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It is well known that $\mathfrak{su}(2)$, the real Lie algebra of traceless skew-Hermitian $2\times 2$ complex matrices, satisfies $\mathfrak{su}(2)_{\mathbb C}\simeq \mathfrak{sl}(2,\mathbb C)$. To see this, it is sufficient to observe that any traceless matrix $A$ can be written as $$A = i\left(\frac{A+A^\dagger}{2i}\right) + \left(\frac{A-A^\dagger}{2}\right),$$ where both components are traceless and skew-Hermitian, and the decomposition is unique.

We also know that $\mathfrak{so}(3)\simeq\mathfrak{su}(2)$, where $\mathfrak{so}(3)$ is the real Lie algebra of traceless skew-orthogonal $3\times 3$ real matrices. This follows from observing that both Lie algebras satisfy the same commutation relations, $[T_i,T_j]=\epsilon_{ijk}T_k$ (or rather, we can always find bases for both spaces satisfying such relations).

This should imply that also $\mathfrak{so}(3)_{\mathbb C}\simeq\mathfrak{sl}(2,\mathbb C)$ (as also mentioned in passing in this answer), but how can I show that this is the case more directly? As far as I understand, this statement should mean that, given any traceless $2\times 2$ complex matrix $A$, there is a bijection sending $A$ to two $3\times 3$ real skew-orthogonal matrices. What is this decomposition?

glS
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  • Is the notation $\mathfrak{su}(2, \mathbb{C})$ used anywhere? – orangeskid Aug 07 '21 at 01:59
  • @orangeskid I think I saw it somewhere, but I guess you are right that is not as standard (and I don't remember where I saw it). It made sense thinking of it as meaning "space of skew-Hermitian 2x2 complex matrices", in analogy with what $\mathfrak{sl}(2,\mathbb C)$ represents. But yea, with su(2) it might be more misleading because the algebra is real – glS Aug 07 '21 at 10:31

3 Answers3

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The most direct way is the straightforward computation of a Lie algebra isomorphism $\phi \colon \mathfrak{sl}_2(\Bbb C)\rightarrow \mathfrak{so}_3(\Bbb C)$. One sets up a matrix $\phi=(a_{ij})$ with parameters $a_{ij}\in \Bbb C$, and then rewrites the identity (using the Lie brackets $[x,y]_1$ and $[x,y]_2$ for the two given algebras) $$ \phi([e_i,e_j]_1)=[\phi(e_i),\phi(e_j)]_2 $$ with respect to the standard basis of $\mathfrak{sl}_2(\Bbb C)$ by polynomial equations for these variables. These equations are quite easy to solve, because the condition $\det(\phi)\neq 0$ is very strong.

In the end we obtain a solution $\phi$, which need not be unique by the way, so the two Lie algebras are isomorphic.

References:

Lie algebra isomorphism between ${\rm sl}(2,{\bf C})$ and ${\bf so}(3,\Bbb C)$

An Explicit Isomorphism Between the Three Dimensional Orthogonal Lie Algebra and the Special Linear Lie Algebra of Dimension $3$

