The associated variety is a closed conical coadjoint-invariant cone

Given: A finite-dimensional complex Lie algebra $\mathfrak{g}$, a two-sided ideal $I⊴U(\mathfrak{g})$, the associated graded ideal $\mathrm{gr}I\subseteq S(\mathfrak{g})=\mathbb{C}[{\mathfrak{g}}^{\ast }]$, and the associated variety $\mathcal{V}(I)=\{\xi \in {\mathfrak{g}}^{\ast }:f(\xi )=0\text{ for all }f\in \mathrm{gr}I\}$.

$\mathcal{V}(I)$ is the zero locus of the family $\mathrm{gr}I$ in the polynomial algebra on ${\mathfrak{g}}^{\ast }$, hence a Zariski closed classical affine algebraic set; $\mathrm{gr}I$ is a graded ideal of $S(\mathfrak{g})$ (The associated graded variety of a two-sided ideal, Classical affine zero loci form the Zariski closed sets, Classical affine algebraic sets, including the empty boundaries).

For every $x\in \mathfrak{g}$, the derivation $D_x$ of $S(\mathfrak{g})$ extending $y\mapsto [x,y]$ satisfies $D_x(\mathrm{gr}I)\subseteq \mathrm{gr}I$; hence $D_x^k(\mathrm{gr}I)\subseteq \mathrm{gr}I$ for all $k\ge 0$ (The adjoint action preserves the associated graded of a two-sided ideal).

Fix a basis of $\mathfrak{g}$ and use its maximum-coordinate norm and the induced submultiplicative operator norm on $\mathrm{End}(\mathfrak{g})$. The exponential series for $-t{\mathrm{ad}}_x$ is dominated on every compact $t$-disc by the scalar series ${\sum }_{k\ge 0}(\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }{\mathrm{ad}}_x\mathrm{\parallel }{)}^k\mathrm{/}k!$, which converges for every $t$ (The exponential series converges absolutely for every real argument); it therefore converges locally uniformly, may be differentiated termwise, and has entire matrix entries (A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation, The sum of a complex power series is analytic throughout its open disc of convergence, Every complex analytic function is holomorphic). Composing those entries with the polynomial $f$ and evaluating at $\xi$ shows $\phi (t)=f(\xi \circ e^{-t{\mathrm{ad}}_x})$ is entire (Complex analytic functions are closed under finite linear combinations, products, quotients with nonzero denominator, and composition). Put $E(t)=e^{-t{\mathrm{ad}}_x}$ and ${\xi }_t=\xi \circ E(t)$. Termwise differentiation gives $E^{\mathrm{\prime }}(t)=-E(t){\mathrm{ad}}_x$. On a linear polynomial $y\in \mathfrak{g}$, this yields $\frac{d}{dt}y({\xi }_t)=-(D_{x}y)({\xi }_t)$; the product and sum rules extend the identity to every polynomial. Iteration gives ${\phi }^{(k)}(0)=(-1{)}^k(D_x^{k}f)(\xi )$ (Linearity, product, reciprocal, and quotient rules for complex derivatives). Absolute convergence permits multiplying the exponential series entrywise; collecting total degree and using the binomial formula gives $E(s)E(t)=E(s+t)$, so $E(0)=1$ and $E(-t)=E(t{)}^{-1}$. Thus the dual maps $\xi \mapsto {\xi }_t$ form the asserted one-parameter group. Finally, an entire function is equal on $\mathbb{C}$ to its Taylor series at $0$, so if all these derivatives vanish then $\phi \equiv 0$ (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain).