I have seen, but don't fully understand, the following statement and sketch proof:

Statement: A codimension $2$ submanifold $C \subset \mathbb{C}^2$ such that $C$ has positive intersection index with every complex line is complex analytic (i.e. every tangent space is complex).

Sketch Proof: Given a point $p \in C$ we can write $C$ as a graph over it's tangent space $T_pC$ in some neighbourhood of $p$. Now if the tangent space is not complex then one can "easily" find a complex line which has negative intersection index with $C$ at $p$. This is a contradiction and hence $T_pM$ is complex.

My problems are as follows:

1) What exactly is meant by the tangent space not being complex here? 2) How does one find this line precisely?

For the first point if we are considering $C$ as a subset in $\mathbb{C}^2$ then I guess we are talking about the tangent space as a plane in $\mathbb{C}^2$ which would necessarily be complex?

For the second point I guess that if we take a non-complex plane a complex plane both in $\mathbb{C}^2$ then their orientations will be oposite and hence the intersection index will be negative?

Any help will be greatly appreciated!

  • $\begingroup$ For $C$ a real submanifold of $B = \Bbb{C}^2$ then $T_pC$ is a real vector space and a subspace of the real vector space $T_p B$. Then $B$ has a complex structure making $T_p B$ a complex vector space. So the question is if $T_p C$ is a complex subspace of $T_pB$. $\endgroup$ – reuns Apr 25 at 19:27
  • $\begingroup$ For starters, that submanifold needs to be oriented. A real 2-dim. subspace is complex iff it is $J$-invariant, where $J$ is the complex structure on $\Bbb C^2$. $\endgroup$ – Ted Shifrin Apr 25 at 19:27
  • $\begingroup$ @reuns Why take smooth charts at all? We’re in $\Bbb C^2$. $\endgroup$ – Ted Shifrin Apr 26 at 3:59
  • $\begingroup$ No, @reuns, as the OP framed the question it is only about working with the tangent space of $C$ as a subspace of $\Bbb C^2$. $\endgroup$ – Ted Shifrin Apr 26 at 16:26
  • $\begingroup$ So that the tangent spaces are complex isn't enough for the manifold to be complex (and to be an analytic variety), we also need the vanishing of the Nijenhuis tensor $\endgroup$ – reuns Apr 27 at 21:13

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