Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 1 / 3

For a Lie group $G$ with identity $e$, its Lie algebra as a set is the tangent space $T_{e}G$:The ambient embedding identifies these tangent vectors with vectors in ${\mathbb{R}}^N$. For a matrix Lie group, the identity is $I$ and the tangent vectors are matrices.

If $\gamma (t)\in {\mathrm{SL}}_n$ with $\gamma (0)=I$, put $A={\gamma }^{\prime }(0)$. The determinant expansion $\det (I+tA+o(t))=1+t\mathrm{tr}A+o(t)$ follows directly from its permutation formula: to first order only the diagonal entries contribute. Since $\det \gamma (t)=1$, differentiation gives $\mathrm{tr}A=0$. ThusIn fact equality holds: if $\mathrm{tr}A=0$, then $\exp (tA)$ has determinant $\exp (t\mathrm{tr}A)=1$ and derivative $A$ at zero.

Take curves $g(s),h(t)$ through the identity with tangent vectors $X,Y$. Form the local group commutator $c(s,t)=g(s)h(t)g(s{)}^{-1}h(t{)}^{-1}$. In a smooth coordinate chart at the identity, its leading mixed term defines the Lie bracket:for a matrix group, where the identity's constant term has zero derivative. More generally take this derivative in the chart with the identity sent to zero. Since the first derivatives vanish on the coordinate axes, changing charts changes the mixed coefficient only by the tangent-space coordinate transformation, so the definition is intrinsic.

For the displayed nilpotent matrices choose $g(s)=I+sX$ and $h(t)=I+tY$, which lie in ${\mathrm{SL}}_2$. Their inverses are $I-sX$ and $I-tY$. Direct multiplication givesIts mixed derivative is thereforeInterchanging the two curves giveswhose mixed derivative is $-H$. Thus $[X,Y]=-[Y,X]$ directly from the curve definition. The general matrix expansion likewise gives the commutator $XY-YX$.

A smooth vector field on a smooth manifold $M$ is a smooth section of its tangent bundle: it assigns $v(p)\in T_{p}M$ smoothly to each point. On a Lie group, it is a left-invariant vector field whenFor $X\in T_{e}G$, define $v_X(g)=(d{L}_g{)}_{e}X$. Smoothness of multiplication makes this a smooth vector field, and $L_g{L}_p=L_{gp}$ plus the chain rule proves left invariance. Every left-invariant vector field arises this way from its value at the identity.

For a matrix Lie group, $v_X(g)=gX$, because the derivative at zero of $g\gamma (t)$ is $gX$. The printed group name in this subpart must be read as ${\mathrm{SL}}_2$, rather than the additive vector space ${\mathfrak{sl}}_2$: it is the matrix-group calculation consistent with the preceding subparts. At the specified matrix,A tangent vector at a nonidentity group element need not itself be trace-free; here $g^{-1}{v}_X(g)=X$ is trace-free, as required. If one instead interpreted the printed ${\mathfrak{sl}}_2$ as its additive Lie group, the field would be the constant field $X$, a different problem.