Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 49 / 1

For a finite-dimensional Lie algebra, write $A_X=\mathrm{ad}X$ for its Adjoint representation. The Killing form is the symmetric bilinear formThe Jacobi identity gives $A_{[X,Y]}=[A_X,A_Y]$. Using the cyclic property of the trace,Thus it is an invariant bilinear form on a Lie algebra. If $[T_a,T_b]=f_{ab}{}^c{T}_c$, the adjoint matrix has entries $(A_{T_a}{)}^c{}_b=f_{ab}{}^c$. Its trace product yields the structure constants of a Lie algebra formulaA semisimple Lie algebra has zero solvable radical, equivalently no nonzero solvable ideal. In characteristic zero, the Cartan criterion for semisimplicity characterizes it by nondegeneracy of its Killing form. A real compact Lie algebra is the Lie algebra of a compact Lie group, equivalently one admitting a positive-definite invariant inner product. For a real semisimple algebra, the compactness criterion from the Killing form says compactness is equivalent to negative-definiteness of $\kappa$. An abelian compact-type algebra has zero Killing form, so this last characterization specifically concerns the semisimple case.

In the ordered basis $(K_1,K_2,J)$, the three Adjoint representation matrices areTheir cross-product traces vanish, while their squared traces are $-2\lambda$, $-2\lambda$ and $-2{\lambda }^2$. ConsequentlyThis is the Killing form for cyclic three-generator brackets with constants $(1,\lambda ,\lambda )$.

For $\lambda >0$, the form is nondegenerate and negative-definite. The algebra is semisimple and compact, isomorphic to the SU(2) Lie algebra. For $\lambda <0$, it is nondegenerate with signature $(2,1)$, so the algebra is semisimple and noncompact, isomorphic to ${\mathfrak{sl}}_2(\mathbb{R})$.

For $\lambda =0$, the only nonzero bracket is $[K_1,K_2]=J$, with $J$ central. This is the Heisenberg Lie algebra, which is nilpotent and neither semisimple nor compact. Compactness is ruled out even though the Killing form is merely degenerate: $A_{K_1}$ is nonzero and nilpotent, whereas in a compact Lie algebra every adjoint map is skew-adjoint for a positive inner product. Skew-adjoint nilpotent operators vanish.

Let $v=(x,y{)}^T$ and $B(\rho )=\left(\begin{array}{cc}\cosh \rho & \sinh \rho \\ \sinh \rho & \cosh \rho \end{array}\right)$. Matrix multiplication givesThus $G$ is a semidirect product of the translation plane by the one-parameter Lorentz boost group. Differentiating left multiplication at the identity gives the left-invariant vector fieldsThey obey $[X_0,X_1]=X_2$, $[X_0,X_2]=X_1$ and $[X_1,X_2]=0$.

Their dual coframe is $d\rho$, $\cosh \rho dx-\sinh \rho dy$, and $-\sinh \rho dx+\cosh \rho dy$. The Haar measure from a left-invariant coframe is their absolute wedge product. Since $\det B(\rho )=1$,One can also check this directly: left multiplication has Jacobian determinant one. Right multiplication has a block triangular Jacobian with determinant one too, so the same measure is both left and right Haar measure.