Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

assume we have a curve X of any genus with one non separating node. Call it $x$. Which is the explicit form of the boundary map $ \mathbb{G}_m\rightarrow H^1(X,\mathcal{O}_X^{*}) $ It seems to me that I have to do the following: given a covering of the curve $\{U_{\alpha}\}$ and $\lambda\in\mathbb{G}_m$ I have to extend locally on the normalization to $\{f_{\alpha}\}$ with $\frac{f_{\alpha}(p)}{f_{\alpha}(q)}=\lambda$, where $p$ and $q$ are the points mapped to the node and then I have to consider the line bundle on $X$ with transition functions $g_{\alpha\beta}=\frac{f_{\alpha}}{f_{\beta}}$. This is correct? How is it described geometrically the action of $\mathbb{G}_m$ on $H^1(X,\mathcal{O}_X^*)$? Why if I have a separating node I get an isomorphism between the Picard groups of $X$ and the one of its normalization?

share|cite|improve this question
up vote 5 down vote accepted

Let $\widetilde{X}$ be the normalization of $X$, and let $x_0$ and $x_1$ be the two points of $\widetilde{X}$ lying above the node $x$.

If $\widetilde{\mathcal L}$ is a line bundle on $\widetilde{X}$ and $\phi$ is an element of Isom$(\widetilde{\mathcal L}_{x_0},\widetilde{\mathcal L}_{x_1})$, then you can make a line bundle $\mathcal L$ on $X$ by identifying the fibres $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$ via $\phi$.

Note that $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$ are both one-dimensional vector spaces, and so the Isom space between them is non-canonically identified with $\mathbb G_m$. More canonically, it is a $\mathbb G_m$-torsor, i.e. we can multiply any isomorphism $\phi$ by an element of $\mathbb G_m$, and this makes the Isom space a principal homogeneous space for $\mathbb G_m$.

Conversely, if $\mathcal L$ is a line bundle on $X$, then its pull-back to $\widetilde{X}$ is a line bundle $\widetilde{L}$, equipped with an isomorphism $\phi$ between $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$. (Because both are canonically identified with the fibre $\mathcal L_x$.)

Thus line bundles on $X$ are the same as pairs $(\widetilde{L},\phi)$ where $\widetilde{L}$ is a line bundle on $\widetilde{X}$ and $\phi$ is an isomorphism between $\widetilde{\mathcal L}_{x_0}$ and $\widetilde{\mathcal L}_{x_1}$.

The $\mathbb G_m$-action is just the action on $\phi$ given by scaling.

The boundary map is given by taking $\widetilde{L} = \mathcal O_{\widetilde{X}}$, the trivial line bundle on $\widetilde{X}$. In this case, the fibres at $x_0$ and $x_1$ are both canonically equal to $k$ (the ground field), and so the Isom space between them is canonically $\mathbb G_m$.

Thus we get $\mathbb G_m$ being canonically identified with those $\mathcal L$ for which $\widetilde{\mathcal L}$ is trivial. (Actually, this identification, and also the $\mathbb G_m$-action, depend on the labelling of $x_0$ and $x_1$. If we switch those two points, it is the same as applying $\lambda \mapsto \lambda^{-1}$ to $\mathbb G_m$.)

If $x$ is a separating node, then the situation is different: the normalization is not connected, it has two pieces, say $X_0$ and $X_1$, each with a unique point $x_0$ and $x_1$ lying over $x$. To get a line bundle on $X$ we have to choose $\mathcal L_0$ and $\mathcal L_1$ on $X_0$ and $X_1$ and then identify the fibre of $\mathcal L_0$ at $x_0$ with the fibre of $\mathcal L_1$ at $x_1$. Again, this identification is only determined up to scaling by an element of $\mathbb G_m$, but if we choose two different identifications, which say differ by a scaling factor of $\lambda$, we nevertheless get the same line bundle on $X$, the reason being that we can apply the automorphism $1 \times \lambda$ to the line bundle $\mathcal L_0 \coprod \mathcal L_1$ on $X_0 \coprod X_1$ (i.e. the automorphism which is $1$ on the first factor and multiplication by $\lambda$ on the second factor), and this will take the first identification to the second.

share|cite|improve this answer

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.