Article ID | Journal | Published Year | Pages | File Type |
---|---|---|---|---|
4657588 | Topology | 2008 | 30 Pages |
Abstract
We prove that 2r+1⋅π∗(Pm(2r))=02r+1⋅π∗(Pm(2r))=0 provided m≥4m≥4 and r≥6r≥6. This is the best possible result. As well, for 2≤r≤52≤r≤5 we obtain upper bounds on the homotopy exponent of Pm(2r)Pm(2r).
Keywords
Related Topics
Physical Sciences and Engineering
Mathematics
Geometry and Topology
Authors
Stephen D. Theriault,