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This is more of a particle physics question than maths. Since $SO(6)$ and $SU(4)$ are isomorphic, how are the fields (say for example scalar fields of ${\mathcal{N}}=4$ Super Yang Mills in $4d$) transforming under 6 dimensional vector representation of $SO(6)$ related to the fields transforming under antisymmetric 6 of $SU(4)$ ?

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The field-theory tag is inappropriate here; removed. These are a different notion of fields. –  Marc van Leeuwen Sep 10 '12 at 11:16

2 Answers 2

I) An existence proof that $SU(4)\cong SPIN(6)$ is e.g. given in the lectures PG course on Spin Geometry taught by José Figueroa-O'Farrill, see Lemma 8.1 in the lecture named Parallel and Killing spinors. In this answer we will concentrate on the explicit realization of such Lie group isomorphism, which OP may find useful to his question about the corresponding Lie algebra representation theory.


$SL(4;\mathbb{C})$ is (the double cover of) the special orthogonal group $SO(6;\mathbb{C})$.

This follows partly because:

  1. There is a bijective isometry from the $6$-dimensional complex vector space $(\mathbb{C}^6,||\cdot||^2)$ $$ \vec{v}~=~(x^1,x^2,x^3,y^1,y^2,y^3), \qquad x^a,y^a\in \mathbb{C},\qquad a\in \{1,2,3\}, \tag{1} $$ endowed with the standard bilinear (as opposed to sesquilinear) form $$||\vec{v}||^2~:=~ \sum_{a,b=1}^3 x^a \eta_{ab} x^b + \sum_{a=1}^3 y^a \eta_{ab} y^b, \qquad\eta_{ab}~=~{\rm diag}(+1,+1,+1), \tag{2} $$ and the space $(so(4,\mathbb{C}),{\rm Pf}(\cdot))$ of antisymmetric complex $4\times 4$ matrices, $$\mathbb{C}^6 ~\cong ~ so(4;\mathbb{C}) ~:=~\{A\in {\rm Mat}_{4\times 4}(\mathbb{C}) \mid A^{t}=-A\}, \tag{3}$$ endowed with the Pfaffian $${\rm Pf}(A)~=~\frac{1}{8}\sum_{\mu,\nu,\lambda,\sigma=1}^4\epsilon_{\mu\nu\lambda\sigma}A^{\mu\nu} A^{\lambda\sigma} ~=~A^{12}A^{34} + A^{31}A^{24} + A^{23}A^{14}~=~||\vec{v}||^2, $$ $$ \epsilon_{1234}~=~1,\tag{4}$$ if we identify

$$ A^{ab}~=~\sum_{c=1}^3\epsilon^{abc} (x_c-iy_c)\qquad \Leftrightarrow\qquad x_a-iy_a ~=~\frac{1}{2}\sum_{b,c=1}^3\epsilon_{abc}A^{bc} ,$$ $$ \epsilon^{123}~=~1~=~\epsilon_{123},\qquad A^{a4}~=~ x^a+iy^a, \qquad a,b,c\in \{1,2,3\}, $$ $$A~=~\left[\begin{array}{cccc} 0& x_3-iy_3 &-x_2+iy_2 & x^1+iy^1 \cr -x_3+iy_3 &0&x_1-iy_1 & x^2+iy^2 \cr x_2-iy_2 &-x_1+iy_1 & 0& x^3+iy^3 \cr -x^1-iy^1&-x^2-iy^2&-x^3-iy^3&0\end{array} \right].\tag{5} $$ We stress again that the bilinear form (2) has indefinite signature because $x^a,y^b \in \mathbb{C}$ are in general complex numbers.

