$A_{\infty}$ singularity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:46:42Z http://mathoverflow.net/feeds/question/98392 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98392/a-infty-singularity $A_{\infty}$ singularity IMeasy 2012-05-30T15:39:01Z 2012-08-25T19:49:37Z <p>What kind of singularity is commonly meant by $A_{\infty}$?</p> http://mathoverflow.net/questions/98392/a-infty-singularity/103648#103648 Answer by Fly by Night for $A_{\infty}$ singularity Fly by Night 2012-08-01T00:52:58Z 2012-08-25T19:49:37Z <p>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.</p>