## 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

.)