Dtyjlui

Let $M$ a manifold and $T_{p}M$ the tangent space at $pin M$. 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 $$begin{cases}dot x=y\ dot y=-xend{cases},$$
what gives $(Asin(t)-Bsin(t),Acos(t)+Bsin(t))$, but how does it work in a more general manifold ? For example what are the integral curves of $$V=xpartial_y-ypartial _x ?$$

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

$TM = displaystylebigcup_{pin M}T_pM$ : 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 $p:TMto M$ that sends any $vin T_xM$ to $x$ (this is well-defined, as given $vin TM$ there is a unique $xin M$ such that $vin T_xM$).

A vector field will be a section of that canonical projection, that is a map $s:Mto TM$ such that $pcirc s = id_M$. Now let's see what this equation means : start from a point $xin M$, 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 $xin M$ 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 $tin (a,b)$, 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 $M=mathbb{R}^2$, $TM cong mathbb{R}^2times mathbb{R}^2$ 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 $mathbb{R^2to R^2}$ (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 :

$begin{cases}dot c_1=c_2\ dot c_2=-c_1end{cases}$

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 $V=xpartial_y - ypartial_x$ becomes $pi_1partial_2 - pi_2partial_1$. 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

$begin{cases}dot c_1=-c_2\ dot c_2=c_1end{cases}$

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 $f partial_1 + gpartial_2$, where $f,g: mathbb{R}^2to mathbb{R}$ 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