Hi Does any one know of a necessary and sufficient condition for a curve to have a singularity of type A_k.

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


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.

A necessary and sufficient condition for a curve to have an A_1 node is the following

df:= (f_x, f_y) = 0

Hessian(f) = non degenerate.

This is essentially the morse lemma.

I know conditions for A_2, A_3, ...... until A_6 node. I was wondering if anyone knew something for a general k.


3 Answers 3


At least over $\mathbb{C}$, there is a simple answer.

A plane curve $f(x,y)=0$ has a singularity of type $A_k$ in $o=(0,0)$ if and only if

  • $o$ 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;

  • the $Milnor$ $number$

$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(f_x, f_y)$

is equal to $k$. Here $\mathcal{O}_{o}$ denotes the ring of convergent power series.

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

  • the corank


is $ \leq 1$;

  • the Milnor number

$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(J_f)$

is equal to $k$.

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.

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.

And yes, there are similar conditions for $D_k$ and $E_6$, $E_7$, $E_8$. Let me state the condition for $D_k$.

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:

  • $f^{(3)}$ factors into at least two different factors and $\mu(f, o)=k$;
  • $f$ is of type $D_k$.

Moreover, $f^{(3)}$ factors into three different factors if and only if $f$ is of type $D_4$.

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.

  • $\begingroup$ Hi Francesco Thank you for your reply. Just two further questions. 1) Is there an explicit expression for the Milnor number in terms of the partial derivatives of f? I am only talking about maps from C^2 to C. 2) Are there similar conditions for other singularities? Such as D_k singularity and E_k singularity? Again, only for maps from C^2 to C. $\endgroup$
    – Ritwik
    Jul 26, 2010 at 2:39
  • $\begingroup$ Hi, I edited the reply to answer your new questions. Best, f. $\endgroup$ Jul 26, 2010 at 9:55

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.


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.

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.

(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.)

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.

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 $X^2 \pm Y^4.$

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 $x_{new}^2 \pm y_{new}^n.$

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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.