What kind of singularity is commonly meant by $A_{\infty}$?
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$\begingroup$ mathoverflow.net/questions/14489/… $\endgroup$– Francesco PolizziCommented May 30, 2012 at 17:01
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$\begingroup$ It would help if you said what type of singularity you think $A_n$ means; Fly by Night's answer below is using a different notion of $A_n$ than the question Francesco links. $\endgroup$– David E SpeyerCommented Aug 1, 2012 at 15:05
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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 $f(x) = \varepsilon_1x_1^2 + \cdots + \varepsilon_{n-1}x_{n-1}^2 \pm x_n^{\mu +1}$. 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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$\begingroup$ @David: They're the same notation. In Francesco's link $n = 2$ and we get the $A_{\mu}$ series given by $f(x) = x_1^2 + x_2^{\mu+1}$, with $A_{\infty}$ being $f(x) = x_1^2$. $\endgroup$ Commented Aug 1, 2012 at 15:13