Next message: Ken Muldrew: "Re: Easy Lunars in 1790"
Walter,
It has to be double checked yet,
but I suspect that in my favorite Volkovyski
example (cited in our paper in Iberoamericana),
all singularities are true (have branches which do not
return when you continue on a Jordan curve),
and there are uncountably many of them.
So they are not K.
Thus it seems that no good conjectures remain.
1. True (in the sense above) cannot be on the boundary
of the completely invariant domain.
2. Direct implies true. (So you fixed the bug in our paper
with Liubich).
3. True does not imply direct. (Simple example).
3. True implies linearly accessible.
4. But not vice versa, by your wonderful example.
5. True can be uncountably many (and thus true does not
imply K).
6. K does not imply true (Take a K-singularity described in Goldberg's
paper. Topologically it is like (sin z)/z).
What else to ask?
Alex
