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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!!

FireFenix777
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2 Answers2

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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.

Marius S.L.
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    Can you help me to calculate the eigen values of $ad(T)$?

    Is the point that $ad(T)$ has a complex eigenvalue whilst $ad(H)$ has all real eigenvalues and those $ad(T)$ can't be mapped onto $ad(H)$??

    Edit: oh, $ad(T)$ will have a complex eigenvalue and thus it can't be diagonalisable over $\mathbb{R}$... I'll see if I can figure out how to calculate it's eigenvalues, thank you.

     – FireFenix777 Sep 20 '20 at 00:18
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    Yeah, I need help calculating the eigenvalues of $ad(T)$, I'd really appreciate it. – FireFenix777 Sep 20 '20 at 00:21
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    The matrix of $\operatorname{ad}T$ is given by the images of the basis vectors $U,V,W$. The first column is $$[T,U]=u[U,U]+v[V,U]+w[W,U]=0-vW+wV=(0,w,-v)^\tau$$ and the others accordingly. I know a complex isomorphism between the Pauli matrices and $U,V,W$. Such an isomorphism has to exist as there is only one three dimensional simple Lie algebra over the complex numbers. Hence the only way that they are different over the reals is because not all eigenvalues are available. – Marius S.L. Sep 20 '20 at 00:30
  • Hmm... So by calculating all the columns of $ad(T)$ it should become obvious that all the eigenvalues are not available? – FireFenix777 Sep 20 '20 at 03:26
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    That's the idea. $\operatorname{ad}T$ is a generic element. Since $\mathfrak{sl}(2,\mathbb{R})$ has an ad-semisimple element $H$, there has to be one, too, in $(\mathbb{R}^3,\times)$ if they were isomorphic. Now we want to show that no combination of $u,v,w$ leads to a diagonalizable (= semisimple) matrix. – Marius S.L. Sep 20 '20 at 11:14
  • Got any tips on how to know if a matrix is NOT diagonalizable?? I've always been a little confused on this, as you have to show that it it not diagonal in EVERY possible basis.... – FireFenix777 Sep 21 '20 at 09:30
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    You can bring a matrix into Jordan normal form where you can see the geometric multiplicities of the eigenvalues. There are also criteria for diagonalizability, like: the minimal polynomial splits into pairwise disjoint linear factors. Some of the many algorithms for matrices are also online available, e.g. https://www.symbolab.com/solver/matrix-calculator or https://www.emathhelp.net/calculators/linear-algebra/ – Marius S.L. Sep 21 '20 at 09:56
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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.

Andreas Cap
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    Cool, this is great, thanks!! – FireFenix777 Sep 20 '20 at 13:54
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    I'm trying to figure out why $B(e_i,e_j)=-2\delta_{i,j}$. I feel like it should always equal zero and can't imagine why a $-2$ would show up... To compute the matrix of $ad(\epsilon_i)$, i would apply $ad(\epsilon_i)$ to the basis element $\epsilon_j$ to get the $j$'th column... It seems like they would all be zero!!! It would be zero if $i \neq j$ and if $i=j$ then $ad(\epsilon_i)(\epsilon_i) = [\epsilon_i,\epsilon_i]=0$ – FireFenix777 Sep 21 '20 at 10:34
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    I have edited the answer to address this. – Andreas Cap Sep 22 '20 at 12:47
  • Answer looks the same to me ?? – FireFenix777 Sep 22 '20 at 12:55
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    I should have said "I'll edit ..." :-) – Andreas Cap Sep 22 '20 at 12:55