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Prove or disprove: Let $L$ be a Lie algebra over any field $F$ with $\dim(L)=\dim(L')=3$. Then there exists a basis $\{x,y,z\}$ of $L$ such that the change of basis matrix from $\{[y,z],[z,x],[x,y]\}$ to $\{x,y,z\}$ is diagonal.

My thinking (Prove): Suppose $$ \begin{bmatrix}[y,z] \\ [z,x] \\ [x,y] \end{bmatrix}= \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}. $$ If $i\neq 0$, let $z'=gx+hy+iz$ then $$ \begin{bmatrix}[y,z'] \\ [z',x] \\ [x,y] \end{bmatrix}= \begin{bmatrix}a'&b'&c'\\d'&e'&f'\\0&0&1\end{bmatrix} \begin{bmatrix}x\\y\\z'\end{bmatrix}. $$ If $c'\neq 0$, let $\alpha=f'/c'$ and $x'=x+\alpha y$ then $$ \begin{bmatrix}[y,z'] \\ [z',x'] \\ [x',y] \end{bmatrix}= \begin{bmatrix}a''&b''&c''\\d''&e''&0\\0&0&1\end{bmatrix} \begin{bmatrix}x'\\y\\z'\end{bmatrix}. $$ $$ \vdots $$ If we can arrive that $$ \begin{bmatrix}[y',z'] \\ [z',x'] \\ [x',y'] \end{bmatrix}= \begin{bmatrix}\theta&0&0\\0&\phi&0\\0&0&1\end{bmatrix} \begin{bmatrix}x'\\y'\\z'\end{bmatrix} $$ then we are done. But it is too tedious to exhaust all cases. I hope there is a smarter approach.

FAX
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1 Answers1

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It essentially goes like this. Since $L=L'$, the Lie algebra $L$ is simple. Simple $3$-dimensional Lie algebras are classified over an arbitrary field, see for example Strade's article on Lie algebras in small dimensions.
Denote the base change matrix by $M_{x,y,z}$. A lengthy calculation with the Jacobi identity shows that $M_{x,y,z}$ is in fact symmetric. It then follows from linear algebra that a basis for $L$ can be chosen so that $M_{x,y,z}$ is diagonal, see for example H. Anton: Elementary Linear Algebra, Eighth Edition, John Wiley & Sons, Inc, page $357$. The case of characteristic $p=2$ is more difficult, because there we cannot diagonalize every symmetric matrix. For more details then see this thesis, page $13-14$.

Dietrich Burde
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  • The symmetry property is amazing, but the diagonalbility of symmetric matrix for any field is not true in general ( https://math.stackexchange.com/questions/2096374 ). – FAX Feb 26 '18 at 13:59
  • But I can still complete it by exhaustion since it becomes simple by the symmetry property, if it is true. Thank a lot ! – FAX Feb 26 '18 at 14:02
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    Yes indeed, see also here. It makes sense to check the linked thesis on this; how they solve this in characteristic $2$. – Dietrich Burde Feb 26 '18 at 14:04