A necessary and sufficient condition for a curve to have an $A_k$ singularity. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:15:19Z http://mathoverflow.net/feeds/question/33236 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33236/a-necessary-and-sufficient-condition-for-a-curve-to-have-an-a-k-singularity A necessary and sufficient condition for a curve to have an $A_k$ singularity. Ritwik 2010-07-24T21:47:50Z 2012-10-03T14:22:17Z <p>Hi Does any one know of a necessary and sufficient condition for a curve to have a singularity of type A_k. </p> <p>More precisely, a curve f=0 has a singularity of type A_k at a point, if there exist local coordinates (x,y), where the function can be written as</p> <p>f(x,y)=x^2+y^{k+1}=0.</p> <p>If you understand what I am talking about, you need not read the rest. But in case you don't follow the question, let me elaborate a bit. </p> <p>A necessary and sufficient condition for a curve to have an A_1 node is the following</p> <p>df:= (f_x, f_y) = 0</p> <p>Hessian(f) = non degenerate. </p> <p>This is essentially the morse lemma.</p> <p>I know conditions for A_2, A_3, ...... until A_6 node. I was wondering if anyone knew something for a general k.</p> http://mathoverflow.net/questions/33236/a-necessary-and-sufficient-condition-for-a-curve-to-have-an-a-k-singularity/33262#33262 Answer by Francesco Polizzi for A necessary and sufficient condition for a curve to have an $A_k$ singularity. Francesco Polizzi 2010-07-25T09:39:42Z 2010-07-26T09:54:16Z <p>At least over $\mathbb{C}$, there is a simple answer. </p> <p>A plane curve $f(x,y)=0$ has a singularity of type $A_k$ in $o=(0,0)$ if and only if</p> <ul> <li><p>$o$ is a $double$ $point$, that is all first partial derivatives of $f$ vanish in $o$ but there is at least one second partial derivative which is not zero;</p></li> <li><p>the $Milnor$ $number$ </p></li> </ul> <p>$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(f_x, f_y)$ </p> <p>is equal to $k$. Here $\mathcal{O}_{o}$ denotes the ring of convergent power series. </p> <p>This can be generalized in higher dimensions. In fact, one proves that a (germ of) complex hypersurface singularity $f(x_1, ...,x_n)=0$ is of type $A_k$ if and only if</p> <ul> <li>the corank </li> </ul> <p>$\textrm{crk}(f):=n-\textrm{rank}(\textrm{Hessian}(f))(o)$ </p> <p>is $\leq 1$;</p> <ul> <li>the Milnor number</li> </ul> <p>$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(J_f)$ </p> <p>is equal to $k$. </p> <p>This follows from a sort of generalized Morse Lemma. See the book GREUEL - LOSSEN - SHUSTIN "Introduction to singularities and deformations" p. 150 for the proof.</p> <p>ADDED TO ANSWER THE COMMENT BELOW. I do not know any explicit expression for the Milnor number, I think that in general you cannot avoid to compute the $\mathbb{C}$-basis for the Milnor algebra. I agree that these computations are tedious by hand, however you can use a Computer Algebra software like SINGULAR (which is free) to do this quickly and easily. </p> <p>And yes, there are similar conditions for $D_k$ and $E_6$, $E_7$, $E_8$. Let me state the condition for $D_k$.</p> <p>Let $f \in \boldsymbol{m}^3 \subset \mathcal{O}_o$ and $k \geq 4$. Denote by $f^{(3)}$ the $3$-jet of $f$. Then the following are equivalent:</p> <ul> <li>$f^{(3)}$ factors into at least two different factors and $\mu(f, o)=k$;</li> <li>$f$ is of type $D_k$.</li> </ul> <p>Moreover, $f^{(3)}$ factors into three different factors if and only if $f$ is of type $D_4$.</p> <p>The conditions for $E_6$, $E_7$, $E_8$ are a bit more complicate and I will not state them here. You will find them in the book of GREUEL, LOSSEN and SHUSTIN, p. 154.</p> http://mathoverflow.net/questions/33236/a-necessary-and-sufficient-condition-for-a-curve-to-have-an-a-k-singularity/36654#36654 Answer by inkspot for A necessary and sufficient condition for a curve to have an $A_k$ singularity. inkspot 2010-08-25T10:36:13Z 2010-08-25T10:36:13Z <p>You can almost settle the issue by counting the number of blow-ups necessary to achieve an embedded resolution: a curve of type $A_{2k}$ or $A_{2k-1}$ requires exactly $k$ blow-ups. Then to distinguish between $2k$ and $2k-1$, look at the singularity that you have after $k-1$ blow-ups and decide whether it is $A_1$ or $A_2$, e.g., by following Francesco's suggestion.