# Decomposition of product of two antisymmetric Lorentz tensors

Suppose I have a tensor $$A_{\mu\nu}$$ in the $$(3,1)\oplus (1,3)$$ representation of the Lorentz group where $$(a,b) =(2s_a+1,2s_b+1)$$. I was wondering on how to decompose explictly in terms of tensors the prouct $$A_{\mu\nu}\otimes A_{\rho\sigma}$$ (where it is the same antisymmetric $$A$$ in the two factors of the product) in a sum of irreducibile representations. If I am not wrong I have that: $$[(3,1)\oplus (1,3)]\otimes [(3,1)\oplus (1,3)] = (5,1)\oplus (1,5)\oplus 2\, (3,3)\oplus [(3,1)\oplus (1,3)]\oplus 2(1,1)$$ However it is not at all clear to me how to translate this into an explicit representation in terms of tensors.

i.e. I would like to do the analogous of:

the product of two vectors $$V_\mu\otimes V_\nu$$ is: $$(2,2)\otimes (2,2) = (3,1)\oplus (1,3)\oplus (3,3)\oplus (1,1)$$ which can be easily written as: $$V_\mu\otimes V_\nu = A_{\mu\nu}+S_{\mu\nu}+\frac{1}{4}g_{\mu\nu}T$$ where $$A_{\mu\nu}$$ is antisymmetric, $$S_{\mu\nu}$$ is symmetric and traceless, while $$T$$ is a scalar times $$g_{\mu\nu}$$ which is Lorentz invariant.

In the case of vectors it is trivial to explictly build these tensors. However, in the case I am asking about, I don't find the procedure to build the analogous objects so obvious or straightforward.

• To get explicit projectors for tensor products of more general representations, arxiv.org/abs/1603.05551 is a good reference. But it would probably be overkill for the two index case. Nov 14, 2021 at 15:30
• You need a shortcut for Appendix VII of Wu-ki Tung's book? Nov 15, 2021 at 18:10
• @CosmasZachos Are you sure about Appendix VII? Nov 18, 2021 at 20:04
• @CosmasZachos Appendix VII is a summary of the unitary (inf.dim) irreps of the restricted Lorentz group, not so obviously related to Young tableaux. Nov 18, 2021 at 20:10
• The combinatorics shadow the tensor decomposition asked for. The OP appears to be asking for projectors. Nov 18, 2021 at 20:13

In general dimension, the decomposition is well understood, it follows from basic Young manipulations. In terms of Dynkin labels

$$(0,1,0,\dots,0)^2=(0,\dots,0)+(2,0,\dots,0)+(0,1,0,\dots,0)+(0,0,0,1,0,\dots,0)+(1,0,1,0,\dots,0)+(0,2,0,\dots,0)$$ of dimension $$\left(\frac12 d(d-1)\right)^2=1+\left(\frac12d(d+1)-1\right)+\frac12d(d-1)+\frac{1}{4!}d(d-1)(d-2)(d-3)+\frac{1}{8}d(d-3)(d-1) (d+2)+\frac{1}{12} d(d-3) (d+1) (d+2)$$ or, in words, a scalar, a rank2 traceless symmetric, a rank2 anti-symmetric, a rank4 anti-symmetric, a rank4 tensor with mixed symmetry (anti with respect to three indices, the fourth one is not), and a rank4 tensor with the symmetries of the Riemann tensor (and traceless).

In $$d=4$$ things are a little bit different because some of these break into self-dual and anti-self-dual parts; this happens to the rank2 anti-symmetric and the Riemann tensor. Also, some rank4 tensors become isomorphic to rank2 tensors, in particular the rank4 anti-symmetric becomes a scalar and the rank4 tensor with mixed symmetry becomes a rank2 traceless symmetric.

So, to summarize, \begin{align} 2\, (3,3)\oplus 2(1,1)&:\text{ two rank2 symmetric tensors}\\ (3,1)\oplus (1,3)&:\text{ dual and anti-self dual parts of a rank2 anti-symmetric tensor}\\ (5,1)\oplus (1,5)&:\text{ dual and anti-self dual parts of a rank4 traceless tensor with the symmetries of the Riemann tensor} \end{align} i.e., schematically $$A^2=2S+A+R$$ where both $$A$$ and $$R$$ can be broken into chiral parts if so desired, and $$S$$ into trace and traceless parts.

