# (Lee 8-12 SM) How do you find a global extension of a vector field defined on an open set?

I have a function $$F: \mathbb{R}^2 \to \mathbb{RP}^2$$ given by $$F(x,y) = [x,y,1]$$, and some vector field $$X$$, and a vector field on $$\mathbb{RP}^2$$ given by $$X(x,y) = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$$.

I want to find a vector field $$Y$$ on $$\mathbb{R}^2$$ that is $$F$$-related to $$X$$.

$$F$$ is not surjective, but nonetheless, I of course need to check how $$Y$$ ought to behave on points that are in $$\mathrm{im \text{ }}X$$.

So, I decided to look at what $$dF$$ pushes vectors to. I chose the chart $$(U, \phi)$$ for $$\mathbb{RP}^2$$ where $$U$$ is the set of points with nonvanishing third entry, and $$\phi([x,y,z]) = (\frac{x}{z}, \frac{y}{z})$$ and found $$dF$$ to be trivial with respect to this coordinate representation.

So, I thought that whatever $$Y$$ should be, it should 'look like $$X$$ in coordinates,' whatever that means.

This is my problem. I do not believe that I understand what I am doing when I define a vector field 'in coordinates.' Let me spew out some words that may or may not clarify my current state of understanding (very little!):

• The above chart is equal to the image of $$F$$. Therefore, I define the vector field $$Y$$ as follows: Given a point $$p \in U$$, look at its coordinate representation $$(x,y)$$. Then, define $$Y(x,y) = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$$ (i.e. exactly what X is).
• This defines a smooth vector field on the image of $$F$$. First of all, it is well defined; I am explicitly choosing a particular chart. Next, it is smooth. If I were to take some other chart instead, the smoothness should fall out of composing diffeomorphisms. (Please, if someone can confirm my understanding here, I will be much obliged. I feel uneasy about doing anything with respect to a particular coordinate representation, even if irrationally)
• Extend this vector field to all of $$\mathbb{RP}^2$$ by some process? Clearly (admittedly, I did not check this part), I cannot simply stipulate that the above characterization $$Y$$ should hold with respect to any chart; it may not be well defined. If the above assumption is correct, what should I do? Define some sort of 'trivial' smooth vector fields on the other charts comprising the atlas, and then paste together with a partition of unity?

Apologies for the menial questions. I do not trust my brain to reason/interpret the textbook correctly when left to my own devices.

• You have a typo. You mean a vector field $Y$ on $\Bbb RP^2$. So what happens to the vector field on $F(\Bbb R^2)$ when you consider a different coordinate patch? If you use coordinates $[1,u,v]$, how are $(x,y)$ and $(u,v)$ related? What is the vector field in the $(u,v)$ coordinates? Commented Apr 4 at 17:52
• Ah I see! I was too pessimistic about whether the obvious extension worked. Commented Apr 5 at 20:50

Let $$\varphi_0 : U_0 \to \mathbb{R}^2$$, $$\varphi_1: U_1 \to \mathbb{R}^2$$, and $$\varphi_2: U_2 \to \mathbb{R}^2$$ be the three standard charts on $$\mathbb{RP}^2$$ (page 6 of Lee). Let's call our coordinates $$[1, u, v]$$, $$[w, 1, z]$$, and $$[x,y,1]$$ respectively.

We will define a global vector field on $$\mathbb{RP}^2$$ by defining local vector fields on each coordinate chart and then gluing them together. We can define local vector fields on the charts of $$\mathbb{RP}^2$$ by defining them as coordinate vector fields on $$\mathbb{R}^2$$ and pulling them back via the $$\varphi_i$$. As you rightly pointed out, since $$F = \varphi_2^{-1}$$ we want to define $$X_2 = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$$ to be our vector field on $$U_2$$ (the image of $$F$$).

Note that $$X_2$$ is not vector field on $$\mathbb{RP}^2$$ as it is not defined at any points of the form $$[X,Y,0] \in \mathbb{RP}^2$$. We want to make sure that our definition of $$X_2$$ can be extended to these other points. We will accomplish this by changing coordinates to the other two charts.

Let's look at $$U_0 \cap U_2$$. Our goal is to rewrite $$X_2$$ in terms of $$(u,v)$$ on this overlap. We have that $$[1,u,v] = [\frac{1}{v}, \frac{u}{v}, 1] = [x, y, 1]$$. Therefore the change of coordinate function is given by $$(\varphi_2\circ\varphi_0^{-1})(u,v) = (\frac{1}{v},\frac{u}{v})$$. Often we write this as $$x = \frac{1}{v}$$ and $$y = \frac{u}{v}$$.