</p> http://mathoverflow.net/questions/33236/a-necessary-and-sufficient-condition-for-a-curve-to-have-an-a-k-singularity/69068#69068 Answer by Fly by Night for A necessary and sufficient condition for a curve to have an $A_k$ singularity. Fly by Night 2011-06-28T23:26:03Z 2011-06-28T23:26:03Z <p>Although the above answer involving the local algebra and the Milnor number is correct, it is often very hard to apply in real situations. Especially if you have a general function with arbitrary coefficients. You can perform a rather messy iterative process to check for an $A_k.$ In general the condition is far too ugly to want to, or be able to, write down.</p> <p>You have a curve in the plane given by $f(x,y) = 0.$ The Taylor series, with respect to $x$ and $y$ is what you're really interested in. Let's assume we are only interested in the origin. If the linear terms vanish then you know that you have a singular point (a critical point of $f$). In that case you consider the quadratic part. If the quadratic part is non-degenerate, i.e. not a perfect square, then you have Morse singularity. These are $\mathscr{A}$-equivalent to $x^2 \pm y^2,$ and give the so-called $A_1^{\pm}$-singularity types.</p> <p>(Notice that $\mathscr{A}$-equivalence has no relevance to the $A$ in $A_k.$ $\mathscr{A}$-equivalence is also called $\mathscr{RL}$-equivalence. You allow diffeomorphic changes of coordinate in the source and target (right and left sides of the commutativity diagram.)</p> <p>If $f$ has a zero linear part and a degenerate quadratic part, we complete the square on the quadratic part. Then take a change of coordinates that turns the quadratic part into $\tilde{x}^2$. The condition for exactly an $A_2$ is that $\tilde{x}$ does not divide the new, post-coordinate change, cubic term. If not then $f$ is $\mathscr{A}$-equivalent to $\tilde{x}^2 + \tilde{y}^3.$ (There is no $\pm$ because $(x,y) \mapsto (x,-y)$ changes the sign of the cubic term.</p> <p>If $\tilde{x}$ does divide the new cubic term, then you can complete the square on the three jet, i.e. on the quadratic and cubic terms as a whole. You take a change of coordinates so that this completed square become, say $X^2$. The condition for an $A_3^{\pm}$ is that $X$ does not divide the new, post-coordinate change, quadric terms. If not then $f$ is $\mathscr{A}$-equivalent to $X^2 \pm Y^4.$</p> <p>In general you follow the same pattern. Complete the square, take a formal power series change of coordinates so that the perfect square becomes $x_{new}^2.$ Check if $x_{new}$ divides the next set of fixed order terms. If not then stop. If $x_{new}$ didn't divide the order $n$-terms then $f$ is $\mathscr{A}$-equivalent to $x_{new}^2 \pm y_{new}^n.$ </p> <p>You just repeat the pattern: Is the quadratic part degenerate? If so then change coordinates (by a formal power series of low, but sufficient order) so that the degenerate part becomes $x_{new}^2 + O(3).$ Check if $x_{new}$ divides the cubic terms. If not then you have $x^2 + y^3.$ If so then complete the square on the new 3-jet and change coordinates so that you have $x_{new}^2 + O(4)$. Does $x_{new}^2$ divide the quartic terms? If not then you have $x^2 \pm y^4.$ If so then complete the square on the new 4-jet and change coordinates. Just keep completing the square, checking divisibility, changing coordinates. </p> <p>The conditions on the coefficients soon spiral out of control. To check an $A_6$ you'll need a computer program. For a general polynomial it's impossible without a computer. (Except for very special cases!) I wrote a program in Maple to calculate the conditions up to $A_k$ once, but the output was so messy then I gave up. Having said that, for an explicit polynomial it's child's play.</p>