2
$\begingroup$

I am reading do Carmo's Riemannian Geometry, and he defines the tangent bundle $TM$ of a differentiable manifold $M$ as $\{ (p, v) : p \in M, v \in T_pM\}.$ That's fine. Next, he defines (on page 25):

A vector field $X$ on a differentiable manifold $M$ is a correspondence that associates to each point $p \in M$ a vector $X(p)$ $\in T_pM$. In terms of mappings, $X$ is a mapping of $M$ into the tangent bundle $TM$. The field is called differentiable if the mapping $X: M \to TM$ is differentiable.

So on one hand, he is saying that $X(p) \in T_p M$, which means that $X(p)$ is some tangent vector $v$. However, he then says that $X$ maps into $TM$, which means that $X(p) = (p, v)$ for some $v \in T_p M$. Am I missing something, or is he contradicting himself?

$\endgroup$
12
  • 3
    $\begingroup$ What he says is fine; $X$ is a map from $M$ to $TM$ which satisfies the property that $X(p) \in TM_p$, the tangent space of $M$ at $m \in M$. When you say $X(p)=v$, that is a point in $T_pM$, but if you view it as a point in $TM$, then one normally write it as $(p,v)$, as otherwise it is not clear what tangent space it is to be understood to lie in. $\endgroup$
    – krm2233
    Commented Sep 27, 2023 at 21:50
  • $\begingroup$ Each point $p$ has its own tangent space $T_p$. This is just saying that a vector field assigns a tangent vector to each point in the manifold. An even more compact way to say it is that a vector field is a section of the tangent bundle. They all mean the same thing as the first definition. $\endgroup$
    – John Douma
    Commented Sep 27, 2023 at 21:52
  • 1
    $\begingroup$ It's a common slight abuse of notation in the field, no surprise you've noticed it if you like formal syntax/langauges (like me). Sometimes geometers write tangent vectors as $v\in T_xM$, sometimes $(p,v)\in TM$, but they refer to the same object. Depends on what you want to stress. The context always makes it clear so you shouldn't have difficulties understanding the material because of it. Don't get too picky about it :) $\endgroup$
    – Al.G.
    Commented Sep 27, 2023 at 21:58
  • 1
    $\begingroup$ @user56202 do Carmo is trying to informally introduce a vector field when he writes "it is a correspondence that associates to each point $p \in M$ a tangent vector $v\in T_pM$." He then gives a formal definition saying it is a function $X\colon M \to TM$ with the property that $X(p) = (p,v) \in T_pM \subseteq TM$. At worst he is using a standard abuse of notation. $\endgroup$
    – krm2233
    Commented Sep 27, 2023 at 22:03
  • 1
    $\begingroup$ @Al.G. Like with a lot of mathematical notation, the standard abuses are context-dependent: if you say "tangent vector $v$" and it is clear from context that you were referring to a vector in a particular tangent space $T_pM$, then writing "$v$" is sufficient. If you said out of the blue, "suppose $v$ is tangent to the sphere $S^2$", then I'd probably ask "tangent where?", and you would have to write $v \in T_pS^2$, or just $(p,v) \in TS^2$. $\endgroup$
    – krm2233
    Commented Sep 27, 2023 at 22:14

1 Answer 1

1
$\begingroup$

Syntactically, if $X:M\to TM$ is a vector field and $p\in M$, then $X(p)=(p,v)$ for some $v\in T_p M$ because $TM$ is a disjoint union of all $T_p M$'s, i.e. formally an object like $\bigcup_p \{p\}\times T_p M$ (but if you write it like this you should specify the proper topology that is meant).

But remember that $X$ is a section, i.e. $\pi \circ X=\textrm{id}$. This means that $X$ always sends $p$ to some pair $(p,\cdot)$, so the first element of it is readily determined. So in some sense it is redundant to write $X(p)=(p,v)\in \{p\}\times T_p M$ and we just write $X(p)=v\in T_p M$, slightly abusing notation for brevity.

It's nice to see someone else got puzzled by this too (like I was in the beginning). Rest assured, it's easy to get over it, as it never really causes any problems.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .