Semi-simple Lie-Algebras

We will always assume that $A$ is a real Lie-algebra admitting an euclidean product $\la.,.\ra$ such that $$ \forall x,y,z\in A:\qquad \la\ad(x)y,z\ra=-\la y,\ad(x)z\ra $$ i.e. $\ad(x)$ is skew-symmetric on $(A,\la.,.\ra)$. Assume further that $A\cap iA=\emptyset$, then by

proposition

The center $Z(A)$ of the Lie-algebra $A$ is an ideal, i.e. $Z(A)$ is invariant under the adjoint representation: indeed if $z\in Z(A)$, then $[x,z]=0$ for all $x\in A$ and we get for any $y\in A$ by Jacobi's identity $$ [x,[y,z]]=-[y,[z,x]]-[z,[x,y]]=0 \quad\mbox{i.e.}\quad \forall y\in A:\quad[y,z]\in Z(A)~. $$ Moreover the orthogonal complement $I^\perp$ of an ideal is again an ideal, this is because if $I$ is $\ad$-invariant, then so is its orthogonal complement $I^\perp$: for all $x\in A$, all $y\in I^\perp$ and all $z\in I$: $$ \la\ad(x)y,z\ra =\la y,\ad(x^*)z\ra =0 \quad\mbox{i.e.}\quad \ad(x)I^\perp\sbe I^\perp~. $$ Moreover $A$ is the Lie-algebra direct sum of $I$ and $I^\perp$, because for all $y\in I^\perp$ and all $z\in I$: $[y,z]\in I\cap I^\perp=\{0\}$.

As $Z(A)^\C=Z(A^\C)$, $A$ is semi-simple iff $A^\C$ is semi-simple. Decomposing $A=Z(A)\oplus Z(A)^\perp$ we assert that $Z(A)^\perp$ is semi-simple, because any point $x$ in the center of $Z(A)^\perp$ satisfies $[x,y]=0$ for all $y\in Z(A)^\perp$ and by definition of the center: $[x,y]=0$ for all $y\in Z(A)$.

$\proof$ We know that $A$ is the direct sum of pairwise orthogonal $\ad$-invariant sub-spaces $A_j$, irreducible under the action of the adjoint representation. As $A_j$ is $\ad$-invariant, $A_j$ is an ideal and irreducibility exactly means that the Lie-Algebra $A_j$ is simple: indeed, for all $k\neq j$: $[A_k,A_j]=\{0\}$ and therefore $A_j$ is irreducible under the action of the adjoint representation iff it is simple as a sub-algebra of $A$. $\eofproof$

Let $X$ be any element in the center of $\sla(n,\C)=\{A\in\Ma(n,\C):\tr A=0\}$, i.e. $[X,Z]=0$ for all $Z\in\sla(n,\C)$, then by

exam

Roots

Cartan-algebras

A sub-algebra $H$ of a semi-simple Lie-algebra $A$ is said to be a Cartan-algebra if the following holds:

For all $x,y\in H$: $[x,y]=0$.
If for some $x\in A$ and all $h\in H$: $[x,h]=0$, then $x\in H$.
For all $h\in H$: $\ad(h)$ is diagonalizable.

I.e. $H$ is a maximal commutative sub-algebra of $A$ and the set of linear mappings $\{ad(h):h\in H\}$ is jointly diagonalizable.
If any two such Cartan sub-algebras of $A$ are isomorphic, the dimension of these algebras is said to be the rank .

Suppose $H$ is a Cartan-algebra of the real semi-simple Lie-algebra $A$. Then $H^\C$ is a Cartan-algebra of the Lie-algebra $A^\C$.

$\proof$ Obviously $A^\C$ is commutative. Assume $z=x+iy\in A^\C$ and $[z,h]=0$ for all $h\in H^\C$. In particular for $h\in H$: $0=[z,h]=[x,h]+i[y,h]$. As $A\cap iA=\{0\}$ we infer that by the maximality of $H$: $x,y\in H$ and thus $z\in H^\C$. Finally, all $x,y\in H$ are diagonalizable and commute. Hence all $x+iy\in H^\C$ are diagonalizable. $\eofproof$

Suppose $H$ is a Cartan sub-algebra of the real semi-simple Lie-algebra $A$. A vector $r\in H\sm\{0\}$ is called a root if $$ \exists x\in A\sm\{0\}\,\forall h\in H:\quad \ad(h)x=\la h,r\ra x~. $$ $x$ is said to be a root vector and the set of all roots is denoted by $R$.

