# Two questions related to $TS^{2}$ as a holomorphic manifold

We consider $TS^{2}$ as a 2 dimensional holomorphic manifold and fix an explicit holomorphic structure on $TS^{2}$ as it is indicated in the answer of Mike Usher to the following question. We have two questions which are not necessarily related to each others(directly).

1)Is there a holomorphic map $q:TS^{2}\to S^{2}$ such that $(TS^{2}, S^{2}, q)$ has an structure of a holomorphic line bundle or at least topological one dim. complex line bundle or two dim real vector bundle?

2) Consider the natural projection $p:TS^{2}\to S^{2}$. By definition, a holomorphic vector field on $S^{2}$ is a holomorphic map $X:S^{2}\to TS^{2}$ with $p\circ X=Id.$ To what extend these holomorphic vector fields have been classified? In particular what can be said about the nature of singularities and closed orbits of a holomorphic vector field? Can we have a singularity with negative index, ex: saddle point? Can we have a limit cycle for a holomorphic vector field? Is the space of holomorphic vector field a Lie algebra, that is closed under the usual Lie bracket?

Note 1: In the above questions, $S^{2}$ is identified with the one dim complex manifold $\mathbb{C}P^{1}$

Note 2: The second question is weakly motivated by the following facts: Let $F:U\to \mathbb{C}$ be a holomorphic map where $U\subset \mathbb{C}$ is an open subset. Then the index of each singular point of the vector field $\dot x=F(x)$ is non negative. Moreover such vector fields have no limit cycle, because $[F,iF]=0$ on the other hand every two commuting flows share on limit cycles, that is every limit cycles of one is invariant under the other ones. The other reason that $\dot x= F(x)$ has no limit cycle is that for each $t$ the flow $\phi_{t}$ is a holomorphic map Now put $T=\text{the period of limit cycle}$. So $\phi_{T}$ as a holomorphic maps has a curve of fixed points. this implies that $\phi_{T}$ is identicaly equal to the identity map. So a closed orbit lies in a band of closed orbit with the same period. so there is no an isolated closed orbit, i.e. limit cycle.

• Careful: if we take the natural orientation on $TS^2$ induced from that of $S^2$, we get $TS^2=O(2)$. But with the opposite orientation we get $O(-2)$. – Ben McKay Oct 26 '14 at 21:00

