$S^3$ is a $3$ -surface as $f(x_1,x_2,y_1,y_2)=x_1^2+x_2^2+y_1^2+y_2^2$ gives $f^{-1}(1)= S^3$ and for every point of $S^3$ is a regular point, i.e., $\nabla f\ne 0 $ at every point of $S^3$.

Since the tangent space to $S^3$ at $p$, denoted $T_pS^3$ equals $\nabla f(p)^\perp$. One can define $X_i(x_1,x_2,x_3,x_4)$ to lie in $\nabla f(p)^\perp, i=1,2,3$.

$\nabla f(p)^\perp$ is of dimension $3$ and hence has $3$ linearly independent vectors in it, but I'm having difficulty in finding these linearly independent vectors. Hit and trial gives: $X_1(x_1,x_2,y_1,y_3)= (y_1,y_2,-x_1,-x_2), X_2(x_1,x_2,y_1,y_3)=(-x_2,x_1,-y_2,y_1), X_3(x_1,x_2,y_1,y_3)=(-y_2,-y_1,x_2,x_1)$ but these are $X_i$'s are not linearly independent when $x_1=0$. How do I get $X_i$'s such that $X_i(*)$ are linearly independent for every $*\in S^3$?


1 Answer 1


Identify $S^3$ with the unit quaternions, which makes them a Lie group under quaternion multiplication. Calculate the tangent space to $1$, which is a $3$ dimensional real vector space, and pick a basis for this space. Now since you're dealing with a Lie group, the pushforward of this basis under the multiplication map $q: S^3 \to S^3$ allows you to extend this vector field smoothly to the whole sphere, and they will be linearly independent everywhere, since the multiplication map is a diffeomorphism.

I don't know what specific fields this will give you when you're done, since I haven't done this calculation in a long time (and also it depends on the choice of basis). If you're still stuck, I'll expand this answer to include a calculation later.

  • $\begingroup$ Thanks a lot for the answer. So far, I haven't studied Lie groups in detail, and I think that these are not expected to be used for answering this question. Is there any other easier way? If not, then I'll just wait till I understand Lie groups in detail so that I can understand this answer. $\endgroup$
    – Koro
    Aug 17, 2023 at 20:37
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    $\begingroup$ All that is needed is the derivative of the left-multiplication map. This hardly requires a detailed study of all the intricacies of Lie group theory. $\endgroup$ Aug 17, 2023 at 20:41
  • $\begingroup$ @Koro You're basically just considering the maps $z \mapsto pz$, where $p \in \{i, j, k\}$. (more generally, choose a basis $\{u_1, u_2, u_3\}$ to the tangent space at $1$, and show the map $X_i: z \in S^3 \mapsto zu_i \in T_z S^3$ is well-defined). It's a direct generalization of the circle case, where multiplication by $i$ (coming from the fact the circle is the unit complex numbers) induces the global vector field to prove parallelizability. $\endgroup$ Aug 17, 2023 at 20:44
  • $\begingroup$ It's probably also worth noting the same technique of multiplying by a basis of unit quaternions extends to the parallelizability of $S^7$ via the octonions, even though the octonions lack a Lie group structure. Proving it works is trickier though, since you have to actually check it works rather than appealing to Lie theory to say we can translate $\endgroup$ Aug 17, 2023 at 20:51
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    $\begingroup$ You wrote ... I think that these are not expected to be used for answering this question... That's what's so exciting about mathematics --- things you don't expect to be relevant turn out to be exactly what you need to give an interesting and enlightening solution to a problem. $\endgroup$
    – Lee Mosher
    Aug 17, 2023 at 21:04

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