Logarithm of complex matrices in holomorphic families

Question

Let $n,k\geq 1$ be integers, let $U \subseteq \mathbb C^n$ be a contractible open subset, and let $f:U\to \mathrm{GL}_k(\mathbb C)$ be a holomorphic function. Does there exist a holomorhpic function $F:U\to \mathrm{M}_k(\mathbb C)$ such that $\exp(F(u))= f(u)$ holds for all $u\in U$?

Here, $\mathrm{M}_k(\mathbb C)$ means complex $k$ by $k$ matrices. The answer is of course "yes" if $k=1$.

As soon as $k\geq 2$, the problem is that for some invertible matrices $A \in \mathrm{GL}_k(\mathbb C)$ the set of matrices $B\in \mathrm{M}_k(\mathbb C)$ with $\exp(B)=A$ is not discrete. This happens for example if $A$ is diagonalisable and has a double eigenvalue. If in the question we require that for all $u\in U$ the eigenvalues of $f(u)$ are pairwise distinct, the answer would again be yes.

Answer 1

The answer is "no" in general. As Denis suspects, the problem is a global one, and it involves matrices with nontrivial Jordan blocks. These have, in a sense, "fewer" logarithms than the commoners. Concretely, I clain that the holomorphic function $$f(z) = \begin{pmatrix} e^{2\pi i z} & 1 \\\\ 0 & 1 \end{pmatrix}$$ has no holomorphic logarithm on $\mathbb C$. If it had one, there would also be a holomorphic square root of $f$ on $\mathbb C$, and not even that exists. Indeed, suppose by contradiction that there was a function $g:\mathbb C \to \mathrm{GL}_2(\mathbb C)$ such that $f(z) = g(z)^2$. The matrix $$f(0) = g(0)^2 = \begin{pmatrix} 1 & 1\\\\ 0 & 1 \end{pmatrix} $$ has only two square roots (a 2-by-2 matrix with distinct eigenvalues has four square roots!) differing by a sign, so we may suppose $$g(0) = \begin{pmatrix} 1 & 1/2 \\\\ 0 & 1 \end{pmatrix}$$ by changing $g$ to $-g$ if necessary. If we move $z$ on the real line from $0$ to $1$, we find by continuity of $g$ $$g(z) = \begin{pmatrix} e^{\pi i z} & (e^{\pi i z}+1)^{-1} \\\\ 0 & 1 \end{pmatrix}$$ and run into a pole as $z$ approaches $1$, end of story.

Answer 2

It might be worth mentioning that there is an analogous problem with $C^{\infty}$ functions, even when all the matrices are diagonalizable.

Let $$F(x,y) = \begin{pmatrix} e^{ix} & 0 \\ 0 & e^{-ix} \end{pmatrix} \begin{pmatrix} \cos y & \sin y \\ - \sin y & \cos y \end{pmatrix}$$

Since $F(x,y)$ is unitary, it is always diagonalizable. We claim that $F$ does not have a smooth logarithm on any open neighborhood of $[0,\pi] \times \{ 0 \}$.

Proof: Let $f(x,y)$ be a logarithm of $F$. First, look at the situation with $y=0$. The continuous $\log$'s of $F(x,0)$ are of the form $\left( \begin{smallmatrix} ix+2k\pi i & 0 \\ 0 & -ix+2 \ell \pi i \end{smallmatrix} \right)$ for integers $k$ and $\ell$. So it is impossible that $f(0,0)$ and $f(\pi,0)$ are both multiplies of the identity. Without loss of generality, assume that $f(\pi, 0)$ is of the form $\left( \begin{smallmatrix} (2a+1) \pi i & 0 \\ 0 & -(2b+1) \pi i \end{smallmatrix} \right)$ with $a \neq b$.

The Jacobian of the exponential map at $\left( \begin{smallmatrix} (2a+1) \pi i & 0 \\ 0 & -(2b+1) \pi i \end{smallmatrix} \right)$ has rank one, and $\partial F/\partial y (\pi, 0)$ is not in the image of that rank one map. So it is impossible for $F$ to equal $e^{f(x,y)}$ for any smooth function $f$ near $(\pi, 0)$. If $f(\pi, 0)$ was a multiple of the identity, then there would be no problem at $(\pi, 0)$, but there would be a problem at $(0,0)$ instead.

Answer 3

I don't think that the multiplicity of eigenvalues be an obstruction. At least, if $g(U)$ is small enough, then such an $F$ exists: first of all, there exists a holomorphic logarithm on the set of all eigenvalues of all elements $u\in U$. Then $F=\log\circ f$ works. For instance, if $f(U)$ is included in the unit ball centered at $I_n$, the branch of the logarithm is the classical one.

If there is an obstruction, it must be a global one, in which the union of spectra of elements of $u$ encircle the origin. I don't have an answer in this case and will continue to think about it.