\exp \left( \begin{smallmatrix}0 & \pi \\ - \pi & 0 \end{smallmatrix} \right)$ if you'd lke smaller entries.

However, I don't see the connection between this and Hall's statement, and I don't know what the best possible constant in Hall's result is. I would consider the possibility that Hall simply wasn't interested in finding the best constant.

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Thanks very much David for your answer. I really appreciate it as I have been stressed out over this problem for a while now! However, how do matrices of the form $\left(\begin{array}{cc} 1 & \epsilon \\ 0 & 1 \end{array}\right)$ give me an example of a matrix $A$ such that $\log A$ is not in the Lie algebra? I presume in your example above that we have $\mathfrak{sl}_2(\Bbb{C})$ as our Lie algebra, but what would be the matrix Lie group in question? –  fpqc Sep 2 '12 at 14:26
    
$\log \begin{pmatrix} 1 & \epsilon \\ 0 & 1 \end{pmatrix}$ is in the Lie algebra, however it is not in a ball (of radius $\pi$, say) around $\begin{pmatrix} 2 \pi i & 0 \\ 0 & - 2\pi i \end{pmatrix}$. This example demonstrates that the exponential of an open ball need not be an open set. As I said, I don't see the connection to Hall's statement. –  David Speyer Sep 2 '12 at 14:33
    
I am wondering now if indeed it is true that if $A \in G$ and $||A - I||<1$ then $\log A$ is in the Lie algebra. I am quite confused now. –  fpqc Sep 2 '12 at 14:35
    
Dear David, I have decided to accept your answer. Though it did not answer my question directly, I felt I learned about some concrete examples through it. Thanks very much. –  fpqc Sep 4 '12 at 22:59
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