Proof. It is obvious that A is a quotient of $U(\mathfrak {g})$ and, in particular, finitely generated.

(A) Recall that the Jacobson radical, $\mathop {\mathrm {Rad}} A$ , of A is the intersection of the kernels of all irreducible representations of A. However, the nilpotent radical $\mathfrak {n}$ of the Lie algebra $\mathfrak {g}$ is the intersection of the kernels of all irreducible representations of $\mathfrak {g}$ . Since irreducible representations of $U(\mathfrak {g})$ are in one-to-one correspondence with irreducible representations of $\mathfrak {g}$ and every irreducible representation of A can be lifted to an irreducible representation of $U(\mathfrak {g})$ , it follows that $\mathfrak {n}\subset \mathop {\mathrm {Rad}} A$ .

(B) The Braun–Kemer–Razmyslov theorem asserts that the Jacobson radical of a finitely generated PI-algebra over a commutative Jacobson ring (in particular, over a field) is nilpotent [Reference Kanel-Belov, Karasik and Rowen18, Theorem 4.0.1, page 149]. This completes the proof.

Proof. Let I be an ideal of an algebra A. Suppose that $A/I$ and I satisfy polynomial identities p and q, respectively. Let n and m be the numbers of variables in p and q, respectively. If $a_{ij}\in A$ ( $i=1,\ldots ,n$ , $j=1,\ldots ,m$ ), then $p(a_{1j},\ldots ,a_{nj})\in I$ for every j because $p(a_{1j}+I,\ldots ,a_{nj}+I)\subset I$ . Hence, $q(p(a_{11},\ldots ,a_{n1}),\ldots ,p(a_{1m},\ldots ,a_{nm}))=0$ . It is easy to see that since p and q are not trivial, this noncommutative polynomial is not trivial. Thus A is a PI-algebra.

Proof of Proposition 3

Denote by S the unital subalgebra of A generated by $\mathfrak {s}$ and suppose that A is a PI-algebra. Then S is a quotient of $U(\mathfrak {s})$ and satisfies a PI. Since we work in characteristic $0$ and $\mathfrak {s}$ is semisimple, we can apply a result of Bahturin, which asserts that every quotient of $U(\mathfrak {s})$ that is a PI-algebra is finite dimensional (see [Reference Bahturin6, Theorem 1 and Corollary]). In particular, S is finite dimensional. However, it follows from Lemma 4 that every element of $\mathfrak {n}$ is nilpotent. The necessity is proved.

Suppose now that S is finite dimensional and every element of $\mathfrak {n}$ is nilpotent. Denote by I the ideal of A generated by $\mathfrak {n}$ . From Lemma 7, it suffices to show that I and $A/I$ are PI-algebras.

Denote by $A_0$ the nonunital subalgebra of A generated by $\mathfrak {n}$ and by $U(\mathfrak {n})_0$ the augmentation ideal of $U(\mathfrak {n})$ . Note that $\mathfrak {n}$ is a nilpotent Lie algebra. So by Lemma 5, there is a $d\in \mathbb {N}$ such that $A_0^d=0$ . Since $\mathfrak {n}$ is a Lie ideal of $\mathfrak {g}$ , it easily follows that $U(\mathfrak {g})U(\mathfrak {n})_0=U(\mathfrak {n})_0 U(\mathfrak {g})$ . Then, $(U(\mathfrak {g})U(\mathfrak {n})_0)^d\subset U(\mathfrak {n})_0^d U(\mathfrak {g})$ and therefore $I^d=A_0^d A=0$ . Hence, I is nilpotent and, in particular, a PI-algebra.

