Lemma 2
For each
d > 0, there is an algebraic morphism
Fd, n :
Q(n + 2)Q(n + 2)
and an algebraic isomorphism
such that
- (i)
- under the identification (by Scholium 1)
Q(n + 2) =
C(
Q'(n),
(1))
we have
Fd, n =
C(
Fd, n - 2,
)
.
- (ii)
-
Fd1, noFd2, n = Fd1d2, n.