Rank and smooth section of a tautological vector bundle on $\mathbb{C}P^{1}$

The tautological vector bundle on $\mathbb{C}P^{1}=S^2$ is given by $L=\{(l,v) \ \vert \ l \subset \mathbb{C}^{2},\ \dim_{\mathbb{C}}l=1,\ v \in l\}$. How do I find it's rank as a real vector bundle; and also, a smooth section $V \colon S^2 \rightarrow L$ having exactly one point $a \in \mathbb{C}P^{1}$ such that $V(a)=0$.

I'm obviously not looking for an out and out answer; hoping to get some hints or big and small nudges to help me.

For the first question, the weakest hint I can think to say pretty much gives away the answer: as a real vector space $\mathbb{C}$ is two dimensional. Rank of a bundle is dimension of the fiber. The fiber is all the vectors $v$ in a line $\ell$, which is by hypothesis 1 (complex) dimensional

For the second question: the are no holomorphic non-zero sections of this bundle. There is a meromorphic section of the form $1/z.$ Maybe this can help construct a smooth section.

Edit: here's an interlude on sections of the tautological bundle $L=\mathcal{O}(-1).$ Now $P^1$ is the set of lines through the origin in $\mathbb{A}^2.$ We may represent a line with homogeneous coordinates like $[Z_0:Z_1],$ with the understanding that $[Z_0:Z_1]=[\lambda Z_0:\lambda Z_1],$ for $\lambda\neq 0.$ If you scale both coordinates you still have the same line.

Now a section of the tautological bundle $\mathcal{O}(-1)$ is a function $s\colon P^1\to \mathcal{O}(-1)$ with $s(\ell)=v,$ choosing a vector from each line $v\in\ell$.

So in homogeneous coordinates, an element $v$ in the line $\ell=[Z_0:Z_1]$ in $\mathbb{A}^2$ is a vector of the form $v=f\cdot(Z_0,Z_1),$ for some scalar $f.$ We write $s([Z_0:Z_1])=f(Z_0,Z_1)(Z_0,Z_1),$ where now we let $f$ be a scalar function depending on the homogeneous coordinates $Z_0$ and $Z_1$. For this function to be well-defined, we must have

$$ s([\lambda Z_0:\lambda Z_1])=f(\lambda Z_0,\lambda Z_1)(\lambda Z_0,\lambda Z_1) = \lambda f(\lambda Z_0,\lambda Z_1)(Z_0,Z_1)\\ =s([Z_0:Z_1])=f(Z_0,Z_1)(Z_0,Z_1) $$ for all $(Z_0,Z_1)$. So $$f(\lambda Z_0,\lambda Z_1)=\frac{1}{\lambda}(Z_0,Z_1).$$

In other words, $f$ is a homogeneous function of degree $-1.$ A generic form of degree $-1$ looks like $f(Z_0,Z_1)=\frac{1}{aZ_0+bZ_1}.$ Which is the reason for the notation $\mathcal{O}(-1)$ (it is the set of homogeneous forms of degree $-1$). So the generic section is

$$ s([Z_0:Z_1])=\frac{1}{aZ_0+bZ_1}(Z_0,Z_1). $$

Up to a change of coordinates, we may as well take it to be

$$ s([Z_0:Z_1])=\frac{1}{Z_0}(Z_0,Z_1). $$

In affine coordinates, we put $z=Z_0/Z_1$ and we have

$$ s(z:1)=\left(1,\frac{1}{z}\right). $$

And this is what I mean when I said the generic section of the tautological bundle looks like $1/z.$

Note in particular that, similar to the function $1/z$, this section has a pole at $z=0$ (or $[0:1]$). Also it has no zeros at all (unlike the function $1/z$, which has a zero at $z=\infty$ (i.e. $[1:0]$)).

It is a consequence of some basic theorems of complex geometry that all holomorphic sections are of this form (or rather, the only holomorphic section is the zero section, and any meromorphic section must have at least one pole, or number of poles minus number of roots is one). In particular, there is no holomorphic everywhere section with exactly one zero.

