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Parsa
Parsa
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By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1} $.

The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.

Read pages 74-75 in Mumford's book very carefully.

By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1} $.

The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.

By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1} $.

The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.

Read pages 74-75 in Mumford's book very carefully.

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Parsa
Parsa
  • 631
  • 3
  • 13

By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1} $.

The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.