  1. There is a group action $\rho: SL(4;\mathbb{C})\times so(4;\mathbb{C}) \to so(4;\mathbb{C})$ given by $$g\quad \mapsto\quad\rho(g)A~:= ~gA g^{t}, \qquad g\in SL(4;\mathbb{C})\subseteq {\rm Mat}_{4\times 4}(\mathbb{C}),$$ $$A\in so(4;\mathbb{C})\subseteq {\rm Mat}_{4\times 4}(\mathbb{C}),\tag{6} $$ which is length preserving $$ {\rm Pf}(\rho(g)A)~=~{\rm Pf}(gAg^t)~=~\det(g){\rm Pf}(A)~=~{\rm Pf}(A),\tag{7}$$ i.e. $g$ is an orthogonal transformation. In other words, there is a Lie group homomorphism
    $$\rho: SL(4,\mathbb{C}) \quad\to\quad O(so(4;\mathbb{C}) ,\mathbb{C})~\cong~ O(6;\mathbb{C}) , \qquad \rho(\pm {\bf 1}_{4 \times 4})~=~{\bf 1}_{so(4;\mathbb{C})}.\tag{8}$$


$SL(4;\mathbb{R})$ is (the double cover of) the restricted split orthogonal group $SO^+(3,3;\mathbb{R})$.

This follows partly from Section II by restricting the coordinates $x^a\in\mathbb{R}$ to real numbers; the coordinates $y^a\in i\mathbb{R}$ to imaginary numbers; and the matrices $A$ in eq. (3) to real matrices $$\mathbb{R}^6 ~\cong ~ so(4,\mathbb{R}) ~:=~\{A\in {\rm Mat}_{4\times 4}(\mathbb{R}) \mid A^{t}=-A\}. \tag{9}$$ In other words, there is a Lie group homomorphism $$\rho: SL(4;\mathbb{R}) \quad\to\quad O(so(4;\mathbb{R}) ;\mathbb{R})~\cong~ O(3,3;\mathbb{R}) , \qquad \rho(\pm {\bf 1}_{4 \times 4})~=~{\bf 1}_{so(4;\mathbb{R})}.\tag{10}$$

IV) Let us now address OP's question (v3).

$SU(4)$ is (the double cover of) the special orthogonal group $SO(6;\mathbb{R})$.

This follows partly because:

  1. The complex $6$-dimensional vector space $${\rm Re}(\mathbb{C}^6)\oplus_{\mathbb{R}} {\rm Im}(\mathbb{C}^6)~=~ \mathbb{C}^6~\cong~so(4;\mathbb{C})~=~so^{-}(4;\mathbb{C}) \oplus_{\mathbb{R}} so^{+}(4;\mathbb{C})\tag{11}$$ splits in two real $6$-dimensional subspaces $$\left.\begin{array}{c} {\rm Im}(\mathbb{C}^6) \cr {\rm Re}(\mathbb{C}^6) \end{array} \right\}~\cong~so^{\pm}(4;\mathbb{C})~:=~ \{ A\in so(4;\mathbb{C}) \mid *A = \pm A^{\dagger} \} \tag{12} $$ via a selfdual/antiselfdual-type condition. Here the Hodge-dual is defined as $$(*A)^{\mu\nu}~:=~ \frac{1}{2}\sum_{\lambda,\sigma=1}^4\epsilon^{\mu\nu\lambda\sigma} A_{\lambda\sigma}, \qquad A_{\lambda\sigma}~:=~\sum_{\mu,\nu=1}^4 \eta_{\lambda\mu}A^{\mu\nu}\eta_{\nu\sigma},$$ $$\mu,\nu,\lambda,\sigma\in \{1,2,3,4\} , \qquad \epsilon^{1234}~=~1,\tag{13} $$ where $$\eta_{\mu\nu}~=~{\rm diag}(+1,+1,+1,+1)\tag{14} $$ is the standard bilinear form on $\mathbb{C}^4$ to raise and lower indices. The Hodge-dual is an involution $$ {*}^2~=~ \det (\eta_{\mu\nu})~=~ 1.\tag{15}$$ We have $$ (*A)^{ab}~=~\sum_{c=1}^3\epsilon^{abc} A_{c4} , \qquad (*A)^{a4}~=~ \frac{1}{2}\sum_{b,c=1}^3\epsilon^{abc} A_{bc}, $$ $$ a,b,c\in \{1,2,3\}, \qquad \epsilon^{123}~=~1.\tag{16}$$

  2. Hence there is a bijective isometry between the positive definite, real $6$-dimensional, Euclidean vector space $({\rm Re}(\mathbb{C}^6),||\cdot||^2)$, and the space $(so^{-}(4,\mathbb{C}),{\rm Pf}(\cdot))$ of "antiselfdual" $4\times 4$ matrices.

  3. The Lie group $SU(4)~\subseteq~ SL(4;\mathbb{C})~\subseteq~{\rm Mat}_{4\times 4}(\mathbb{C})$ is a subgroup of $SL(4;\mathbb{C})$. Hence we can consider the restriction of the group action $\rho: SL(4;\mathbb{C})\times so(4,\mathbb{C}) \to so(4;\mathbb{C})$ to just $SU(4)$.

  4. The two subspaces $so^{\pm}(4;\mathbb{C})$ are invariant under the (restricted) group action $\rho: SU(4)\times so(4;\mathbb{C}) \to so(4;\mathbb{C})$, $$ *(\rho(g)A) ~=~\pm (\rho(g)A)^{\dagger}, \qquad g\in SU(4), \qquad A\in so^{\pm}(4;\mathbb{C}).\tag{17} $$

  5. In conclusion, there is a Lie group homomorphism $$\rho: SU(4) \quad\to\quad O(so^{-}(4;\mathbb{C}) ,\mathbb{R})~\cong~ O({\rm Re}(\mathbb{C}^6) ;\mathbb{R})~\cong~ O(6;\mathbb{R}) ,$$ $$\rho(\pm {\bf 1}_{4 \times 4})~=~{\bf 1}_{so^{-}(4;\mathbb{C})}.\tag{18}$$


$SU(2,2)$ is (the double cover of) the Lie group $SO^+(2,4;\mathbb{R})$.

This follows partly from Section IV by changing the signatures of the bilinear form $$ \eta_{ab}~=~{\rm diag}(-1,-1,+1), \qquad \eta_{\mu\nu}~=~{\rm diag}(-1,-1,+1,+1).\tag{19} $$

VI) Since this post is about (local) isomorphisms of 15-dimensional simple Lie groups, let us for completeness mention the following.

$SL(2;\mathbb{H})$ is (the double cover of) the restricted Lorentz group $SO^+(1,5;\mathbb{R})$.

This follows partly because:

  1. There is a star algebra monomorphism $\Phi:\mathbb{H}\to {\rm Mat}_{2\times 2}(\mathbb{C})$ given by $$\mathbb{H}~\ni~ q~=~q^0+iq^3 +jq^2+kq^1~=~\alpha +\beta j $$ $$~~\stackrel{\Phi}{\mapsto}~~ \begin{pmatrix} \alpha & \beta \cr -\bar{\beta} & \bar{\alpha} \end{pmatrix} ~=~ q^0{\bf 1}_{2\times 2}+i\sum_{a=1}^3 q^a \sigma_a~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}),$$ $$\alpha~\equiv~ q^0+iq^3~\in~\mathbb{C}, \qquad \beta~\equiv~ q^2+iq^1~\in~\mathbb{C},\tag{20}$$ Define an involution $(\cdot)^t: \mathbb{H}\to\mathbb{H}$ as
    $$q^t~:=~q|_{q^2\to-q^2}~:=~q^0+iq^3 -jq^2+kq^1, \qquad q~\in~\mathbb{H},\tag{21}$$ which we will call the "transposed" quaternion. Note that $$ \Phi(q)^{\dagger}~=~\Phi(\bar{q})\quad\text{and}\quad\Phi(q)^t~=~\Phi(q^t), \qquad q~\in~\mathbb{H}. \tag{22}$$

  2. The star algebra monomorphism (20) is extended in the natural way to a star algebra monomorphism
    $\Phi:{\rm Mat}_{2\times 2}(\mathbb{H})\to {\rm Mat}_{4\times 4}(\mathbb{C})$ (which we also call $\Phi$) such that $$ \Phi(m)^{\dagger}~=~\Phi(m^{\dagger})\quad\text{and}\quad\Phi(m)^t~=~\Phi(m^t), \qquad m~\in~{\rm Mat}_{2\times 2}(\mathbb{H}). \tag{23}$$

  3. There is no canonical notion of a quaternionic determinant, so we define the special linear group of quaternionic $2\times 2$ matrices as $$SL(2;\mathbb{H})~:=~\Phi^{-1}(SL(4;\mathbb{C})). \tag{24}$$

