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My question concerns the lifting of degree $0$ map from $S^3$ to itself.
Let us suppose that all maps are smooth here.

Looking at $S^3$ as the space of unit quaternions, one way to define degree is the following $$\deg(u)=\frac{-1}{12\pi^2}\int_{S^3} \Re((u^{-1}du)^3).$$ We can check that if $u=\exp(\phi)$ for $\phi: S^3 \rightarrow \Im \mathbb H$ smooth then the degree is zero. Does the converse holds?
If I have $u$ a degree $0$ map from $S^3$ to itself, can I find a smooth map $\phi: S^3 \rightarrow\Im \mathbb H$ such that $u=\exp(\phi)$?

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  • $\begingroup$ What is "con" in "$u$ a degree 0 con map"? $\endgroup$
    – LSpice
    Feb 27 at 22:43
  • $\begingroup$ Are both minus signs preceding the integral intentional? And does u^(-1) denote a) the inverse function of u, or b) the inverse quaternion? $\endgroup$ Feb 28 at 0:09
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    $\begingroup$ @LSpice: it must mean "continuous" $\endgroup$ Feb 28 at 2:21

2 Answers 2

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No. The reason this holds in the complex case (any degree $0$ map $S^1 \to S^1$ factors through $\exp: \mathbb{R}i\to S^1$) is that in that case the exponential map is a covering map, in particular a local homeomorphism. This tells you that there is never a local problem when constructing a lift, and you only need things to be compatible globally, which is where the degree condition comes in.

In the quaternion case, the exponential map is not a local homeomorphism: it collapses the sphere of radius $\pi k$ in $\Im \mathbb{H}$ to the element $(-1)^k\in S^3$, for each $k$. Now take a geodesic path $I\subseteq S^3$ joining $-1$ and $1$. There exists a degree $0$ map $f: S^3 \to S^3$ which is the identity in an open neighbourhood of $I$ (by adding suitable local degrees away from $I$). I claim this map cannot lift through $\Im \mathbb{H}$. Indeed, if it does, it would need to take $-1$ and $1$ to spheres of radius $k\pi$ and $(k+1)\pi$ in $\Im\mathbb{H}$ for some $k\in \mathbb{Z}$, by looking at how the lift needs to vary along $I$. But then at least one of $k,k+1$ is nonzero, say $k$. Now I claim there is simply no section to the map $\exp: \Im\mathbb{H}\to S^3$, restricted to a neighbourhood of the sphere of radius $k\pi$ on the left and a neighbourhood of $(-1)^k$ on the right. Indeed, a continuous section would take $(-1)^k$ to a single point in the sphere of radius $k\pi$, but since the (restricted) map is a homeomorphism away from the sphere of radius $k\pi$, looking at a sequence converging to any other point contradicts continuity of the section.

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    $\begingroup$ You can even do this with just a neighborhood of $-1$ since zero isn't odd so $-1$ can't lift to $0$. $\endgroup$
    – Will Sawin
    Feb 28 at 0:19
  • $\begingroup$ Ah, good point! $\endgroup$ Feb 28 at 3:55
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In fact I get the full answer. Since the degree determines the homotopy because $\pi_3(S^3)=\mathbb Z$. Then a degree zero map is homotopic to a constant. You can find a sequence of homotopy $H_1,\dots, H_n$ such that each $H_i$ has support in the sphere minus a point and $H_1\circ\dots\circ H_n$ defines the full homotpy. Then you can lift each $H_i$ and finally the initial map is equal to some $\exp(\phi_1)\dots \exp(\phi_n)$. But now since the quaternions don't commute the product of exponential is not an exponential. So the fundamental reason why the result is not true is not the fact that the exponential is not a covering but the fact that product of exponential is not an exponential. Thanks every body for your help.

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