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:

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

  • $\begingroup$ When you plug in $R=1+\Delta R+O(\Delta R^2)$ into condition 2, you get that $\Delta R$ is traceless. Can you explain how you get that 6-term identity? $\endgroup$ – Pavel Safronov Jan 23 '12 at 0:51
  • $\begingroup$ You say "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". Attention, Your premise is wrong: if $R$ is sufficiently close to $I$ then we have just $\det(R)\neq 0$. Therefore your conclusion is not correct. We need to be careful. $\endgroup$ – Giuseppe Jan 23 '12 at 9:02
  • $\begingroup$ Pointed out this misunderstanding, I would ask: Have you tried to work out the whole thing intrinsecally? Probably yes, so you surely know that $\det:(GL(n),\cdot)\to\(\mathbb{R}^\times,\cdot)$ is a Lie group morphism whose induced Lie algebra morphism $\det_\ast:\mathfrak{gl}(n)\to\mathbb{R}$ is exactly the trace. $\endgroup$ – Giuseppe Jan 23 '12 at 9:16
  • $\begingroup$ Infact $\textrm{trace}:\mathfrak{gl}(n)\to\mathbb{R}$ is the unique linear functional on $\mathfrak{gl}(n)$ such that $\textrm{trace}(A.B-B.A)=0,\ \forall A,B\in\mathfrak{gl}(n)$. $\endgroup$ – Giuseppe Jan 23 '12 at 10:39

The tensor $\iota^{abijk}$ is defined so that it is $ab$-antisymmetric; it is a difference of the 3 terms with $a$ in $\delta$ and 3 terms with $b$ in $\delta$. The antisymmetry means $$\iota^{abijk}=-\iota^{baijk}$$ In the same way, one may show that the tensor is fully antisymmetric in $ijk$; the permutation of any pair of indices among the last 3 changes the sign, too. The antisymmetry in 3 indices may be shown by an antisymmetry with respect to two permutations; or antisymmetry with respect to 1 permutation and invariance under the cyclical permutation of $ijk$ but I have checked it.

In $SO(3)$, every tensor that is antisymmetric in 3 indices has to be a multiple of $\epsilon_{ijk}$; it only contains one independent component. It follows that $$\iota^{abijk}=T^{ab} \epsilon^{ijk}$$ for some tensor $T^{ab}$. However, we know that $T^{ab}$ is antisymmetric, so it carries the same information as a vector $v^c$, $$ T^{ab} = \epsilon^{abc} v^c $$ but we also know that $v^c$ is $SO(3)$-invariant so it must vanish because $0$ is the only rotationally invariant vector. Alternatively, one could compute $v^c$ out of contractions of your tensor $\iota$ in various ways, more precisely by $$v^c = K \epsilon^{abc} \epsilon^{ijk} \iota^{abijk} $$ where $K$ is a real nonzero coefficient I could easily determine and one would get zero by "brute but straightforward" calculations. So we have $T^{ab}=0$ and therefore $\iota^{abijk}=0$.

There are various tricks used above, and others. Some of them generalize, some of them need to be enriched by other tricks in the case of other groups and other tensors etc. It's hard to summarize "all tricks" that are enough to prove similar identities because two identities are, in general, qualitatively different. If you specified a class of identities to be discussed, it could be discussed, but you haven't.

But yes, the generalization to $SO(N)$ would work as well. One could still factorize $\iota$; now, the 3-index factor would have to be proportional to the structure constants of the group. The right value of the determinant of an $SO(N)$ matrix isn't an extra continuous condition so it must tautologically hold and indeed, $T^{ab}$ would vanish for any $SO(N)$. That's really because $SO(N)$ has no nonzero invariant antisymmetric tensors with two indices.

Taking the last method, one could talk about invariant tensors of any kind. In $SO(3)$, by definition, you said it's the group where $\delta$ and $\epsilon$ are the only two invariant "independent" tensors. So all other invariant tensors must really be their functions. In this case, one may argue that the 5-index object $\iota$ is a polynomial in $\delta$ and $\epsilon$ and because $5=2+3$ is the only way to divide the indices, it must be of the form $\epsilon \delta$. However, one may also use the $ab$, $ijk$ antisymmetry of $\iota$ to show that the coefficients of all terms like $\epsilon\delta$ cancel.


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