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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.

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Perhaps you know it already, but rationally connected varieties have been well studied, I think. –  Hailong Dao Nov 26 '12 at 22:06
I'm almost sure the the answer for 1 is positive. I'll try to write an argument. –  Rami Nov 26 '12 at 22:07
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. –  Rami Nov 26 '12 at 22:14
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. –  Allen Knutson Nov 27 '12 at 2:39
The answer to question 1 is Lemma 3.9 here arxiv.org/abs/1208.4055 –  Frank Nov 28 '12 at 12:53
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2 Answers

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 $V$.

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Why should that irreducible curve be smooth? –  Will Sawin Nov 27 '12 at 6:39
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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.).

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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 $V$? –  Tom Goodwillie Nov 28 '12 at 1:22
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