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Let $X$ be a smooth, rationally connected variety over $\mathbb{C}$. A priori, this means that for any two points $p,q\in X$, there is a rational curve $C$ containing $p$ and $q$. Can $C$ be taken to be smooth?

adrido
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  • Yes. Look in Koll'ar's book Rational Curves on Algebraic Varieties for all kinds of results like this. What you want is somewhere in Chapter IV. – Schemer Sep 07 '15 at 20:02
  • Nice question ! – Georges Elencwajg Sep 07 '15 at 20:13
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    @GeorgesElencwajg:Dear Georges, I believe Theorem IV.3.9 gives an affirmative answer in the case where X is projective and dimX>2. – adrido Sep 08 '15 at 01:08
  • @Relapsarian: thanks for the reference! it helped. – adrido Sep 08 '15 at 01:45
  • Dear @adrido: you are absolutely right. Thanks for your precise reference. – Georges Elencwajg Sep 08 '15 at 05:53
  • Dear @adrido, did you notice that Theorem IV.3.9 does not answer your question? It only says that in a rationally connected smooth variety $X$ of dimension $\geq 3$ you can join two points by a smooth rational curve provided both points are in some mysterious open subset $X^0\subset X$ of the variety. – Georges Elencwajg Sep 08 '15 at 06:26
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    @GeorgesElencwajg: Dear Georges, you are right, it does not give an answer for general $X$. There are two theorems in the version I have: http://link.springer.com/book/10.1007%2F978-3-662-03276-3. There is theorem IV.3.9.4 which does not assume properness and requires the points to lie on some largest open set. There is also IV.3.9 which assumes properness but requires no further assumption on where the points are. – adrido Sep 08 '15 at 12:41
  • Dear adrido: but this is fantatic! Actually most of the varieties studied in algebraic geometry are complete and theorem IV (3;9;3) on page 203 indeed gives a great answer to your very interesting question for those. I'm learning a lot from you, thanks a lot ! Unfortunately I can't upvote your question more than once :-) – Georges Elencwajg Sep 08 '15 at 13:04
  • Dear adrido and @GeorgesElencwajg: sorry for being sloppy in my original comment. I was assuming we were talking about proper varieties, but now I see that was not mentioned in the question. – Schemer Sep 08 '15 at 13:09
  • @GeorgesElencwajg: Dear Georges, you are very welcome! But it's really me who should thank you! After all, I've been learning from your posts ever since I joined MSE. – adrido Sep 08 '15 at 13:32

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No: an amusing counterexample is obtained by taking for $X$ a rational curve singular at $p$ !

EDIT
This is a counterexample with $X$ singular (I had forgotten that adrido required a smooth variety).
However, in the positive direction, there also exist singular varieties such that two arbitrary points $p,q$ can always be joined by a smooth rational curve
A quadratic conic hypersurface $X\subset \mathbb A^n$ with apex $O$ is such an example:
Given two points $p,q\in X$, they are either aligned with $O$ and thus can be joined by a ruling (=line through $O$) of $X$, or else there exists a plane $P$ containing $p$ and $q$ but not $O$ and thus $P\cap X$ is a smooth conic joining $p$ and $q$.

Georges Elencwajg
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  • maybe i'm dumb but if $X$ is singular at $p$ then $X$ is not smooth ? – mercio Sep 07 '15 at 19:53
  • You are not dumb at all: I misread the question, forgetting that $X$ is required to be smooth. I'll delete my answer soon unless I find a correct argument. – Georges Elencwajg Sep 07 '15 at 19:57
  • @GeorgesElencwajg: I find the positive result very useful! what properties do you require of a singular $X$ so that there exists a smooth curve through any of its two points? – adrido Sep 07 '15 at 20:43
  • Dear adrido, unfortunately I don't know: I just wrote down the first example that came to my mind, because I love this sort of geometry that would have been familiar to Euclid ! – Georges Elencwajg Sep 07 '15 at 21:08