Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $SU(4)$ ?

share|improve this question
1  
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

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 $\mathfrak{so}(6)$.

share|improve this answer

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 Lie group $SO(6;\mathbb{C})$.

This follows partly because:

  1. There is a bijective isometry from the $6$-dimensional complex vector space $(\mathbb{C}^6,||\cdot||^2)$ $$ \vec{v}~=~(x_1,x_2,x_3,y_1,y_2,y_3), \qquad x_a,y_a\in \mathbb{C},\qquad a\in \{1,2,3\}, $$ endowed with the standard bilinear (as opposed to sequilinear) form $$|| \vec{v}||^2~:=~ \sum_{a=1}^3 x_a^2 + \sum_{a=1}^3 y_a^2, $$ and the space $(so(4,\mathbb{C}),{\rm Pf}(\cdot))$ of antisymmetric complex $4\times 4$ matrices, $$\mathbb{C}^6 ~\cong ~ so(4,\mathbb{C}) ~:=~\{A\in {\rm Mat}_{4\times 4}(\mathbb{C}) \mid A^{t}=-A\}, $$ endowed with the Pfaffian $${\rm Pf}(A)~=~\frac{1}{8}\sum_{\lambda,\sigma=1}^4\epsilon_{\mu\nu\lambda\sigma}A_{\mu\nu} A_{\lambda\sigma} ~=~A_{12}A_{34} + A_{31}A_{24} + A_{23}A_{14}~=~|| \vec{v}||^2, \qquad\epsilon_{1234}~=~1,$$ if we identify $$ A_{a4}~=~ x_a+iy_a, \qquad a,b,c\in \{1,2,3\}, $$ $$ A_{ab}~=~\sum_{c=1}^3\epsilon_{abc} (x_c-iy_c)\qquad \Leftrightarrow\qquad x_a-iy_a ~=~\frac{1}{2}\sum_{b,c=1}^3\epsilon_{abc}A_{bc} ,\qquad\epsilon_{123}~=~1,$$ $$A~=~\left[\begin{array}{cccc} 0& x_3-iy_3 &-x_2+iy_2 & x_1+iy_1 \cr -x_3+iy_3 &0&x_1-iy_1 & x_2+iy_2 \cr x_2-iy_2 &-x_1+iy_1 & 0& x_3+iy_3 \cr -x_1-iy_1&-x_2-iy_2&-x_3-iy_3&0\end{array} \right]. $$

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

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

This follows partly because:

  1. The complex $6$-dimensional vector space $${\rm Re}(\mathbb{C}^6)\oplus_{\mathbb{R}} {\rm Im}(\mathbb{C}^6)~=~ \mathbb{C}^6~\cong~so(4,\mathbb{C})~=~so^{-}(4,\mathbb{C}) \oplus_{\mathbb{R}} so^{+}(4,\mathbb{C})$$ splits in two real $6$-dimensional subspaces $$\left.\begin{array}{c} {\rm Im}(\mathbb{C}^6) \cr {\rm Re}(\mathbb{C}^6) \end{array} \right\}~\cong~so^{\pm}(4,\mathbb{C})~:=~ \{ A\in so(4,\mathbb{C}) \mid *A = \pm A^{\dagger} \} $$ via a selfdual/antiselfdual-type condition. Here the Hodge-dual is defined as $$(*A)_{\mu\nu}~:=~ \frac{1}{2}\sum_{\lambda,\sigma=1}^4\epsilon_{\mu\nu\lambda\sigma} A_{\lambda\sigma}, \qquad \mu,\nu,\lambda,\sigma\in \{1,2,3,4\} , \qquad {*}^2~=~ {\bf 1},$$ $$ (*A)_{ab}~=~\sum_{c=1}^3\epsilon_{abc} A_{c4} , \qquad (*A)_{a4}~=~ \frac{1}{2}\sum_{b,c=1}^3\epsilon_{abc} A_{bc}, \qquad a,b,c\in \{1,2,3\}.$$

  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 $\rho: SU(4)\times so(4,\mathbb{C}) \to so(4,\mathbb{C})$, $$ *(\rho(g)A) ~=~\pm (\rho(g)A)^{\dagger}, \qquad g\in SU(4), \qquad A\in so^{\pm}(4,\mathbb{C}).$$

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

share|improve this answer
    
Correction to the answer(v2): sequilinear should be sesquilinear. –  Qmechanic Dec 3 '12 at 23:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.