Dietrich Burde
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  • you are essentially saying to map $A$ to its skew-Hermitian components in $\mathfrak{su}(2)$, and then exploit the isomorphism $\mathfrak{su}(2)\simeq \mathfrak{so}(3)$ to obtain the final decomposition. While this is fine, I'm not sure how to relate this with it being a "complexification". In other words, I understand this as a (real) Lie algebra isomorphism between $\mathfrak{sl}(2,\mathbb C)$ and $\mathfrak{so}(3)\oplus \mathfrak{so}(3)$, but doesn't talking about "complexification" here also amount to something more? – glS Oct 15 '20 at 10:19
  • No, I am not referring to skew-Hermitian or real case at all. I simply take your question without knowing this. I just do a computation over the complex numbers. So we compute an isomorphism between these two complex Lie algebras, over the complex numbers. The $a_{ij}$ are complex numbers. – Dietrich Burde Oct 15 '20 at 10:20
  • I see. I still don't quite understand how you build $\phi$ though. What do you mean with "rewrite the identity" here? – glS Oct 15 '20 at 10:31
  • For example, since $[e_1,e_2]=e_3$ we must have $[\phi(e_1),\phi(e_2)]=\phi(e_3)$ and we know that $\phi(e_1)=a_{11}f_1+a_{21}f_2+a_{31}f_3$, where $f_1,f_2,f_3$ is the basis of the second Lie algebra. Do this also for $\phi(e_2)$ and $\phi(e_3)$. Then this gives equations for the coefficients, because two vectors are equal iff their coefficients for the basis elements are equal. – Dietrich Burde Oct 15 '20 at 10:33
  • See the answers given at this post for more details of a direct computation in the real case. We can do the same in the complex case, but the answer then will be positive. – Dietrich Burde Oct 15 '20 at 13:31
  • to be clear, in your example if $[e_1,e_2]=e_3$ that means that $e_i\in\mathfrak{so}(3)$, and thus $\phi:\mathfrak{so}(3){\mathbb C}\to\mathfrak{sl}(2;\mathbb C)$ (i.e. the inverse direction of that in the answer), correct? For the calculation, I'll have to write $\phi(e_i)=a{ji} f_j$ and thus the commutation rules of $\mathfrak{so}(3)$ translate into the conditions $a_{ki}a_{\ell j}[f_k,f_\ell]=\epsilon_{ijm}a_{nm}f_n$. I should then use the commutation relations between the $f_i\in\mathfrak{sl}(2)$ to obtain the form of $a_{ij}$, i.e. $\phi$. – glS Oct 30 '20 at 11:03
  • No, I wanted to start with $\mathfrak{sl}(2)$, having basis $(e_1,e_2,e_3)$ with $[e_1,e_2]=e_3$, $[e_3,e_1]=2e_1$ and $[e_3,e_2]=-2e_2$, as said in my answer. But of course, it doesn't matter. You can also proceed like you did. – Dietrich Burde Oct 30 '20 at 11:57
  • going back to this, I tried to sketch the calculation explicitly in my answer. I get 9 conditions plus the condition on the determinant, and I know a solution to these constraints from the known isomorphism $\mathfrak{sl}(2,\mathbb C)\simeq\mathfrak{su}(2)_{\Bbb C}$. However, I'm not really sure how I could have got to a solution without using this. Is there a general way to approach this sort of system of polynomials? I can see the conditions can be combined to given conditions on the determinant, but how does one go from that to explicit solutions? – glS Aug 07 '21 at 00:49
  • @glS By explicitly solving the equations, first the "easy ones" (substitution of a variable), and then the other ones, possibly computing a Gröbner basis. But actually, I never used Gröbner, but solved quadratic equations in the variables with a resultant. – Dietrich Burde Aug 07 '21 at 09:58
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In general, for every semi-simple Lie algebra $\frak{g}$ (finite dimensional, over a field of char $0$) we have an imbedding $$ad\colon {\frak g} \to so({\frak g}, k)$$ where $k$ is the Killing form on $g$ (non-degenerate).

In our case ${\frak g}= sl(2, \mathbb{C})$ over the field $\mathbb{C}$, any two non-degenerate bilinear symmetric form over $g$ are isomorphic. Moreover, $sl(2, \mathbb{C})$ and $so(3, \mathbb{C})$ have the same dimension. So the above map is an isomorphism.

orangeskid
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General approach

The general idea is to find the conditions for a linear map $\phi:\mathfrak g\to\mathfrak h$ to be an invertible Lie algebra homomorphism, and thus such that $\phi([g_1,g_2])=[\phi(g_1),\phi(g_2)]$ for all $g_1,g_2\in\mathfrak g$. Here $\mathfrak g,\mathfrak h$ are generic Lie algebras over some as-of-yet unspecified field. We must clearly assume them to have the same dimension if we are to hope for $\phi$ to be invertible.

Let $\{e_i\}\subset\mathfrak g,\{t_i\}\subset\mathfrak h$ be bases for $\mathfrak g,\mathfrak h$, respectively, and write the corresponding structure coefficients as $$[e_i,e_j] = {c_{ij}}^k e_k, \qquad [t_i,t_j] = {d_{ij}}^k t_k.$$ Let us also write the coefficients of $\phi$ as $\phi(e_i)=\sum_j \phi_{ji}t_j$. We thus must have $$ \phi([e_i,e_j]) = \sum_k {c_{ij}}^k \phi(e_k) = \sum_{k\ell} {c_{ij}}^k \phi_{\ell k} t_\ell \\ = [\phi(e_i),\phi(e_j)] = \sum_{mn} \phi_{mi}\phi_{nj} [t_m,t_n] = \sum_{mn\ell} \phi_{mi}\phi_{nj} {d_{mn}}^\ell t_\ell. $$ In summary, we have the conditions $$\sum_k {c_{ij}}^k \phi_{\ell k} = \sum_{mn} \phi_{mi} \phi_{nj} {d_{mn}}^\ell,$$ for all $i,j,\ell$.

