For $\mu \ge 0,$ the $A_{\mu}$ series of map germs $f : (\mathbb{R}^n,0) \to (\mathbb{R},0)$ is, for $\varepsilon_i = \pm 1$, given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2 \pm \varepsilon_nx_n^{\mu x_n^{\mu +1}$. These all have algebraically isolated singularities at $0 \in \mathbb{R}^n$. I this setting, finite Milnor number is equivalent to the singularity being isolated. For $\mu = \infty$, the $A_{\infty}$ singularity is non-isolated, and is given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2$. Notice the lack of $x_n$. If you're working over $\mathbb{C}$ then drop all of the $\pm$s.
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For $\mu \ge 0,$ the $A_{\mu}$ series of map germs $f : (\mathbb{R}^n,0) \to (\mathbb{R},0)$ is, for $\varepsilon_i = \pm 1$, given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2 \pm \varepsilon_nx_n^{\mu +1}$. These all have algebraically isolated singularities at $0 \in \mathbb{R}^n$. I this setting, finite Milnor number is equivalent to the singularity being isolated. For $\mu = \infty$, the $A_{\infty}$ singularity is non-isolated, and is given by $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2$. Notice the lack of $x_n$. If you're working over $\mathbb{C}$ then drop all of the $\pm$s. |
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