Timeline for Smooth map homotopic to Lie group homomorphism
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2020 at 1:51 | vote | accept | Hang | ||
| Jul 7, 2020 at 1:48 | comment | added | Igor Belegradek | @Hang If $X$, $K$ are connected CW complexes and $K$ has contractible universal cover, then the set of homotopy classes $[X,K]$ of maps from $X$ to $K$ is bijective to the set of conjugacy classes of the induced $\pi_1$-homomorphism (e.g. Spanier, "Algebraic Topology", Ch.11, Section1, Theorem 11). Thus $[T^n, T^n]$ is bijective to the set of endomorphisms of $\mathbb Z^n$. Those are $n\times n$ matrices over $\mathbb Z$. Any such matrix (as a map of $\mathbb R^n$) descends to a self-map of an $n$-torus. | |
| Jul 7, 2020 at 0:29 | comment | added | Hang | Thank you for the great answer! Could you explain more about the self-map of torus? I'm not familiar with obstruction theory. Or, any reference is also good enough. | |
| Jul 6, 2020 at 23:45 | comment | added | Igor Belegradek | In fact, compact connected simple Lie groups have (see en.wikipedia.org/wiki/List_of_simple_Lie_groups) finite fundamental group and hence their self-coverings are diffeomorphisms. Thus in the above answer one can drop the assumption ``$G$ is simply-connected''. | |
| Jul 6, 2020 at 21:06 | history | edited | Igor Belegradek | CC BY-SA 4.0 |
edited body
|
| Jul 6, 2020 at 20:38 | history | undeleted | Igor Belegradek | ||
| Jul 6, 2020 at 20:36 | history | deleted | Igor Belegradek | via Vote | |
| Jul 6, 2020 at 20:35 | history | answered | Igor Belegradek | CC BY-SA 4.0 |