# 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 $V=\partial _x$?

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.

$$TM = \displaystyle\bigcup_{p\in M}T_pM$$ : 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

• Thank you for your answer and sorry to not answer earlier (it take time for me to understand every thing). So several questions 1) If you have a vector field $V$ and a path $c:(a,b)\to M$, for $t$ fixed there are two vector tangent at $c(t)$ that are interesting : $c'(t)$ or $V(c(t))$. Why only these two are interesting ? There are many different tangent vector at $c(t)$. 2) Why $(x,y)$ is not a good notation for coordinate ? Because in fact this suggest that $x(x,y)=x$ and $y(x,y)=y$ ? (which not good for notation). I will have other question, but I have to think a bit before :) – user623855 Jan 12 at 15:40