MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.