Coding Theory 15: Ruled Surface Code Parameters

Coding theory / 15

Ruled surface code parameters

Ruled surfaces give surface codes where length and distance can be estimated fiber by fiber.

Let \(X=\mathbb P(E)\to C\) be a ruled surface over \(\mathbb F_q\). Suppose \(C\) has \(\gamma\) rational points. Each rational fiber has \(q+1\) rational points, so the natural evaluation set has length

\[n=(q+1)\gamma.\]

For divisors \(D=aC_0+bf\), sections restrict to degree \(a\) polynomials on each fiber. Hansen bound then yields distance estimates involving the number of fibers contained in a zero divisor and the intersection degree on the remaining fibers.

Example 1: product of two projective lines

For \(X=\mathbb P^1\times\mathbb P^1\), with \(D=aF_1+bF_2\) and \(0\le a,b<q+1\), the code from all rational points has

\[n=(q+1)^2,\qquad k=(a+1)(b+1),\qquad d\ge(q+1-a)(q+1-b).\]

This is the product Reed-Solomon behavior.

Example 2: why the fiber bound has this shape

For the chosen evaluation ruling, a section can vanish identically on at most \(b\) evaluation fibers. On every other evaluation fiber it has at most \(a\) zeros. Thus the zero count is at most \(b(q+1)+(q+1-b)a\), giving the distance bound above.

General ruled-surface bound

Write a divisor class as \(D=aC_0+bf\). Let \(m\) be the maximum number of rational fibers that can be components of a zero divisor in the linear system. In the non-ample case treated in Chapter 6 this is expressed by a number of the form \(m=a(\lceil\kappa\rceil-e)+b\), while in the ample case it is governed by \(m=b-ae\). Here \(\kappa\) is the positivity threshold from the ruled-surface ample/nef criterion; when \(e\ge0\) in the decomposable non-ample case, \(\kappa=e\) and the formula reduces to \(m=b\). The fiberwise Hansen estimate gives the bound

\[d\ge n-(\gamma-m)a-(q+1)m,\]

under the theorem hypotheses

\[m<\gamma,\qquad \gamma(q+1)\ge (\gamma-m)a+(q+1)m.\]

The second inequality is the numerical condition used in the theorem. For the clean injective-evaluation conclusion used in parameter computations, the strict form

\[(q+1)\gamma>(\gamma-m)a+(q+1)m\]

gives a positive distance lower bound and hence no nonzero section can vanish on all evaluation points. At equality the distance estimate only gives \(d\ge0\), so injectivity should be checked separately or obtained from the sharper hypotheses of the particular family. This is the common source of the formulas for rational and elliptic ruled surfaces.

Proof audit

Apply Hansen’s bound to the rational fibers \(f_1,\ldots,f_\gamma\). Each has at most \(q+1\) rational points and \(D\cdot f_i=a\). If at most \(m\) fibers are components of a zero divisor, then a nonzero section can vanish at no more than

\[m(q+1)+(\gamma-m)a\]

evaluation points. Hence

\[d\ge (q+1)\gamma-m(q+1)-(\gamma-m)a=n-(\gamma-m)a-(q+1)m.\]

The number \(m\) comes from intersecting the zero divisor with a nef divisor: in the non-ample case Chapter 6 uses a nef class governed by \(\kappa\), giving \(m=a(\lceil\kappa\rceil-e)+b\); in the ample case \(C_0\) is nef and gives \(m=b-ae\).

Rational ruled surfaces

Over \(\mathbb P^1\), rank-two bundles split, so rational ruled surfaces are controlled by decomposable bundles. This makes dimension computations explicit and connects the resulting codes to product-type evaluation codes. Chapter 6 compares their relative parameters with the Gilbert-Varshamov benchmark and with product Reed-Solomon intuition.

Example 3: decomposable bundle over \(\mathbb P^1\)

If \(E=\mathcal O\oplus\mathcal O(-e)\) with \(e\ge0\), then

\[S^a(E)=\bigoplus_{j=0}^a \mathcal O(-ej).\]

For \(0\le a<q+1\) and \(0\le b<q+1\), the rational ruled-surface code has

\[n=(q+1)^2,\qquad d\ge(q+1-a)(q+1-b),\]

and

\[k=\sum_{0\le j\le a,\; ej\le b}(b-ej+1).\]

For \(e=0\) this becomes \(k=(a+1)(b+1)\), the product case.

The dimension formula is just cohomology on \(\mathbb P^1\):

\[H^0\bigl(\mathbb P^1,\mathcal O(b-ej)\bigr)\]

has dimension \(b-ej+1\) when \(b-ej\ge0\) and dimension \(0\) otherwise. Summing over the symmetric-power decomposition gives the displayed value of \(k\).

Example 4: rate-distance tradeoff on the product

Taking larger \(a\) and \(b\) raises \(k=(a+1)(b+1)\) but lowers \(d\ge(q+1-a)(q+1-b)\). The two ruling directions make the tradeoff rectangular rather than one-dimensional.