As we assume that for a real Lie-algebra $A$ all mappings $\ad(x)$, $x\in A$, are skew-symmetric, all eigen-value of $\ad(x)$ are purely imaginary. Moreover $\la.,.\ra$ is real on $H\times H$ and therefore any root $r$ must be in $iH\sbe H^\C$. Hence for any root $r$ the linear mapping $$ \ad(r):A^\C\rar A^\C \quad\mbox{is self-adjoint.} $$ Given any $r\in A$ we define the vector-space $A_r$ to be the space of all $x\in A$ such that for all $h\in H$: $\ad(h)x=\la h,r\ra x$, i.e. \begin{equation}\label{rooeq1}\tag{ROO1} A_r\colon=\bigcap_{h\in H}\ker(\ad(h)-\la h,r\ra)~. \end{equation} For $r=0$ we conclude by the maximality of $H$: $A_0=H$ and for $r\notin R\cup\{0\}$ we have $A_r=\{0\}$. Since all $\ad(h)$, $h\in H^\C$ are jointly diagonalizable, we get a decomposition $$ A^\C=H^\C\oplus\bigoplus_{r\in R}A_r^\C $$ As $H^\C$ is spanned by its self-adjoint elements, i.e. elements $h$ satisfying $h^*=h$, we may assume that this decomposition is orthogonal.

1. For all $r,s\in A$ we have $[A_r,A_s]\sbe A_{r+s}$. 2. If $z\in A_r$, then $z^*\in A_{-r}$. 3. $\lhull{R}=H$.

$\proof$ 1. Cf. e.g. lemma. Suppose $x\in A_r$ and $y\in A_s$; we have to verify that for all $h\in H$: $[h,[x,y]]=\la h,r+s\ra[x,y]$. By Jacobi`s identity we have: $$ [h,[x,y]] =-[x,[y,h]]-[y,[h,x]]~. $$ Now $[h,x]=\la h,r\ra x$ and $[h,y]=\la h,s\ra y$ and thus $$ [h,[x,y]] =[x,\la h,s\ra y]-[y,\la h,r\ra x] =\la h,s+r\ra [x,y]~. $$ 2. Suppose $z=x+iy$ and $[h,x+iy]=\la h,r\ra(x+iy)$. As $(\l z)^*=\bar\l z^*=\bar\l(-x+iy)$ and $\la h,r\ra$ is purely imaginary we get: $$ -\la h,r\ra(-x+iy) \cl{\la h,r\ra}(x+iy)^* =[h,x+iy]^* =([h,x]+i[h,y])^* =-[h,x]+i[h,y] =[h,-x+iy]~. $$ This shows that $-r$ is a root with root-vector $z^*$.
3. If $\lhull{R}\neq H$ then there is some normed vector $h\in H$ such that for all $r\in R$: $\la h,r\ra=0$; since $r$ is a root we must have for all $h_1\in H$ and all $x_r\in A_r$: $[h_1,x_r]=\la h_1,r\ra x_r$, in particular: $[h,A_r]=0$. As $[h,A_0]=0$ we infer that $h$ commutes with all $x\in A$, i.e. $h$ is in the center of $A$. On the other hand $A$ is semi-simple, i.e. its center is trivial and thus $h=0$. $\eofproof$

Sub-algebras isomorphic to $\sla(2,\C)$

Given a root $r\in H$ and a root vector $x\in A_r$, we will show that the sub-algebra generated by $r,x$ and $x^*$ is isomorphic to $\sla(2)$. By definition we have $[r,x]=\Vert r\Vert^2 x$ and $[r,x^*]=-\Vert r\Vert^2 x^*$ and thus both $[r,x]$ and $[r,x]$ lie in this sub-algebra, but what about $[x,x^*]$?

For all $x\in A_r$, all $y\in A_{-r}$ we have: $[x,y]=\la y,x^*\ra r$.

