I am working through the exercises in "Lie Groups, Lie Algebras, and Representations" – Hall and can't complete exercise 11 of chapter 3. My aim was to demonstrate that there does not exist a vector space isomorphism $A$ between the two spaces that also preserves the commutator.
$$[AX, AY] = A[X, Y]$$
To this end I computed the following commutation relations on bases for two spaces.
For the $\mathfrak{su}(2)$ basis matrices $e_1, e_2, e_3$ it holds that
$$[e_1, e_2] = 2e_3 \,\,\,\,\,\, [e_1, e_3] = -2e_2 \,\,\,\,\,\, [e_2, e_3] = 2e_1$$
For the $\mathfrak{sl}(2, \mathbb{R})$ basis matrices $f_1, f_2, f_3$ it holds that
$$[f_1, f_2] = 2f_2 \,\,\,\,\,\, [f_1, f_3] = -2f_3 \,\,\,\,\,\, [f_2, f_3] = f_1$$
It is clear that for the linear bijection $(e_1, e_2, e_3) \mapsto (f_1, f_2, f_3)$ would not preserve the relationships, nor would a permutation of the target matrices. However, I need to show no invertible matrix satisfies
$$[AX, AY] = A[X, Y]$$
So from there I began to derive equations for the elements of $A$. They are ugly expressions in terms of the sub-determinants of the $A$ matrix, and given them I can't think of a way to conclude $A$ cannot exist. Is there an easier way to finish the proof than to derive the equations for $A$?
Note: I have looked up solutions for this problem and the only technique I see hinted at is to consider Killing forms (which have not yet been covered in this book).
Best Answer
Your approach works without problems, if you write the condition $[Ax,Ay]=A[x,y]$ for all $x,y$ in terms of the $9$ coefficients of the matrix $A$. The polynomial equations in these $9$ unknowns over $\mathbb{R}$ quickly yield $\det(A)=0$, a contradiction.
Another elementary argument is the following. $\mathfrak{sl}(2,\mathbb{R})$ has a $2$-dimensional subalgebra, e.g., $\mathfrak{a}=\langle f_1,f_2\rangle$, but $\mathfrak{su}(2)$ has no $2$-dimensional subalgebra. Hence they cannot be isomorphic.