Find parametrization for a possible "ruled" surface in $\mathbb R^4$ Let us endow $\mathbb R^4$ with a group law $\cdot$ such that the dilations $\delta_\lambda:(\mathbb R^4,\cdot)\to (\mathbb R^4,\cdot), (x_1,x_2,x_3,x_4)\mapsto (\lambda x_1,\lambda x_2,\lambda^2 x_3,\lambda^3 x_4)$, are automorphisms of the group for each $\lambda \in \mathbb R$.
Now, let $S\subseteq \mathbb R^4$ be an hypersurface given by $$S=\{(x_1,x_2,x_3,x_4)\in \mathbb R^4: f(x_1,x_2,x_3,x_4):=\frac{1}{6}x_2(x_1^2+x_2^2)-\frac{1}{2}x_1 x_3+x_4=0\}.$$
It is easy to see that $S$ is dilation invariant, i.e. $\delta_\lambda(S)=S \,\forall \lambda \in \mathbb R$. Since $S$ has this property, I was trying to see if it is possible to write $S$ as a "ruled" surface, that is writing it with a parametric representation of the form $$S(u_1,u_2,u_3)=p(u_1,u_2)+u_3 r(u_1,u_2) \quad (*).$$
The important thing here is that we have to remember that the dilations we are working with aren't the usual Euclidean ones, so we would like that the parametric representation of $S$ had the following property: for each fixed $(u_1,u_2)$, $u_3\mapsto u_3 r(u_1,u_2)$ is a line which is a subgroup of $(\mathbb R^4,\cdot)$ and it is invariant under dilations of the group, that is the image of $u_3\mapsto u_3 r(u_1,u_2)$ is $\delta_\lambda$-homogeneous subgroup of topological dimension 1.
Now, we know that the only $\delta_\lambda$-homogeneous subgroup of topological dimension 1 in $(\mathbb R^4,\cdot,\delta_\lambda)$ are of the form:


*

*$\{(\alpha t,\beta t,0,0):t\in \mathbb R\}, \alpha,\beta$ fixed;

*$\{(0,0, t,0):t\in \mathbb R\}$;

*$\{(0,0,0,t):t\in \mathbb R\}$.


My question is: is it possible to find a parametric representation of $S$ of this form? I've tried to write $S$ as the image of $$(s_1,s_2,s_3)\mapsto (s_1,s_2,s_3,-\frac{1}{6}s_2(s_1^2+s_2^2)+\frac{1}{2}s_1 s_3),$$ but it doesn't seem possible to manipulate this parametrization in order to obtain a parametric representation of the form $(*).$ Am I wrong? Also, one can explicit $x_3$ in function of the other variables, but also using the parametrization with $x_3$ explicit I can't arrive to an expression of the form $(*)$. 
Do you see a way to manage this problem? Or do you think it is not possible to do something like that? 
 A: Not entirely sure I understand your goal, but this is too long for a comment.
For each $(x_{1}, x_{2}, x_{3}, x_{4})$, the curve
$$
\lambda \mapsto (\lambda x_{1}, \lambda x_{2}, \lambda^{2} x_{3}, \lambda^{3} x_{4})
\tag{1}
$$
is a twisted cubic transverse to each of the hyperplanes $\{x_{i} = 1\}$. The hypersurface $S$ may therefore be parametrized by parametrizing the $2$-dimensional intersection $S \cap \{x_{i} = 1\}$ (for any convenient $i$), and then using $\lambda$ in (1) as third parameter.
For example, the surface $S \cap \{x_{3} = 1\}$, which has equation
$$
\tfrac{1}{6} x_{2}(x_{1}^{2} + x_{2}^{2}) - \tfrac{1}{2} x_{1} + x_{4} = 0,
$$
may be parametrized by
$$
(u_{1}, u_{2}) \mapsto \bigl[u_{1}, u_{2}, 1, \tfrac{1}{2} u_{1} - \tfrac{1}{6} u_{2}(u_{1}^{2} + u_{2}^{2})\bigr].
$$
The hypersurface $S$ may therefore be parametrized by
$$
(u_{1}, u_{2}, \lambda) \mapsto \bigl[\lambda u_{1}, \lambda u_{2}, \lambda^{2}, \lambda^{3}\bigl(\tfrac{1}{2} u_{1} - \tfrac{1}{6} u_{2}(u_{1}^{2} + u_{2}^{2})\bigr)\bigr].
$$
Is this the type of parametrization you want?
