Kouznetsov Wrote:Your figure looks similar to Fig.3 by Kneser.
Ya, now that I go again through his paper do find out what he really does, I see that his picture is indeed the upper half of mine. I will start a thread about the Kneser construction, step by step how I understand it. I think there maybe broad interest in it, especially because not everyone interested in his paper is able to read German.
Quote:bo198214 Wrote:We call a function \( \text{slog} \) defined on the domain \( H \) a super logarithm iff \( \text{sexp}(\text{slog}(z))=z \) for each \( z\in H \) (but not necessarily \( \text{slog}(\text{sexp}(z)) \) for each \( z\in\text{slog}(H) \)).Why "non necessarily"?
\( H=\mathrm{sexp}(D) \) ?
\( D=\mathrm{slog}(H) \) ?
Well if you define \( \text{sexp} \) on \( \mathbb{C}\setminus\{z:\Re(z)\le -2\} \) then I guess the image of \( \text{sexp} \) is whole \( \mathbb{C} \). While \( \text{slog} \) has a singularity at \( L \).
I mean you do it the same for the logarithm. You chose a branch for the logarithm and then \( \exp(\log(z))=z \) for all complex \( z \). But of course \( \log(\exp(2\pi i))=\log(1)=0\neq 2\pi i \). A branching similar to the logarithm at 0 takes place at \( L \) and \( L^\ast \) for the \( \text{slog} \).
