The rotation group SO(3) can be viewed as the group that preserves our old friends the delta tensor $\delta^{ab}$ and $\epsilon^{abc}$ (the totally antisymmetric tensor). In equations, this says:

- $R^i_{ a}R^j_{ b} \delta^{ab} = \delta^{ij}$, a.k.a. $RR^T = I$.
and
- $R^i_{ a}R^j_{ b}R^k_c \epsilon^{abc} = \epsilon^{ijk}$, a.k.a. $Det(R) = 1$.

When we derive the fact that the Lie algebra by using R infinitesimally different from I, $R = I + \delta R$, the first condition gives us

1B $\delta R_{ab} = -\delta R_{ba}$.

This is all very familiar, the antisymmetric matrices. Since we are near the identity matrix, the determinant of $R$ is automatically one, and we don't need to plug $R = I + \delta R$ into condition 2.

But if we do, and combine with 1B, we derive an identity between $\delta$ and $\epsilon$ that looks like this:
$\iota^{abijk} = \delta^{ak} \epsilon^{ijb}+\delta^{aj} \epsilon^{ibk}+\delta^{ai} \epsilon^{bjk} -\delta^{bk} \epsilon^{ija}-\delta^{bj} \epsilon^{iak}-\delta^{bi} \epsilon^{ajk}=0$. (I take the big expression to be the definition of $\iota$.)

This identity seems to work numerically.
The issue is that when computing explicit commutators for spinors in the Lorentz algebra in QFT, this identity keeps popping up in various forms (typically making it hard to simplify an answer into a desired form until you recognize it.)

The question is whether there is a good way to understand this identity and others that might arise with the Lorentz group invariant tensors $\epsilon ^{\mu \nu \rho \sigma}$, or even with more general Lie groups. I've looked at the diagrammatic form of the identity and didn't see any enlightenment there (although it helped me prove to myself that it should be true numerically.)

Edit: For reasons of avoiding circular logic, in the following I'll assume that some group is defined by $\delta$ and $\epsilon$ being invariant tensors, but I don't know which tensors they are, i.e. they are not necessarily kronecker delta and the totally antisymmetric tensors. I define the lower index version $R_{ab}$ of a group element $R^a_b$ using the $\delta$ tensor. I think I can do this consistently, and that these are the only facts about $\epsilon$ and $\delta$ I have actually used.

Explicitly, plugging into 2 gives $\delta^i_a \delta^j_b \delta R^k_c \epsilon^{abc} + \delta^i_a \delta^k_c \delta R^j_b \epsilon^{abc} +\delta^k_c \delta^j_b \delta R^i_a \epsilon^{abc} = 0$.
Factoring out the $\delta R$ gives
1C:
$\delta R_{ab} (\delta^{ak} \epsilon^{ijb}+\delta^{aj} \epsilon^{ibk}+\delta^{ai} \epsilon^{bjk})=0$.
However by condition 1B, $\delta R_{ab}$ is antisymmetric. Only the anti-symmetric part of the thing multiplying $\delta R_{ab}$ contributes. Therefore we get
$\delta R_{ab} \iota^{abcde} =0.$ Logically, this could be a further condition on $\delta R$, but for SO(3) at least it seems that $\iota^{abcde}$ is just always $0$, which I argued for above based on the fact that condition 2 is equivalent to $Det(R) = 1$. I do not know what happens in general.

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