But your question didn't ask for holomorphic sections. It asked for smooth sections. Smooth sections are much less constrained, so there is some hope of finding one with no poles and exactly one zero. One way to construct a smooth section is using a partition of unity.

Given an open cover $\mathcal{U}$ of a compact or even paracompact space, we can find a collection of functions $\{\rho_U\}_{U\in\mathcal{U}}$ such that for every point $x\in X$ we have only finitely many $U$ with $\rho_U(x)\neq 0$, and $\sum_{U\in\mathcal{U}}\rho_U(x)=1,$ and $\text{supp} (\rho_U)\subseteq U.$

For $P^1$, our partition of unity can be very explicit. Put $$\rho(z)=\begin{cases}\exp{\left(\frac{1}{1-1/\lvert z\rvert^2}\right)} & \lvert z\rvert < 1\\0 & \lvert z\rvert \geq 1\end{cases}.$$

This function has the property that $\rho(0)=1$, $\rho(z)=0$ for $\lvert z\rvert\geq 1$, and it is smooth everywhere. It is extremely not holomorphic, however. The complement $(1-\rho)$ is also smooth and highly non-holomorphic, but instead vanishes at $z=0$, and equals $1$ for $\lvert z\rvert\geq 1$.

Given sections of any bundles over the two affine patches of $P^1$, $s_1$ and $s_2$, we can hopefully construct locally, where whatever bundle is trivial, we can use these functions to construct a global section $s_1\rho+s_2(1-\rho).$

In our case, it's pretty straightforward. Our answer is just $V=(1-\rho)s$. Because this section no longer has a pole at $z=0$ because of the zero of the mollifier bump function at $z=0.$ And $(1-\rho)$ approaches $0$ exponentially, far faster than $1/z$ approaches $\infty$, so $(1-\rho)$ will dominate $s$ in the product, so $V$ will still have a zero at $z=0.$ And since $s$ has no zero anywhere else, and $(1-\rho)>0$ away from $z=0$, therefore neither does $V.$ These are some facts you can check yourself using basic calculus.

So the final answer is

$$ V(z)=(1-\rho(z))s(z)=\begin{cases}\left(1,\frac{1}{z}\right)\left(1-\exp{\left(\frac{1}{1-1/\lvert z\rvert^2}\right)}\right) & \lvert z\rvert < 1\\\left(1,\frac{1}{z}\right) & \lvert z\rvert \geq 1\end{cases}. $$

There is exactly one point where $V$ vanishes, which is $z=0.$

There is another way to get a section of the tautological line bundle $L=\mathcal{O}(-1).$

When a finite dimensional vector space $V$ is endowed with an a nondegenerate bilinear form (such as a real inner product $g$), this induces a natural isomorphism between $V$ and its dual space $V^*$, given by $v\mapsto g(\cdot,v).$

As an aside, this is sometimes called the musical isomorphism because when you write your vectors in a basis using the Einstein summation convention, vector components have superscript indices, dual vector components have subscript indices, and the isomorphism and its inverse can be viewed as raising and lowering the indices, using the metric tensor. There is a somewhat whimsical convention to use the musical symbols $\sharp$ and $\flat$ for the operator that raises and lowers indices, respectively (since in music they denote raising and lowering pitch).

So anyway, any inner product space is isomorphic to its dual. For vector bundles, this isomorphism applies fiber-wise, so any vector bundle with inner product is also isomorphic to its dual bundle.

And while the tautological line bundle $L=\mathcal{O}(-1)$ has no holomorphic global sections (only meromorphic), its dual bundle is $L^*=\mathcal{O}(1)$, the space of linear forms on $\mathbb{C}^2,$ which does have global sections.

Wait a second, how can it be that $\mathcal{O}(-1)$ has no global sections, but $\mathcal{O}(1)$ has many, if the two bundles are isomorphic??

