How should I show that the Lie algebra so(6) of SO(6) is isomorphic to the Lie algebra su(4) of SU(4)?

Question

As far as I can see, an isomorphism of Lie algebras is a bijective map which preserves the Lie bracket.

I need to show that $\mathfrak{so}(6)$ (the Lie algebra of SO(6)) is isomorphic to the $\mathfrak{su}(4)$ (the Lie algebra of SO(4)). I know that $\mathfrak{so}(6)$ is the set of 6x6 real antisymmetric matrices and $\mathfrak{su}(4)$ is the set of 4x4 anti Hermitian matrices. Both types of matrices have 15 real independent components. Is this enough to say that both are isomorphic to $\mathbb{R}^{15}$? Since the Lie bracket of $\mathbb{R}^{15}$ is $[x,y]=0$, the preservation of the Lie bracket under the maps appears to be trivial.

At first I hoped that this would be enough to prove that $\mathfrak{su}(4)$ and $\mathfrak{so}(6)$ were isomorphic, but I don't think the map from $\mathfrak{su}(4)$ to $\mathbb{R}^{15}$ to $\mathfrak{so}(6)$ would preserve the Lie bracket. Am I right in saying that Lie algebra homomorphism need not be transitive, i.e. $\psi : \mathscr{A} \rightarrow \mathscr{B}, \ \phi : \mathscr{B} \rightarrow \mathscr{C}$ Lie algebra homomorphisms need not imply $\phi \circ \psi : \mathscr{A} \rightarrow \mathscr{C}$ a Lie algebra homomorphism.

Getting back to the original problem, how should I show that $\mathfrak{su}(4)$ and $\mathfrak{so}(6)$ are isomorphic? I suppose I need to explicitly find a map between them and show that it's an isomorphism? Is there a somewhat general way of finding such a map?

This is my first time posting here, so sorry if my question is a little bit long winded! Thanks, Alex

Answer 1

We have the isomorphism of Lie groups $SO(6) \simeq SU(4)/\{± id\}$, because $SU(4)$ acts on $\Lambda^2 (\mathbb{C}^4)$ with an invariant orthogonal structure given by a choice of an element of $\Lambda^2 (\mathbb{C}^4)^*$. Then it follows that both Lie algebras are isomorphic. Alternatelvely, one can write down bases for both Lie algebras and indeed construct explicitly a linear isomorphism (this is better not to do by hand, but with some computer algebra system like Magma).

Answer 2

As you suspect, neither $\mathfrak{su}(4)$ nor $\mathfrak{so}(6)$ is isomoprhic to $\mathbb R^{15}$ as a Lie algebra, exactly because the first two have nontrivial brackets but the bracket on the last is zero.

Note that $\mathfrak{su}(4)$ is defined in terms of its action on $\mathbb C^4$, and $\mathfrak{so}(6)$ is defined in terms of its action on $\mathbb R^6$. So the best way to show that $\mathfrak{su}(4)$ are $\mathfrak{so}(6)$ is to make the first act on $\mathbb R^6$ or the second on $\mathbb C^4$, and then to check that these actions are the ones desired.

I find the former easier, since it extends to an action of $\mathrm{SU}(4)$ on $\mathbb R^6$, whereas $\mathrm{SO}(6)$ does not act in the desired way on $\mathbb C^4$. The trick is to notice that $\binom42 = 6$. We take the $\mathrm{SU}(4)$ action on $\mathbb C^4$, and use it to act on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $g(v\wedge w) = gv \wedge gw$ for $g\in \mathrm{SU}(4)$ and $v,w \in \mathbb C^4$; the infinitesimal version of this is $x(v\wedge w) = xv \wedge w + v \wedge xw$ for $x\in \mathfrak{su}(4)$.

Of course, the action of $\mathrm{SU}(4)$ on $\mathbb C^4$ extends to an action of $\mathrm{SL}(4,\mathbb C)$. So I will temporarily work with it.

Define a pairing $\langle,\rangle$ on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $\langle v_1\wedge w_1, v_2\wedge w_2 \rangle = \det(v_1,w_1,v_2,w_2)$, where $(v_1,w_1,v_2,w_2)$ denotes the matrix with rows $v_1,w_1,v_2,w_2$ — that this is well-defined follows from standard facts about the determinant. By definition, $\mathrm{SL}(4,\mathbb C)$ consists of all $\mathbb C$-linear automorphisms of $\mathbb C^4$ that preserve the determinant, and therefore the $\mathrm{SL}(4,\mathbb C)$ action on $\mathbb C^6$ preserves this pairing.

