- \phi \colon U\to V} , we say that U and V are
biholomorphically equivalent or that they are
biholomorphic. If n = 1 , {\displaystyle n=1,}
every simply...
- n>1} , open
balls and open
polydiscs are not
biholomorphically equivalent, that is,
there is no
biholomorphic mapping between the two. This was
proven by...
-
between charts are
biholomorphic,
complex manifolds are, in particular,
smooth and
canonically oriented (not just orientable: a
biholomorphic map to (a subset...
-
around every point on the
sphere there is a
neighborhood that can be
biholomorphically identified with C {\displaystyle \mathbf {C} } . On the
other hand...
-
theorem says that a
simply connected domain in the
complex plane is "
biholomorphically equivalent" (i.e.
there is a
bijection between them that is holomorphic...
-
considered as a
Riemann surface, the open unit disk is
isomorphic ("
biholomorphic", or "conformally equivalent") to the
upper half-plane, and the two...
- curvature. Then the
universal cover of M, with its
natural metric, is
biholomorphically isometric to the
metric product of
complex Euclidean space, with some...
- of G {\displaystyle G} as a
neighbourhood of 0 {\displaystyle 0} ).
Biholomorphic mapping –
Bijective holomorphic function with a
holomorphic inversePages...
- Thus,
under this definition, a map is
conformal if and only if it is
biholomorphic. The two
definitions for
conformal maps are not equivalent.
Being one-to-one...
-
isometry group. In the
category of
Riemann surfaces, an
automorphism is a
biholomorphic map (also
called a
conformal map), from a
surface to itself. For example...
