This question is related to this question: "Solutions of equations characterizing a complex structure." Where, here we suppose the Euclidean space instead of Sphere and the following equations happen if and only if the almost complex structure $J_{\delta ,\beta}$ (defined as the introduced structure in this question) is a complex structure on $T\mathbb{R}^{2n}$(the tangent space of $\mathbb{R}^{2n}$).

Let $\mathbb{R}^{2n}$ be the Euclidean $2n$ dimensional space and $(x^1,...,x^n,y^1,...,y^n)$ be its coordinate system. Suppose $u=\frac{\beta}{\delta}$ and $v=\frac{1}{\delta}$ where $\beta , \delta$ are real-valued functions on $\mathbb{R}^{2n}$. Is there any other solutions except constant functions satisfying the following Equations? $$\frac{\partial v}{\partial x^l}+ v\frac{\partial u}{\partial y^l} +u\frac{\partial v}{\partial y^l}=0,\hspace{1cm} \forall l, \hspace{1cm} 1‎\leq l‎\leq n$$ and $$\frac{\partial u}{\partial x^l}- v\frac{\partial v}{\partial y^l} +u\frac{\partial u}{\partial y^l}=0,\hspace{1cm} \forall l, \hspace{1cm} 1‎\leq l‎\leq n$$


1 Answer 1


The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial y^\ell}\left(\mathrm{d} y^\ell - z\,\mathrm{d}x^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-x^\ell\frac{\partial z}{\partial y^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial y^\ell}\,\mathrm{d}\left(y^\ell - z\,x^\ell\right). \tag3 $$ Now, defining functions $w^\ell = y^\ell - z\,x^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z - \bar z=0$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now, $\Phi$ maps the hyperplane $z-\bar z=0$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs: One is the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + x\cdot x}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - (x\cdot y)^2\,}}{1 + x\cdot x}, $$ and the other is the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering more generic hyperquadrics with no real points, though this is far from exhausting all of the global solutions.)

  • $\begingroup$ I think the variable $z=u+iv$ does a role as well as $i=\sqrt{-1}$. Is it right? I'm confused by this. $\endgroup$ Apr 1, 2016 at 17:17
  • $\begingroup$ @Solar: Sorry, I don't understand. What do you mean by 'the variable $z = u + i\,v$ does a role'? What is 'it' in "Is it right?"? I'm confused by your comment. Certainly, by $i$, I mean $\sqrt{-1}$. $\endgroup$ Apr 1, 2016 at 17:37
  • $\begingroup$ I checked the Example but the stated $u(x,y)$ and $v(x,y)$ in the Example don't satisfy in the sufficient and necessary Equations for the integrability of $J_{\delta,\beta}$. $\endgroup$ Apr 4, 2016 at 14:55
  • $\begingroup$ @Solar: Then you have a mistake in your equations, because the $u(x,y)$ and $v(x,y)$ as given in the Example do satisfy your given equations. You do understand that, in my notation, I am using $x = (x^1,\ldots,x^n)$ and $y = (y^1,\ldots,y^n)$ as vectors in $\mathbb{R}^n$, right? $\endgroup$ Apr 4, 2016 at 16:08
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    $\begingroup$ @Solar: I did the same thing, but using Maple to check the calculations, and I found that the pair $\bigl(u(x,y),v(x,y)\bigr)$ does solve the equation, so I don't know what you are doing wrong. $\endgroup$ Apr 4, 2016 at 17:43

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