$\proof$ By proposition: $[A_r,A_{-r}]\sbe A_0=H$ and thus $[x,y]\in H$. Next we notice that $\ad(x)^*=\ad(x^*)$ and by proposition for all $h\in H$: $\ad(h)x^*=\la h,-r\ra x^*$; therefore: \begin{equation}\label{rooeq2}\tag{ROO2} \la h,[x,y]\ra =\la h,\ad(x)y\ra =\la\ad(x^*)h,y\ra =-\la\ad(h)x^*,y\ra =\la\la h,r\ra x^*,y\ra =\la h,r\ra\la x^*,y\ra~. \end{equation} Hence for $h\perp r$ we also have $[x,y]\perp h$; as $[x,y]$ is in $H$ it follows that $[x,y]=cr$ for some $c\in\C$. Putting $h=r$ in \eqref{rooeq2} we get: $$ \bar c\Vert r\Vert^2 =\la r,[x,y]\ra =\Vert r\Vert^2\la x^*,y\ra, \quad\mbox{i.e.}\quad \bar c=\la x^*,y\ra~. $$ $\eofproof$

For all roots $r$, all $x\in A_r\sm\{0\}$ and all $y\in A_{-r}\sm\{0\}$ the sub-space $\lhull{r,x,y}$ is a sub-algebra satisfying $$ [r,x]=\Vert r\Vert^2 x,\quad [r,y]=-\Vert r\Vert^2 y,\quad [x,y]=\la y,x^*\ra r~. $$ As $r,x,y$ are eigen-vectors of the self-adjoint operator $\ad(r):A\rar A$ with eigen-values $0,\Vert r\Vert^2$ and $-\Vert r\Vert^2$, they must also be linearly independent. To verify that this sub-algebra is isomorphic to $\sla(2,\C)$ we are looking for scalars $\a,\b,\g$, such that $h_r\colon=\a r$, $x_r\colon=\b x$, $y_r\colon=\g x^*$ satisfy the commutation relations: \begin{equation}\label{rooeq3}\tag{ROO3} [h_r,x_r]=2x_r,\quad [h_r,y_r]=-2y_r,\quad [x_r,y_r]=h_r~. \end{equation} This implies that: \begin{eqnarray*} [h_r,x_r]&=&\a\b[r,x]=\a\b\Vert r\Vert^2 x=\a \Vert r\Vert^2x_r\\ [h_r,y_r]&=&-\a\g[r,x^*]=-\a\g\Vert r\Vert^2 x^*=-\a\Vert r\Vert^2y_r\\ [x_r,y_r]&=&\b\g[x,x^*]=\b\g c_rr=\b\g c_rh_r/\a~. \end{eqnarray*} Thus the numbers $\a,\b,\g$ simply need to satisfy: $\a\Vert r\Vert^2=2$ and $\b\g c_r/\a=1$, e.g.: $$ \a=\frac2{\Vert r\Vert^2} \quad\mbox{and}\quad \b=\g=\sqrt{\frac{\a}{c_r}}=\sqrt{\frac{2}{c_r\Vert r\Vert^2}}~. $$ The vector $r^\vee\colon=h_r\colon=2r/\Vert r\Vert^2$ is called the co-root of the root $r$ and the vectors $x_r\in A_r$ and $y_r\in A_{-r}$ satisfy \eqref{rooeq3} and: \begin{equation}\label{rooeq4}\tag{ROO4} y_r=x_r^*\quad\mbox{and}\quad \la r,h_r\ra=2~. \end{equation} Suppose $s=cr$ is another root for some $c\in\C\sm\{0\}$, then $h_s=2cr/\Vert cr\Vert^2=h_r/c$. Now $\ad:\lhull{h_r,x_r,x_r^*}\rar\Hom(A)$ is a representation and \begin{equation}\label{rooeq5}\tag{ROO5} \ad(h_r)x_s =\ad(ch_s)x_s =2cx_s \end{equation} i.e. $x_s$ is an eigen-vector of $\ad(h_r)$ with eigen-value $2c$. By proposition: $2c\in\Z$. Exchanging the roles of $r$ and $s$, we get: $2/c\in\Z$. Hence $c\in\{\pm1/2,\pm1,\pm2\}$ and the only possible roots are $\pm r/2,\pm r,\pm2r$. Choosing among these the one having minimal norm and renaming this to be $r$, we conclude that only $\pm r,\pm2r$ are possible roots. So let $B$ be the sub-space generated by $h_r$, $A_r$, $A_{2r}$, $A_{-r}$ and $A_{-2r}$. Then by proposition and lemma $B$ is in fact a sub-algebra and therefore $$ \ad:\lhull{h_r,x_r,x_r^*}\rar\Hom(B) $$ is a representation. Of course $B$ contains the sub-space $\lhull{h_r,x_r,x_r^*}$ invariant under this representation. The orthogonal complement $Z$ of $\lhull{h_r,x_r,x_r^*}$ in $B$ is also invariant. What are the eigen-values of the self-adjoint operator $\ad(h_r):Z\rar Z$? As a self-adjoint mapping on $B$ $\ad(h_r)$ has the eigen-values $0,\pm2$ and by \eqref{rooeq5} possibly $\pm4$ and the only eigen-vector with eigen-value $0$ is $h_r$. The self-adjoint mapping $$ \ad(h_r):\lhull{h_r,x_r,x_r^*}\rar\lhull{h_r,x_r,x_r^*} $$ has the eigen-values $-2,0,2$. Now, if $Z\neq\{0\}$, then $0\neq x_{2r}\in B$ and by \eqref{rooeq5}: $\ad(h_r)x_{2r}=4x_{2r}$. Again by proposition this implies that $\ad(h_r):Z\rar Z$ must have the eigen-values $\pm4,\pm2$ and $0$. However, the only eigen-vector of $\ad(h_r):B\rar B$ with eigen-value $0$ is $h_r$. Therefore $Z=\{0\}$, i.e. $B=\lhull{h_r,x_r,x_r^*}$, $A_r=\lhull{x_r}$ and $A_{-r}=\lhull{x_r^*}$.

