# Solutions of equations characterizing a complex structure

Let $(S^n,g)$ denote the unit $n$-sphere endowed with its induced metric $g$ from its embedding into $\mathbb{R}^{n+1}$. The Levi-Civita connection of $g$ induces a splitting of the tangent bundle of $\pi:TS^n\to S^n$ into horizontal and vertical parts as $TTS^n = V\oplus H$, where each summand is canonically isomorphic to $\pi^*(TS^n)$. Thus, every vector field $X$ on $S^n$ can be uniquely lifted to $TS^n$ as either a horizontal vector field $X^h$ or a vertical vector field $X^v$.

Given a triple of (smooth) real-valued functions $(\alpha,\delta,\beta)$ on $TS^n$ satisfying $\alpha \delta - \beta ^2 = 1$, one can define an almost-complex structure $J_{\delta, \beta}$ on $TS^n$ by the conditions $$J_{\delta , \beta}(X^h)=\beta X^h + \alpha X^v,\\ J_{\delta , \beta}(X^v)=-\beta X^v - \delta X^h,$$ where $X^h$ and $X^v$ are the horizontal and vertical lifts of any vector field $X$ on $S^n$.

Question: What are the possibilities for $(\alpha,\beta,\delta)$ if we require that $J_{\delta,\beta}$ be integrable?

Remarks:

(1) I am particularly interested in integrable $J_{\delta,\beta}$ for which $\beta$ is not constant. I am also interested in understanding the integrable $J_{\delta,\beta}$ in which $(\alpha,\beta,\delta)$ are only defined on some open subset of $TS^n$.

(2) I already know some local solutions that take a special form with respect to conformal coordinates on $(S^n,g)$: If $x = (x_1,\ldots,x_n)$ are local conformal coordinates on $U\subset S^n$, so that $g = \lambda^2(x) \sum _{i=1}^n dx_i ‎\otimes dx_i$, let $y_i:TU\to\mathbb{R}$ defined by $y_i(v) = \mathrm{d}x_i(v)$ be the associated tangential coordinates. Then one can compute that a basis for the $(1,0)$-forms for $J_{\delta,\beta}$ on $TU$ are given by $$\zeta_k = \mathrm{d}y_k + y_j(\delta_{jk}\,\mu_l\,\mathrm{d}x_l +\mu_j\,\mathrm{d}x_k-\mu_k\,\mathrm{d}x_j) - z\, \mathrm{d}x_k$$ where the summation convention is assumed and $$\mu_j = \frac{\partial (\log\lambda)}{\partial x_j}\qquad\text{and}\qquad z = \frac{i+\beta}{\delta}.$$ Then $J_{\delta,\beta}$ is integrable if and only if $$\mathrm{d}\zeta_k\equiv0\mod \zeta_1,\ldots,\zeta_n\,.$$ When $n>1$, this works out to be $2n$ first-order partial differential equations for $z$, so the system is overdetermined (and nonlinear). I already know the solutions when one supposes that the function $z$ is assumed to be a function of $E(x,y)=\lambda ^2(x)\sum _{i=1}^n (y_i)^2$. I want to know whether there are any solutions that are not of this form.

Addendum: I have some questions on Robert Bryant's answer:

