Math 8205
Spring 1998
Hint for Part b. of Exercise 6
There are two cases to consider:
-
dim(Sec X) = 2r + 1
-
dim(Sec X) = 2r
In the first case, it would be nice to show that we can choose
L so that each point of L lies on at most
finitely many secant lines of X. One way to accomplish
this is to consider the closed subset Z
Sec X, defined to be the closure of the following subset:
Z0 = {z :
z
Sec X
and z lies on infinitely secant lines <p,q>,
p
X
and q
X}.
Specifically, one can estimate the dimension of Z, showing that
dim Z
2r
- 1.
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