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Consider an elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers  (the field under consideration is $\mathbb{C}$,$X$,$C$ are smooth and projective ). Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1-forms are pull backs of those from the base curve?.

Consider an elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has a reduced singular fiber apart from possibly multiple fibers  (the field under consideration is $\mathbb{C}$,$X$,$C$ are smooth and projective ). Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1-forms are pull backs of those from the base curve?.

Consider an algebraic elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers (the field under consideration is $\mathbb{C}$).Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1-forms are pull backs of those from the base curve?.

Consider an elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers (the field under consideration is $\mathbb{C}$,$X$,$C$ are smooth and projective ).  Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1-forms are pull backs of those from the base curve?.

Consider an algebraic elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers (the field under consideration is $\mathbb{C}$). Now Hodge theory gives us \begin{equation} q(X) = dim_{\mathbb{C}}(H^{0}(X,\Omega^1_X)) = dim_{\mathbb{C}}(H^1(X,\mathcal{O}_X)) \end{equation} On the other hand the $dim_{\mathbb{C}}(H^1(X,\mathcal{O}_X))$ can be seen to be equal to the genus $g$ of $C$. We also have $dim_{\mathbb{C}}H^0(C,K_C) = g$ and hence the canonical inclusion map of sheaves \begin{equation} 0 \rightarrow f^{*}K_c \rightarrow \Omega^1_X \implies H^{0}(C,K_c) = H^{0}(X,\Omega^1_X). \end{equation} Now what I seek is to understand this "equality" in a more geometric way. Is there a better way to see (apart from the above method) to see that all the 1 forms in fact come from the base curve?.

Consider an algebraic elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers (the field under consideration is $\mathbb{C}$).Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1-forms are pull backs of those from the base curve?.