1. Are our leaves of foliation $‎\lbrace(u,v)‎\rbrace \times H_+$?
2. What is the proof of the proposition in the answer?
3. Why the functions $\delta, \beta$ of the associated complex structure $J_{\delta, \beta}$ (as the associated complex structure for the equation $Z.Z+1=0$) are not functions of $E(x,y)=\lambda^2 \sum_{i=1}^n (y^i)^2$? And are there other types of solutions?
• @Solar: I noticed that you have not responded to the answer that I posted. Do you need more details, have doubts or objections, etc.? Let me know and I will do my best to answer them. Feb 13, 2016 at 0:17
• @Solar: Answers to your questions: (1) Yes. (2) I have entered the proof as Remark 2 below. (3) Actually, for the hypersurface $Z{\cdot}Z+1 = 0$, the functions $\delta$ and $\beta$ are functions of the type you mention (i.e., they are functions of $|v|^2$). However, for the generic hyperquadric that I wrote down, they are not. This actually follows from Remark 1, since the form for $E(x,y)$ that you wrote down is not invariant under the action of the affine group of $\mathbb{R}^{n+1}$, but you can just take an explicit example when $n=1$ and check this directly. Feb 15, 2016 at 15:49
• @Solar: By the way, I just realized and feel that I should point out that there is something wrong with your explicit equations for $u$ and $v$ in the coordinate system $(x,y)$. These are not the differential equations $(u,v)$ must satisfy in order for $J_{\delta,\beta}$ be integrable. The correct equations when $n>1$ (as I show in my answer) are $2n$ in number, but you have written down $3n$ equations, which is too many. Also, it is clearly not right for $n=1$ because, when $n=1$ all $J_{\delta,\beta}$ are integrable complex structures. You need to fix this, because it misleads readers. Feb 16, 2016 at 13:15
• @Solar: You have improved your question because now you are correctly stating that the number of (real) equations on $z$ is $2n$ instead of $3n$, but you still don't have them right because, for example, $z = i$ satisfies your equations, but the corresponding almost complex structure $J_{1,0}$ is not integrable. You seem to be having problems working things out in coordinates. Perhaps you are confusing projective coordinates (which carry geodesics to straight lines) with conformal coordinates. Would you like me to try editing it? Feb 20, 2016 at 4:11
• @RobertBryant Dear Prof. Bryant, I get more than what I expected and I am grateful for your complete answers and explanations. Mar 31, 2016 at 12:40

## 1 Answer

The answer is 'Yes, there are many other solutions, even global ones, and, when $n>1$, the local solutions depend on one holomorphic function of n complex variables.' (Of course, when $n=1$, $J_{\delta,\beta}$ is always integrable.) For a geometric context of this question, see Remark 3 below.

Briefly, here is a description of how they can be understood: Let $\mathbb{R}^{n+1}$ be given its standard inner product (and extend it complex linearly to a complex inner product on $\mathbb{C}^{n+1}$, which will be used below). Then $$S^n = \bigl\{ u\in\mathbb{R}^{n+1}\ \bigl|\ u{\cdot}u = 1\ \bigr\}$$ and $$TS^n = \bigl\{ (u,v)\in\mathbb{R}^{n+1}{\times}\mathbb{R}^{n+1} \ \bigl|\ u{\cdot}u = 1,\ u{\cdot}v = 0\ \bigr\}.$$

Let $H_+ = \{ a+ib\ |\ b>0\ \}\subset\mathbb{C}$ be the upper half-line in $\mathbb{C}$. Define a mapping $$\Phi:TS^n\times H_+\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$$ by $\Phi\bigl((u,v),z\bigr) = v - z\,u$. The mapping $\Phi$ is easily seen to be a diffeomorphism (one-to-one, onto, and smoothly invertible). One can think of $\Phi$ as establishing a smooth (though not holomorphic) foliation of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ by half-lines, one for each point of $TS^n$.

Now, given real-valued functions $(\alpha,\beta,\delta)$ on an open set $U\subset TS^n$ defining an almost complex structure $J_{\delta,\beta}$ on $U$ as per the OP's description, one has $\alpha\delta-\beta^2 = 1$, so without loss of generality, one can replace $J_{\delta,\beta}$ by $-J_{\delta,\beta}$, so one can assume that $\alpha$ and $\delta$ are positive. Set $z = (\beta{+}i)/\delta = \alpha/(\beta{-}i):U\to H_+$. (One way to interpret this is that $TS^n\times H_+$ can be thought of as pairs $\bigl((u,v),z\bigr)$ consisting of a point $(u,v)$ of $TS^n$ and a complex structure (parametrized by $z\in H_+$) on the vector space $T_{(u,v)}TS^n$. The map $\Phi$ is just an identification of this $H_+$-bundle over $TS^n$ with the complex manifold $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$.)

Then one has the following result, easily proved by a moving frames calculation (see Remark 2 below):

Proposition: When $n>1$, the almost complex structure $J_{\delta,\beta}$ on $U\subset TS^n$ is integrable if and only if the image of the mapping $$\Phi_z = \Phi\bigl((u,v),z(u,v)\bigr):U\to \mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$$ is a holomorphic hypersurface in $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$. Moreover, in this case, the map $\Phi_z$ is a holomorphic embedding of $(U,J_{\delta,\beta})$ into $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$.