On the other hand, being reductive, $\mathfrak {g}/\mathfrak {n}$ is a direct sum of a semisimple summand and an abelian summand $\mathfrak {a}$ [Reference Dixmier9, Proposition 1.7.3, page 27]. Note that the restriction of the map $\mathfrak {g}\to \mathfrak {g}/\mathfrak {n}$ to $\mathfrak {s}$ is injective and so we can identify the semisimple summand in $\mathfrak {g}/\mathfrak {n}$ with $\mathfrak {s}$ . Consider the naturally defined homomorphisms $S\to A\to A/I$ and $U(\mathfrak {a})\to U(\mathfrak {g}/\mathfrak {n})\to A/I$ . Their ranges commute since $U(\mathfrak {s})$ and $U(\mathfrak {a})$ commute in $U(\mathfrak {g}/\mathfrak {n})$ and the diagram

is commutative. By the universal property of tensor products of associative algebras, we have the induced homomorphism $S\otimes U(\mathfrak {a})\to A/I$ . This homomorphism is surjective because $A/I$ and $S\otimes U(\mathfrak {a})$ are quotients of $U(\mathfrak {g})/(U(\mathfrak {g})U(\mathfrak {n})_0)$ and $U(\mathfrak {s})\otimes U(\mathfrak {a})$ , respectively, and the last two algebras are isomorphic being isomorphic to $U(\mathfrak {g}/\mathfrak {n})$ . By Regev’s theorem [Reference Kanel-Belov, Karasik and Rowen18, Theorem 3.4.7, page 138], the class of PI-algebras is stable under tensor products. Thus, $S\otimes U(\mathfrak {a})$ is a PI-algebra since it is the tensor product of a finite-dimensional and abelian algebra. Being a quotient of a PI-algebra, $A/I$ also satisfies a PI. This completes the proof of the sufficiency.

Proof. Note that $\mathfrak {n}=[\mathfrak {g},\mathfrak {r}]$ , where $\mathfrak {r}$ is the solvable radical of $\mathfrak {g}$ [Reference Dixmier9, Proposition 1.7.1, page 26]. Turovskii’s theorem [Reference Turovskii and Maksudov24] asserts that $[\mathfrak {g},\mathfrak {h}]\subset \mathop {\mathrm {Rad}} B$ for every solvable ideal $\mathfrak {h}$ of $\mathfrak {g}$ (for a proof, see [Reference Beltiţă and Şabac7, Section 24, Theorem 1, page 130]). By putting $\mathfrak {h}=\mathfrak {r}$ , we have the result.

Proof. Put $r\!:=e^b-1$ . If $b^d=0$ for some $d\in \mathbb {N}$ , then $r=b+\cdots + b^{d-1}/(d-1)!$ and therefore $r^d=0$ .

Suppose now that there is $d\in \mathbb {N}$ such that $r^d=0$ . Since b is topologically nilpotent, its spectrum is $\{0\}$ . Let f be the function on $\mathbb {C}$ such that $f(z)=(e^z-1)/z$ when $z\ne 0$ and $f(0)=1$ . Since f is holomorphic, it follows from the spectral mapping theorem [Reference Helemskii15, Theorem 2.2.23] that the spectrum of $f(b)$ is $\{1\}$ and therefore $f(b)$ is invertible. Since $bf(b)=e^b-1=r$ and $r^d=0$ , we have $b^df(b)^d=0$ . Since $f(b)$ is invertible, $b^d=0$ .

Proof. If (1) holds, then $1+r$ is generalised scalar [Reference Laursen and Neumann20, Theorem 1.5.12, page 66], that is, admits a $C^\infty $ -functional calculus on $\mathbb {C}$ . (This result and the proposition cited below are stated for operators but, in fact, the arguments for them use only the Banach algebra structure.) Shifting by a number does not change the property of being generalised scalar. Thus, r is both topologically nilpotent and generalised scalar. It follows from [Reference Laursen and Neumann20, Proposition 1.5.10, page 64] that r is nilpotent.

However, if r is nilpotent, then we immediately have the desired estimate for positive k with constants $C>0$ and $\alpha>0$ depending only on the degree of nilpotency of r, say d, and $\|r\|$ . To prove it for negative k, note that $(1+r)^{-1}=1+r'$ for some nilpotent $r'$ whose degree of nilpotency and norm depend only on d and $\|r\|$ . It follows that (1) holds with C and $\alpha $ also depending only on d and $\|r\|$ .

