Prop 2.50. Let \(\varphi:A\to B\) a filtered algebra homomorphism. Then
\(\mathrm{gr}\varphi\) injective \(\Rightarrow \varphi\) injective.
\(\mathrm{gr}\varphi\) surjective \(\Rightarrow \varphi\) surjective.
Suppose \(\varphi(A_k)=\varphi(A)\cap B_k\) for all \(k\). Then \(\varphi\) injective / surjective \(\Leftrightarrow \mathrm{gr}\varphi\) injective / surjective.
Let \(I\subset A\) be a left ideal, define
\[\mathrm{gr}I = \bigoplus_{k=0}^\infty \frac{I\cap A_k}{I\cap A_{k-1}} \subset \mathrm{gr}A.\]
\(\mathrm{gr}I\) is a left ideal of \(\mathrm{gr}A\).
\(\mathrm{gr}\) preserves inclusions of left ideals.
If \(I\subset J\) and \(\mathrm{gr}I = \mathrm{gr}J\), then \(I=J\).
From this we can prove \(\mathcal{U}(L)\) is Noetherian, since \(\mathrm{gr}\mathcal{U}(L) \cong S(L)\) is Noetherian by Hilbert’s basis theorem.
Consider the universal enveloping algebra of a semisimple Lie algebra \(\mathfrak{g}\) over \(\mathbb{C}\). To describe its center, consider
\[\mathrm{Aut}(\mathfrak{g}) = \{\phi \in \mathrm{GL}(\mathfrak{g}) : \phi([x,y]) = [\phi(x),\phi(y)] \text{ for all } x,y\in \mathfrak{g}\}.\]
Lemma (Nilpotent adjoint generates automorphism). For \(x\in \mathfrak{g}\) such that \(\mathrm{ad}x\) is nilpotent, then \(\exp(\mathrm{ad}x) \in \mathrm{Aut}(\mathfrak{g})\).
To prove this, the essential part is to verify that
\[\exp(\mathrm{ad}x)([y,z]) = [\exp(\mathrm{ad}x)(y), \exp(\mathrm{ad}x)(z)].\]
And we can start by showing
\[\mathrm{ad}x ([y,z]) = [\mathrm{ad}x(y), z] + [y, \mathrm{ad}x(z)]\]
following from the Jacobi identity.
From here note by binomial expansion we have
\[\mathrm{ad}^n x ([y,z]) = \sum_{k=0}^n \binom{n}{k} [\mathrm{ad}^k x (y), \mathrm{ad}^{n-k} x (z)].\]
Expanding inside the exponential series, we get the desired equality.
\(G\) acts on \(T(\mathfrak{g})\) by acting on each component. This induces an action on \(\mathcal{U}(\mathfrak{g})\) since the generator relations are preserved.
It’s infinitesimal version is \(\mathfrak{g}\) acting by adjoint on itself, giving action on \(T(\mathfrak{g})\) as
\[y\cdot (x_1 \otimes \cdots \otimes x_n) = \sum_{i=1}^n x_1 \otimes \cdots \otimes [y,x_i] \otimes \cdots \otimes x_n.\]
This induces an action on \(\mathcal{U}(\mathfrak{g})\) as well, which actually simplifies to
\[y \cdot u = yu - uy.\]