Here we are visualizing a quartic surface containing a line in the spherical geometry of $\mathbb{RP}^3$.

In the space of all quartic surfaces it is exceedingly rare that one contains a line, specifically it occurs with measure zero. For an explantion see here

We are further refining our interest in quartic surfaces defined by two cubic curves.

Note that if a plane contains a line $L$ that lines a quartic surface, then the intersection of that plane with the quartic surface will be a quartic curve containing a line, meaning the remaining curve will be of degree at most 3 (allowing for multiplicity of $L$).

Without loss of generality let $L = \{ x_0 = x_1 = 0 \}$. Given two cubic curves, $p(x_0,x_2, x_3)$ and $q(x_1, x_2, x_3)$ we define the quartic surface

which vanishes on $L$ since then $x_0=x_1=0$. When $x_0=0$ we recover $f= x_1 q$ and when $x_1 = 0$ we recover $f = x_0 p$. So these defining curves live on the surface in orthogonal planes. If we consider the whole family of planes containing $L$ we get a corresponding smooth family of cubic curves. This is an example of (a real slice of) an elliptic K3 surface.

Why do most quartics not contain a line?

The set of all quartic surfaces is itself a space, and the set of all quartic surfaces containing a line is a subspace. Both spaces have a dimension and the dimension of quartics containing a line is 1 dimension lower than the dimension of the full space of quartics. You would say that the space of quartics containing a line has codimension 1.

We will proceed using an argument similar to one presented in Section 1.2 of (Hassett 2016).

Since we are considering surfaces in $\mathbb P^3$ our quartic surface is defined by a homogenous polynomial $f$ of degree 3 in $\mathbb C[x_0, x_1, x_2, x_3]$.

This polynomial can be written as a linear combination of terms, monomial terms that look like $x_0^{e_0} x_1^{e_1} x_2^{e_2} x_3 {e^3}$ for $e_i \geq 0$ and $e_0 + e_1 + e_2 + e_3 = 4$.

How many such polynomials are there? The answer is 35.

To prove this we can use the “stars and bars” argument. We are trying to split the degree $d$ between the values $e_0, e_1, e_2, e_3$. In general distributing $d$ identical items between $n$ bins. Any way of distribution can be represented as a string of $d$ *’s representing value and $n-1$ |’s seperating the bins. For example the string *|**|*| corresponds to the monomial $x_0 x_1^2 x_2$. In total the string will always be length $d+n-1$ and each string is uniquely defined by a choice of where to put the stars. So the number of possible strings and hence the number of monomials is ${d+n-1}\choose d$. In our case of $d=4$ and $n=4$ we get 35.

Now consider the projective line

If we restrict $f$ to $L$ all the terms with $x_2$ or $x_3$ vanish and we get

Now if $L$ is contained in the surface that means that $f$ vanishes on $L$. So $f\mid _L$ must be the zero polynomial and all those coefficients must be zero.

So asking that the surface contains $L$ is a 5 equation constraint in the coefficient space meaning the space of quartics containing $L$ is degree 30.

But we asked about the space of quartics contianing any line, not $L$ specifically. To answer this we have to “quotient” both spaces.

The group $GL(4)$ of 4x4 matrices acts on our affine space $\mathbb C^4$ in the natural way. Considering the set of orbits of quartics under this $GL(4)$ action gives us a way to think about a quartic up to its embedding in the space.

We will also consider the orbit of quartics containing $L$ under the action of $G$ where $G$ is defined

the $GL(4)$-stabilizer subgroup of $L$. By the way we’ve constructed $L$, this condition is equivalent to asking that $M$ is a matrix whose bottom left 2x2 submatrix is filled with zeros. The rest of the matrix is completely unconstrained which makes $G$ a 12-dimensional space, 4 less than the 16-dimensional space of $GL(4)$.

There is a natural injection from the set of orbits of quartics with $L$ to the set of orbits of quartics. Yet we compute the dimension of the space of quartics up to embedding as 35 - 16 = 19 and the dimension of the space of quartics with $L$ as 30 - 12 = 18.

And we are done.

This same story can be repeated with replacing quartic with cubic or quadric and the dimensions will line up differently. More to come on this soon.

References

Hassett, Brendan. 2016. “Cubic Fourfolds, K3 Surfaces, and Rationality Questions.” arXiv:1601.05501. Version 2. Preprint, arXiv, July 17. https://doi.org/10.48550/arXiv.1601.05501.