- that ζ and η are
cohomologous to each other.
Exact forms are
sometimes said to be
cohomologous to zero. The set of all
forms cohomologous to a
given form...
-
forms on a manifold. One
classifies two
closed forms α, β ∈ Ωk(M) as
cohomologous if they
differ by an
exact form, that is, if α − β is exact. This classification...
-
implies that, in practice, one
studies only
classes of
bialgebras that are
cohomologous to a Lie
bialgebra on a coboundary. They are also
called Poisson-Hopf...
- U\subset X} , and by the Poincaré
lemma any Kähler form will
locally be
cohomologous to zero. Thus the
local Kähler
potential ρ {\displaystyle \rho } is the...
- the real line.
First observe that any 0 {\displaystyle 0} -chain is
cohomologous to 0 {\displaystyle 0} .
Since this
reduces to the case of a
point p...
- ^{2}(M)} a
family of
symplectic form on M {\displaystyle M}
which are
cohomologous, i.e. the
deRham cohomology class [ ω t ] ∈ H d R 2 ( M ) {\displaystyle...
- h ) {\displaystyle c^{\prime }(g,h)=f(gh)f(g)^{-1}f(h)^{-1}c(g,h)}
cohomologous to c. Thus L
defines a
unique class in H2(G, F∗). This
class might not...
-
apply the same
techniques to
symplectic forms,
thereby proving that a
cohomologous family of
symplectic forms are
related to one
another by diffeomorphisms:...
-
related as the
hatted and non-hatted
versions of ω
above are said to be
cohomologous. They
belong to the same
second cohomology class, i.e. they are represented...
- the
first example of
symplectic forms on a
closed manifold that are
cohomologous but not
diffeomorphic and also
classified the
rational and
ruled symplectic...
