よって、

0 → H^0(X, R) → H^0(X, Ω^0) → H^0(X, dΩ^0)

H^0(X, Ω^0) → H^0(X, dΩ^0) → H^1(X, R) → 0

0 → H^p(X, dΩ^k) → H^(p+1)(X, dΩ^(k-1)) → 0 (p≧1)

が完全になる