Now we know how to replace $$x$$ and $$y$$ with $$u$$ and $$v$$, but we also need to change coordinates in the tangent vectors. This is covered in Chapter 3 of Lee in the section on changing coordinates (see Example 3.16). We compute that \begin{align*} \frac{\partial}{\partial x} &= -\frac{y}{x^2} \frac{\partial}{\partial u} - \frac{1}{x^2} \frac{\partial}{\partial v}\\ &= -uv \frac{\partial}{\partial u} - v^2 \frac{\partial}{\partial v} \end{align*} and \begin{align*} \frac{\partial}{\partial y} &= \frac{1}{x} \frac{\partial}{\partial u}\\ &= v \frac{\partial}{\partial u}. \end{align*}

Therefore, on $$U_0 \cap U_2$$ we can express $$X_2$$ as \begin{align*} X_2 &= \frac{1}{v}\left(v\frac{\partial}{\partial u}\right) - \frac{u}{v}\left(-uv \frac{\partial}{\partial u} - v^2 \frac{\partial}{\partial v}\right)\\ &= (1+u^2)\frac{\partial}{\partial u} + uv \frac{\partial}{\partial v} \end{align*}

Here is the key step: we define a coordinate vector field on all of $$U_0$$ by $$X_0 = (1+u^2)\frac{\partial}{\partial u} + uv \frac{\partial}{\partial v}.$$

By construction, $$X_0$$ agrees with $$X_2$$ on $$U_0 \cap U_2$$. All of the coordinate functions are smooth in the definitions of $$X_0$$ and $$X_2$$, so $$X_0$$ is a smooth vector field on $$U_0$$ and $$X_2$$ is a smooth vector field on $$U_2$$ by Proposition 8.1 of Lee. Since $$X_0$$ and $$X_2$$ agree on $$U_0 \cap U_2$$, we can glue these smooth vector fields together to get a smooth vector field on $$U_0 \cup U_2$$. The reason we can glue follows from the definition of a vector field as a section of $$\pi: TM \to M$$ (sections that agree on the intersection of their domains glue to give a section on the union of their domains).

I'll leave it to you to follow the same procedure to construct a vector bundle on $$U_1$$ that agrees with $$X_2$$ on $$U_1 \cap U_2$$ and check that the coordinate functions are smooth. This allows you to know that $$X_2$$ really does extend smoothly to all of $$\mathbb{RP}^2$$.

More generally, one can construct a vector field on a smooth manifold $$M$$ by defining coordinate vector fields on each chart of $$M$$ and then checking that we can glue these locally defined coordinate vector fields into a global vector field. One can check that this is equivalent to defining a section of $$\pi: TM \to M$$. By 'checking that we can glue these local sections', what I mean is that if $$\varphi: U \to \mathbb{R}^n$$ and $$\psi: V \to \mathbb{R}^n$$ are two charts of $$M$$ with coordinates $$(x^1, \ldots, x^n)$$ and $$(y^1, \ldots, y^n)$$ respectively, and if $$X_p = X^i(p) \frac{\partial}{\partial x^i}\bigg|_{\varphi(p)}, \ \ \ p \in U$$ and $$Y_p = Y^i(p) \frac{\partial}{\partial y^i}\bigg|_{\psi(p)}, \ \ \ p \in V$$ are the coordinate vector fields we defined on $$U$$ and $$V$$, then on $$U \cap V$$ changing coordinates will not change the vector field. Formally this means that $$X_p = Y_p$$ for all $$p \in U \cap V$$. We check this by using the change of coordinates formula for tangent vectors (page 63 of Lee) $$\frac{\partial}{\partial x^i}\bigg|_{\varphi(p)} = \frac{\partial y^j}{\partial x^i}(\varphi(p))\frac{\partial}{\partial y^j}\bigg|_{\psi(p)},$$ where $$\displaystyle \frac{\partial y^j}{\partial x^i}(\varphi(p))$$ is obtained from the Jacobian of the transition function $$\psi \circ \varphi^{-1}$$. So we are trying to verify that $$X^i(p)\frac{\partial}{\partial x^i}\bigg|_{\varphi(p)} = X^i(p)\frac{\partial y^j}{\partial x^i}(\varphi(p))\frac{\partial}{\partial y^j}\bigg|_{\psi(p)} = Y^j(p)\frac{\partial}{\partial y^j}\bigg|_{\psi(p)}.$$ This is precisely what we checked above to guarantee that the local sections glue: $$Y^j = X^i \frac{\partial y^j}{\partial x^i}.$$