We know that for special orthogonal groups $SO(N)$ there exists invariant tensors (invariant under the group action). These are $\delta_{ij}$ and the totally anti-symmetric $\epsilon_{m_1,m_2,...m_N}$ tensor.

Similarly for $SU(N)$ the invariant tensors are $\delta_k^i$, $\epsilon_{m_1,m_2,...m_N}$ and $\epsilon^{m_1,m_2,...m_N}$ ($\delta_k^i$ is an invariant tensor of $U(N)$ too but not so for the $\epsilon$'s).

These objects are very useful in constructing singlets out of objects transforming under representations of $SO(N)$ or $SU(N)$.

Question 1: Are there such tensors for the Symplectic group and Exceptional groups? I am particularly interested in the groups $Sp(2N)$ and $E_7$. Is there a systematic method of obtaining the same?

Question 2. This question is just for the sake of curiosity. Can we also find invariant tensors for supergroups like $OSp(4|\mathcal{N})$ or $SU(2,2|\mathcal{N}/2)$ that appears in numerous $\mathcal{N}$-extended supersymmetric field theories?


1 Answer 1


Here is a partial answer: Define $Sp(2N,\mathbb{R})$ as the group of matrices $S$ such that $S \cdot\Omega\cdot S^T=\Omega$ where $\Omega_{ij}$ is a non-degenerate anti-symmetric matrix. Then $\Omega_{ij}$ is an invariant tensor similar to the Kronecker delta for orthogonal transformations. I don't think there are any more (not 100% sure).

For $E_7$: $E_7$ may be defined to be the group that preserves a antisymmetric second-rank tensor $g_{\mu\nu}$ and a totally symmetric fourth-rank tensor $f_{\mu\nu\rho\sigma}$ with $\mu,\nu,\rho,\sigma=1,2,\ldots, 56$ (i.e., the 56 dimensional representation is the defining representation). In more detail: $$ g_{\mu_1\mu_2} = (S_{\mu_1})^{\nu_1}\ (S_{\mu_2})^{\nu_2}\ g_{\nu_1\nu_2} \ , $$ with a similar one for the other tensor.

The $E_7$ description is described in P. Cvitanovic's Group Theory book. You can take a look there for the supersymmetric groups for which he has an interesting approach.

  • $\begingroup$ Yes of course! This comes by definition. Are are there any other? $\endgroup$
    – Orbifold
    Sep 7, 2014 at 8:12
  • $\begingroup$ @Orbifold I have updated my answer to include the $E_7$ case. I didn't know that answer on the top of my head and had to look it up. $\endgroup$
    – suresh
    Sep 7, 2014 at 10:08

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