Past exam of the lie Groups and Differentiable Manifolds course of the University of Oxford / 2008 / A / 2

The matrix exponential series differentiates term by term on bounded intervals and gives $(e^{tX}{)}^{\prime }=X{e}^{tX}$ and $e^{0X}=I$, proving existence. If $a$ is another differentiable solution, thenIts initial value is $I$, so $e^{-tX}a(t)=I$, giving $a(t)=e^{tX}$ uniquely.

The exponential satisfies the initial conditions and $e^{(s+t)X}=e^{sX}{e}^{tX}$ because the two matrix exponents commute. Conversely, differentiating $a(s+t)=a(s)a(t)$ with respect to $s$ at zero gives $a^{\prime }(t)=Xa(t)$. Even if only differentiability at zero was initially specified, the identity $a(t+h)=a(h)a(t)$ supplies differentiability at every $t$. The uniqueness result just proved yields $\boxed{a(t)=e^{tX}}$.

For fixed $X\in L(G)$, set $b(t)=\varphi (e^{tX})$. The homomorphism property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\varphi )(X)=D{\varphi }_I(X)$. The one-parameter subgroup uniqueness theorem givesFor the Adjoint representation, ${\mathrm{Ad}}_{g}Y=gY{g}^{-1}$. ThusThe differential $L(\varphi )$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\varphi$.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.