Lie groups and Lie algebras

Group multiplication is nonlinear, so we cannot simply add two group elements. But the bracket $[X,Y] = XY - YX$ measures the *failure of $G$ to be commutative* to second order, and remarkably that failure is itself linear — it lives in the tangent space. This is what lets us replace hard nonlinear questions about $G$ (does it commute? what are its subgroups?) with linear-algebra questions about $\mathfrak{g}$ (does the bracket vanish? what are its ideals?), which is exactly why Lie theory is so powerful.

Step 1 (setup). Let $X \in \mathfrak{g}$, $Y \in \mathfrak{g}$ come from curves $\alpha(t), \beta(t) \in G$ with $\alpha(0)=\beta(0)=I$, $\alpha'(0)=X$, $\beta'(0)=Y$; to first order $\alpha(t) \approx I + tX$ and $\beta(t) \approx I + tY$.

Step 2 (the commutator curve). Define $\gamma(t) = e^{tX} e^{tY} e^{-tX} e^{-tY}$, which lies in $G$ because $G$ is a group and $\gamma(0) = I$. Expanding each factor to second order in $t$ (using $e^{sZ} \approx I + sZ + \tfrac12 s^2 Z^2$) and multiplying out gives $\gamma(t) = e^{tX} e^{tY} e^{-tX} e^{-tY} = I + t^2[X,Y] + O(t^3)$; the linear terms in $t$ cancel exactly because it is a commutator $g h g^{-1} h^{-1}$, leaving a quadratic leading term equal to $XY - YX$.

Step 3 (extracting the tangent vector). Reparametrize by $s = t^2$ and set $\sigma(s) = \gamma(\sqrt{s})$ for $s \ge 0$; this is a smooth curve in $G$ with $\sigma(0) = I$ and $\sigma'(0) = [X,Y]$ by Step 2. Since $\sigma$ is a curve through the identity of $G$, its velocity vector $[X,Y]$ lies in $T_e G = \mathfrak{g}$ by definition of the tangent space. Hence $[X,Y] \in \mathfrak{g}$.

Step 4 (Jacobi identity). Direct algebraic expansion of $[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0$ using $[X,Y] = XY - YX$ shows every term of the form $XYZ$ appears exactly twice with opposite signs and cancels, so the identity holds automatically for any associative matrix product — no extra geometric input is needed once the bracket is $XY-YX$. Together with bilinearity and antisymmetry $[X,Y]=-[Y,X]$ (immediate from the formula), this confirms $(\mathfrak{g}, [\cdot,\cdot])$ is a Lie algebra.

This identification is exactly why angular velocity in physics is a $3$-vector: the "instantaneous rotation rate" of a rigid body lives in $\mathfrak{so}(3)$, and the isomorphism with $(\mathbb{R}^3,\times)$ is why $\vec\omega \times \vec r$ gives the velocity of a point $\vec r$ on a spinning body.

Step 1 (basis). Every skew-symmetric $X \in \mathfrak{so}(3)$ has the form $X_u = \begin{pmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{pmatrix}$ for a unique $u=(u_1,u_2,u_3) \in \mathbb{R}^3$ (three free entries above the diagonal determine the rest), and $u \mapsto X_u$ is manifestly linear and bijective, so it is a linear isomorphism of $3$-dimensional vector spaces.

Step 2 (action on vectors). A direct computation shows $X_u v = u \times v$ for every $v \in \mathbb{R}^3$, i.e. $X_u$ acts on $\mathbb{R}^3$ exactly as "cross with $u$".

Step 3 (bracket matches cross product). Using Step 2 twice, for any $v$: $[X_u,X_v]w = X_u(X_v w) - X_v(X_u w) = u\times(v\times w) - v\times(u\times w)$. Applying the vector triple-product identity $a\times(b\times c) = b(a\cdot c) - c(a\cdot b)$ to both terms and simplifying, the right side collapses to $(u\times v)\times w = X_{u\times v} w$ for every $w$, hence $[X_u,X_v] = X_{u\times v}$ as operators — exactly $\mathfrak{so}(3) \cong (\mathbb{R}^3, \times), \qquad [X_u, X_v] = X_{u \times v}$.

Step 4 (conclusion). Since $u \mapsto X_u$ is a linear bijection that turns the cross product into the matrix bracket, it is a Lie algebra isomorphism $(\mathbb{R}^3,\times) \cong \mathfrak{so}(3)$, as claimed.