Specific case $\mathfrak{sl}(2,\mathbb C)\to\mathfrak{so}(3,\mathbb C)$

We are looking for a linear operator $\phi:\mathfrak{sl}(2,\mathbb C)\to\mathfrak{so}(3,\mathbb C)$ that preserves the Lie structure.

The (complex) Lie algebra $\mathfrak{sl}(2,\mathbb C)$ is generated by three basis elements, $e_1,e_2,e_3$, such that $$[e_3,e_1]=2e_1,\qquad [e_3,e_2]=-2e_2, \qquad [e_1,e_2]=e_3.$$

For the complex Lie algebra $\mathfrak{so}(3,\mathbb C)$ we have the basis $\{t_1,t_2,t_3\}$ with $[t_i,t_j]=\epsilon_{ijk}t_k$.

The conditions found above thus become here: $$ 2\phi_{k1} = \sum_{ij} \phi_{i3}\phi_{j1} \epsilon_{ijk}, \qquad 2\phi_{k2} = -\sum_{ij}\phi_{i3}\phi_{j2} \epsilon_{ijk}, \\ \phi_{k3} = \sum_{ij}\phi_{i1}\phi_{j2} \epsilon_{ijk}. $$

More explicitly, this is the system of equations: $$ 2\phi_{11} = \phi_{23}\phi_{31} - \phi_{33}\phi_{21}, \quad 2\phi_{21} = \phi_{33}\phi_{11} - \phi_{13}\phi_{31}, \quad 2\phi_{31} = \phi_{13}\phi_{21} - \phi_{23}\phi_{11}, \\ 2\phi_{12} = \phi_{33}\phi_{22} - \phi_{23}\phi_{32}, \quad 2\phi_{22} = \phi_{13}\phi_{32} - \phi_{33}\phi_{12}, \quad 2\phi_{32} = \phi_{23}\phi_{12} - \phi_{13}\phi_{22}, \\ \phi_{13} = \phi_{21}\phi_{32} - \phi_{31}\phi_{22}, \quad \phi_{23} = \phi_{31}\phi_{12} - \phi_{32}\phi_{11}, \quad \phi_{33} = \phi_{11}\phi_{22} - \phi_{21}\phi_{12}. $$ It remains to find a solution for these $9$ equation, that also satisfies the invertibility condition: $\det\phi\neq0$.

A known solution

Without attempting to solve these equations head on, there is one solution we already know. Indeed, we can easily verify the relations $$[2i t_3, i(t_1\pm i t_2)] = \pm 2i (t_1\pm it_2), \qquad [i(t_1+it_2),i(t_1-it_2)]= 2i t_3,$$ which implement the isomorphism $\mathfrak{sl}(2,\mathbb C)\to\mathfrak{su}(2)_{\Bbb C}$. This means that $$\phi = \begin{pmatrix} i & i & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 2i \end{pmatrix}$$ is one of the isomorphisms we've been looking for. One can then readily verify that this $\phi$ satisfies the nine conditions given above, and that $\det\phi=-4\neq0$.

Head-on solution to the equations

Well, I suppose I wanted to hurt myself a little today, so I went on and worked out the landscape of possible solutions to the 9 equations above, with the help of Mathematica.

  1. $\phi_{11}=\phi_{21}=0$ implies $\phi=0$.

  2. $\phi_{11}=0, \phi_{21}\neq0$ implies $$\phi_{13} = \pm2 i, \qquad 2\phi_{31}=\pm 2i \phi_{21}.$$ Moreover, we have the conditions $$\phi_{12}^2\phi_{21}^2=4+4\phi_{21}\phi_{22}, \qquad \pm 2i\phi_{12}\phi_{21}=2\phi_{23}, \\ \pm 2i (\phi_{12}^2\phi_{21}-2\phi_{22})=4\phi_{32}, \qquad \phi_{21}\phi_{12}+\phi_{33}=0.$$ Using these we get the matrices $$\begin{pmatrix}0 & \phi_{12} & \pm 2i \\ \phi_{21} & \phi_{22} & \pm 2i \phi_{12}\phi_{21} \\ \pm i\phi_{21} & \mp \frac{i}{2} (\phi_{12}^2\phi_{21}-2\phi_{22}) & -\phi_{12}\phi_{21}\end{pmatrix},$$ with the additional condition $$\phi_{12}^2\phi_{21}^2 = 4 + 4\phi_{21}\phi_{22}.$$ The determinant of this matrix reads $$\det(\phi) = -2 \phi_{12}^2\phi_{21}^2 + 4 \phi_{21}\phi_{22},$$ and we thus have an invertible matrix.