Proof of Theorem 1

We show that (1) $\Leftrightarrow $ (2a) and (2a) $\Rightarrow $ (2b) $\Rightarrow $ (3b) $\Rightarrow $ (4) $\Rightarrow $ (3a) $\Rightarrow $ (2a). Let $\mathfrak {s}$ be a Levi subalgebra of $\mathfrak {g}$ , A the associative unital subalgebra of B generated by $\mathfrak {g}$ and $A_0$ the nonunital subalgebra of A generated by $\mathfrak {n}$ .

(1) $\Leftrightarrow $ (2a). Since $\mathfrak {s}$ is semisimple, it follows from a result of Taylor [Reference Taylor23] (see also [Reference Beltiţă and Şabac7] or [Reference Aristov2]) that the image of $U(\mathfrak {s})$ in a Banach algebra is finite dimensional. Being dense in B, the subalgebra A satisfies a PI if and only if so does B. Applying Proposition 3 to A, we conclude that (1) and (2a) are equivalent.

(2a) $\Rightarrow $ (2b). Applying Lemma 5 to $A_0$ shows that A is nilpotent when every element of $\mathfrak {n}$ is nilpotent.

(2b) $\Rightarrow $ (3b). Suppose that $A_0$ is nilpotent, that is, there is $d\in \mathbb {N}$ such that $A_0^d=0$ . In particular, every b in $\mathfrak {n}$ is of degree of nilpotency at most d and so is $e^b-1$ by Lemma 9.

(3b) $\Rightarrow $ (4). Let d be a positive integer such that $e^b-1$ is nilpotent of degree at most d for every $b\in \mathfrak {n}$ . Applying the implication (2) $\Rightarrow $ (1) in Lemma 10 shows that there are constants $C>0$ and $\alpha>0$ such that

(4.1) $$ \begin{align} \|e^{kb}\|\le C(1+ |k|)^\alpha\quad \text{for}\ b\in\mathfrak{n}\ \text{and}\ k\in\mathbb{Z} \end{align} $$

when $\|e^b-1\|\le 1$ . Therefore, there is $\varepsilon>0$ such that (4.1) holds when $\|b\|\le \varepsilon $ .

Now fix a nonzero element b of $\mathfrak {n}$ . Take $k\in \mathbb {N}$ such that $(k-1)\varepsilon \le \|b\|\le k\varepsilon $ and put $b'\!:=k^{-1}b$ . Then $\|b'\|\le \varepsilon $ and

(4.2) $$ \begin{align} \|e^{b}\|=\|e^{kb'}\|\le C(1+ k)^\alpha. \end{align} $$

Since $k\le 1+\varepsilon ^{-1}\|b\|$ , we have

$$ \begin{align*} 1+ k\le 2+ \varepsilon^{-1}\|b\|\le 2 \bigg(1+ \frac{\varepsilon^{-1}}2\bigg) (1+ \|b\|). \end{align*} $$

Combining this with (4.2) shows that there is $C'>0$ such that $\|e^b\|\le C' (1+\|b\|)^\alpha $ and $C'$ is independent in b.

(4) $\Rightarrow $ (3a). Suppose that for some $C>0$ and $\alpha>0$ , the inequality $\|e^{b}\|\le C (1+\|b\|)^\alpha $ holds when $b\in \mathfrak {n}$ . We claim that $r\!:=e^b-1$ is topologically nilpotent whenever $b\in \mathfrak {n}$ . Indeed, Lemma 8 implies that $b\in \mathop {\mathrm {Rad}} B$ . Note that $r=\sum _{n=1}^{\infty }{b^n}/{n!}$ and $\mathop {\mathrm {Rad}} B$ is closed. So r is also contained in $\mathop {\mathrm {Rad}} B$ . Being an element of the radical of a Banach algebra, r is topologically nilpotent.

Further, it follows from the assumption that

$$ \begin{align*} \|(1+r)^k\|=\|e^{kb}\|\le C (1+\|kb\|)^\alpha\le C (1+\|b\|)^\alpha (1+|k|)^\alpha \end{align*} $$

for every $k\in \mathbb {Z}$ . So by the implication (1) $\Rightarrow $ (2) in Lemma 10, r is nilpotent.