Note that this Proposition implies that, if $X\subset \mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ is a nonsingular holomorphic hypersurface that is transverse to the half-line foliation determined by $\Phi$ and intersects each such half-line in at most one point, then $X$ is the image of $\Phi_z$ for some $z:U\to H_+$ where $U\subset TS^n$ is an open subset, and thus, writing $z=(\beta{+}i)/\delta$ for real-valued functions $(\delta,\beta)$ on $U$ (with $\delta>0$), the almost complex structure $J_{\delta,\beta}$ is integrable on $U\subset TS^n$.

Thus, when $n>1$, the space of local solutions is identified with the space of holomorphic hypersurfaces in $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ satisfying some open conditions, which, locally, are 'parametrized' by one holomorphic function of $n$ complex variables. (As already remarked, when $n=1$, $J_{\delta,\beta}$ is always integrable.)

Now, there are many holomorphic hypersurfaces in $\mathbb{C}^{n+1}\setminus \mathbb{R}^{n+1}$ that satisfy these conditions. For example, the hyperquadric in $\mathbb{C}^{n+1}$ defined by $Z{\cdot}Z +1 = 0$ has no real points and meets each half-line of the $\Phi$-foliation transversely in a unique point, and this gives a solution of the kind that the OP has already found. More generally, consider a hyperquadric of the form $$Z{\cdot} AZ + B{\cdot} Z + 1 = 0,$$ where $A$ is a complex symmetric $(n{+}1)$-by-$(n{+}1)$ matrix and $B\in \mathbb{C}^{n+1}$ is a complex vector. The conditions needed to guarantee that this hyperquadric has no real points and meets each half-line of $\Phi$ transversely in a single point are open conditions on the pair $(A,B)$, and this set is not empty because it contains the above example.

This already works out to provide a space of globally defined complex structures on $TS^n$ that has (real) dimension $(n{+}1)(n{+}4)$, which is far more than the dimension of the space of solutions that the OP mentioned as already known. The generic one is not of the kind mentioned by the OP.

Remark 1: It turns out that this problem has a much larger symmetry group than one might, at first, suspect. It is well-known that the tangent bundle of $S^n$ can be regarded as the space of oriented lines in $\mathbb{R}^{n+1}$; simply let $L(u,v)\subset\mathbb{R}^{n+1}$ be the oriented line through $v$ with direction $u$. Then the complexification of $L(u,v)$ as a line in $\mathbb{C}^{n+1}$ is divided into two (complex) half-lines, a positive half and and a negative half. The above map $\Phi$ simply gives this foliation of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ explicitly. All of this is equivariant under the group $G$ of affine transformations of $\mathbb{R}^{n+1}$, a Lie group of dimension $(n{+}1)(n{+}2)$. It turns out that the entire problem is invariant under this group. However, even under this larger group, the space of hyperquadric solutions listed above is not homogeneous (i.e., they are not all equivalent) since the dimension of this space is $(n{+}1)(n{+}4)$.

Remark 2: The following is the moving frames calculation I used to prove the Proposition. While it can be done as a bundle calculation without taking a local moving frame, choosing a local moving frame reduces the number of 'auxiliary variables' and so may be easier to understand for the novice. The usual summation convention for repeated indices in a term will be assumed in what follows.

Let $U\subset S^n\subset\mathbb{R}^{n+1}$ be any proper open subset and let $e_0:U\to \mathbb{R}^{n+1}$ denote the (vector-valued) inclusion mapping. Let $e_1,\ldots,e_n:U\to\mathbb{R}^{n+1}$ be any (smooth) orthonormal tangential frame field extending $e_0$, i.e., $e_a\cdot e_b = \delta_{ab}$ for $0\le a,b\le n$.

There exist smooth $1$-forms $\omega_i$ and $\omega_{ij}=-\omega_{ji}$ (where $1\le i,j\le n$) on $U$ satisfying the structure equations $$\mathrm{d}e_0 = e_i\,\omega_i \qquad \text{and}\qquad \mathrm{d}e_i = -e_0\,\omega_i + e_j\,\omega_{ji}\,.$$ The $n$ $1$-forms $\omega_i$ are linearly independent on $U$ and, in fact, form a basis for the $1$-forms on $U$. These forms satisfy the structure equations $$\mathrm{d}\omega_i = -\omega_{ij}\wedge\omega_j \qquad \text{and}\qquad \mathrm{d}\omega_{ij} = -\omega_{ik}\wedge\omega_{kj} + \omega_i\wedge\omega_j\,.$$

