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

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 2

Here I provide an explicit correspondence between the basis matrices of $su(4)$ and $so(6)$, and hence prove the isomorphism.

Let me first introduce my notations. The $su(4)$ algebra is spanned by 15 traceless Hermitian $4\times4$ matrices, which can be constructed by the Kronecker product (tensor product) of two Pauli matrices. Let us denote the Pauli matrices as $\sigma^1$, $\sigma^2$ and $\sigma^3$, and also introduce $\sigma^0$ to be the $2\times2$ identity matrix, then all the $4\times 4$ Hermitian matrices can be decomposed onto $\sigma^{\mu\nu}\equiv\sigma^\mu\otimes\sigma^\nu$ (with $\mu,\nu=0,1,2,3$). The traceless condition rules out $\sigma^{00}$, leaving 15 matrices as the basis of $su(4)$. On the other hand, the $so(6)$ algebra is spanned by 15 antisymmetric real $6\times6$ matrices. Let $A^{ij}$ be a $6\times6$ matrix which has a $+1$ at row-$i$ column-$j$ and a $-1$ at row-$j$ column-$i$ and zero elsewhere (note that $A^{ij}=-A^{ji}$). The 15 subsets $\{i,j\}$ in the set $\{1,\cdots,6\}$ correspond to the 15 antisymmetric $A^{ij}$ matrices, spanning the $so(6)$ algebra.

The following table lists a possible one-to-one mapping between $\sigma^{\mu\nu}$ and $A^{ij}$ up to an overall factor. (More precisely, $\sigma^{\mu\nu}\leftrightarrow A^{ij}$ means $\sigma^{\mu\nu}/(2\mathrm{i})=A^{ij}$ here.) $$ \begin{array}{cccc} & \sigma^{01}\leftrightarrow A^{24} & \sigma^{02}\leftrightarrow A^{46} & \sigma^{03}\leftrightarrow A^{26} \\ \sigma^{10}\leftrightarrow A^{15} & \sigma^{11}\leftrightarrow A^{36} & \sigma^{12}\leftrightarrow A^{32} & \sigma^{13}\leftrightarrow A^{43} \\ \sigma^{20}\leftrightarrow A^{13} & \sigma^{21}\leftrightarrow A^{65} & \sigma^{22}\leftrightarrow A^{25} & \sigma^{23}\leftrightarrow A^{54} \\ \sigma^{30}\leftrightarrow A^{35} & \sigma^{31}\leftrightarrow A^{61} & \sigma^{32}\leftrightarrow A^{21} & \sigma^{33}\leftrightarrow A^{14} \\ \end{array} $$ It can be verified that the Lie bracket is preserved under this mapping.

The above mapping is found by making use of the spinor representation of the $SO(6)$ group. It is known that the $SO(6)$ group has an 8-dimensional spinor representation, which can be further split into two conjugate 4-dimensional spinor representations, as $8=4\oplus\bar{4}$. The action of $SO(6)$ in one of the 4-dimensional spinor representation is identical to the action of $SU(4)$ in its fundamental representation upto a $\pm1$ sign, thus making a connection between the two Lie groups, and hence between their Lie algebras.

The spinor representation of $SO(6)$ can be derived from the representation of the real Clifford algebra $Cl_{0,6}\cong M_8(\mathbb{R})$ (which is 8-dimensional). For example, we may choose the following representation for the 6 generators $\gamma^i$ ($i=1,\cdots,6$) of $Cl_{0,6}$: $$\gamma^1=\sigma^{112},\gamma^2=\sigma^{120},\gamma^3=\sigma^{132},\gamma^4=\sigma^{321},\gamma^5=\sigma^{302},\gamma^6=\sigma^{323},$$ where $\sigma^{\mu\nu\lambda}\equiv\sigma^\mu\otimes\sigma^\nu\otimes\sigma^\lambda$ denotes an $8\times8$ Hermitian matrix. Then the generators of $SO(6)$ can be constructed from $$ A^{ij}=\frac{1}{2}[\gamma^i,\gamma^j]. $$ This gives the 8-dimensional spinor representation of $SO(6)$, such as $A^{24}=-i\sigma^{201}$, $A^{46}=-i\sigma^{002}$ etc. The representation splits in the $Cl_{0,6}$ pseudo-scalar $\gamma^7\equiv\prod_{i=1}^6\gamma^i=\sigma^{200}$ diagonal basis into the left-handed ($\gamma^7=+1$) and the right-handed ($\gamma^7=+1$) spinors, both are 4-dimensional spinors. Suppose we take the left-handed spinor representation, it would correspond to the projection by removing the first index of Pauli matrices: $\sigma^{\mu\nu\lambda}\to\sigma^{\nu\lambda}$ (which will not affect the Lie bracket as $\mu$ only takes two values $\mu=0,2$ and hence will not contribute to the Lie bracket). By this we established the correspondence $A^{24}\leftrightarrow\sigma^{01}$, $A^{46}\leftrightarrow\sigma^{02}$ etc, as concluded in the above table. Note that this mapping is not unique, that a different choice of the Clifford algebra $Cl_{0,6}$ representation will lead to a different mapping between $su(4)$ and $so(6)$.

Answer 3

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 4

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).

Answer 5

$\mathrm{SO}(6)$ and $\mathrm{SU}(4)$ both have 15 generators, so there is a natural isomorphism between them. $\mathrm{SU}(4)$ is in fact the universal covering group of $\mathrm{SO}(6)$.