So, this is an exercise. But from math.stackexchange I have been suggested to post this question here.

To find the Thom-Boardman stratification of the smooth map $f(x,y,a,b,c,d)=x^2y+y^3+a(x^2+y^2)+bx+cy$, where $a,b,c$ are parameters.

As I have seen, this is also known as the $D_4$ singularity.

I don't know if the concept of Thom-Boardman symbols is well-known by the math community. I am trying to follow "Singularities of Differentiable Maps: Volume 1: The Classification of Critical Points Caustics, Wave Fronts". In few words, the set $\Sigma^{i_1}(f)$ consists of all points at which the kernel of $Df$ has dimension $i_1$. Then $\Sigma^{i_1,i_2}(f)=\Sigma^{i_2}(\Sigma^{i_1}(f))$ and so on.

So I started by

$$\dfrac{\partial f}{\partial x}=2xy+2ax+b, \; \dfrac{\partial f}{\partial y}=x^2+3y^2+2ay+c.$$

This means that the critical set of $f$ is the set of points $$\left\{(x,y,a,b,c)\in\mathbb R^5 \, |\, 2xy+2ax+b=0 \text{ and } 3y^2+2ay+c=0 \right\}.$$ Which I "see" as a $3$-dimensional manifold in $\mathbb R^5$. So according to the book, this is $\Sigma^2(f)$.

Next step is to find the critical set of $\Sigma^2(f)$. So we have

$$D\Sigma^2(f)= \begin{bmatrix} 2y+2a & 2x\\ 2x & 6y+2a \end{bmatrix}, $$

and then $|D\Sigma^2(f)|=12y^2+16ay+4a-4x^2$. So,

$$\Sigma^{2,2}(f)=\left\{ (x,y,a,b,c)\in\mathbb \Sigma^2(f) \, |\, 12y^2+16ay+4a-4x^2=0\right\}.$$

Since $\Sigma^2(f)$ is $3$-dimensional, $\Sigma^2\Sigma^2(f)$ as written above is $2$-dimensional. Following the same ideas I obtain

$$\Sigma^{2,2,2}(f)=\left\{ (x,y,a,b,c)\in\mathbb \Sigma^{2,2}(f)\,|\, x=0 \text{ and } 3y=-2a \right\}, $$

which is a line. But then, I was expecting to find the origin in the next step. After all, the origin is a singular point of $f$ and I am trying to compute the symbol of $f$ at $0$. However, the next derivative is constant, therefore it corresponds to the symbol $\Sigma^{2,2,2,0}$. So although the origin is an element of $\Sigma^{2,2,2}(f)$ it is not a strata??? I was expecting that the origin was a $0-$dimensional strata with symbol $\Sigma^{2,2,2,2,0}(f)$ but ...

Maybe it is something I am misunderstanding. Could anyone please help me? Am I really doing something wrong?