Define functions $v_i:TU\to\mathbb{R}$ by $v_i(u,v) = e_i(u){\cdot}v$ for $1\le i\le n$, so that $v = e_i(u)\,v_i$ for all $(u,v)\in TU$ and define $1$-forms $\eta_i$ on $TU$ by $\eta_i = \mathrm{d}v_i + v_j\,\omega_{ij}$. Then the $2n$ $1$-forms $\omega_i$, $\eta_i$ are a basis of the $1$-forms on $TU$. Moreover, they have the property that a vector $w\in T(TU)$ is tangent to the fibers of the basepoint projection $TU\to U$ if and only if $\omega_i(w) = 0$ while $w$ is tangent to the horizontal plane field on $TU$ if and only if $\eta_i(w) = 0$.

Now, the OP defines, for any (say, smooth) triple of functions $(\alpha,\beta,\delta)$ on $TU$ satisfying $\alpha\delta-\beta^2 = 1$, an almost-complex structure $J_{\delta,\beta}:T(TU)\to T(TU)$ by the rules $$\omega_k\circ J_{\delta,\beta} = \beta\,\omega_k - \delta\,\eta_k \quad\text{and}\quad \eta_k\circ J_{\delta,\beta} = -\beta\,\eta_k + \alpha\,\omega_k\,.$$ It is easy to check that this does define an almost-complex structure and that a basis for the $(1,0)$-forms on $TU$ with respect to this almost-complex structure is given by $$\zeta_k = \eta_k - z\,\omega_k$$ where $z = (\beta{+}i)/\delta = \alpha/(\beta{-}i)$ has nonzero imaginary part (in particular, $z$ is nowhere vanishing). Note that the $\mathbb{C}$-valued $1$-forms $\zeta_k$ and $\overline{\zeta_k}$ for $1\le k\le n$ provide a basis for the $1$-forms on $TU$.

Since $J_{-\delta,-\beta} = - J_{\delta,\beta}$, one can always reduce to a case in which $\delta>0$ (and, of course, $\alpha>0$) when investigating integrability. For simplicity, I will assume this from now on, which is equivalent to the condition that $z$ have positive imaginary part.

Now, the condition that $J_{\delta,\beta}$ be integrable is the condition that the ideal generated by $\zeta_1,\ldots,\zeta_n$ be closed under exterior differentiation, i.e., that $$\mathrm{d}\zeta_k \equiv 0\ \text{modulo}\ \zeta_1,\ldots,\zeta_n\,.$$ Using the structure equations, one calculates $$\mathrm{d}\zeta_k = \zeta_j\wedge(\omega_{kj} + (v_j/z)\,\omega_k) - \tfrac{1}{2z}\,\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) \wedge \omega_k\,.$$ Since $\omega_k = (\zeta_k-\overline{\zeta_k})/(\bar z - z)$, this yields $$\mathrm{d}\zeta_k \equiv \frac{\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2)}{2z(\bar z - z)}\wedge \overline{\zeta_k}\ \ \text{modulo}\ \zeta_1,\ldots,\zeta_n\,.$$ Thus (since $n>1$), $J_{\delta,\beta}$ is integrable if and only if $$\mathrm{d}(z^2{+}{v_1}^2{+}\cdots {+}{v_n}^2) \wedge\zeta_1\wedge\cdots\wedge\zeta_n = 0. \tag1$$

On the other hand, consider the mapping $\Phi_z:U\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ defined by $$\Phi_z = v - z u = v_1\,e_1 + v_2\,e_2+ \cdots +v_n\,e_n - z\,e_0.$$ Again, using the structure equations, one computes $$\mathrm{d}\Phi_z = -e_0\,\tfrac{1}{2z}\,\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) + (e_k{+}(v_k/z)\,e_0)\,\zeta_k\,.$$ Write $\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) = p_k\,\zeta_k + q_k\,\overline{\zeta_k}$ for some functions $p_k$ and $q_k$. It follows that the map $\Phi_z:U\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ has a complex linear differential (when $U$ is endowed with the almost-complex structure $J_{\delta,\beta}$) if and only if $q_k = 0$ for all $k$, which, by (1), holds if and only if $J_{\delta,\beta}$ is integrable.

