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Let $D$ be the open unit ball of $\mathbb{C}^{n}$, and $\tilde D$ be the submanifold of $D \times \mathbb CP^{n - 1}$ given by $\{(z, l) \in U \times \mathbb CP^{n - 1} | z^{i}l^{j} = z^{j} l^{i}\}$ where $[l_1: \ldots l_n]$ are the homogeneous coordinates on $\mathbb CP^{n - 1}$. Denote the projection $(z, l) \to z$ by $\pi$. We can cover $\tilde D$ by $\tilde D_{i} = \{ l^{i} \neq 0\}$ with local coodinates $\tilde z(i)^{j} = \frac{l^{j}}{l^{i}} = \frac{z^{j}}{z^{i}}$ for $j \neq i$ and $\tilde z(i)^{i} = z^{i}$. Denote $E := \pi^{-1}(0)$. The claims is that $E$ on $\tilde D_{i}$ is given by $\tilde z(i)^{i} = 0$.

Basically, $E \cap \tilde D_{i} = \{(0, l) \in D \times \mathbb CP^{n - 1} | l_{i} \neq 0\}$, but $\tilde z(i)^{i} = 1$ by definition. How would $\tilde D_{i}$ ever be given by $\tilde z(i)^{i} = 0$?

Adam
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  • Something is odd about the notation: There are only $n-1$ non-constant $\tilde{z}(i)^{j}$, but $\tilde{D}$ is $n$-dimensional. Be that as it may, the geometric idea is you're taking lines through the origin of $D$ but "indexing" the origins of distinct lines by points of the projective space. The exceptional divisor $E$ is the set of origins. (The diagrams here may be helpful.) – Andrew D. Hwang Apr 05 '21 at 01:50
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    You are right. I overlooked that part and now it all makes sense. Thank you. – Adam Apr 05 '21 at 04:32

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