Brian Bi
$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref{(\ref{#1})}$

## Section 2.9. Lie algebras

Parentheticals to remark 2.9.4 Fix $$a, b \in A$$ and regard $$g$$ as a function of $$t$$ only so that we have \begin{equation*} \frac{\d}{\d t} (g(t)(ab)) = g'(t)(ab) \end{equation*} Then \begin{align*} g'(t)(ab) &= \frac{\d}{\d t} (g(t)(ab)) \\ &= \frac{\d}{\d t} (g(t)(a)g(t)(b)) \\ &= \frac{\d}{\d t}(g(t)(a)) \cdot g(t)(b) + g(t)(a) \frac{\d}{\d t}(g(t)(b)) \\ &= g'(t)(a) g(t)(b) + g(t)(a) g'(t)(b) \end{align*} At $$t = 0$$ we have $$g(t) = \operatorname{Id}$$ and $$g'(t) = D$$ so \begin{equation*} D(ab) = D(a)b + aD(b) \end{equation*}

For the other direction, we first observe that $$D^n(ab) = \sum_{i=0}^n \binom{n}{i} D^i(a) D^{n-i}(b)$$, which is easily proven by induction on $$n$$. Then it becomes clear that \begin{align*} e^{tD}(ab) &= ab + tD(ab) + \frac{t^2}{2!} D^2(ab) + \frac{t^3}{3!} D^3(ab) + \ldots \\ &= \sum_{i=0}^\infty \left[\frac{t^i}{i!} \sum_{j=0}^i \binom{i}{j} D^j(a)D^{i-j}(b)\right] \\ &= \sum_{j=0}^\infty \sum_{i=j}^\infty \frac{t^i}{i!} \binom{i}{j} D^j(a)D^{i-j}(b) \\ &= \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{t^{j+k}}{(j+k)!} \binom{j+k}{j} D^j(a) D^k(b) \\ &= \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{t^{j+k}}{j!k!} D^j(a) D^k(b) \\ &= \left[\sum_{j=0}^\infty \frac{t^j}{j!} D^j(a) \sum_{k=0}^\infty \frac{t^k}{k!} D^k(b)\right] \\ &= e^{tD}(a) e^{tD}(b) \end{align*} so $$e^{tD}$$ is an algebra homomorphism, which is an isomorphism since it has an inverse given by $$e^{-tD}$$.

Exercise 2.9.5 The Lie algebra of vectors in $$\mathbb{R}^3$$ with the cross product is isomorphic to the matrix Lie algebra $$\mathfrak{so}(3)$$ if we identify $$(x, y, z)$$ with the skew-symmetric matrix \begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix} Verification is tedious but straightforward.

Exercise 2.9.11 If $$x \in \mathfrak{g}$$, and $$x = \sum_i c_i x_i$$ with $$c_i \in K$$, then we can identify $$x$$ with the element $$\sum_i c_i x_i$$ in $$\mathcal{U}(\mathfrak{g})$$. Call this $$\iota(x)$$. A representation $$\rho$$ of $$\mathfrak{g}$$ assigns a linear operator to each $$x \in \mathfrak{g}$$ that preserves linearity and the Lie bracket, so that $$\rho([x_i, x_j]) = [\rho(x_i), \rho(x_j)]$$. If we define $$\rho'$$ on the image of $$\iota$$ (that is, the copy of $$\mathfrak{g}$$ in $$\mathcal{U}(\mathfrak{g})$$) such that $$\rho'(\iota(x)) = \rho(x)$$, the defining relations of $$\mathcal{U}(\mathfrak{g})$$ imply that $$\rho'(\iota(x_i)\iota(x_j) - \iota(x_j)\iota(x_i)) = \sum_k c^k_{ij} \rho(\iota(x_k))$$, so $$\rho'$$ can be uniquely extended to a representation of $$\mathcal{U}(\mathfrak{g})$$. In a very similar way, if we have a representation on the universal enveloping algebra, its restriction to the image of the injection map $$\iota$$ induces a representation of the Lie algebra. (We do not give a formal proof since the question says explain rather than prove or show.)