7
$\begingroup$

This question is motivated by the question Path Connectedness of Varieties and some funny little theorem I was trying to prove. Let $X$ be a (quasiprojective smooth connected) algebraic variety over an algebraically closed field of an arbitrary characteristic. We know from the answer above that any two points of $X$ can be connected by a curve. Can we control its genus?

More precisely, we say that the two points $x$ and $y$ are $n$-path-connected, if there is a sequence of smooth curves from $x$ to $y$ such that every curve has genus less or equal to $n$. It is an equivalence relations and I am interested in its equivalence classes that are reasonably to be called $n$-path-connected components. Let me ask three precise questions.

  1. Is it true that there exists $n$ such that $X$ is $n$-path-connected?
  2. How do you find such minimal $n$?
  3. Is there an algorithm/method for computing/describing $n$-path-connected components of $X$?

PS A curve of genus $g$ is an instructive example. It is $g$-path-connected but its $(g-1)$-path-connected components are points.

$\endgroup$
7
  • $\begingroup$ Perhaps you know it already, but rationally connected varieties have been well studied, I think. $\endgroup$ Nov 26, 2012 at 22:06
  • $\begingroup$ I'm almost sure the the answer for 1 is positive. I'll try to write an argument. $\endgroup$
    – Rami
    Nov 26, 2012 at 22:07
  • $\begingroup$ Hailong Dao, are you implying that the minimal number $n$ is a bi-rational invariant? It make sense. If it so it gives a good strategy for 2 and probably 3, in view of the en.wikipedia.org/wiki/Minimal_model_program. $\endgroup$
    – Rami
    Nov 26, 2012 at 22:14
  • $\begingroup$ It seems like one should be able to pass to the quotient where two points that are $0$-connected are identified. I'm running into the problem that that is not quite an algebraic equivalence relation, if I recall correctly. $\endgroup$ Nov 27, 2012 at 2:39
  • 1
    $\begingroup$ The answer to question 1 is Lemma 3.9 here arxiv.org/abs/1208.4055 $\endgroup$
    – Frank
    Nov 28, 2012 at 12:53

2 Answers 2

4
$\begingroup$

Here's an idea for the first question. I'm really thinking over $\mathbb C$. Maybe someone who knows enough can make it into a proof. I think that if $X$ has degree $D$ then $n$ can be taken to be $\frac{(D-1)(D-2)}{2}$, the genus of a smooth plane curve of degree $D$. The idea is to show that each $n$-path-component is open.

Let $X$ be $d$-dimensional. Let $a\in X$ be a point. Choose a linear projection $\pi$ to $\mathbb P^{d+1}$ such that it embeds a neighborhood $U$ of $a$, and such that $U=\pi^{-1}(\pi(U))$. By intersecting $\pi(X)$ with $2$-dimensional planes through $a$ we see that for any point $b\in X$ sufficiently close to $a$ there is an irreducible curve of genus at most $n$ mapping into $\pi(X)$ such that $\pi(a)$ and $\pi(b)$ are in its image. Lift so that the curve maps to $X$.

Edit: This is wrong. I forgot that we were seeking smooth curves in a smooth variety.

Edit: Let's start again. Taking a cue from Misha's answer: Suppose $V$ is quasiprojective, smooth, and connected, and of degree $d$. Let $p$ and $q$ be distinct points in $V$. If $dim(V)>1$, then there is a hyperplane $H$ through $p$ and $q$ that is transverse to $V$. (I'll discuss this claim below.) Now $V\cap H$ is smooth and of dimension $d-1$ (by transversality) and connected (by Lefschetz, since $dim(V)>1$) and again of degree $d$. Repeat until you have reduced the dimension to one. Now you have a smooth curve of degree $d$ in some projective space, so its genus is at most $\frac{(d-1)(d-2)}{2}$.

Now, why did that hyperplane exist? Let $L$ be the line determined by $p$ and $q$, and let $P$ be the projective space of all hyperplanes containing $L$. To make $H\in P$ transverse to $V$ at $p$, we just need to avoid a proper closed subspace of $P$, a projective space of codimension $dim (V)$, or $dim(V)-1$ if $L$ is tangent to $V$ at $p$. Likewise to make $H\in P$ transverse to $V$ at $p$ we just need to avoid another such subspace of $P$. Let $U\subset P$ be the remaining open subset. To make $H\in U$ transverse to $V-\lbrace p,q\rbrace$, consider the space of all pairs $(H,x)$ with $H\in U$ and $x\in H-\lbrace p,q\rbrace$. The projection $(H,x)\mapsto x$ to the complement of $\lbrace p,q\rbrace$ is a submersion, so the inverse image of $V-\lbrace p,q\rbrace$ is smooth. The projection $(H,x)\mapsto H$ of this smooth thing to $U$ is transverse to some point $H$. This $H$ is then transverse to $V$.

$\endgroup$
1
  • $\begingroup$ Why should that irreducible curve be smooth? $\endgroup$
    – Will Sawin
    Nov 27, 2012 at 6:39
2
$\begingroup$

Here is a variation on Tom's argument (for the 1st question) that works: I will also work over ${\mathbb C}$, although, I do not think it is important. I will be proving that $n$ can be bounded depending only on dimension of ambient space and degree $d$ of the subvariety $V\subset P^N$. The proof is by induction on dimension $N$ of the ambient projective space. Everything is clear, if $N=1$ or if $dim(V)=1$. Suppose that we have a bound $n(d,N-1)$; consider a smooth subvariety $V\subset P^N$. Now, given any pair of points $p, q\in V$, I can find a projective hyperplane $H\subset P^N$ intersecting $V$ transversally (algebraic geometers would call it Bertini's transversality, topologists would call it Sard's theorem). By Lefschetz, if $dim(V)\ge 2$, then $W=V\cap H$ is connected. For two generic projective hyperplanes $H_p, H_q$ through $p, q$, intersections $W_p=H_p\cap W, W_q=H_q\cap W$ are nonempty, smooth and connected. Now, we are done by induction applied to $W, W_p, W_q$ (connect $p$ to $x\in W\cap W_p$, then connect $x$ to $y\in W\cap W_q$, etc.).

$\endgroup$
1
  • $\begingroup$ Oh. The point I was missing was that $V\cap H$ must be connected. But can't you simplify your argument further and just find an $H$ through $p$ and $q$ that is transverse to $V$? $\endgroup$ Nov 28, 2012 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.