If $V$ is a vector field, how to interpret it when it's written in the tangent space ? For example, what are the integral curves of $V=\partial _x$?

Question

Let $M$ a manifold and $T_{p}M$ the tangent space at $p\in M$. I'm not sure how to interpret a vector field in $M$. I know that a vector field $V$ is in $TM$ and $V(p)\in T_pM$. In $\mathbb R^n$ a vector field can be written as $F(x_1,...,x_n)= (f_1(x_1,...,x_n),...,f_n(x_1,...,x_n))$ where $f_i$ are scalar field, but in a general manifold I'm a bit confused. For example in $\mathbb R^2$, the integral curves of $f(x,y)=(y,-x)$ are the solution of the system $$\begin{cases}\dot x=y\\ \dot y=-x\end{cases},$$ what gives $(A\sin(t)-B\sin(t),A\cos(t)+B\sin(t))$, but how does it work in a more general manifold ? For example what are the integral curves of $$V=x\partial_y-y\partial _x \ \ ?$$

It's probably very easy but I'm not used to see vector field as derivatives functions.

Answer 1

$TM = \displaystyle\bigcup_{p\in M}T_pM$ : the tangent bundle is the union of the tangent spaces, with a suitable differential structure.

This union is disjoint, and so there is a canonical projection $p:TM\to M$ that sends any $v\in T_xM$ to $x$ (this is well-defined, as given $v\in TM$ there is a unique $x\in M$ such that $v\in T_xM$).

A vector field will be a section of that canonical projection, that is a map $s:M\to TM$ such that $p\circ s = id_M$. Now let's see what this equation means : start from a point $x\in M$, apply $s$. You get a vector $s(x)$. The equation tells you that $p(s(x))= x$. But recall the definition of $p$: this means that $s(x) \in T_xM$.

So a vector field on $M$ is just a map $s$ that assigns to each point $x\in M$ a vector $s(x)$ in the tangent space at $x$.

Now if you have a vector field $V$ and a path $c: (a,b)\to M$ you can look at two things: for each $t$ you have two tangent vectors at $c(t)$ that are interesting; one is $V(c(t))$ and the other is $d_tc(1) = c'(t)$ . This one comes from the fact that for each $t\in (a,b)$, there is a canonical identification between $T_t(a,b)$ and $\mathbb{R}$, so we can canonically see $d_tc: \mathbb{R}\to T_{c(t)}M$, so $d_tc(1)$ makes sense.

To get a bit of a feel of what this does, you may look at your example, where the fact that $f$ is a vector field is a bit hidden. Indeed, when $M=\mathbb{R}^2$, $TM \cong \mathbb{R}^2\times \mathbb{R}^2$ where $p$ becomes the projection on the first two coordinates, so that (it's illuminating to try and see why for yourself) a vector field on $\mathbb{R}^2$ is essentially the same as a map $\mathbb{R^2\to R^2}$ (note: the codomain of this map should be seen as the last two coordinates of the tangent bundle, i.e. if $f$ is such a vector field you should think of $(x,y)$ as a point, but $f(x,y)$ as a vector).

So if you have a path $c:(a,b)\to \mathbb{R}^2$, $c(t) = (c_1(t),c_2(t))$, then our two vectors of interest at the time $t$ are $c'(t) = (c_1'(t),c_2'(t))$ (which you should think of as a vector) and $f(c(t))= (c_2(t), -c_1(t))$.

So now if you are on a general manifold with a vector space $V$, you may want to look at paths where the two "vectors of interest" are the same at each time: you're looking for $c$ such that $V(c(t))= d_tc(1)$ for all $t$. This is just the analog of a differential equation, for instance in your example, you get $(c_1'(t),c_2'(t)) = (c_2(t),-c_1(t))$, so in other words :

$\begin{cases}\dot c_1=c_2\\ \dot c_2=-c_1\end{cases}$

which is exactly your equation (with $x=c_1, y=c_2$).

So in general, looking for such a $c$ is the same thing as trying to solve a differential equation; and a total solution is what you call an "integral curve".

Now on $\mathbb{R}^2$, you have specific vector fields, $\partial_x$ and $\partial_y$ - but I don't like those notations; let me write them $\partial_1$ and $\partial_2$ instead. These generalize to $\mathbb{R}^n$ with $\partial_1,...,\partial_n$. What they do is very simple : they're constant vector fields, so $\partial_1 (x,y) = (1,0)$ and $\partial_2 (x,y) = (0,1)$.

Then you also have specific functions on $\mathbb{R}^2$, the coordinate functions (which again generalize to $\mathbb{R}^n$), which you denoted by $x,y$, but I still don't like those notations, so let me denote them by $\pi_1,\pi_2$.They just pick out the right coordinate : $\pi_1(x,y) = x, \pi_2(x,y) = y$ (see why I don't like the notation $x,y$ ? )

So with my notations, your vector field $V=x\partial_y - y\partial_x$ becomes $\pi_1\partial_2 - \pi_2\partial_1$. And now if $c:(a,b)\to \mathbb{R^2}$, $c=(c_1,c_2)$ is a path, for it to satisfy the differential equation associated to $V$ means that $V(c(t))=(c_1'(t),c_2'(t))$ for all $t$; but $V(c(t)) = \pi_1(c_1(t),c_2(t)) \partial_2(c(t)) - \pi_2(c_1(t),c_2(t)) \partial_1(c(t)) = c_1(t) (0,1) - c_2(t) (1,0) = (-c_2(t), c_1(t))$, so the equation simply becomes

$\begin{cases}\dot c_1=-c_2\\ \dot c_2=c_1\end{cases}$

So it's almost the same as your example.

It turns out that the tangent bundle of $\mathbb{R}^2$ is very simple so that a vector field in the form is always of the form $f \partial_1 + g\partial_2$, where $f,g: \mathbb{R}^2\to \mathbb{R}$ are $C^\infty$ (or $C^k$, depending on the regularity you want to impose on the vector fields), but for more general manifolds you have more complicated vector fields