Well a real inner product is nondegenerate bilinear form, but $\mathcal{O}_{P^1(\mathbb{C})}(-1)$ is a complex vector bundle. An inner product on a complex vector space is not bilinear, but rather sesquilinear or hermitian, meaning linear in one argument, and conjugate-linear in the other. (This is required to make the associated norm real and positive definite). A nondegenerate hermitian form $h$ induces a natural isomorphism $V\overset{\cong}\to \bar V^*$ given by $v\mapsto h(\cdot,v)$, or alternately $\bar V\overset{\cong}\to V^*$ given by $v\mapsto h(v,\cdot)$ (where we adopt the convention that $h$ be conjugate-linear in its first argument, linear in its second). That is a way to say that the isomorphism between $V$ and $V^*$ induced by $h$ is conjugate-linear.

So $\mathcal{O}(1)$ and $\mathcal{O}(-1)$ are isomorphic, but the isomorphism is conjugate-linear. That is not an isomorphism in the category of holomorphic vector bundles. But it is an isomorphism of the underlying real vector bundles. The image of a global section of $\mathcal{O}(1)$ under this isomorphism will be a global section of $\mathcal{O}(-1).$ It just won't be holomorphic. In fact it will be antiholomorphic. Which means it will still be continuous and smooth.

Enough with the generalities and background theory, let's compute the section.

So $\mathcal{O}(1)$ is the space of linear functionals on lines in $\mathbb{C}^2$. A generic linear functional is a map like $f(Z_0,Z_1)=aZ_0+bZ_1.$ We may as well take our functional to be $f(Z_0,Z_1)=Z_0.$ This is a global section of $\mathcal{O}(1)\to P^1(\mathbb{C})$ because for each line $\ell\in P^1(\mathbb{C})$, $f$ restricts to a linear function on $\ell$, which is also the fiber over $\ell$ in $\mathcal{O}(-1).$ In other words, $f$ is an element of $\ell^*$, hence an element of $\mathcal{O}(1)$ in the fiber over $\ell.$

And $\mathcal{O}(-1)$ inherits its hermitian inner product from $\mathbb{C}^2$ (see for example John Ma's answer here or Gunnar's). We have a section of $\mathcal{O}(-1)$ given by $s([Z_0:Z_1])=\frac{1}{Z_0}(Z_0,Z_1)$. The inner product is $h(s,s)=\frac{\lvert Z_0\rvert^2+\lvert Z_1\rvert^2}{\lvert Z_0\rvert^2}.$

There is a corresponding inner product $\tilde{h}$ on $\mathcal{O}(1)$, given by $\tilde{h}(f,g)=h(u,v),$ where $u$ is the vector such that $h(u,\cdot)=g$, and $v$ such that $h(\cdot,v)=f.$

In our case, $f(Z_0,Z_1)=Z_0$, and so $f(s)=1.$ What vector $\lambda s$ will satisfy $f(s)=h(\lambda s,s)$? Let's solve for $\lambda.$

$$1=f(s)=h(\lambda s,s)=\lambda\frac{\lvert Z_0\rvert^2+\lvert Z_1\rvert^2}{\lvert Z_0\rvert^2}$$

hence

$$\lambda=\frac{\lvert Z_0\rvert^2}{\lvert Z_0\rvert^2+\lvert Z_1\rvert^2}.$$

Thus the section we seek, the image of the linear functional $(Z_0,Z_1)\mapsto Z_0$ under the isomorphism between a bundle and its dual, is $$V=\lambda s=\left(\frac{\lvert Z_0\rvert^2}{\lvert Z_0\rvert^2+\lvert Z_1\rvert^2}\right)\frac{1}{Z_0}(Z_0,Z_1)=\frac{ \bar{Z_0}}{\lvert Z_0\rvert^2+\lvert Z_1\rvert^2}(Z_0,Z_1).$$

In the affine coordinate $z=Z_0/Z_1$ it becomes $$V=\lambda s=\bar{z}\frac{(z,1)}{1+\lvert z\rvert^2}=\frac{(\lvert z\rvert^2,\bar{z})}{1+\lvert z\rvert^2}.$$

Note that at $z=0$, this section $V$ has a zero, and as $z\to \infty$, $\lambda s\to (1,0).$ Thus $V$ has exactly one zero, and no poles, and is smooth everywhere (but not holomorphic).