• As a fun fact, the fact that Riemann appears in the decomposition of two copies of an anti-symmetric tensor is what's behind the double copy. Nov 15, 2021 at 20:30
• Thank you for the answer. However it is not clear to me how do you build these objects explicitly
– Fra
Nov 16, 2021 at 9:24
• It is the same idea as in the OP. First you write $A_{\mu\nu}B_{\rho\sigma}$. Then you break into symmetric and anti-symmetric parts $\sim A_{\mu\nu}B_{\rho\sigma}+A_{\mu\sigma}B_{\rho\nu}+A_{\mu\rho}B_{\nu\sigma}+\cdots$. Then subtract all the traces. Have you done this exercise for two vectors instead of two tensors? $u_\mu v_\nu=\frac12(u_\mu u_\nu+u_\nu u_\mu)+\frac12(u_\mu u_\nu-u_\nu u_\mu)$, which gives you the symmetric and anti-symmetric. The tensor case is analogous, except that there are more terms. Nov 16, 2021 at 21:56
• I tried, can't interpret some terms, like the four indices symmetrized pieces
– Fra
Nov 17, 2021 at 17:47
• Please with all these hints it seems like you are playing some sick game. If you have a clear answer or suggestion I am all ears
– Fra
Nov 19, 2021 at 10:19
1. The complexified proper Lorentz group $$SO(1,3;\mathbb{C})$$ has double cover $$SL(2,\mathbb{C})_L\times SL(2,\mathbb{C})_R$$, so a 4-vector representation $${\bf 4}~\cong~ {\bf 2}~\otimes~ \bar{\bf 2}~\equiv~ ({\bf 2}, \bar{\bf 2}),\tag{1}$$ i.e. $$V^{\mu}~\cong~ V^{\alpha\dot{\alpha}},\tag{2}$$ where $$\mu=0,1,2,3$$ is a Lorentz index, and where $$\alpha=1,2$$ is a left Weyl spinor index. The upshot is that we can trade 1 Lorentz index $$\mu$$ for 1 undotted spinor index $$\alpha$$ and 1 dotted spinor index $$\dot{\alpha}$$. (The intertwiners in eq. (2) are Pauli matrices. See e.g. this related Phys.SE post.)

2. Similarly, \begin{align}\bigwedge\!{}^2 {\bf 4}~\cong~& {\rm Sym}^2{\bf 2}~\otimes~ \bigwedge\!{}^2 \bar{\bf 2} ~\oplus~ \bigwedge\!{}^2{\bf 2}~\otimes~ {\rm Sym}^2\bar{\bf 2}\cr ~\cong~& ({\bf 3},\bar{\bf 1}) ~\oplus~ ({\bf 1}, \bar{\bf 3}), \end{align}\tag{3} i.e. an antisymmetric matrix contains 2 undotted and 2 dotted spinor indices: \begin{align} A^{\mu_1\mu_2}~\cong~& A^{\alpha_1\dot{\alpha}_1,\alpha_2\dot{\alpha}_2} \cr ~=~& A^{\alpha_1\alpha_2} \epsilon^{\dot{\alpha}_1\dot{\alpha}_2}+ \epsilon^{\alpha_1\alpha_2} A^{\dot{\alpha}_1\dot{\alpha}_2}\cr ~=~& A^I \epsilon^{\dot{\alpha}_1\dot{\alpha}_2}+ \epsilon^{\alpha_1\alpha_2} A^{\dot{I}}.\end{align}\tag{4} Here $$I=(\alpha_1\alpha_2)$$ is a symmetric double-index that can take 3 values: $$(11), \quad(12)=(21), \quad\text{and}\quad(22).\tag{5}$$

