Show that every locally free sheaf of finite rank on \(\PP^1\) is isomorphic to a direct sum of invertible sheaves.
Hint: Choose a subinvertible sheaf of maximal degree, and use induction on the rank.
Solution
Let \(\mathcal E\) be locally free of rank \(r\) on \(\PP^1\).
1. There is a largest integer \(a\) such that\[ H^0\qty(\PP^1,\mathcal E(-a))\ne0. \]
For \(N\gg0\), Serre’s theorem makes \(\mathcal E(N)\) globally generated, so it has a nonzero section. Thus the displayed set of integers is nonempty.
For \(a\gg0\), Serre duality on \(\PP^1\) gives \[ H^0\qty(\PP^1,\mathcal E(-a))^\vee \cong H^1\qty(\PP^1,\mathcal E^\vee(a-2)). \] The group on the right vanishes for \(a\gg0\) by Serre vanishing. Hence the set is bounded above and has a largest element \(a\).
2. A nonzero section of \(\mathcal E(-a)\) yields an exact sequence\[ 0\longrightarrow\OO_{\PP^1}(a) \longrightarrow\mathcal E \longrightarrow\mathcal F \longrightarrow0 \]with \(\mathcal F\) locally free of rank \(r-1\).
A nonzero section gives an injective map \[ \OO_{\PP^1}(a)\longrightarrow\mathcal E. \] Saturate its image. On the nonsingular curve \(\PP^1\), the saturation is an invertible subsheaf, hence isomorphic to \(\OO_{\PP^1}(a+m)\) for some \(m\ge0\), where \(m\) is the degree of the zero divisor removed by saturation. If \(m>0\), then \[ H^0\qty(\PP^1,\mathcal E(-a-m))\ne0, \] contradicting maximality of \(a\). Thus \(m=0\): the original image is already saturated. Its quotient is torsion-free on a nonsingular curve and therefore locally free, of rank \(r-1\).
3. Assume inductively that every locally free sheaf of rank \(r-1\) on\(\PP^1\) splits. Then\[ \mathcal F\cong\bigoplus_{j=1}^{r-1}\OO_{\PP^1}(b_j) \]for some integers \(b_j\).
2. The rank-one base case is immediate because every invertiblesheaf on \(\PP^1\) is \(\OO(n)\) for some \(n\in\ZZ\).
4. Every summand in step 3 satisfies\[ b_j\le a. \]
Twist the exact sequence of step 2 by \(\OO(-a-1)\): \[ 0\longrightarrow\OO(-1) \longrightarrow\mathcal E(-a-1) \longrightarrow\mathcal F(-a-1) \longrightarrow0. \] The projective-line cohomology calculation [@Har10a, Theorem III.5.1] gives \[ H^1(\PP^1,\OO(-1))=0. \] Hence the map on global sections \[ H^0\qty(\PP^1,\mathcal E(-a-1)) \longrightarrow H^0\qty(\PP^1,\mathcal F(-a-1)) \] is surjective.
If some \(b_j\ge a+1\), then \[ H^0\qty(\PP^1,\OO(b_j-a-1))\ne0, \] so \(H^0(\PP^1,\mathcal F(-a-1))\ne0\). Surjectivity would then give \[ H^0\qty(\PP^1,\mathcal E(-a-1))\ne0, \] contradicting the maximality of \(a\). Therefore every \(b_j\le a\).
5. The exact sequence in step 2 splits.
Its extension class lies in \[ \Ext^1\qty(\mathcal F,\OO(a)). \] Using step 3 and the extension calculation PR-ET5PQ, \[ \begin{aligned} \Ext^1\qty(\mathcal F,\OO(a)) &\cong \bigoplus_{j=1}^{r-1} \Ext^1\qty(\OO(b_j),\OO(a))\\ &\cong \bigoplus_{j=1}^{r-1} H^1\qty(\PP^1,\OO(a-b_j)). \end{aligned} \] By step 4, \(a-b_j\ge0\). Theorem III.5.1 gives \[ H^1\qty(\PP^1,\OO(a-b_j))=0 \] for every \(j\). Hence the entire Ext group vanishes and the sequence splits.
6. Therefore\[ \boxed{ \mathcal E \cong \OO_{\PP^1}(a) \oplus \bigoplus_{j=1}^{r-1}\OO_{\PP^1}(b_j).} \]
Step 5 splits the exact sequence of step 2, and step 3 splits its quotient. This proves the rank-\(r\) case from the rank-\((r-1)\) case. Induction from rank one proves the theorem for every finite rank.
7. Q.E.D.
Steps <1>1–<1>6 prove that every locally free sheaf of finite rank on \(\PP^1\) is a direct sum of invertible sheaves.