Milnor lemma 2 pg 34 "Any orientation preserving diffeomorphism f on $R^m$ is smoothly homotopic to the identity"

So he proves that $f\simeq df_0$ ,which he says is clearly homotopic to the identity. Can you explain me why?

Here I found two explanations I don't understand: 1) $Gl^{+}(m,\mathbb{R})$ is path connected. Why is $df_0 \in Gl^{+}(m,\mathbb{R})$? What prevents $df_{0}\in Gl^{-}(m,\mathbb{R})$?

2) $df_0$ is isomorphic everywhere and thus(why?) isotopic to identity.



First, $f$ is orientation-preserving. Second, $GL(n)^+$ is path-connected (e.g., use the $QR$ decomposition).

  • $\begingroup$ Thanks, f is o.p. $\Rightarrow sign(det(df_{0}))=1>0\Rightarrow df_{0}\in GL^{+}(n)$ $\endgroup$ – TKM Feb 18 '14 at 4:55
  • $\begingroup$ Nice answer Ted; so the isotopy in question is defined by $(x,t)\mapsto\gamma(t)x$, where $\gamma$ is a smooth path joining $df_0$ to $\mathrm{id}_{\mathbf{R}^n}$? @TKM: this answer fully answers your question, and is certainly more appropriate than the other one. Please consider accepting it. $\endgroup$ – user135041 May 4 '14 at 17:01

This is not really an answer to your specific question, but another way of proving the lemma.

Since $f$ is an orientation preserving diffeomorphism on $\mathbb R^m$ it must have degree 1. By the Hopf degree theorem (p. 51 in Milnor), it is smoothly homotopic to the identity.

  • 3
    $\begingroup$ Oh come on! A cannon to kill a fly! (Not to mention almost surely circular reasoning.) $\endgroup$ – Ted Shifrin Feb 18 '14 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.