# Finding 3 linearly independent tangent vector fields on $S^3$

$$S^3$$ 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 $$*\in S^3$$?

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.
• @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. Aug 17, 2023 at 20:44
• 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 Aug 17, 2023 at 20:51