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I am trying to prove that the lie Algebras $\mathfrak{so} \left( 3; 1 \right)$ and $\mathfrak{sl} \left( 2, \mathbb{C} \right)$, viewed as a real Lie algebra, as isomorphic. To do so, I have considered the following basis for $\mathfrak{so} \left( 3; 1 \right)$:

$$E_1 = \left[ \begin{matrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], E_2 = \left[ \begin{matrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], E_3 = \left[ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} \right], E_4 = \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right], E_5 = \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{matrix} \right], E_6 = \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix} \right].$$

And for $\mathfrak{sl} \left( 2, \mathbb{C} \right)$, viewed as a real vector space, I have considered the following basis:

$$H_1 = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right], H_2 = \left[ \begin{matrix} \iota & 0 \\ 0 & - \iota \end{matrix} \right], X_1 = \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right], X_2 = \left[ \begin{matrix} 0 & \iota \\ 0 & 0 \end{matrix} \right], Y_1 = \left[ \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right], Y_2 = \left[ \begin{matrix} 0 & 0 \\ \iota & 0 \end{matrix} \right].$$

With this, we have the following commutation relations:

$$[E_1, E_2] = E_4, [E_1, E_3] = E_5, [E_1,E_4] = -E_2, [E_1, E_5] = -E_3, [E_1, E_6] = 0, [E_2, E_3] = E_6, [E_2, E_4] = E_1, [E_2, E_5] = 0, [E_2, E_6] = -E_3, [E_3, E_4] = 0, [E_3, E_5] = -E_1, [E_3, E_6] = -E_2, [E_4, E_5] = E_6, [E_4, E_6] = -E_5, [E_5, E_6] = -E_4,$$

and

$$[H_1, H_2] = 0, [H_1, X_1] = 2X_1, [H_1, X_2] = 2X_2, [H_1, Y_2] = -2Y_1, [H_1, Y_2] = -2Y_2, [H_2, X_1] = 2X_2, [H_2, X_2] = -2X_1, [H_2, Y_1] = -2Y_2, [H_2, Y_1] = 2Y_1, [X_1, X_2] = 0, [X_1, Y_1] = H_1, [X_1, Y_2] = H_2, [X_2, Y_1] = H_2, [X_2, Y_2] = - H_1, [Y_1, Y_2] = 0.$$

With only this knowledge, I want to construct a Lie algebra isomorphism $\phi : \mathfrak{so} \left( 3; 1 \right) \rightarrow \mathfrak{sl} \left( 2, \mathbb{C} \right)$, where the codomain is viewed as a real Lie algebra.

THe problem is that if we look at the commutation relations, we cannot let any of the $E_i$'s to be $H_1$ because none of them preserve any of the other vectors. That is to say, if we let, for instance $\phi \left( E_2 \right) = H_1$, then all the commutation relations with $E_2$ give some other vector, i.e., $[E_2, E_j] = E_k$, where $k \neq j$ (or, it is zero). But [H_1, X_1] is a multiple of $X_1$. So if $\phi \left( E_j \right) = X_1$ for some $j$, then we must have $\phi \left( \left[ E_2, E_j \right] \right) = \left[ \phi \left( E_2 \right), \phi \left( E_j \right) \right]$, which is a multiple of $X_1$. Then, we would get $\phi \left( E_k \right) = X_1$, which cannot be possible unless $j = k$ (because we want $\phi$ to be an isomorphism).

So, can we conclude that $\mathfrak{so} \left( 3; 1 \right)$ is not isomorphic to $\mathfrak{sl} \left( 2, \mathbb{C} \right)$ viewed as a real Lie algebra? Or am I doing something wrong? The book "Lie Groups, Lie Algebras and Representations" by Hall claims that they are isomorphic in an exercise of chapter $7$.

Aniruddha Deshmukh
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    There is a typo in $E_5$. The matrix $[E_1,E_3]$ is different from your $E_5$ (it should be a symmetric matrix). Also, you have to consider that $\phi(E_i)$ is an arbitrary linear combination of the six basis vectors of $\mathfrak{sl}_2(\Bbb C)$, viewed as real Lie algebra. You cannot assume that $\phi(E_j)=X_1$, for example. – Dietrich Burde Jun 02 '21 at 10:21
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    That's not a good basis to use for $\mathfrak{sl} \left( 2, \mathbb{C} \right)$ for this purpose. Take instead the Pauli spin matrices, and the Pauli spin matrices times $i$. The Pauli spin matrices times $i$ are antihermitian and obey the same commutation relations as the 3 Euclidean infintesimal rotations $E_1, E_2, E_4$ in your notation (up to scalar). Then the Pauli spin matrices (hermitian) have the same commutation relations as $E_3, E_5, E_6$. –  Jun 02 '21 at 10:27
  • @DietrichBurde, I have made the edits. – Aniruddha Deshmukh Jun 03 '21 at 04:46
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    Yes, good. But the problem remains. An isomorphism need not map an $E_j$ to a basis vector. We only know that, say, $\phi(E_1)=a_1X_1+a_2X_2+b_1Y_1+b_2Y_2+c_1H_1+c_2H_2$. – Dietrich Burde Jun 05 '21 at 08:14
  • Continuing the discussion, is it possible that $\mathfrak{so} \left( 3;1 \right)_{\mathbb{C}}$ (the complexification of $\mathfrak{so} \left( 3;1 \right)$) is isomorphic to $\mathfrak{sl} \left( 2, \mathbb{C} \right) \oplus \mathfrak{sl} \left( 2, \mathbb{C} \right)$? I think it is not. Because the former is only $6$ dimensional, while the latter is $9$ dimensional. – Aniruddha Deshmukh Jun 07 '21 at 04:09

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