For each root $r$, the only multiples of $r$, that are roots are $r$ and $-r$. 2. For each root $r$: $\dim A_r=1$ and thus $\dim A=\dim H+|R|$. 3. For all roots $r,s$ we have $$ \la h_r,s\ra=\frac{2\la s,r\ra}{\Vert r\Vert^2}\in\Z~. $$

$\proof$ We only have to verify the third statement. The adjoint action restricted to the algebra $\lhull{h_r,x_r,x_r^*}$ gives a representation of this algebra in $\Hom(A)$. If $y$ is a root vector for the root $s$, then in particular $[h_r,y]=\la h_r,s\ra y$, i.e. $y$ is an eigen-vector for $\ad|\lhull{h_r,x_r,x_r^*}$. By proposition this implies that $\la h_r,s\ra\in\Z$. $\eofproof$

The Weyl group

The roots of $\sla(3,\C)$ are given by $\pm H_j$, $j=1,2,3$, and the Weyl group of $\sla(3,\C)$ turned out to be the group generated by the reflections about the lines orthogonal to the roots - cf. subsection.

The Weyl group $W(A)$ of a semi-simple Lie-algebra $A$ is the group generated by the reflections about hyperspaces orthogonal to $r\in R$: $$ R_r:H\rar H,\quad R_r(h)\colon=h-\frac{2\la h,r\ra}{\Vert r\Vert^2}r~. $$

