Exercise 31, § 2

a) Let A be a normed algebra, and let $x, y$ be two elements of A such that $xy-yx=y$. Prove that $y^n= 0$ for $n >2\|x\|$. (Use the formula $xy^n-y^nx=ny^n$).

b) Let $\mathfrak{L}$ be a finite-dimensional Lie algebra over $\mathbf{C}$. Prove that, if $\mathfrak{L}$ is not nilpotent, it contains two non-zero elements $x, y$ such that $[x, y] =y$. (Use the fact that there exists an $x\in \mathfrak{L}$ such that ad($x$) is not nilpotent.)

c) Let $\mathfrak{L}$ be a finite-dimensional real or complex Lie algebra such that the enveloping algebra of $\mathfrak{L}$ possesses a norm compatible with its algebra structure. Prove that $\mathfrak{L}$ is nilpotent. (Use a) and b).)

d) Let V be a finite-dimensional real or complex vector space, $\mathfrak{L}$ a Lie subalgebra of $\mathfrak{g}l(V)$ consisting of nilpotent endomorphisms, and $x_1, . . . , x_n$ elements of $\mathfrak{L}$. Let $\alpha >0$. Prove that there exists a Hilbert space structure on V such that $\|x_i\|\leqslant \alpha$ for $1\leqslant i\leqslant n$.

e) Let $\mathfrak{L}$ be a finite-dimensional real or complex nilpotent Lie algebra, and U its enveloping algebra. Prove that there exists on U a norm compatible with its algebra structure. (Using the method of LIE, I, §3, exerc. 5 and §7, exerc. 3, show that there exists a family $(\pi_{\lambda})$ of representations of $\mathfrak{L}$ in finite-dimensional spaces $V_{\lambda}$, having the following property: for every $u\in U$, there exists a $\lambda$ such that $\pi_{\lambda}(u)\not= 0$, and $\pi_{\lambda}(\mathfrak{L})$ consists for every $\lambda$ of nilpotent endomorphisms. By endowing each $V_{\lambda}$ with a Hilbert space structure provided by d), one obtains in the Hilbert space V Hilbert sum of the $V_{\lambda}$ a representation $\pi$ of $\mathfrak{L}$ by continuous endomorphisms, whence an injective morphism $\varphi$ of U into $\mathscr{L}(V)$. Put $\|u\|_U=\|\varphi (u)\|$ for every $u\in U$).

f) Let A be a complex unital normed algebra $x$ and $y$ elements of A. If $A\not=\{0\}$, one has $xy-yx\not= 1$. (First proof: one has Sp$'_A(xy) =$ Sp$'_A(yx)$; if $xy=yx+ 1$, then Sp$_A(xy)$ is deduced from Sp$_A(yx)$ by the translation $z\mapsto z+ 1$; deduce that Sp$_A(xy)$ is unbounded, which is absurd. Second proof: if $[x, y] = 1$, one has $[xy, y] =y$, whence $y^n= 0$ for $n$ sufficiently large according to a) ; as $[x, y^p] =py^{p-1}$ deduce that $y= 0$).