  3. If $\phi_{11}\neq0$ and $1+\phi_{11}\phi_{12}\neq0$, then $$\begin{pmatrix} \phi_{11} & \phi_{12} & \pm2i\sqrt{1+\phi_{11}\phi_{12}} \\ \phi_{21} & \phi_{22} & \frac{\phi_{13}(\phi_{12}\phi_{21}+\phi_{11}\phi_{22})}{2(1+\phi_{11}\phi_{12})} \\ \frac12(\phi_{13}\phi_{21}-\phi_{11}\phi_{23}) & \frac12(\phi_{12}\phi_{23}-\phi_{13}\phi_{22}) & \frac12(\phi_{11}\phi_{22}-\phi_{12}\phi_{21}) \end{pmatrix},$$ where in the third row $\phi_{13}$ and $\phi_{23}$ are to be replaced with their explicit expressions (I didn't do it here to keep the expression from exploding). The final expression thus depends on the variables $\phi_{11},\phi_{12},\phi_{21},\phi_{22}$. The additional condition on this matrix is $$2\sqrt{(1+\phi_{11}\phi_{12})(\phi_{11}^2+\phi_{21}^2)} = \pm [\phi_{11}^2\phi_{22} - \phi_{21}(2+\phi_{11}\phi_{12})].$$ There is probably a way to see the compatibility of this condition with the determinant condition, but the equation get a little bit too unwieldy for me.

  4. If $\phi_{11}\neq0$ and $1+\phi_{11}\phi_{12}=0$ (note that this is the case for the solution given at the beginning), we find $$\begin{pmatrix} \phi_{11} & \phi_{12}& 0 \\ \phi_{21} & \phi_{12}^2\phi_{21} & \pm2i\sqrt{1+\phi_{12}^2\phi_{21}^2} \\ \mp i \phi_{11}\sqrt{1+\phi_{12}^2\phi_{21}^2} & \pm i\phi_{12} \sqrt{1+\phi_{12}^2\phi_{21}^2} & -2 \phi_{12}\phi_{21} \end{pmatrix},$$ whose determinant is $4\phi_{12}\phi_{11}$, which is consistent with our original solution, choosing $\phi_{11}=\phi_{12}=i$ and $\phi_{21}=-1$. We also see that this is a valid solution whenever $\phi_{12},\phi_{11}\neq0$.

An example of a class of solutions generalising the one we gave at the beginning is obtained fixing $\phi_{11}=\phi_{12}=i$ but leaving free $\phi_{21}$: $$\begin{pmatrix} i & i & 0 \\ \phi_{21} & -\phi_{21} & \pm 2i\sqrt{1-\phi_{21}^2} \\ \pm \sqrt{1-\phi_{21}^2} & \mp \sqrt{1-\phi_{21}^2} & -2i \phi_{21} \end{pmatrix}.$$ For example, if $\phi_{21}=0$ we get $$\phi = \begin{pmatrix} i & i & 0 \\ 0 & 0 & \pm 2i \\ \pm 1 & \mp 1 & 0 \end{pmatrix}.$$

Mathematica code

The snippet I used to make exploring the solutions easier is

expr\[Phi][k_, 1] := 1/2 Sum[
    \[Phi][i, 3] \[Phi][j, 1] LeviCivitaTensor[3][[i, j, k]],
    {i, 1, 3}, {j, 1, 3}
    ];
expr\[Phi][k_, 2] := -(1/2) Sum[
    \[Phi][i, 3] \[Phi][j, 2] LeviCivitaTensor[3][[i, j, k]],
    {i, 1, 3}, {j, 1, 3}
    ];
expr\[Phi][k_, 3] := Sum[
   \[Phi][i, 1] \[Phi][j, 2] LeviCivitaTensor[3][[i, j, k]],
   {i, 1, 3}, {j, 1, 3}
   ];

Reduce[
  Flatten@Table[
    [Phi][i, j] == expr[Phi][i, j],
    {i, 3}, {j, 3}
    ],
  Flatten@Array[[Phi]@## &, {3, 3}]
  ] // FullSimplify

and then adding assumptions as appropriate.

glS
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