Dimension of moduli of lines on quadric

Question

What is the dimension of the moduli space of lines on a general quadric hypersurface in $\mathbb{P}^n$? Maybe the question is quite trivial, but different intuitive approaches (à la Italian algebraic geometry) yield different answers.

Answer 1

I'm not sure what you mean when you say that different approaches give different answers. As Fredrik Meyer says, the answer is $2n-5$: here are a some different arguements that prove it.

By the way, note that all smooth quadrics are isomorphic, so you can weaken the condition "general" to "smooth".

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Method 1 (prehistoric): First fix a point $p$ on the quadric. Any line through $p$ lies in the projective tangent space $\mathbf T_p Q$.The intersection $\mathbf T_p Q \cap Q$ is a cone over a smooth quadric $Q'$ of dimension $n-3$, so the lines on $Q$ through $p$ are in bijection with the points of $Q'$. Varying $p$, we get a space of dimension $(n-1)+(2n-3)=2n-4$. But we overcounted: for each line, there is a 1-parameter family of points $p$ for which it appears. So the correct count is $2n-4-1=2n-5$.

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Method 2 (classical): The space of quadrics in $\mathbf P^n$ is $\mathbf P^N$ where $N={n+2 \choose 2}-1$. The space of lines in $\mathbf P^n$ is the Grassmannian $\mathbf G(1,n)$, which has dimension $2(n-1)$.

Now in the product $\mathbf G(1,n) \times \mathbf P^N$, consider the incidence correspondence $$ I := \left\{ (l,Q) \mid l \subset Q \right\}. $$

We will see below that $I$ is irreducible.

The second projection $\pi_2: I \rightarrow \mathbf P^N$ is surjective, because every quadric contains at least one line. So the general fibre of $\pi_2$ has dimension $\operatorname{dim} I - N$.

On the other hand, vanishing on a fixed line imposes 3 linear conditions on quadrics, so the first projection $\pi_1: I \rightarrow \mathbf G(1,n)$ has fibres which are projective spaces of dimension $N-3$. (In fact this shows that $I$ is a projective bundle over $\mathbf G(1,n)$, hence irreducible as claimed.) So $\operatorname{dim} I = 2(n-1)+N-3$.

Combining the last two paragraphs we see that the space of lines on a general quadric, hence any smooth quadric, has dimension $2(n-1)+N-3-N=2n-5$.

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Method 3 (modern): Let $E$ be the tautologial rank-2 bundle on $\mathbf G(1,n)$. Then $V = \operatorname{Sym}^2 E^*$ is a bundle of rank 3, and $Q$ defines a section $\sigma$ of $V$ (fibrewise, by restricting $Q$ to each line.) The space of lines on $Q$ is then the zero-locus $Z(\sigma)$ of the section $\sigma$. Checking that a general such $\sigma$ intersects the zero-section transversely, we get that $Z(\sigma)$ has codimension equal to the rank of $V$, that is 3. So $Z(\sigma)$ has dimension $2(n-1)-3=2n-5$ again.

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Comments: The first method seems historically plausible, but it isn't rigorous as written. Moreover, it seems special to quadrics (a singular hypersurface of higher degree need not be a cone); by contrast, the other two methods work for general hypersurfaces of any degree. (In this context general really is necessary, not just smooth.)

The second method is more or less complete, as long as you accept that incidence correspondences are projective varieties.

The third method is trickier (the claim about transverality has to be proved) but it has a big advantage: when the expected dimension $2n-d-3$ is zero, this method actually tells you how many points the zero-locus contains (by calculating the top Chern class of the bundle). You can use this to prove that a general cubic has 27 lines, or that a general quintic threefold has 2875 lines.

You can read the paper "Galois Groups of Enumerative Problems" by Joe Harris if you want to see how to tackle the transversality question. There is also the paper "Rational curves on smooth cubic hypersurfaces" by Coskun and Starr that shows how to extend results of this kind to all (not just general) smooth cubic hypersurfaces.