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I'm trying to make sense of what restricting a vector field means.

More specifically, if we have $S^3$ as a submanifold of $\mathbb{R}^4$ and the vector field, say, $X=(x_4,-x_2,x_3,-x_1)$, what does $X\restriction{S^3}$ mean?

My confusion lies in the fact that $T_p(S^3)$ is not the same thing as $T_p(\mathbb{R^3})$, even though the former can be viewed as a subspace (via $dj$, the differential of the inclusion map) of the latter. $X$ maps each point $p$ of $\mathbb{R}^3$ to a tangent vector in $T_p(\mathbb{R^3})$, so it's not clear to me how this can be made into a mapping from $S^3$ into $T_p(S^3)$.

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Short answer: (in general) you can't.

Long answer: vector fields can be pushed forward and differential forms pulled back. Given an embedding of one manifold into another: $f: \Sigma \to M$, you can map vector fields on $\Sigma$ to vector fields on (the image of $\Sigma$ in) $M$ using the differential map.

$$ df(p) : T_p\Sigma \to T_{f(p)}M $$

and you can map one-forms on $M$ to one-forms on $\Sigma$

$$ f^*(p): T_{f(p)}^*M \to T_p^*\Sigma $$

by $(f^*\omega)(v) = \omega(df(v))$.

So given the inclusion map $\iota: \mathbb{S}^3\to \mathbb{R}^4$ you cannot canonically map vector fields on $\mathbb{R}^4$ to $\mathbb{S}^3$.


There are, however, ways to get around this problem.

  1. Since $f$ is an embedding, we have that $df$ is injective, and hence a bijection between $T_p\Sigma$ and its image. So if you have a priori a vector field $v$ which lives in the image of $T_p\Sigma$ under the map $df$, you can of course "restrict" it to $T_p\Sigma$. (A linear map that is injective but not surjective is not invertible in general, but it can be inverted if you restrict the codomain to its image.) This is precisely the specific case in your question.
  2. Another case where you can do such a thing is when you have a Riemannian structure on $\Sigma$ and $M$, and $f$ is an isometric embedding. Then you can exploit the now canonical identification between tangent and cotangent vectors to "pull back" a vector field. This is in fact equivalent to first taking the orthogonal projection of a vector field on $M$ to the (image of the) tangent space $T\Sigma$ using the Riemannian metric, and then using point 1 above.

So for your specific question:

Fixing a point $p$, at most one of $x_1, x_2, x_3, x_4$ is degenerate, so the rest can form a coordinate system in a neighborhood of $p$. Assume that they are $x_1, x_2, x_3$. The map is $$ (x_1, x_2, x_3) \mapsto (x_1,x_2,x_3, \sqrt{1 - x_1^2 - x_2^2 - x_3^2})$$ so the differential is $$ d\iota(x_1,x_2,x_3): \sum y^i \partial_i \mapsto \sum_1^3 y^i\partial_i - \frac{\sum_1^3 y^ix_i}{\sqrt{1 - x_1^2 - x_2^2 - x_3^2}}\partial_4 $$ which implies that the image of the differential map is precisely the subspace of $T\mathbb{R}^4$ that is in the kernel of $\sum_{i = 1}^4x_i dx_i$. Since $(x_4, -x_3, x_2, -x_1)$ does satisfy this condition, along points on the image of $\mathbb{S}^3$ it can be written as the image of a vector field on $\mathbb{S}^3$ under the inclusion map.

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