In the previous discussion of the Weyl group of $\sla(3,\C)$ a representation of $\SU(3)$ had been our starting point. In proposition we verified that the linear map $\Ad(g):\sla(3,\C)\rar\sla(3,\C)$, $X\mapsto gXg^{-1}$ maps root vectors for the root $s\in{\cal H}$ - now we denote this space by $H$ - to root vectors for the root $\Ad(g)s$, provided ${\cal H}$ is $\Ad(g)$ invariant, i.e. $g\in N({\cal H})$. Inspired by this result and the fact that for any linear operator $W\in\Hom(E)$: $\Ad(e^W)=e^{\ad(W)}$, cf. subsection, a substitute for the mapping $\Ad(g)$ will be a mapping of type $e^{\ad(x)}\in\Hom(A)$.
Now we are given the reflection $R_r$, mapping the root $s$ to $R_r(s)$ - which we`d like to prove to be a root as well - and we need to find a linear mapping (actually an isomorphism) $S_r^{-1}:A\rar A$ such that for all root vectors $x$ for the root $s$: $$ \ad(h)(S_r^{-1}(x))=\la h,R_r(s)\ra S_r^{-1}(x), $$ i.e. $S_r^{-1}(x)$ is a root vector for the root $R_r(s)$. Now $$ \la h,R_r(s)\ra S_r^{-1}(x) =\la R_r(h),s\ra S_r^{-1}(x) =S_r^{-1}(\la R_r(h),s\ra x) =S_r^{-1}\ad(R_r(h))x $$ which evidently shows $R_r(s)$ to be a root with root vector $S_r^{-1}(x)$ provided $S_r$ meets the following condition: \begin{equation}\label{rooeq6}\tag{ROO6} \forall h\in H:\qquad S_r\ad(h)S_r^{-1}=\ad(R_r(h))~. \end{equation}

For any representation $\psi:A\rar\Hom(E)$ in a finite dimensional space $E$: $$ \forall x,y,z\in A:\quad e^{\psi(x)}\psi(y)e^{-\psi(x)}=\psi(e^{\ad(x)}y) \quad\mbox{and}\quad e^{\psi(z)}e^{\psi(x)}\psi(y)e^{-\psi(x)}e^{-\psi(z)} =\psi(e^{\ad(z)}e^{\ad(x)}y) $$

$\proof$ By subsection we have for any linear operator $W\in\Hom(E)$: $\Ad(e^W)=e^{\ad(W)}$, i.e. $\Ad(e^{\psi(x)})\psi(y)=e^{\ad(\psi(x))}\psi(y)$. As $\psi:A\rar\Hom(E)$ is an algebra homomorphism, i.e. $\ad(\psi(x))\psi(y)=\psi(\ad(x)y)$, we also have: $$ e^{\ad(\psi(x))}\psi(y)=\psi(e^{\ad(x)}y) $$ It follows that $$ \forall x,y\in A:\quad e^{\psi(x)}\psi(y)e^{-\psi(x)} =\Ad(e^{\psi(x)})\psi(y) =e^{\ad(\psi(x))}\psi(y) =\psi(e^{\ad(x)}y)~. $$ and applying the formula twice: $$ e^{\psi(z)}e^{\psi(x)}\psi(y)e^{-\psi(x)}e^{-\psi(z)} =e^{\psi(z)}\psi(e^{\ad(x)}y)e^{-\psi(z)} =\psi(e^{\ad(z)}e^{\ad(x)}y) $$ $\eofproof$

