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Let $\theta \in \mathbb{C}$ be a fixed complex number. The submersion $f:\mathbb{C}^{2}\to \mathbb{C}\; \text{with}\; f(x,y)=y-\theta x$ defines a complex foliation on $\mathbb{C}^{2}$. Consider the quotient foliation on each of the $\mathbb{C}^{2}/\mathbb{Z}^{2}$ or $\mathbb{C}^{2}/\mathbb{Z}^{4}$.(Here we consider the natural and obvious action of either$\mathbb{Z}^{2}$ or $\mathbb{Z}^{4}$ on $\mathbb{C}^{2})$.

To what extend these foliations are classified up to topological equivalent(in term of $\theta$).

In particular what is the structure of this complex foliation for real rational $\theta$? What about irrational $\theta$?

Motivated by irrational rotation algebras, what type of $C^{*}$ algebras would appear, in this way?(The foliation $C^{*}$ algebra of this complex foliation)

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Ali Taghavi
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  • Could you please describe what are those "obvious" actions ? I can think of different ones, and don't know which is the one you refer to. – Loïc Teyssier Jun 19 '15 at 13:10
  • @LoïcTeyssier The action is by translation. $\mathbb{Z}^{2}$ on $\mathbb{C}^{2}$ and $\mathbb{Z}^{4}$ on $\mathbb{C}^{2}\simeq \mathbb{R}^{4}$. What type of other action you are thinking about? – Ali Taghavi Jun 19 '15 at 13:16
  • You have a multiplicative action $x\mapsto x\alpha$ for $\alpha\neq 0$, for instance, and the quotient is a torus if $\alpha$ is not real. Anyway, I understood that you were speaking about translation, but there still is an ambiguity, as the action of $\mathbb Z^2$ on $\mathbb C$ usually depends on a complex non-real parameter $\tau$.So I take it that $\tau=i$ here. – Loïc Teyssier Jun 19 '15 at 13:21
  • @LoïcTeyssier I do not see the role of $\mathbb{Z}$ in the multiplicative action you mentioned. Could you please more explain? For the same reason I do not understand what was the ambiguity? The action of $\mathbb{Z}^{4}$ on $\mathbb{C}^{2}$ is $(n_{1},n_{2},n_{3},n_{4}).(x+yi,z+iw)=x+n_{1}+(y+n_{2})i, z+n_{3}+(w+n_{4})i$ – Ali Taghavi Jun 19 '15 at 13:38
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    Well… for $n\in\mathbb Z$ you have a diffeomorphism $x\mapsto x\alpha^n$. Similarly the additive $\mathbb Z^2$-action is given by $(n,m)\cdot x=x+n+m\tau$, which corresponds to $\tau=i$ in your example. – Loïc Teyssier Jun 19 '15 at 13:52
  • Anyway, the question I understand here is to describe the set of $\theta$ for which the induced foliations are topologically equivalent, right? – Loïc Teyssier Jun 19 '15 at 13:54
  • @LoïcTeyssier Yes, exactly. This is already done for the real case: That is one replace $\mathbb{C}^{2}$ by $\mathbb{R}^{2}$ and consider $\theta \in \mathbb{R}$. Please see http://mathoverflow.net/questions/178073/the-kronecker-foliation-or-a-kronecker-foliation – Ali Taghavi Jun 19 '15 at 14:02
  • @LoïcTeyssier the action $x\mapsto x\alpha^{n}$ is on $\mathbb{C}$ or $\mathbb{C} \setminus {0}$(with quotient=torus)? – Ali Taghavi Jun 19 '15 at 14:15
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    Yes, obviously you're right, the torus is the quotient of $\mathbb C^*$. – Loïc Teyssier Jun 19 '15 at 14:24
  • @LoïcTeyssier So it would be interesting to find explicitly, the foliation of $\mathbb{C}\setminus {0}$ which quotient is the $\theta$-kronecer foliation. – Ali Taghavi Jun 19 '15 at 14:33
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    The additive and multiplicative actions are related through the exponential mapping. – Loïc Teyssier Jun 19 '15 at 14:34
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    What do you mean by topologically equivalent? Are you calling two complex foliations (foliations of a complex manifold by complex submanifolds) equivalent if they are related by a diffemorphism, with no regard for the complex structure? – Tom Goodwillie Jun 19 '15 at 15:59
  • @TomGoodwillie In my question I am interested in topological classification, So I mean " By homeomorphism". However, maybe it is more reasonable to compare two holomorphic foliations with holomorphic object. But I am motivated by real case which says two Kronecer foliation $\theta_{1}$ and $\theta_{2}$ are not even topological equivalent if the slops are not in the same orbit of action $Sl_{2} \mathbb{Z}: a\theta+b/c\theta+d$ – Ali Taghavi Jun 19 '15 at 16:11
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    OK, but if the equivalence relation has nothing to do with complex structure then I would suggest forgetting about complex structure. Any vector subspace of $\mathbb R^n$ determines a foliation of $\mathbb R^n/\mathbb Z^n$. You could think about the case $n=3$ before the case $n=4$. – Tom Goodwillie Jun 19 '15 at 20:27
  • @TomGoodwillie Prof. Goodwillie I really thank you very much for your very helpful comment. I understand from your comment that : we have a family of one dim foliations $\mathcal{F}_{\theta}, \theta \in \mathbb{R}P^{2}$ (or 2 dimensional foliation for $\theta \in G(2,3)\simeq \mathbb{R}P^{3}$). Now the question is that:"What is the topolgical classification of such foliations in term of parameter $\theta$". It would be interesting that, the action of some group(what group?) on parameter space determines this classification. Thanks again for your very interesting comment. – Ali Taghavi Jun 19 '15 at 21:04
  • I correct: $G(2,3)\simeq \mathbb{R}P^{2}$. – Ali Taghavi Jun 19 '15 at 21:13

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