Is the intersection number on $S^k$ is always zero? Choose $X$ a compact submanifold of $S^k$, and $Z$ a closed submanifold of complementary dimension, viewing $I_2(X,Z) = I_2(i, Z)$ where $i: X \hookrightarrow Y$ is the inclusion. Then $X,Z$ can always be homotopped to be disjoint?
Asked
Active
Viewed 293 times
2
-
No, the intersections aren't always zero. For example, you can choose $Z$ to be zero dimensional and $X=S^k$. – Ryan Budney Jan 16 '14 at 06:37
-
Thank you Ryan - how about when $X,Z$ are not finite set? Is the reason stands? Thanks @RyanBudney – 1LiterTears Jan 16 '14 at 06:44
-
1Yes, if they're manifolds of dimension strictly between $0$ and $k$ then for example studiousus's answer applies. You could also argue the two manifolds can be isotoped to be disjoint, because the sphere is the one-point compactification of Euclidean space, which is contractible. – Ryan Budney Jan 16 '14 at 06:50
1 Answers
2
Yes, intersection number is zero unless either X or Z is a finite set. This follows from vanishing of homology groups of the sphere except in dimension 0 and k. You can also homotope the submanifolds to be disjoint. Hint: first do this for $R^k$ instead of the sphere.
Moishe Kohan
- 97,719
-
Dear studiousus - thank you very much. Is my claim right, though? Thank you! – 1LiterTears Jan 16 '14 at 06:10
-
Yes, that's what I said in my answer: True if $dim(X) dim(Z)\ne 0$, false if $dim(X)dim(Z)=0$. – Moishe Kohan Jan 16 '14 at 08:57