6
$\begingroup$

It is easy to see that within the disk algebra $A(D)$ $$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 1&1\\z&1+z \end{pmatrix} $$ is a product of two exponential matrices. Is $\Delta(z)$ itself an exponential matrix of a holomorphic matrix $M(z)$ ? I don't think so, but could not come up with a proof. Is there a general method to deal with these questions?

$\endgroup$
1
  • $\begingroup$ See again my answer, the upper-left entries lacked the correct sign. $\endgroup$ May 2, 2016 at 12:38

3 Answers 3

2
$\begingroup$

$\Delta(z)$ does not have a nonpositive real eigenvalue for $|z| < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic functional calculus produces the desired analytic logarithm $M(z)$ for $|z| < 4$.

$\endgroup$
3
$\begingroup$

It seems that $\Delta(z)$ is the exponential of an holomorphic $M(z)$. Using the eigenvalues and eigenvectors of $\Delta$, I find $$M(z)=\frac\mu{\sqrt{z(z+4)}}\begin{pmatrix} -z & 2 \\ 2z & z \end{pmatrix},$$ where $\mu$ is the solution of $$\sinh\mu=\frac12\sqrt{z(z+4)}.$$ Because $\sinh^{-1}$ is an odd holomorphic function, the quotient $\frac\mu{\sqrt{z(z+4)}}$ can be chosen holomorphic, at least in a domain where $$\left|\frac14z(z+4)\right|<1,$$ because the convergence radius of the power series $$\sinh^{-1} x=x-\frac1{2\cdot3}x^3+\frac{1\cdot3}{2\cdot4\cdot5}x^5+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot7}x^7+\cdots$$ equals $1$. To have a positive answer to the question, we need to prove that $\sinh^{-1}$ extends holomorphically to a slightly larger domain, containing the image of $D$ under the multi-valued map $$z\mapsto\frac12\sqrt{z(z+4)}\,.$$ A good news is that this image does not reach the singularities $\pm i$ of the series above.

$\endgroup$
2
$\begingroup$

To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac12\sum_{n=1}^\infty(-1)^{n-1}\frac{(n-1)!}{(2 n-1)\text{!!}}\left(\frac z2\right)^{n-1}\begin{pmatrix}-z&2\\2z&z\end{pmatrix} $$ which (I think) boils down to the same $M(z)$ as in another answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.