  4. The set of "antisymmetric" quaternionic $2\times 2$ matrices is $$so(2;\mathbb{H})~:=~\left\{ m\in{\rm Mat}_{2\times 2}(\mathbb{H}) \mid m^t=-m \right\}$$ $$~=~\left\{ \begin{pmatrix} rj & -q^t \cr q & sj \end{pmatrix} \in{\rm Mat}_{2\times 2}(\mathbb{H}) \mid r,s\in\mathbb{R}, ~q\in\mathbb{H} \right\}~=~\Phi^{-1}(so(4;\mathbb{C})), \tag{25} $$ cf. the above definition (21) of "transposed" quaternion.

  5. Comparing with the bilinear form (2) from Section II, we identify one real variable $$ x^3~=~\frac{r+s}{2}~\in~\mathbb{R} \tag{26} $$ and five imaginary variables $$ y^3~=~ \frac{i(r-s)}{2},\quad x^1~=~-iq^1,\quad y^1~=~-iq^2,\quad x^2~=~iq^3,\quad y^2~=~iq^0, \tag{27} $$ In other words, the bilinear form (2) restricts to $$ ||\vec{v}||^2~=~ rs-|q|^2,\qquad r,s~\in~\mathbb{R}, \qquad q~\in~\mathbb{H} ,\tag{28} $$ with signature (1,5).

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Physicists use dimension to denote representations of particular simple Lie-group/Lie-algebra, and mathematicians use weights, because finite-dimensional irreducible representations are all highest-weight modules.

$\mathfrak{su}(4)$ is a rank 3 Lie algebra of $A$ series, i.e. $A_3$. Denote its roots as $\alpha_1$, $\alpha_2$, and $\alpha_3$, and the corresponding weights $\omega_1$, $\omega_2$ and $\omega_3$. Lowest weight representations $\Lambda_{\omega_1}$ and $\Lambda_{\omega_3}$ are 4 dimensional, and $\Lambda_{\omega_2}$ is 6-dimensional. Weyl dimensions formula gives dimension of the representation $\Lambda_{n_1 \omega_1 + n_2 \omega_2 + n_3 \omega_3}$: $$ \dim \Lambda_{n_1 \omega_1 + n_2 \omega_2 + n_3 \omega_3} = \\ \frac{1}{12} \left(n_1+1\right) \left(n_2+1\right) \left(n_3+1\right)\left(n_1+n_2+2\right) \left(n_2+n_3+2\right) \left(n_1+n_2+n_3+3\right) $$ $\Lambda_{\omega_2}$ is the only 6-dimensional representation of $\mathfrak{su}(4)$. Highest weight modules are stable under the action of the Weyl group $\mathcal{W}_{\mathfrak{su}(4)} \simeq S_3$, thus they can be constructed as an orbit of the highest weight under the action of Weyl group: $$ \begin{eqnarray} \Lambda_{\omega_1} &=& \{ | \omega_1 \rangle, | \omega_1-\alpha_1 \rangle, | \omega_1 -\alpha_1 - \alpha_2 \rangle, | \omega_1 - \alpha_1-\alpha_2-\alpha_3 \rangle \} \\ \Lambda_{\omega_3} &=& \{ | \omega_3 \rangle, | \omega_3-\alpha_3 \rangle, | \omega_3 -\alpha_3 - \alpha_2 \rangle, | \omega_3 - \alpha_1-\alpha_2-\alpha_3 \rangle \} \\ \Lambda_{\omega_2} &=& \{ | \omega_2 \rangle, | \omega_2 - \alpha_2 \rangle, | \omega_2 - \alpha_2 - \alpha_3 \rangle, | \omega_2 - \alpha_1 - \alpha_2 \rangle , \\ &\phantom{=}& \phantom{-} | \omega_2 - \alpha_1 - \alpha_2 - \alpha_3 \rangle, | \omega_2 - \alpha_1 - 2 \alpha_2 - \alpha_3 \rangle \} \end{eqnarray} $$ $\mathcal{W}_{\mathfrak{su}(4)}$ is generated by three refections $\mathcal{W}_{\mathfrak{su}(4)} = \langle w_{\alpha_1}, w_{\alpha_2}, w_{\alpha_3}\rangle$.

