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For vector fields: 3 reasons why I think vector fields are injective

  1. (This is from the undergraduate point of view) For a vector field $X: \mathbb R^n \to \mathbb R^{2n}$ (my undergraduate calculus textbook doesn't give a range for "vector fields", but we know from differential geometry that $\mathbb R^n$ is parallelisable tangent bundle, i.e. $T\mathbb R^n$ is isomorphic as a tangent bundle to $\mathbb R^n \times \mathbb R^n$) , I really don't see a way, geometrically, we have the same $2n$-vector $X_p = X_q$ for different $n$-points $p, q \in \mathbb R^n$, since $X_p$ is supposed to emanate from $p$.

The next two use differential geometry: Let $M$ be a smooth manifold.

  1. Let $M$ have tangent bundle $\pi: TM \to M$. Consider a vector field, which need not be smooth, $X: M \to TM$, we have $\pi \circ X$ as the identity of $M$, which is bijective and thus injective. By this, $X$ is injective.

  2. Alternatively, I think we don't need any (explicit) notion of tangent bundle. Let $X$ be a vector field on $M$ (a map whose domain is $M$ and such that for each $p$ in $M$, $X_p \in T_pM$). Let $p,q \in M$. We have $X_p \in T_pM$ and $X_q \in T_qM$. We don't really have that $T_qM$ and $T_pM$ intersect anywhere as sets even though as $\mathbb R-$vector spaces they are isomorphic to $\mathbb R^n$, where $n$ is the dimension of $M$. Saying that $X_p$ and $X_q$ are equal implies $T_pM$ and $T_qM$ intersect somewhere (namely at $X_p$), which is nonsensical unless $T_pM = T_qM$, which is also nonsensical unless $p=q$.

For vector bundles in general: Are sections (need not be smooth) injective? I think indeed so generalising from the case of tangent bundles.

Thanks in advance!

Ekhin Taylor R. Wilson
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1 Answers1

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If you think of a vector field on $\Bbb R^n$ as a map $f\colon\Bbb R^n\to\Bbb R^n$, of course it need not be injective (the $0$-vector field is the easiest counterexample). If, however, you think of a vector field as a section of $T\Bbb R^n$, then, as you already proved, it's injective. We can see this directly because we're then looking at the map $\sigma\colon\Bbb R^n\to\Bbb R^{2n}$ given by $\sigma(x)=(x,f(x))$, and the identity map in the first coordinate makes the mapping injective.

Ted Shifrin
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  • You said $f\colon\Bbb R^n\to\Bbb R^n$ and not $f\colon\Bbb R^n\to\Bbb R^{2n}$. So the undergraduate vector field is more of $f\colon\Bbb R^n\to\Bbb R^n$ which is different from the one in differential geometry which is more of $f\colon\Bbb R^n\to\Bbb R^{2n}$? – Ekhin Taylor R. Wilson Nov 06 '19 at 00:31
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    As I already said, the vector field in differential topology is of the form $\sigma(x)=(x,f(x))$, i.e., the graph of the calculus student's vector field (in order to keep track of the base point). – Ted Shifrin Nov 06 '19 at 00:40
  • I assume your answer to my question in comment is 'yes'. – Ekhin Taylor R. Wilson Nov 18 '19 at 16:43
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    I still won't think of a section of the trivial bundle as a map $\Bbb R^n\to\Bbb R^{2n}$, because it is specifically the graph of a function. Of course, as such, it is a special mapping as you specify. – Ted Shifrin Nov 18 '19 at 16:45