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This is an exercise from Gromov's Partial differential relations. (page 5)

Let $V$ and $V'$ be two closed complex submanifolds in $\mathbb{C}^N$ of complimentory dimension. Prove that $V$ and $V'$ intersect if the following sets are compact for all k.

$V_k = \{(v,v') \subset V\times V'|dist(v,v') \leq k$}.

I was looking for a differential geometry approach to solving this problem along the lines of Theorem 2 of Frankel(1961) but anything would do.

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By compactness, there is a pair of points at the minimal distance. On the other hand, any positive distance can be made smaller (e.g., cf. the proof of Lefschetz theorem in Milnor's "Morse theory": the critical points of the distance function to a complex manifold always have positive index).

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    $\begingroup$ I am confused. Take the lines $x=0$ and $x=1$ in $\mathbb{C}^2$. There is no compactness of course, but still the distance function has minimum value $1$. Why does your argument not apply? $\endgroup$ – abx Jan 14 '14 at 11:33
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    $\begingroup$ It's not an argument, it's a hint; check Milnor's book. Your function is constant, hence no compactness. Basically, it's like the maximum principle (which can be restated as the minimum principle in the absence of zeroes): a function (in an appropriate class) is either constant or takes maximum on the boundary (which is empty in our case). But I'm too old to work out the technical details :) $\endgroup$ – Alex Degtyarev Jan 14 '14 at 11:40
  • $\begingroup$ OK, thanks (I am not young either…). $\endgroup$ – abx Jan 14 '14 at 12:35
  • $\begingroup$ @AlexDegtyarev You mentioned On the other hand, any positive distance can be made smaller and you gave a reference to Milnor's Morse. But after skimming through the proof of the Lefschetz theorem, there is no mention of the distance function having a positive index. There is a bound though, and that is what is used in that book. Also in the answer there is no mention of the fact that $V$ and $V'$ have complementary dimension. $\endgroup$ – bellztoll Jan 16 '14 at 8:30

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