Let $M$ be a manifold with the property that $f^{*}(TM)$ is isomorphic to TM, for every diffeomorphism $f$ on $M$. Does this imply that $M$ is parallelizable?
$\begingroup$
$\endgroup$
3
asked Dec 19, 2013 at 15:01
-
$\begingroup$ Doesn't every diffeomorphism $f : M \to N$ induce an isomorphism $df : TM \to f^\ast TN$? $\endgroup$– Ricardo AndradeCommented Dec 19, 2013 at 15:12
-
$\begingroup$ $df$ cover $f$, but an isomorphism of two vector bundles on $M$ must cover $Id_{M}$ $\endgroup$– Ali TaghaviCommented Dec 19, 2013 at 15:15
-
1$\begingroup$ Since $df : TM \to TN$ covers $f$, it induces a map $TM \to f^\ast TN$ over $\mathrm{id}_M$, which I also called $df$ above. $\endgroup$– Ricardo AndradeCommented Dec 19, 2013 at 15:19
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Isn't the 2-dimensional sphere a counterexample? If $f$ has degree 1, then it's homotopic to the identity, so $f^*(TM)\cong TM$. If $f$ has degree $-1$, then it's homotopic to the antipode map $a$, so $f^*(TM)\cong a^*(TM)\cong TM$, where the last $\cong$ is evident if we embed the sphere in the standard way in $\mathbb R^3$ so that the tangent spaces at a point and its antipode are the same as vector spaces. Since a diffeomorphism must have degree $\pm1$, we have $f^*(TM)\cong TM$ in all cases. But even-dimensional spheres don't admit even one nowhere-vanishing vector field, because they have non-zero Euler characteristic, so they are far from parallelizable.
answered Dec 19, 2013 at 15:06
-
$\begingroup$ why $S^{2}$ is a counterexample? $\endgroup$ Commented Dec 19, 2013 at 15:12
-
$\begingroup$ @AliTaghavi I edited a proof into my answer. $\endgroup$ Commented Dec 19, 2013 at 15:23