Pull-back of line bundles under the normalization map.

Question

Let $\Gamma\subset\mathbb{P}^2$ be a singular plane curve and let us consider $f:C\to\Gamma$ the normalization map. Given $L\in\mathrm{Pic}(\Gamma)$, is it always true that $h^0(\Gamma,L)=h^0(C, f^*L)$? If not, can you exhibit a counterexample?

Answer 1

Alternately, for any such $\Gamma$ and $C$, we always have a short exact sequence $$0 \to O_{\Gamma} \to f_* O_C \to F \to 0,$$ where $F$ is a finite length $\mathcal{O}_{\Gamma}$-module supported where $\Gamma$ is non-smooth.

Twisting by a very positive (ie, high multiple of an ample) Cartier divisor $L$ and taking cohomology we get $$ 0 \to H^0(\Gamma, O_{\Gamma}(L)) \to H^0(C, O_{C}(f^*L)) \to H^0(\Gamma, F(L)) \to H^1(\Gamma, O_{\Gamma}(L))$$

Now, $H^0(\Gamma, F(L)) \neq 0$ but $H^1(\Gamma, O_{\Gamma}(L)) = 0$ by Serre vanishing. And so you have a counter-example.

Answer 2

Here is a concrete counterexample.

Let $X$ be a degree $4$ plane curve with a unique node $r \in X$. The normalization $f \colon \tilde{X} \to X$ is a genus $2$ curve, hence hyperelliptic. Let $p, q \in \tilde{X}$ be two points that are conjugate under the hyperelliptic involution and do not map to $r$ under the normalization map. Considering $p$ and $q$ as points of $X$, I claim that the line bundle $\mathcal{L} := \mathcal{O}_{X}(p+q)$ has the desired properties. Indeed, I claim that we have $h^{0}(X, \mathcal{L})=1$, but $h^{0}(\tilde{X}, f^{*}(\mathcal{L}))=2$.

For the second equality, just observe that the line bundle $f^{*}(\mathcal{L})$ is just the line bundle associated to a Cartier divisor given by a conjugate pair of points. In other words, this line bundle is the pullback of $\mathcal{O}(1)$ on $\mathbb{P}^1$ under the degree $2$ map $\tilde{X} \to \mathbb{P}^1$. This has a $2$-dimensional space of global sections, coming from the global sections over $\mathbb{P}^1$.

The first equality can be seen using Riemann-Roch. Examining the Riemann-Roch formula, we see this equality is equivalent to the equality $h^{0}(X, \omega(-p-q))=1$, where $\omega$ is the dualizing sheaf. But $\omega$ is just the restriction of $\mathcal{O}(1)$ on $\mathbb{P}^2$ (adjunction!), so the vector space in question is just homogeneous polynomials in $X, Y, Z$ of degree $1$ that vanish at two distinct points. Linear algebra shows that this is a $1$-dimensional space.

Answer 3

A trivial counterexample is given by a reducible plane curve, and the structure sheaf.

Here's another hands on attempt for an irreducible curve. Think of a line bundle on Gamma as obtained from a line bundle on C, by identifying the fibers over identified points of C, via a non zero scalar multiple. Then starting from the trivial line bundle on C, we can identify the fibers over two distinct points which are identified on Gamma, by multiplication say by 2.

then the resulting line bundle on Gamma pulls back to the trivial bundle on C, all of whose sections are constant. But no non zero constant section of the trivial bundle on C can glue together by multiplication by 2, to give a section on Gamma.

Thus there are non trivial line bundles on Gamma with no non zero sections, but whose pull backs to C are the trivial bundle with non zero constant sections.

This should be essentially the same as the long exact sequence answer above.

It is precisely the existence of such non trivial bundles on say a nodal Gamma that pull back to trivial bundles on C that give rise to the structure of the jacobian of C as an "extension" of the Jacobian of Gamma, by copies of the multiplicative group of the scalar field.

I hope this is ok. One could consult Serre's Algebraic groups and class fields, for a discussion of generalized Jacobians on irreducible singular curves, in terms of their normalizations.