# Lie algebra "generated" by matrix-valued curve?

Let $A(t)$ be a $n\times n$-matrix-valued continuous (plus possibly other niceness conditions; see below) curve, with the matrix entries being complex in general. If I am not mistaken, $A(t)$ generates a minimal Lie algebra $\mathfrak{g}$ of matrices, in the sense that $\mathfrak{g}$ is the intersection of all $n\times n$ matrix Lie algebras containing $A(t)$ for all $t$.

Now consider the equation $$\frac{d U(t,t_0)}{d t}=A(t) U(t,t_0)$$ with initial condition $$U(t_0,t_0)=\mathbf{1}_n$$ where $\mathbf{1}_n$ is the $n \times n$ unit matrix.

I am interested in the matrix logarithms of $U(t,t_0)$ for arbitrarily large $t$. For $t$ close to $t_0$ at least one of them lies in $\mathfrak{g}$ because the condition $\int_{t_0}^t \|A(s)\| ds < \pi$ is satisfied -as long as $A(t)$ is nice- and thus the Magnus series for the logarithm converges. For large $t$ I have no reason to expect that the Magnus series will continue to converge; however $U(t,t_0)$ has at least one logarithm, because the matrix exponential is surjective when considering matrices with complex-valued entries. My question is this:

Where do the matrix logarithms of $U(t,t_0)$ lie? Is it possible that they all lie outside $\mathfrak{g}$?

I would also appreciate it if anyone knows any good references dealing with these equations. Thank you!

EDIT: Robert Bryant gave an example of a $U$ whose logarithm lies outside $\mathfrak{g}$. However one may extend $\mathfrak{g}$ by adding the unit matrix, so that that logarithm is now included. Would this $\mathfrak{g'}$ include at least one of the logarithms of $U(t,t_0)$ in general?

• I don't understand your notation $U(t,t_0)$, and I'm sure you don't mean to have $U(t,t_0) = \mathbf{1}_n$, which doesn't make sense. Don't you just mean $U'(t) = A(t)U(t)$ with initial condition $U(t_0)=\mathbf{1}_n$? Apr 9, 2012 at 1:06
• Sorry, I was being sloppy. I meant $U(t_0,t_0)=\mathbf{1}_n$. Fixing... Apr 9, 2012 at 1:14
• Still, what is the point of having two arguments for $U$? Apr 9, 2012 at 1:15
• The point of using $U(t,t_0)$ is that the matrices $U(t_1,t_2)$ possess the semigroup property $U(t,t_0)=U(t,t_1)U(t_1,t_0)$. Apr 9, 2012 at 1:19
• But, if you take my $1$-variable solution, your $2$-variable solution is expressed as $U(t,t_0) = U(t)U(t_0)^{-1}$, so there's no need for a $2$-variable expression. Apr 9, 2012 at 1:45

## 1 Answer

It can't always be done. Take $n=2$, and start with a smooth curve $V:[0,1]\to SL(2,\mathbb{C})$ such that $V(0) = \mathbb{1}_2$ and $$V(1) = \pmatrix{-1&1\\\\ 0&-1}.$$

Set $A(t) = V'(t)V(t)^{-1}$. Then $A:[0,1]\to{\frak{sl}}(2,\mathbb{C})$, and it satisfies $V'(t) = A(t)V(t)$ with $V(0) = \mathbb{1}_2$, but $V(1)$ is not the square of any element of $SL(2,\mathbb{C})$ and hence is not the exponential of any element of ${\frak{g}} = {\frak{sl}}(2,\mathbb{C})$.

Added after the question was edited: Giving yourself just the extra 'room' of adding the center of ${\frak{gl}}(n,\mathbb{C})$ to your algebra $\frak{g}$ won't do it. Here's a counterexample: Consider $SL(2,\mathbb{C})$ as a subgroup of $GL(3,\mathbb{C})$ in the obvious way, and let $W:[0,1]\to GL(3,\mathbb{C})$ be defined by $$W(t) = \pmatrix{V(t)&0\\\\ 0&1},$$ where $V$ is as above. Set $A(t) = W'(t)W(t)^{-1}$. Then $A:[0,1]\to {\frak{g}} = {\frak{sl}}(2,\mathbb{C})\subset {\frak{gl}}(3,\mathbb{C})$, and it satisfies $W'(t) = A(t)W(t)$ with $W(0) = \mathbb{1}_3$. But it is easy to see that $W(1)$ is not the exponential of any element of ${\frak{g}'} = \mathbb{C}\cdot\mathbb{1}_3\oplus {\frak{sl}}(2,\mathbb{C}) = \mathbb{C}\cdot\mathbb{1}_3\oplus {\frak{g}}$.

• That is correct, of course. (One of) the logarithms of V(1) as given above is $i \pi\mathbf{1}_2$−[0 1;0 0] (That last one is the $2\times 2$ upper triangular nilpotent matrix with -1 on the superdiagonal). If I were to enlarge $\mathfrak{g}$ to $\mathfrak{g'}$ by including the unit matrix, would I then get a logarithm in $\mathfrak{g'}$? (I will edit the question to include this tomorrow) Apr 9, 2012 at 2:20
• That's what I was looking for. Thank you very much for your time! Apr 9, 2012 at 22:16