Coding Theory 16: Dimensions on Ruled Surfaces

Coding theory / 16

Dimensions on ruled surfaces

The dimension of a ruled-surface code is a cohomology problem on the base curve.

For \(X=\mathbb P(E)\xrightarrow{\pi}C\) and a divisor corresponding to \(\mathcal O_X(a)\otimes\pi^*\mathcal O_C(bP_0)\), the projection formula gives

\[H^0(X,\mathcal L(D))\simeq H^0(C,S^a(E)\otimes\mathcal O_C(bP_0)).\]

Thus code dimension reduces to vector-bundle cohomology on \(C\). Riemann-Roch for vector bundles gives

\[h^0(C,F)-h^1(C,F)=\deg F+\operatorname{rank}(F)(1-g).\]

The hard part is controlling \(h^1\).

Example 1: decomposable case

If \(E=L_1\oplus L_2\), then

\[S^a(E)=\bigoplus_{i=0}^a L_1^i\otimes L_2^{a-i}.\]

The dimension becomes a sum of line-bundle dimensions on \(C\).

Example 2: product case again

For \(E=\mathcal O\oplus\mathcal O\) over \(\mathbb P^1\), \(S^a(E)\simeq\mathcal O^{a+1}\). Tensoring with \(\mathcal O(b)\) gives \(h^0=(a+1)(b+1)\).

Riemann-Roch for symmetric powers

For a rank-two vector bundle \(E\) of degree \(d\) on a genus \(g\) curve, the symmetric power \(S^a(E)\) has rank \(a+1\) and degree \(da(a+1)/2\). Therefore

\[h^0(C,S^a(E)\otimes\mathcal O_C(bP_0))-h^1(C,S^a(E)\otimes\mathcal O_C(bP_0)) =(a+1)\left(b+{da\over2}+1-g\right).\]

This is the numerical backbone of the dimension formula. Every exact dimension statement in Chapter 6 is obtained by proving that the corresponding \(h^1\) term vanishes or by computing its contribution.

Vanishing criteria

The reference gives sufficient conditions forcing \(h^1(C,S^a(E)\otimes\mathcal O_C(bP_0))=0\). One route uses global generation after twisting; another uses stability and semistability.

A vector bundle \(F\) is stable if every proper subbundle \(F_0\) satisfies

\[{\deg F_0\over\operatorname{rank}F_0}< {\deg F\over\operatorname{rank}F}.\]

Semistability replaces \(<\) by \(\le\). Large enough degree for a semistable bundle implies generation by global sections or vanishing of \(H^1\).

Example 3: line bundles are stable

A line bundle has no nonzero proper subbundle of smaller positive rank, so it is stable. This is why the decomposable line-bundle case is easier than the indecomposable vector-bundle case.

Example 4: why \(h^1\) changes the code dimension

If Riemann-Roch predicts \(h^0-h^1=20\), then the code dimension is exactly \(20\) only when \(h^1=0\). If \(h^1=3\), then \(h^0=23\). Lower bounds can be strict, which is precisely the issue in the degree-one elliptic ruled case.