But the pairing is nondegenerate, also by standard facts about the determinant. Over $\mathbb C$, a vector space has a unique-up-to-isomorphism nondegenerate pairing. It follows that the action of $\mathrm{SL}(4,\mathbb C)$ on $\mathbb C^6$ factors through the action of $\mathrm{SO}(\mathbb C,6)$, where $\mathrm{SO}(\mathbb C,6)$ is the group of complex matrices preserving the pairing $\langle,\rangle$ (isomorphic to any other copy of such a group).

So, we have constructed a homomorphism $\mathrm{SU}(4) \to \mathrm{SL}(4,\mathbb C) \to \mathrm{SO}(6,\mathbb C)$. But the domain $\mathrm{SU}(4)$ is a compact group, and so its image must be compact (since the homomorphism is a continuous map of manifolds). It is a fact (but I don't remember how easy it is to prove) that every compact subgroup of $\mathrm{SO}(6,\mathbb C)$ is contained within a conjugate of $\mathrm{SO}(6,\mathbb R)$. We therefore get a map $\mathrm{SU}(4) \to \mathrm{SO}(6,\mathbb R)$.

Finally, it is not difficult to check that the action of $\mathfrak{sl}(4)$ on $\mathbb C^6$ constructed above has trivial kernel. Indeed, suppose that $x \in \mathfrak{sl}(4)$ acts trivially. Choose the standard basis $e_1,\dots,e_4$ of $\mathbb C^4$; it induces a basis $e_{12},e_{13},\dots,e_{34}$ on $\mathbb C^6$, where $e_{ii'} = e_i \wedge e_{i'}$ for $i<i'$. If the $(i,j)$th matrix entry for $x$ was $x_i^j$, so that $x(e_i) = \sum_j x_i^j e_j$ then $x(e_{ii'}) = x(e_i)\wedge e_{i'} + e_i \wedge x(e_{i'}) = \sum_j x_i^j e_j \wedge e_{i'} + \sum_{j'} x_{i'}^{j'} e_i \wedge e_{j'}$. For $x$ to act by zero, this sum would have to be zero for all values of $i,i'$. But since we are in four dimensions, for any $i,i'$, there is a $j \neq i,i'$, whence $e_j \wedge e_{i'}$ is independent of $e_i \wedge e_{j'}$ for any $j'$. Thus the only way for $x$ to act as $0$ on $\mathbb C^6$ is if $x_i^j = 0$ for all $i,j$.

Therefore the map $\mathfrak{su}(4) \to \mathfrak{so}(6)$ constructed above has trivial kernel. Since it is between two Lie algebras of the same (finite) dimension, it therefore must be an isomorphism.

Answer 3

This is just an addendum to @Everett You's answer. Indeed, I would like an isomorphism that is easier to remember, so here's how I would write it. Let $ij\equiv i\wedge j$ be the anti-symmetric matrix with $-1$ in the entry $(i,j)$ and $+1$ in the symmetric entry $(j,i)$. Let me also use $1',2',3'$ to reprsent $4,5,6$. Then the following table denote the isomorphism

\begin{array}{c|cccc} & 0 & 1 & 2 & 3\\\hline 0 & & 23 & 31 & 12 \\ 1 & 2'3'& 1'1 & 1'2 & 1'3\\ 2 & 3'1'& 2'1 & 2'2 & 2'3\\ 3 & 1'2'& 3'1 & 3'2 & 3'3 \end{array}

Here is how to read this table. The row $\mu$ and column $\nu$ denotes $\sigma^{\mu\nu}=\sigma^\mu\otimes \sigma^\nu$ in $i \mathfrak{su}(4)$ and the entry in row $\mu$, column $\nu$ is the corresponding anti-symmetric matrix $i\wedge j$, i.e., $\sigma^{\mu \nu} \leftrightarrow i\wedge j$.

In full rigor, the Lie algebra isomorphism should be between the real Lie algebra $\mathfrak{su}(4)$ spanned by $s^{\mu \nu} \equiv\sigma^{\mu \nu}/{2i}$ and the real Lie algebra spanned by $i\wedge j$. Then $s^{0,1} \leftrightarrow 2\wedge 3$ (as an example).