# The Mumford line bundle of $(-1)^* L$

Let $X$ be an abelian variety over a field $k$, $L$ a line bundle on $X$.

Let $\varphi_L : X \to X^t$ be the morphism obtained by considering the Mumford line bundle $\Lambda (L) = m^*L \otimes p_1 ^* L^{-1} \otimes p_2 ^* L^{-1}$ on $X \times X$ as a family of line bundles on the first coordinate parameterized by the second coordinate.

There is a claim that

Direct computation yields $\varphi_{(-1)^* L} (x) = - \varphi _L(x)$ for every $L$ and $x$.

From the formula for $\varphi _L$ and $\Lambda (L)$, I see no reason why this should be true. Moreover, there seems to be an easy counterexample: namely take $L$ to be a symmetric line bundle, i.e. such that $(-1)^* L \cong L$, then this would imply that $\varphi_L = 0$ is constant zero.

• This is not true, since $\varphi_L$ only depends on the numerical class of $L$ in the Néron-Severi group, and $(-1)^*$ acts trivially on numerical classes. What is true is that $\varphi_{L^{-1}}(x)=-\varphi_L(x)$. – rfauffar Jun 8 '16 at 13:41
• Do you mean to say that actually this morphism is the same as $\varphi_L$? – Future Jun 10 '16 at 0:04
• Yes, that is what I'm saying. – rfauffar Jun 12 '16 at 20:46

As you and rfauffar say, the claim is not right. First of all, $T_x^*(-1)^*L \simeq (-1)^*T_{-x}L$. Further, if $M\in X^t={\rm Pic }^0 X$, $(-1)^*M\simeq M^{-1}$ (See Mumford AVs, p 75). The homomorphism $\phi_L$ takes values in $X^t={\rm Pic}^0 X$ (of course!). Therefore (equalities and minus signs in $X^t$) $$\phi_{(-1)^*L}(x) = (-1)^* \left( T_{-x}L\otimes L^{-1}\right)= (-1)\ \phi_L(-x) = (-1)\ (-1)\ \phi_L(x) = \phi_L(x).$$