This discussion seems to involve two holomorphic structures on the smooth manifold $T S^2$.
Structure 1 Identify $S^2$ with $\mathbb{CP}^1$. Place a complex structure on the real tangent bundle to $\mathbb{CP}^1$ by using the complex structure on $\mathbb{CP}^1$. This is the structure Alex Degtyarev's answer refers to. As Alex says, this is the line bundle $\mathcal{O}(2)$ and has a three dimensional vector space of holomorphic sections. If we write $z$ for a coordinate on $\mathbb{CP}^1$, they are $\frac{\partial}{\partial z}$, $z \frac{\partial}{\partial z}$ and $z^2 \frac{\partial}{\partial z}$. They form a Lie algebra isomorphic to $\mathfrak{sl}_2(\mathbb{C})$. You can see this by direct computation, but a better method is to notice that they are the infinitesimal version of the $SL_2(\mathbb{C})$ action on $\mathbb{CP}^1$. This, the orbit structure is described by one parameter subgroups of $SL_2(\mathbb{C})$ and should come down to some computations with Jordan form.
Structure 2 As in your previous question, identify $TS^2$ with $\{ (x,y,z) \in \mathbb{C}^3 : x^2 + y^2 + z^2 = 1 \}$. This can't be the same as structure $1$ because $\mathbb{CP}^1$ doesn't embed holomorphically into $\mathbb{C}^3$, so there is no place for the zero section to land. Let's call this complex manifold $X$.
There is a holomorphic map $X \to \mathbb{CP}^1$, sending $(x,y,z)$ to whichever of $(x+iy:1-z)$ and $(1+z:x-iy)$ is well defined. (If both are well defined then the equation $(x+iy)(x-iy) = (1+z)(1-z)$ shows they are equal.) This is an affine bundle, meaning that the fibers are $\mathbb{C}$'s and the transition maps are in $\mathbb{C}^{\times} \ltimes \mathbb{C}$. But it doesn't have any sections at all, since $\mathbb{CP}^1$ doesn't embed in $\mathbb{C}^3$, so it isn't a line bundle.
Moreover, you can turn an affine bundle into a line bundle by applying the quotient map $\mathbb{C}^{\times} \ltimes \mathbb{C} \to \mathbb{C}^{\times}$ to the transition functions. (The resulting line bundle is diffeomorphic to the original affine bundle, since all affine bundles have sections in the smooth category.) But I get that the result is $\mathcal{O}(-2) \cong T^{\ast} S^2$, not $T S^2$. Of course, those are the same as smooth manifolds, but they are different as holomorphic manifolds. (As symplectic manifolds, I don't know.)
• what is the explicit vector space structure on each fiber when you define $[x+yi,1-z]? what is the trivializations ? – Ali Taghavi Oct 28 '14 at 16:53 • It was interesting for me that the formula which you mentioned is similar to representation of all projections in$M_{2}(\mathbb{C})$in the matrix form $$\begin{pmatrix} 1-z&x+yi\\x-yi&1+z\end{pmatrix}$$$x,y,z$are real with$x^{2}+y^{2}+z^{2}=1$. – Ali Taghavi Oct 28 '14 at 16:59$TS^2\to S^2$has Euler class$2$; ergo, the line bundle in question is$\mathcal{O}_{\Bbb P^1}(2)$. It's holomorphic sections form a vector space of dimension$3$; they can be thought of as degree$\le2$polynomials on$\Bbb C$. Added in proof: All this is independent of the holomorphic structure on$TS^2$assuming that there is a fibration$TS^2\to S^2$. Indeed, over$\Bbb P^1$, a line bundle is determined by its$\text{degree}=c_1=e$(Euler class), and the latter is the self-intersection of the zero section, which, in turn, is the generator of$H_2(TS^2)=\Bbb Z$. So, everything is in the topology of the oriented manifold$TS^2$. • thanks for the answer. could you please more explain, with some references please? what is "ergo"? what is the algebraic or holomorphic projecting map$g$? Are you considering the first question? – Ali Taghavi Oct 26 '14 at 18:49 • "ergo" is the Latin for "hence". No, I'm not considering the first question, since you are using$q$and$p$, implying that the two are unrelated. However, assuming that you do have a fibration and that$S^2\subset TS^2$is its zero section, the Eiler class (${}=c_1$in this dimension) is merely the self-intersection$[S^2]^2$, so it's independent of the structures involved. (In fact, you don't even need to assume that$S^2$is the zero section, as it's the generator of$H_2(TS^2)$and the square of this class depends on nothing but the orientation.) – Alex Degtyarev Oct 26 '14 at 18:56 • Sorry, concerning your other questions: I take for the projection just the natural projection, but this is irrelevant (see above). For characteristic classes, the keywords are "obstruction theory"; for the dimension, the name would be Riemann--Roch theorem, even though this is an overkill. (So I'm not including this into the answer :) – Alex Degtyarev Oct 26 '14 at 19:07 • I'm not good at that. As I said, the topological part is the obstruction theory (any textbook in algebraic topology that covers this subject): in dimension$\dim_{\Bbb R}=2\dim_{\Bbb C}=2$, both$c_1$and$e$are obstructions to the same thing, viz. existence of a section; hence, they are equal. I do not have a formal education in algebraic geometry; everything about$\Bbb P^1$is very classical but, from the modern point of view, it's Hodge theory ($h^1=\frac12b_1=0$, hence holomorphic classification is the same as topological for line bundles) plus Riemann--Roch for the dimension. – Alex Degtyarev Oct 26 '14 at 19:16 • For the dimension, you can also use common sense: a bundle is determined by its degree$d$, hence sections are merely polynomials of degree$\le d$. All this is covered (I hope) in Hirzebruch's "Topological methods in algebraic geometry", but I'm sure that algebraic geometers will name lots of much more elementary texts about$\Bbb P^1\$. – Alex Degtyarev Oct 26 '14 at 19:18