# Geometric interpretation of the trace of a linear system in Hartshorne's Algebraic Geometry

In Hartshorne's Algebraic geometry, page 158 gives the definition of the trace of $\mathfrak d$ on $Y$, where $i: \, Y \to X$ is a closed immersion of nonsingular projective varieties over an algebraically closed field $k$ and $\mathfrak d$ is a linear system on $X$. Let $\mathfrak d$ corresponds to an invertible sheaf $\mathscr L$ on $X$ and a sub-vector space $V \subset \Gamma(X, \mathscr L)$ over $k$. Then the trace $\mathfrak d \big |_Y$ of $\mathfrak d$ on $Y$ is defined as the linear system corresponding to $i^* \mathscr L$ and the image of $V$ under the natural map $\Gamma(X, \mathscr L) \to \Gamma (Y, i^* \mathscr L)$ which comes from the natural map $\mathscr L \to i_* i^* \mathscr L$.

I have some difficulty to understand its geometrical interpretation of $\mathfrak d \big |_Y$. The book says $\mathfrak d \big |_Y$ consists of all divisor $D\cdot Y$ where $D \in \mathfrak d$ is a divisor whose support does not contain $Y$. I do not understand why it is true. I really appreciate if someone could show me some details on it. One more question is: if the support of a divisor corresponding to $s \in \mathfrak d$ does contain $Y$, I think the image of $s$ under the map $\Gamma(X, \mathscr L) \to \Gamma (Y, i^* \mathscr L)$ should be $0$. I do not know if it is true or not.

Thank you very much.

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## 1 Answer

Usually, in order to have a better geometrical intuition, it is useful to identify an invertible sheaf $\mathscr{L}$ with a line bundle on $X$.

Then, if $i \colon Y \to X$ is a closed immersion of smooth projective varieties, $i^* \mathscr{L}$ is (essentially by definition) the restriction of the line bundle $\mathcal{L}$ to $Y$, and the natural map $\Gamma(X, \, \mathscr{L}) \to \Gamma(Y, \, i^*\mathscr{L})$ is given by the restriction of global sections. In particular, the zero set of any section $\sigma \in V \subset \Gamma(X, \, \mathscr{L})$ gives a divisor on $Y$ belonging to the trace $\mathfrak{d}|_Y,$ where $\mathfrak{d}$ is the linear system on $X$ corresponding to $V$.

Your last guess is true. Indeed, there is a short exact sequence of coherent sheaves on $X$ $$0 \longrightarrow \mathscr{L} \otimes \mathscr{I}_Y \longrightarrow \mathscr{L} \longrightarrow i_* i^*\mathscr{L} \longrightarrow 0$$ which induces, passing to global sections, $$0 \longrightarrow \Gamma(X, \,\mathscr{L} \otimes \mathscr{I}_Y) \longrightarrow \Gamma(X, \,\mathscr{L}) \longrightarrow \Gamma(Y,\,i^*\mathscr{L}).$$ So the kernel of the restriction map $\Gamma(X, \,\mathscr{L}) \longrightarrow \Gamma(Y,\,i^*\mathscr{L})$ is precisely $\Gamma(X, \,\mathscr{L} \otimes \mathscr{I}_Y)$, that is the subvector space of global sections $\sigma$ which are identically zero on $Y$. This means that the corresponding divisor $s=\textrm{div}(\sigma)$ contains $Y$ in its support.

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Thank you Francesco for your response. Could you please correct me if my understanding your explanation is wrong: That any section $\sigma$ in $\Gamma(X, \mathscr L \otimes \mathscr I_Y)$ is zero on $Y$ means the image of $\sigma$ is $0$ under the natural map $\Gamma(X, \mathscr L \otimes \mathscr I_Y) \to \Gamma(Y, i^*(\mathscr L \otimes \mathscr I_Y))$, which is true basically because $I \otimes_A A/I =0$ for any ideal $I$ in a commutative ring $A$. – user41541 Oct 21 '13 at 0:52