# pullback of poincare bundle

Let $X$ be a smooth curve and ${x,y}$ are two smooth points. Let $J_X$ be the jacobian of X i.e, the variety which parametrizes the degree 0 line bundles on X and let $\mathcal{P}$ is the poincare bundle over $X\times J_X$. We denote by $\mathcal{P}_x$ and $\mathcal{P}_y$ the restrictions of $\mathcal{P}$ along $x\times J_X$ and $y\times J_X$ respectively. There is an isomorphism $\phi_{y-x}:J_X\rightarrow J_X$ defined by $L\mapsto L\otimes \mathcal{O}_X(y-x)$. My question is: Is $\phi_{y-x}^*\mathcal{P}_x\cong \mathcal{P}_y$?

• This might depend on what you call the Poincaré bundle. For the usual definition, $\mathcal{P}_x$ is algebraically equivalent to zero, and therefore invariant by translation. – abx Oct 1 '17 at 9:35
• @abx why is $\mathcal{P}_x$ algebraically equivalent to 0? – Kumar Oct 1 '17 at 9:57
• @Kumar. What is your definition of $J_X$? Does it have an origin? If it has an origin $0$, what is the invertible sheaf $\mathcal{P}|_{X\times \{0\}}$? Is that sheaf algebraically equivalent to $0$? At any rate, if you are not defining $J_X$ to represent isomorphism classes of invertible sheaves that are algebraically equivalent to zero, then you need to specify what is your definition of $J_X$. – Jason Starr Oct 1 '17 at 10:08
• @Jason Starr $J_X$ is the jacobian of X (degree 0 line bundles)...sorry i forgot to mention....but then why $\mathcal{P}_x$ algebrically equivalent to 0 in this case? – Kumar Oct 1 '17 at 10:13
• @Kumar. There is no single definition of $\mathcal{P}$. However, for the most common normalizations of $\mathcal{P}$, also$\mathcal{P}|_{\{x\}\times J_X}$ is algebraically equivalent to zero. For instance, one common normalization is to choose one rational point $x_0$ of $X$ and impose that $\mathcal{P}|_{\{x_0\} \times J_X}$ is the structure sheaf on $J_X$. Since $X$ is connected, it then follows that $\mathcal{P}|_{\{x\}\times J_X}$ is algebraically equivalent to zero for every $x$. – Jason Starr Oct 1 '17 at 10:27

I am just writing the comments as an answer. First of all, the essential point made by abx is that for an invertible sheaf $\mathcal{L}$ on an Abelian variety, if $\mathcal{L}$ has degree $0$, then the pullback of $\mathcal{L}$ by each translation is isomorphic to $\mathcal{L}$. For one reference, this follows from Facts (i) and (iv) on pp. 75-76 of the following.

MR2514037 (2010e:14040)
Mumford, David
Abelian varieties.
With appendices by C. P. Ramanujam and Yuri Manin.
Corrected reprint of the second (1974) edition.
Tata Institute of Fundamental Research Studies in Mathematics, 5.
Published for the Tata Institute of Fundamental Research, Bombay;
by Hindustan Book Agency, New Delhi, 2008. xii+263 pp.

Mumford uses a definition of $\text{Pic}^0$ that might not agree with every definition. By (iv), for any other definition of the degree of an invertible sheaf, so long as it is compatible with pullback by the "multiplication-by-$n$" maps on the Abelian variety, it follows that "degree zero" implies that the invertible sheaf is in $\text{Pic}^0$ with Mumford's definition. Then (i) implies that the invertible sheaf is stable for translations.

It is a matter of convention how to define a Poincare sheaf, and different authors follow different conventions. In the sources that I know, every Poincare sheaf on $X \times \text{Pic}^0(X)$ is normalized so that its restrictions to fibers $\{x\}\times \text{Pic}^0(X)$ are each invertible sheaves of degree $0$, i.e., parameterized by $\text{Pic}^0$.

• thanks...just another question if I replace $J^0_X$ by $J^1_X$ , still $\phi_{y-x}^*\mathcal{P}_x\neq \mathcal{P}_y$....because an isomorphism $\phi$ of $J^1_X$ can always be induced by an isomorphism of $J^0_X$.. – user111251 Oct 1 '17 at 17:34
• @ZSun. I do not understand your question. – Jason Starr Oct 1 '17 at 17:50
• $J^1_X$ is the moduli of line bundles of degree 1. Let $\mathcal{P}$ be the poincare bundle on $J^1_X$. Define an isomorphism $\phi_{y-x}:J^1_X\rightarrow J^1_X$ defined by $L\mapsto L\otimes \mathcal{O}_X(y-x)$. My question is: Is $\phi_{y-x}^*\mathcal{P}_x\cong \mathcal{P}_y$? .....$x$ and $y$ are not the point $x_0$. – user111251 Oct 1 '17 at 18:40
• @ZSun. What is your definition of the Poincare sheaf in that case? If you "normalize" so that $\mathcal{P}_{\{x_0\}\times \text{Pic}^d(X)}$ is algebraically equivalent to zero, then for every $x\in X$, also $\mathcal{P}_{\{x_0\}\times \text{Pic}^d(X)}$ is algebraically equivalent to zero. In that case, these invertible sheaves are invariant under pullback on $\text{Pic}^d(X)$ by translation by a rational point of $\text{Pic}^0(X)$. – Jason Starr Oct 1 '17 at 18:46