Let $M$ be a manifold and $X,Y$ be vector fields on $M$. The bracket $[X,Y]:=XY-YX$ is a vector field when $X,Y$ are smooth, but why is $XY$ not a vector field when $X,Y$ are smooth?

By definition, a smooth vector field takes a smooth map on $M$ to a smooth map on $M$. So if $f$ is a smooth map on $M$ then $XY(f)$ is $X(Y(f))$ and by hypothesis $Y(f)$ is a smooth map on $M$ so that $X(Y(f))$ is a smooth map on $M$ as well.

  • 2
    $\begingroup$ Is "takes a smooth map to a smooth map" the complete definition of vector field? $\endgroup$ – Hagen von Eitzen Apr 23 '19 at 17:35
  • $\begingroup$ By definition a smooth vector field is a function from $M$ to the set of tangents, and each tangent by definition is a real function on the collection of smooth maps on $M$. @HagenvonEitzen $\endgroup$ – user555729 Apr 23 '19 at 17:38
  • $\begingroup$ In short, $XYf$ involves the 2nd derivative of $f$. $\endgroup$ – Yuval Apr 23 '19 at 17:38
  • $\begingroup$ I think my problem is I don’t know the definition of the product in $XY$, @YuDing $\endgroup$ – user555729 Apr 23 '19 at 17:44
  • $\begingroup$ @User12239 We can say XY is a second order differential operator. YX is also a second order differential operator, but [X, Y]=XY-YX turns out to be a first order differential operator! $\endgroup$ – Yuval Apr 23 '19 at 17:48

You seem to have some confusion about the definition of a vector field. There are two equivalent ways to define vector fields on a manifold $M$.

  1. A vector field is a section of the tangent bundle $TM$. i.e. a smooth map $X\colon M \to TM$ with the property that $pr\circ X = id_M$, where $pr\colon TM \to M$ is projection. (Note, this is not precisely the same as the comment you write above or what is written in your question and I believe it is the source of your confusion).

  2. A vector field is a linear map $X\colon C^\infty(M) \to C^\infty(M)$ with the property that it is a derivation, i.e. $X(fg) = fX(g) + X(f)g$ for all $f,g \in C^\infty(M)$.

To answer your question of why for two vector fields $X$ and $Y$, $XY$ is not a vector field, you should check one of the two definitions. The comment by @Hagen von Eitzen hints to use the second definition (which should be the easier definition to use for this question).

In particular, the product of vector fields can be defined two ways as well, depending on which definition of vector fields you are using. What you wrote in your post, that $XY(f) = X(Y(f))$ makes sense using definition 2: you are composing two linear maps, $$ C^\infty(M) \rightarrow^Y C^\infty(M) \rightarrow^X C^\infty(M).$$ $$ f \, \mapsto\, Y(f) \, \mapsto \,X(Y(f))$$ Now, to see if this linear map fits definition 2 of a vector field you need to check that...

If you want to translate to definition 1, you have to do a little bit of work.

  • $\begingroup$ Thanks I will review the definition, I didn’t have in mind it is a derivation. Yes it is a derivation because tangents are derivations. $\endgroup$ – user555729 Apr 23 '19 at 18:06
  • $\begingroup$ It is very useful to know both definitions and why they are equivalent. After answering your question using definition 2, I recommend trying to answer it with definition 1 as well. $\endgroup$ – AnonymousCoward Apr 23 '19 at 18:11
  • $\begingroup$ The book that I’m reading hasn’t yet defined the tangent bundle although I know a little about it. Once my book reaches that section I’ll check the equivalence of these two definitions. $\endgroup$ – user555729 Apr 23 '19 at 18:14
  • $\begingroup$ Which book is that? $\endgroup$ – AnonymousCoward Apr 23 '19 at 20:10
  • $\begingroup$ It’s a classic by Bishop $\endgroup$ – user555729 Apr 23 '19 at 20:43

$XY$ won't necessarily be a derivation on $M$. For example take $M=\mathbb{R}$, $X=Y=\frac{\partial}{\partial x}$ and $f=g=id_{\mathbb{R}}$. Then $XY(fg)(x)=2$ but $g(x)XY(f)(x)+f(x)XY(g)(x)=0$ for all $x\in\mathbb{R}$.

  • $\begingroup$ Yes I had forgotten that the definition I had in my mind will imply the property of being a derivation $\endgroup$ – user555729 Apr 23 '19 at 22:29

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