3. So \begin{align} \bigwedge\!{}^2 {\bf 4}~\otimes~\bigwedge\!{}^2 {\bf 4} ~\cong~& ({\bf 3}\otimes {\bf 3},\bar{\bf 1}) ~\oplus~ ({\bf 1}, \bar{\bf 3}\otimes \bar{\bf 3}) ~\oplus~ 2({\bf 3}, \bar{\bf 3}) . \end{align}\tag{6} Here the corresponding tensor$$^1$$ \begin{align} M^{\mu_1\mu_2,\nu_1\nu_2} ~\cong~&M^{\alpha_1\dot{\alpha}_1\alpha_2\dot{\alpha}_2,\beta_1\dot{\beta}_1\beta_2\dot{\beta}_2}\cr ~=~&M^{IJ}\epsilon^{\dot{\alpha}_1\dot{\alpha}_2}\epsilon^{\dot{\beta}_1\dot{\beta}_2} +M^{\dot{I}\dot{J}}\epsilon^{\alpha_1\alpha_2}\epsilon^{\beta_1\beta_2}\cr &M^{I\dot{J}}\epsilon^{\dot{\alpha}_1\dot{\alpha}_2}\epsilon^{\beta_1\beta_2} +M^{\dot{I}J}\epsilon^{\alpha_1\alpha_2}\epsilon^{\dot{\beta}_1\dot{\beta}_2} \end{align}\tag{7} on the LHS of eq. (6) contains 4 undotted and 4 dotted spinor indices, and it decomposes according to the RHS of eq. (6).

4. Worth mentioning is the further decomposition of $${\bf 3}~\otimes~ {\bf 3}~\cong~{\bf 5} ~\oplus~{\bf 3} ~\oplus~{\bf 1} \tag{8}$$ into irreducible representations. Here the corresponding tensor $$M^{IJ}~=~M_{(S)}^{IJ}+\epsilon^{IJK}M^{(A)}_K+g^{IJ}M\tag{9}$$ on the LHS of eq. (8) contains 4 undotted spinor indices, and it decomposes in a traceless symmetric part, an antisymmetric part, and a trace part, cf. the RHS of eq. (8).

--

$$^1$$ In eq. (7), OP is actually only interested in tensors of the special form $$M^{\mu_1\mu_2,\nu_1\nu_2}~=~A^{\mu_1\mu_2}A^{\nu_1\nu_2}.\tag{10}$$

The map between vector and bispinor indices is $$$$V_{\alpha \dot{\alpha}} = (\sigma^\mu)_{\alpha\dot{\alpha}} V_\mu, \quad V_\mu = -\frac{1}{2} (\bar{\sigma}_\mu)^{\dot{\alpha}\alpha} V_{\alpha \dot{\alpha}}$$$$ but we can derive a useful decomposition by just keeping this formalism in the back of our minds and not going into the details. By the rules of $$\mathfrak{su}(2)$$, your tensor product is built out of \begin{align} (3,1) \otimes (1,3) &= (1,3) \otimes (3,1) = (3,3) \\ (3,1) \otimes (3,1) &= (5,1) \oplus (3,1) \oplus (1,1) \\ (1,3) \otimes (1,3) &= (1,5) \oplus (1,3) \oplus (1,1) \end{align} where $$(3,1)$$ and $$(1,3)$$ are the self-dual and anti-self-dual components respectively. $$$$A^{\pm}_{\mu\nu} = \frac{1}{2} \left ( A_{\mu\nu} \pm \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} A_{\rho\sigma} \right )$$$$

Singlet

Now what is a singlet we can write just out of $$A_{\mu\nu}$$? The most obvious one is $$A_{\kappa\lambda} A_{\kappa\lambda}$$. It turns out that this singlet, times the invariant tensor $$\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}$$ is all we need because the analogous operation performed on the self-dual and anti-self-dual halves yields \begin{align} (\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}) A^\pm_{\kappa\lambda} A^\pm_{\kappa\lambda} = (\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}) \left [ A_{\kappa\lambda} A_{\kappa\lambda} \pm \frac{1}{2} \epsilon_{\kappa\lambda\omega\tau} A_{\kappa\lambda} A_{\omega\tau} \right ]. \end{align} These two terms form the $$2(1, 1)$$ part of your tensor product. But by the symmetries of the problem, it makes sense that they would appear with equal coefficients so that the epsilons cancel out.

