MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 3 added 18 characters in body

The two maps are naturally one the "dual" of the other. Let me give an immediate proof when $k=\mathbf{C}$.

Over the complex numbers one has $$J(C)=H^0(C, \Omega)^*/H_1(C, Omega^1_C)^*/H_1(C, \mathbf{Z}),$$ hence $T_0J(C)=H^0(C, \Omega)^*$.

It follows that Omega^1_C)^*$.

Therefore the automorphism $\sigma \colon C \to C$ induces an automorphism $\bar{\sigma} \colon H^0(C, \Omega)^* Omega^1_C)^* \to H^0(C, \Omega)^*$. Omega^1_C)^*$.

Dualizing it, we obtain the automorphism $\bar{\sigma}^* \colon H^0(C, \Omega) Omega^1_C) \to H^0(C, \Omega).$Omega^1_C).$
show/hide this revision's text 2 edited body

The two maps are naturally one the "dual" of the other. Let me give an immediate proof when $k=\mathbb{C}$.k=\mathbf{C}$.

Over the complex numbers one has $$J(C)=H^0(C, \Omega)^*/H_1(C, \mathbf{Z}),$$ hence $T_0J(C)=H^0(C, \Omega)^*$.

It follows that the automorphism $\sigma \colon C \to C$ induces an automorphism $\bar{\sigma} \colon H^0(C, \Omega)^* \to H^0(C, \Omega)^*$.

Dualizing it, we obtain the automorphism $\bar{\sigma}^* \colon H^0(C, \Omega) \to H^0(C, \Omega).$
show/hide this revision's text 1

The two maps are naturally one the "dual" of the other. Let me give an immediate proof when $k=\mathbb{C}$.

Over the complex numbers one has $$J(C)=H^0(C, \Omega)^*/H_1(C, \mathbf{Z}),$$ hence $T_0J(C)=H^0(C, \Omega)^*$.

It follows that the automorphism $\sigma \colon C \to C$ induces an automorphism $\bar{\sigma} \colon H^0(C, \Omega)^* \to H^0(C, \Omega)^*$.

Dualizing it, we obtain the automorphism $\bar{\sigma}^* \colon H^0(C, \Omega) \to H^0(C, \Omega).$