Furthermore, the image of the tangent space at a point of $U$ is the set of vectors of the form $$-e_0\,\tfrac{1}{2z}\,(p_k\,a_k + q_k\,\overline{a_k}) + (e_k{+}(v_k/z)\,e_0)\,a_k$$ where the $a_k$ are complex numbers. But this is a complex subspace of $\mathbb{C}^{n+1}$ if and only if $q_k=0$ for all $1\le k\le n$, i.e., if and only if (1) holds. Thus, the image $\Phi_z(U)$ is a complex submanifold of $\mathbb{C}^{n+1}$ (of complex dimension $n$) if and only if (1) holds. Thus, the Proposition is proved.

Remark 3: This question can be placed in a larger context that partly explains its interest.
The underlying geometry is that of a $2n$-manifold $M^{2n}$ endowed with a splitting of its tangent bundle $TM = S_1\oplus S_2$ into two $n$-dimensional subbundles $S_1$ and $S_2$ together with a bundle isomorphism $\iota:S_1\to S_2$. Given such a structure on $M$ a family of almost-complex structures can be defined on $M$ as in the OP's question: For each mapping $h = (\alpha,\beta,\delta):M\to\mathbb{R}^3$ satisfying $\alpha\delta-\beta^2 = 1$, one has an almost-complex structure $J_h$ that satisfies $$J_h(s) = \beta\,s + \alpha\,\iota(s) \quad\text{and}\quad J_h\bigl(\iota(s)\bigl) = -\beta\,\iota(s) + \delta\,s$$ for any section $s$ of $S_1$.

This can be understood more geometrically in terms of a somewhat weaker structure, namely, a pair of bundles $P\to M$ and $S\to M$ where $P\to M$ is an orientable $2$-plane bundle and $S\to M$ is an $n$-plane bundle, together with a vector bundle isomorphism $\iota:TM\to P\otimes S$.
(Such a structure is sometimes called a(n) (oriented) Segre structure of type $(2,n)$.) Then to each reduction of structure of $P$ from an oriented real $2$-plane bundle to a complex line bundle $L$, one associates a corresponding almost-complex structure on $M$ using the isomorphism $\iota:TM\to L\otimes_{\mathbb{R}}S$. Such reductions of structure of $P$ are given by sections of an associated bundle $J(P)\to M$ whose fiber is $\mathrm{GL}(2,\mathbb{R})/\mathrm{GL}(1,\mathbb{C})$, a homogeneous space that is topologically and geometrically the hyperboloid of $2$-sheets in $\mathbb{R}^3$.

Then one finds that the data $(M,P,S,\iota)$ can be used to define a natural almost-complex structure on $J(P)$ such that a section of $J(P)$ represents an integrable almost-complex structure if and only if the image of the section is an almost-complex submanifold of $J(P)$. Usually, this almost-complex structure on $J(P)$ is not, itself, integrable, but this is determined by the torsion and curvature of the natural $G$-structure associated to the data $(M,P,S,\iota)$.

In the particular case of the OP's problem, the underlying data $(M,P,S,\iota)$ determine a flat $G$-structure, and the almost-complex structure on $J(P)$ is integrable. Indeed, $J(P)$ is the disjoint union of two copies of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ as complex manifolds.

• Dear Professor Bryant: the OP says s/he is unable to comment, so s/he added some questions for you at the end of his/her post. Feb 15, 2016 at 15:08
• @RobertBryant Dear Professor Bryant, could you please write the calculation of $\mathrm{d}\Phi_z = -e_0\,\tfrac{1}{2z}\,\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) + (e_k{+}v_k\,e_0)\,\zeta_k\,$ in more details? I can not get what you have wrote before. Oct 12, 2016 at 8:41
• @Baghban: Oh, sorry, there is a typo in that formula; the term $e_k+v_k\,e_0$ should have been $e_k+(v_k/z)\,e_0$. I have edited the answer to correct the typo (and a following typo in a line below that). Oct 12, 2016 at 11:48
• @RobertBryant Thank you so much. Dear Professor Bryant, I would be grateful if you could check the formula of $\mathrm{d}\zeta_k$. Unfortunately, I got a non-similar formula for that. $\mathrm{d}\zeta_k= \zeta_j \wedge \omega _{kj} -(v_s\omega _s +\mathrm{d}z)\wedge \omega _k$. Oct 12, 2016 at 15:40
• @Baghban: You are right that there were also typos in the formulae for $\mathrm{d}\zeta_k$. I have fixed those now; they did not affect the argument. Oct 12, 2016 at 16:53