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In a paper the author lists, without justification, generators for a Lie algebra. I would be grateful if someone could justify these choices and perhaps suggest how I might have found them for myself.

Consider the three-dimensional affine space with a chosen coordinate system $(x, y, z)$ and the seven-dimensional group, $\text{Aff}_7$ of orientation-preserving linear transformations of the space that map the half-space $z \ge 0$ into itself.

The space of all $k$-jets of surfaces tangent to the plane $z = 0$ at the point $(0, 0, 0)$ is called the small space of $k$-jets and is denoted by $J_k.$

Consider the fibration $\pi : J_3 \twoheadrightarrow J_2$ where, for homogeneous polynomials $P_k$ of degree $k$, we have $\pi(P_2+P_3)=P_2$. Let $e$ be the fibre $\pi^{-1}(x^2+y^2) = x^2 + y^2 + P_3$. Let $E$ be the maximum subgroup of $\text{Aff}_7$ under which the fiber $e$ is invariant. The author goes on to say that a basis for the Lie algebra of $E$ has the basis: $$\begin{eqnarray*} v_1 &=& z\frac{\partial}{\partial x} \\ \\ v_2 &=& z\frac{\partial}{\partial y} \\ \\ v_3 &=& -\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+2z\frac{\partial}{\partial z}\right) \\ \\ v_4 &=& x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x} \end{eqnarray*}$$ As I mentioned above: I would be grateful if someone could justify these choices and perhaps suggest how I might have found them for myself.

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I would try the following: observe that $E$ can also be described as the subgroup of $\text{Aff}_7$ under which the pullback of the polynomial $p:=z-x^2-y^2$ vanishes up to oder 3 along the zero set $z-x^2-y^2=0$. This translates into: $p$ is pulled back to an element in the ideal generated by $p$ up to terms of oder 3 or more (so formally an element of the ideal $(p)$ in the quotient ring of polynomials modulo ideal $(x^3,y^3,z^3)$). If we let $V$ denote the Lie algebra of $\text{Aff}_7$ and $W$ the Lie Algebra of $E$, then in terms of Lie algebras the above means that $W$ consists of those derivations $v\in V$ which applied to $p$ give an element in $(p)$ mod $(x^3,y^3,z^3)$. Now the problem should reduce to a linear algebra problem by noting that $V$ has generators $z\partial_z, x\partial_x, y\partial_x, z\partial_x, x\partial_y, y\partial_y, z\partial_y$ and the ideal $(p)$ mod $(x^3,y^3,z^3)$ has a vector basis $p,z\cdot p,x\cdot p,y\cdot p$. This might need some expanding. I haven't done these computations but checked that the basis given in your question satisfies $v_1(p)=2xp+P_3$, $v_2(p)=2yp+P_3$, $v_3(p)=-2p$ and $v_4(p)=0$.

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