Multiplier ideal

Remarks

The condition does not depend on the coordinates, since a change of coordinates multiplies $dV$ by a smooth positive function. The stalks are ideals, so $\mathcal{I}(\varphi)$ is an ideal sheaf; Nadel's theorem says it is coherent. Example: $X = \C$, $\varphi = c \log\abs{z}$ with $c > 0$. In polar coordinates, $\int_{\abs{z} < r} \abs{z}^{2k} \abs{z}^{-2c} \, dV = 2\pi \int_0^r t^{2k - 2c + 1} \, dt$, which is finite exactly when $k > c - 1$. So the stalk at $0$ is $(z^{\floor{c}})$; for example $\mathcal{I}(\log\abs{z})_0 = (z)$. Non-example: for $\varphi = \log\abs{z}$, the germ $f = 1$ is not in the stalk at $0$, since $\int \abs{z}^{-2} \, dV$ diverges near $0$. If $\varphi$ is locally bounded, then $\mathcal{I}(\varphi) = \mathcal{O}_X$. Algebraic form: for an ideal $\mathfrak{a} = (g_1, \dots, g_k)$ and a real $c > 0$, $\mathcal{J}(c \cdot \mathfrak{a}) = \mathcal{I}\left(\tfrac{c}{2} \log \sum_i \abs{g_i}^2\right)$, the germs with $\abs{f}^2 / (\sum_i \abs{g_i}^2)^c$ locally integrable. Some authors use the weight $e^{-\varphi}$ instead of $e^{-2\varphi}$.