We are going to apply the previous lemma to the representation $\ad:\lhull{h_r,x_r,y_r}\rar A$. Can we find a vector $v\in\lhull{h_r,x_r,y_r}$ such that $S_r=e^{\ad(v)}$? By \eqref{rooeq6} and lemma $v$ must satisfy \begin{equation}\label{rooeq7}\tag{ROO7} \forall h\in H:\qquad \ad(e^{\ad(v)}h)=\ad(R_r(h))~. \end{equation} Assume $S_r=e^{\ad(v)}$, for arbitrary $v\in\lhull{h_r,x_r,y_r}$. If $h\perp r$ then $[h,h_r]=0$ and $[h,x_r]=\la h,r\ra x_r=0=\la h,-r\ra y_r=[h,y_r]$ and thus $[h,v]=0$. Hence $e^{\ad(v)}h=h$; on the other hand $R_r(h)=h$ and we conclude that \eqref{rooeq7} holds for all $v\in\lhull{h_r,x_r,y_r}$ and all $h\perp r$. Thus our only requirement \eqref{rooeq7} is a consequence of \begin{equation}\label{rooeq8}\tag{ROO8} e^{\ad(v)}h_r=-h_r~. \end{equation} This is just a matter of solving an equation in the three dimensional space $\lhull{h_r,x_r,y_r}$. The matrices of $\ad(h_r)$, $\ad(x_r)$ and $\ad(y_r)$ with respect to the basis $h_r,x_r,y_r$ are easily calculated: $$ \left(\begin{array}{ccc} 0&0&1\\ 0&2&0\\ 0&0&-2 \end{array}\right), \left(\begin{array}{ccc} 0&0&1\\ -2&0&0\\ 0&0&0 \end{array}\right), \left(\begin{array}{ccc} 0&-1&0\\ 0&0&0\\ 2&0&0 \end{array}\right) $$ Now put $v=\a h_r+\b x_r+\g y_r$, then \eqref{rooeq8} is equivalent to the matrix equation $$ \exp\left(\begin{array}{ccc} 0&-\g&\b\\ -2\b&2\a&0\\ 2\g&0&-2a \end{array}\right) \left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right) = \left(\begin{array}{c} -1\\ 0\\ 0 \end{array}\right) $$ This equation doesn`t have a solution if we assume only one of the coefficients $\a,\b$ or $\g$ differs from zero. So let`s assume $\a=0$ but $\b,\g\neq0$; putting $\b\g=p^2$ the matrix equation comes down to two equations: $$ e^{-2p}(e^{4p}+1)=-2\quad\mbox{and}\quad ae^{-2p}(e^{4p}-1)=0~. $$ Thus $p\in\{i\pi/2,i\pi,3i\pi/2\}$ and $e^{-2p}=-1$, i.e. $p=i\pi/2$ and we choose $\b=\pi/2$ and $\g=-\pi/2$; for this choice we get $$ \exp\left(\begin{array}{ccc} 0&\frac12\pi&\frac12\pi\\ -\pi&0&0\\ -\pi&0&0 \end{array}\right) =\left(\begin{array}{ccc} -1&0&0\\ 0&0&-1\\ 0&-1&0 \end{array}\right) $$ Thus we finally found our operator $S_r$ \begin{equation}\label{rooeq9}\tag{ROO9} S_r=e^{\ad(v)}=e^{\frac12\pi\ad(x_r-y_r)}~. \end{equation} On $\lhull{h_r,x_r,y_r}$ this mapping is given by $$ h_r\mapsto-h_r, x_r\mapsto-y_r, y_r\mapsto-x_r~. $$

The mapping $\ad(x_r-y_r)$ is skew symmetric on the whole space $A$ and therefore $S_r\in\Hom(A)$ is an isometry - actually on $\lhull{h_r,x_r,y_r}$ it`s a rotation about the axis defined by the vector $x_r-y_r$ with angle $\pi\norm{x_r-y_r}/2$.

The Weyl group $W(A)$ is finite and for each root $r$, the mapping $R_r$ maps the set of roots onto itself. Hence the set of roots is invariant under the Weyl group $W(A)$. Moreover for each root vector $x$ with root $s$ the vector $e^{-\frac12\pi\ad(x_r-y_r)}x$ is a root vector with root $R_r(s)$.

$\proof$ We are left proving finiteness of the Weyl group: As any element $g$ of the Weyl group is a linear transformation on $H$ and the set of roots $R$ spans the Cartan algebra $H$, $g$ is determined by its action on $R$. But $g$ simply permutes the roots. Hence $W$ is isomorphic to a subgroup of the group of permutations of $R$. $\eofproof$

Remember, if $\ad(h_r)$ is self-adjoint, then the vectors $h_r,x_r,y_r$ are pairwise orthogonal but not orthonormal, for $\norm{h_r}\neq1$ in general. The vectors $x_r$ and $y_r$ can always be choosen to be normalized and thus both $h_r/\norm{h_r}=r/\Vert r\Vert,x_r,y_r$ and $$ b_0\colon=ih_r/\norm{h_r}=r/\Vert r\Vert^2,\quad b_1\colon=(x_r-y_r)/\sqrt2,\quad b_2\colon=i(x_r+y_r)/\sqrt2 $$ form orthonormal bases of $\lhull{h_r,x_r,y_r}$. The advantage of the latter over the former: they are skew symmetric and thus the linear mappings $\ad(b_j)$, $j=0,1,2$, are isometries.

Compute the matrix of $\ad(\a b_0+\b b_1+\g b_2)$ with respect to the basis $b_0,b_1,b_2$.

Other choices of $v$ are possible, e.g. $v=\frac12i\pi(x_r+y_r)$. For this choice we get: $$ \exp\left(\begin{array}{ccc} 0&-\frac12i\pi&\frac12i\pi\\ -i\pi&0&0\\ i\pi&0&0 \end{array}\right) =\left(\begin{array}{ccc} -1&0&0\\ 0&0&1\\ 0&1&0 \end{array}\right) $$

Employing the second formula in lemma we may look for the linear operator $S_r$ in the group generated by the set $\{e^{\psi(x)}:x\in\lhull{h_r,x_r,y_r}\}$. Verify that the matrices of the linear operators $e^{\ad(sx_r)}$ and $e^{\ad(ty_r)}$ with respect to the basis $h_r,x_r,y_r$ are given by: $$ \left(\begin{array}{ccc} 1&0&s\\ -2s&1&-s^2\\ 0&0&1 \end{array}\right), \left(\begin{array}{ccc} 1&-t&0\\ 0&1&0\\ 2t&-t^2&1 \end{array}\right) $$ Show that $e^{\psi(x_r)}e^{-\psi(y_r)}e^{\psi(x_r)}$ and $e^{\psi(y_r)}e^{-\psi(x_r)}e^{\psi(y_r)}$ are possible choices for $S_r$.

Suppose $\b,\g\in\C$. Find all eigen-values and eigen-vectors of the matrix $$ \left(\begin{array}{ccc} 0&-\g&\b\\ -2\b&0&0\\ 2\g&0&0 \end{array}\right) $$

The characteristic polynomial: $p(z)=(4\b\g-z^2)z$. Hence the eigen-values are $0$, $-2\sqrt{\b\g}$ and $2\sqrt{\b\g}$. The corresponding eigen-vectors are the columns of the matrix: $$ \left(\begin{array}{ccc} 0&\sqrt{\b\g}&\sqrt{\b\g}\\ \b&\b&-\b\\ \g&-\g&\g \end{array}\right) $$

Suppose $\a,\b,\g\in\C$. Find all eigen-values and eigen-vectors of the matrix $$ \left(\begin{array}{ccc} 0&-\g&\b\\ -2\b&2\a&0\\ 2\g&0&-2\a \end{array}\right) $$

For any Lie algebra $A$ define the Killing form $B:A\times A\rar\bK$ by $$ B(x,y)\colon=\tr(\ad(x)\ad(y))~. $$ 1. Show that $B$ is symmetric and $B(\ad(x)y,z)=-B(y,\ad(x)z)$. 2. If $A$ is semi-simple than $B(x,y)$ is not degenerated, i.e. for all $x\neq0$ there is some $y$ such that $B(x,y)\neq0$. Moreover, if $\ad(x)$ is skew adjoint, then $-B(x,y)$ is euclidean. 3. If $A$ is nilpotent, then for all $x,y\in A$: $B(x,y)=0$.

1. By Jacobi`s identity we have: $\ad([x,y])=\ad(x)\ad(y)-\ad(y)\ad(x)$ and thus $$ \ad(\ad(x)y)\ad(z) =\ad([x,y])\ad(z) =\ad(x)\ad(y)\ad(z)-\ad(y)\ad(x)\ad(z) $$ and similarly: $$ \ad(y)\ad(\ad(x)z) =\ad(y)\ad(x)\ad(z)-\ad(y)\ad(z)\ad(x) $$ It follows that $$ \tr(\ad(\ad(x)y)\ad(z)+\ad(y)\ad(\ad(x)z)) =\tr(\ad(x)\ad(y)\ad(z)-\ad(y)\ad(z)\ad(x)) =0~. $$ 2. Assume $\ad(x)^*=\ad(x^*)$, then $B(x,x^*)=\tr(\ad(x)\ad(x)^*)$, which vanishes iff $\ad(x)=0$, i.e. iff $x$ is in the center of $A$. If in addition $\ad(x)^*=-ad(x)$, then $-B(x,y)=\tr(\ad(x)\ad(x)^*)\geq0$. 3. As $A$ is nilpotent there is some $n\in\N$ such that for all $x_1,\ldots,x_n\in A$: $\ad(x_1)\cdots\ad(x_n)=0$. Hence the operator $\ad(x)\ad(y)$ is nilpotent, which implies that its only eigen-value is $0$, in particular $\tr(\ad(x)\ad(y))=0$.