Symmetric tensor

Another thing we can do to $$A_{\mu\nu}A_{\rho\sigma}$$ is contract two indices and tracelessly symmetrize the other two. Putting the original two indices back with a delta, this leads to $$$$\left [ \delta_{\nu\rho} \left ( A_{\mu\lambda}A_{\sigma\lambda} - \frac{1}{4}\delta_{\mu\sigma} A_{\kappa\lambda}A_{\kappa\lambda} \right ) - (\mu \leftrightarrow \nu) \right ] - (\rho \leftrightarrow \sigma).$$$$ This is again all we need because it turns out the above is equal to $$$$\left [ \delta_{\nu\rho} \left ( A^+_{\{ \mu\lambda}A^-_{\sigma \} \lambda} - \frac{1}{4}\delta_{\mu\sigma} A^+_{\kappa\lambda}A^-_{\kappa\lambda} \right ) - (\mu \leftrightarrow \nu) \right ] - (\rho \leftrightarrow \sigma).$$$$ If we didn't have identical tensors, this would produce the two terms in $$2(3, 3)$$. But since we cannot do $$A^+B^-$$ and $$A^-B^+$$ anymore, they collapse to just one.

Anti-symmetric tensor

This one is zero for identical tensors because $$(1, 3)$$ is in the alternating product of $$(1, 3) \otimes (1, 3)$$. In terms of indices, it means we cannot make a symmetric tensor out of $$A^+_{\alpha\beta} A^+_{\gamma\delta}$$ because $$\epsilon^{\beta\gamma}$$ is our only choice for how to contract them.

The rest

We are now at the point where we can write \begin{align} A_{\mu\nu}A_{\rho\sigma} = \left [ A_{\mu\nu}A_{\rho\sigma} - x(\delta_{\nu\rho}A_{\mu\lambda}A_{\sigma\lambda} - \delta_{\mu\rho}A_{\nu\lambda}A_{\sigma\lambda} - \delta_{\nu\sigma}A_{\mu\lambda}A_{\rho\lambda} + \delta_{\mu\sigma}A_{\nu\lambda}A_{\rho\lambda}) - (x/2 + y)(\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}) A_{\kappa\lambda} A_{\kappa\lambda} \right ] + x \left [ \delta_{\nu\rho}A_{\mu\lambda}A_{\sigma\lambda} - \delta_{\mu\rho}A_{\nu\lambda}A_{\sigma\lambda} - \delta_{\nu\sigma}A_{\mu\lambda}A_{\rho\lambda} + \delta_{\mu\sigma}A_{\nu\lambda}A_{\rho\lambda} + \frac{1}{2} (\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}) A_{\kappa\lambda} A_{\kappa\lambda} \right ] + y(\delta_{\mu\rho} \delta_{\nu\sigma} - \delta_{\mu\sigma} \delta_{\nu\rho}) A_{\kappa\lambda} A_{\kappa\lambda}. \end{align} I think it is not too bad now to solve for $$x$$ and $$y$$ by imposing that the first piece in square brackets have the index symmetries of the Riemann tensor. As an update, it looks like tracelessness alone is enough to impose this, leading to $$x = -\frac{1}{2}$$ and $$y = \frac{1}{12}$$. The non-trivial symmetries then come in purely as a sanity check.

• Thank you so much for the answer. I have a quesiton however, as you define it the Riemann tensor doen't have the correct symmetry properties it seems
– Fra
Nov 22, 2021 at 16:04
• The properties should match those of the Riemann tensor minus the traces. By this I mean the 2x2 Young diagram. Perhaps one will also need to separate $(1, 5)$ from $(5, 1)$ before this is manifest. Nov 22, 2021 at 16:10
• It doen't satisfy the bianchi identity
– Fra
Nov 22, 2021 at 16:10
• Are you sure 2x2 Young diagrams need to satisfy the Bianchi identity? It seems like a fact about geometry instead of representations. Nov 22, 2021 at 16:39
• hmm you are right I think. I have one last doubt: why $(3,1)\oplus (1,3)$ is zero?
– Fra
Nov 22, 2021 at 16:59