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Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove. Edit: This looks like some work with homological algebra.

Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove.

Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove.

Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove. Edit: This looks like some work with homological algebra.

Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to show.

Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove.