Consider the Lie Algebra $L=(\mathbb{R}^3,\times)$. Show $\nexists x \in L$ s.t. $ad(x)$ is diagonalisable Consider the Lie Algebra $L=(\mathbb{R}^3,\times)$. Show $\nexists x \in L$ s.t. $ad(x)$ is diagonalisable
I'm trying to show that the 2 dimensional special linear Lie Algebra $sl(n,\mathbb{R})$ is not isomorphic to $L=(\mathbb{R}^3,\times)$ where $\times$ is the cross product (which is antisymmetric and satisfies the Jacobi identity, and thus turns $\mathbb{R}^3$ into a Lie algebra). I have been provided with a hint, which is to show that $\nexists x \in L$ s.t. $ad(x)$ is diagonalisable.
Ontop of showing that there can not be such an element in $L$, I'm trying to understand why this would show that the Lie algebras are not isomorphic. To this extent, my idea is to consider the standard presentation of $sl(n,\mathbb{R})$, which includes a matrix with only diagonal elements, oft called $h$. Then I could try to show that $ad(h)$ is also diagonalisable, then use some arguement about how isomorphisms map diagonalisable elements to diagonalisable elements.
Anyway... If anyone could help me out with the parts I'm stuck on and perhaps crystallize some my ideas for me it'd be greatly appreciated.. Thanks yall!!
 A: The two Lie algebras are isomorphic over $\mathbb{C}$, but not over $\mathbb{R}$. The multiplications in $\mathfrak{sl}(2,\mathbb{R})$ are $[H,X]=2X,[H,Y]=-2Y$ and $[X,Y]=H.$ Here we have $\operatorname{ad}H=\operatorname{diag}(0,2,-2).$
Now the multiplications in $(\mathbb{R}^3,\times)$ are given by $[U,V]=W,[V,W]=U,[W,U]=V.$ If there was an isomorphism, then there would exist a vector $T:=uU+vV+wW$ which maps onto $H.$ Then $\operatorname{ad}T$ maps on $\operatorname{ad}H.$ You can calculate the eigenvalues of $\operatorname{ad}T$ which should include a complex one.
A: There is a neat way to prove this which doesn't need the relation to $\mathfrak{sl}(2,\mathbb R)$. Consider$B(x,y):=tr(ad(x)\circ ad(y))$, which evidently defines an $\mathbb R$-valued bilinear form. Now if $ad(x)$ would be diagonalizable, then using an eigenbasis, you conclude that $B(x,x)\geq 0$ (since we are working over $\mathbb R$). But you can easily compute the matrix of $B$ with respect to the standard basis explicitly and conclude that $B(e_i,e_j)=-2\delta_{ij}$, which shows that $B$ is negative definite.
This is a special instance of the Killing form, whose signature can be used to distinguish different real forms of a complex semisimple Lie algebra in general.
Edit (addressing the comment below): To compute $B$, you have to write the matrix corresponding to $ad(e_i)$. The $j$th column of this is obtained by exanding $ad(e_i)(e_j)=e_i\times e_j$ in terms of the $e$'s. For example $ad(e_1)$ maps $e_1$ to $0$, $e_2$ to $e_3$ and $e_3$ to $-e_2$, so this gives $\begin{pmatrix} 0 & 0 & 0\\ 0& 0 &-1\\ 0 & 1 & 0\end{pmatrix}$. Squaring this, you get a diagonal matrix with entries $0$, $-1$, and $-1$, which has trace $-2$, so $B(e_1,e_1)=-2$. The remaining values for $B(e_i,e_j)$ are computed similarly.