(3a) $\Rightarrow $ (2a). Suppose that $e^b-1$ is nilpotent for every $b\in \mathfrak {n}$ . Then it follows immediately from Lemma 9 that b is nilpotent.

This completes the proof of Theorem 1.

Proof of Theorem 2

To deduce Theorem 2 from Theorem 1, note that every Lie-algebra homomorphism $\phi \!:\mathfrak {g}_1\to \mathfrak {g}_2$ maps the nilpotent radical of $\mathfrak {g}_1$ into the nilpotent radical of $\mathfrak {g}_2$ . If, in addition, $\phi $ is surjective, then so is the Lie-algebra homomorphism $\mathfrak {r}_1\to \mathfrak {r}_2$ of the solvable radicals (see, for example, [Reference Aristov3, Lemma 4.10]). Since the nilpotent radicals of $\mathfrak {g}_1$ and $\mathfrak {g}_2$ coincide respectively with $[\mathfrak {g}_1,\mathfrak {r}_1]$ and $[\mathfrak {g}_2,\mathfrak {r}_1]$ (see the reference in the proof of Lemma 8), we have a surjective Lie-algebra homomorphism of the nilpotent radicals. Thus, (1), (2a), (2b), (3a) and (3b) are equivalent since so are the corresponding conditions in Theorem 1.

The proof of the implication (4) $\Rightarrow $ (3a) is a slight modification of the argument for the corresponding implication in Theorem 1. First note that there is $K>0$ such that

(4.3) $$ \begin{align} \|\theta(\eta)\|\le K|\eta|\quad\text{for every}\ \eta\in \mathfrak{n}. \end{align} $$

Then writing $\theta (\eta )$ instead of b and using (4.3), we deduce from the inequality $\|e^{\theta (\eta )}\|\le C (1+|\eta |)^\alpha $ for $\eta \in \mathfrak {n}$ that

$$ \begin{align*} \|(1+r)^k\|\le C (1+|\eta|)^\alpha (1+|k|)^\alpha \end{align*} $$

and then apply Lemma 10.

Finally, if (1) holds, then so does (4) in Theorem 1 for elements b of the form $\theta (\eta )$ with $\eta \in \mathfrak {n}$ . It follows from (4.3) that (4) in Theorem 2 is also satisfied.

Proof. Since $\mathfrak {g}$ is reductive, the nilpotent radical is trivial. Thus by Theorem 2, every completion of $U(\mathfrak {g})$ with respect to a submultiplicative prenorm satisfies a PI.

Proof. There is a basis $e_1,e_2$ in $\mathfrak {g}$ such that $[e_1,e_2]=e_2$ . Note that $\mathfrak {n}=[\mathfrak {g},\mathfrak {g}]=\mathbb {C} e_2$ . Let B be a Banach-algebra completion of $U(\mathfrak {g})$ and $\pi $ the corresponding Lie-algebra homomorphism from $\mathfrak {g}$ to B. It is not hard to see that $\pi (e_2)$ is nilpotent (see, for example, [Reference Pirkovskii22, Example 5.1]). Then B satisfies a PI by Theorem 2.

Proof. Since $\mathfrak {g}$ is nonabelian and nilpotent, there is a sequence $(\theta _n)$ of representations of $U(\mathfrak {g})$ on (finite-dimensional) Hilbert spaces such that the sequence $(d_n)$ of the corresponding degrees of nilpotency of $\theta _n(\mathfrak {n})$ is unbounded (see, for example, [Reference Aristov3, Proposition 4.14 and Lemma 4.16]). Using renormalisation, we can assume that the set $\{\|\theta _n(e_j)\|\}$ , where $e_1,\ldots ,e_k$ are algebraic generators of $\mathfrak {g}$ , is bounded for every j. Then the infinite sum $\theta \!:=\oplus _{n=1}^\infty \theta _n$ is a well-defined representation of $U(\mathfrak {g})$ on a Hilbert space. Since for every $p\in \mathbb {N}_+$ there are $n\in \mathbb {N}_+$ and $\eta \in \mathfrak {n}$ such that $\theta _n(\eta )^p\ne 0$ , the completion of the range of $\theta $ does not satisfy a PI by Theorem 2.