Roots $\alpha_1$ and $\alpha_3$ generate sub-algebra $h = \mathfrak{su}(2)\oplus \mathfrak{su}(2) \subset \mathfrak{su}(4)$. Fundamental irreducible representations $\Lambda_{\omega_1}$, $\Lambda_{\omega_2}$, $\Lambda_{\omega_3}$ of $\mathfrak{su}(4)$ are reducible under $h$. The decomposition of $\Lambda_\omega$ into irreducible $h$-modules can be obtained by considering the orbit of the highest weight $\omega$ under action of $\mathcal{W}_h = \langle w_{\alpha_1}, w_{\alpha_3} \rangle$: $$ \begin{eqnarray} \Lambda_{\omega_1}^{\mathfrak{su}(4)} &=& \Lambda_{\omega_1 \oplus 0}^{\mathfrak{su}(2)} \oplus \Lambda_{0 \oplus \omega_3}^{\mathfrak{su}(2)} = \{ | \omega_1 \rangle, | \omega_1 - \alpha_1 \rangle \} \oplus \{ | \omega_1 -\alpha_1 - \alpha_2 \rangle, | \omega_1 - \alpha_1-\alpha_2-\alpha_3 \rangle \} \\ \Lambda_{\omega_3}^{\mathfrak{su}(4)} &=& \Lambda_{\omega_3 \oplus 0}^{\mathfrak{su}(2)} \oplus \Lambda_{0 \oplus \omega_1}^{\mathfrak{su}(2)} = \{ | \omega_3 \rangle, | \omega_3-\alpha_3 \rangle \} \oplus \{ | \omega_3 -\alpha_3 - \alpha_2 \rangle, | \omega_3 - \alpha_1-\alpha_2-\alpha_3 \rangle \} \\ \Lambda_{\omega_2}^{\mathfrak{su}(4)} &=& \Lambda_{0 \oplus 0}^{\mathfrak{su}(2)} \oplus \Lambda_{0 \oplus 0}^{\mathfrak{su}(2)} \oplus \Lambda_{3 \omega}^{\mathfrak{su}(2)} = \{ | \omega_2 \rangle \} \oplus \{ | \omega_2 - \alpha_1 - 2 \alpha_2 - \alpha_3 \rangle \} \oplus \\ &\phantom{= }& \{ | \omega_2 - \alpha_2 \rangle, | \omega_2 - \alpha_2 - \alpha_3 \rangle, | \omega_2 - \alpha_1 - \alpha_2 \rangle , | \omega_2 - \alpha_1 - \alpha_2 - \alpha_3 \rangle \} \end{eqnarray} $$ The 4-dimensional module decomposes with respect to $\langle w_{\alpha_1}\rangle$ as: $$ \{ | \omega_2 - \alpha_2 \rangle, | \omega_2 - \alpha_1 - \alpha_2 \rangle \} \oplus \{ | \omega_2 - \alpha_2 - \alpha_3 \rangle, | \omega_2 - \alpha_1 - \alpha_2 - \alpha_3 \rangle \} $$ and with respect to $\langle w_{\alpha_3} \rangle$ as: $$ \{ | \omega_2 - \alpha_2 \rangle, | \omega_2 - \alpha_2 - \alpha_3 \rangle \} \oplus \{ | \omega_2 - \alpha_1 - \alpha_2 \rangle, | \omega_2 - \alpha_1 - \alpha_2 - \alpha_3 \rangle \} $$

This analysis tells us how the six-dimensional representation $\Lambda_{\omega_2}$ is constructed. The scalar components correspond to anti-symmetric tensors of each $\mathfrak{su}(2)$ and the four-dimensional component corresponds to direct product of fundamental representations. Each of the element of the above vector spaces corresponds to a field of the super Yang-Mills theory.

Now to make the connection to $SO(6)$, you need to know the isomorphism with $SU(4)$, i.e. how Cartan-Weyl roots of $\mathfrak{so}(6)$ relate to those of $\mathfrak{su}(4)$. Then you need to find the isomorphism between $\Lambda_{\omega_2}$ constructed above, and the $6$-dimensional representation of $\mathfrak{so}(6)$.

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