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

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

• you need to find a basis of $\mathfrak s\mathfrak u(4)$ and define the (iso)morphism on it, sending each basis element to a basis element of $\mathfrak s\mathfrak 0(6)$. Then extend the morphism linearly to the whole $\mathfrak s\mathfrak u(4)$ and check compatibility w.r.t. the commutators. This is enough. In summary, the biggest problems are to find the basis and to make the basis correspond in the "right" order. Nov 17, 2013 at 20:11
• maybe this can help math.stackexchange.com/questions/423419/… Nov 17, 2013 at 20:17
• Thanks - I was thinking of proceeding like that, only I was not too keen on it due to the large number of basis vectors involved. I'll give it a go though!
– user109506
Nov 18, 2013 at 1:21

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.

• Just two typos : 1.) 'So, we have constructed a homomorphism $SU(4) \to SL(4;\mathbb{C}) \to SO(6;\mathbb{C})$ and 2.) 'induces a basis .. on $\mathbb{C}^6$ Nov 18, 2013 at 15:07
• Just a question : I do not know what is an isomorphism between two different pairings and how the map factors from the action of $SO(6;C)$, is there some universal property in action here ? Nov 18, 2013 at 15:09
• By the way, I had asked a similar question here. having read your answer, I think now I can try the other isomorphism proof that I was looking for. But could you please give me some reference for exceptional isomorphisms between low rank Lie algebras and also for understanding the uniqueness of pairings ? Thanks a lot ! Nov 18, 2013 at 15:15
• @user90041 fixed the typos, thanks! As for your second comment, the Gram–Schmidt process finds, for any finite-dimensional complex vector space $V$ with nondegenerate inner product, an orthonormal basis, thereby determining an isomorphism $\mathrm{SO}(V) \cong \mathrm{SO}(\dim V)$. The isomorphism is not unique, but different isomorphisms constructed in this way are conjugate. ... Nov 18, 2013 at 20:17
• ... In general, the Gram–Schmidt process requires taking a square root. So over $\mathbb R$, it classifies inner product by their signature (dimension of maximal positive-definite subspace, dimension of maximal negative-definite subspace). This gives the different real forms $\mathrm{SO}(p,q)$. Nov 18, 2013 at 20:17

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 $$so(6)$$.

• Hi, do you have some nice reference? Feb 16, 2016 at 12:12
• @Marion Because I couldn't find a reference to this explicit correspondence, I can only derive it myself and present the result here. Feb 17, 2016 at 17:32
• @EverettYou I think there may be a typo, i.e., you need $\sigma^{\mu\nu}/(2i)\leftrightarrow A^{ij}$ for the Lie brackets to be equal. Jan 28 at 22:29
• @AndrewYuan Thanks, I updated the answer with a comment of this factor. Jan 29 at 10:48

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).

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).

$\mathrm{SO}(6)$ 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 $\mathrm{SO}(6)$.

• Having the same dimension doesn't mean there's an isomorphism between Lie groups / algebras. Jul 21, 2016 at 23:52