Suppose we would like to anti-symmetrize a tensor $$T^{\mu_1, \mu_2,\ldots, \mu_n} = G^{[\mu_1, \mu_2,\ldots, \mu_r]} H^{[\mu_{r+1},\ldots, \mu_n]},$$ where $G$ and $H$ are anti-symmetric. One could do this iteratively by applying the anti-symmetrizer $$\Lambda_{1,2,\cdots, n} = \frac{1}{n}\left(1-\sum_{i=1}^{n-1} P_{i,n}\right)\Lambda_{1,2,\cdots, n-1}, \quad \Lambda_{1,2} = \frac{1 - P_{1,2}}{2}$$ where $P_{i,n}$ swaps the indices $i$ and $n$, for instance $$P_{1,2} T^{\mu_{1} \mu_2} = T^{\mu_2 \mu_1}$$ however this method makes no use of the fact that $G$ and $H$ are already anti-symmetrized. My question is whether there exists a simplified anti-symmtrizer which anti-symmetrizes $T$ with a minimal amount of operations.

One immediate simplification is that we know that $\Lambda_{1,\ldots,r}$ will have no effect on the tensor $G$ and thus no effect on $T$ either. And something similar for $H$.

One idea is that this problem is somehow related to finding all partitions like $\{1,\ldots, r\}$ and $\{r+1,\ldots, n\}$ via swapping elements between them, but I cannot figure out what the correct algorithm should be. I do however know that there are a total of $\binom{n}{r}$ such partitions meaning that we should have a sum of that many products of operators in the end. In the end, this problem is related to a few problems in both relativity and fermionic statistics, but I cannot seem to find a solution anywhere.

  • $\begingroup$ To antisymmetrize $T$ you can use the generalized Kronecker Delta en.wikipedia.org/wiki/… $\endgroup$ – Ryan Parikh Nov 22 '20 at 16:59
  • $\begingroup$ Your suggestion amount to using the anti-symmetrizer, which is already mentioned in the question. The question is how to make use of the parts which are already anti-symmetrized to reduce the expression in the end. $\endgroup$ – Jan Cillié Louw Nov 22 '20 at 18:15

One can think of this problem as all different swaps with the components of the two tensors. With this in mind, we start by defining the index ordered tensors $$G^{\{\mu_1,\ldots, \mu_r\}} = G^{[\mu_{j_1},\ldots, \mu_{j_r}]},$$ where $(j_1,\ldots, j_r)$ is a permutation of $(1,\ldots,r)$ such that $$\mu_{j_1}< \mu_{j_2} <\ldots < \mu_{j_r}.$$ This definition is important since we then not longer have to worry about the ordering in a particular anti-symmetrized tensor, for instance $$G^{\{\mu_1, \mu_2, \ldots\}} = G^{\{\mu_2, \mu_1, \ldots\}}.$$ Then the anti-symmetrizing process is done with $$T^{[\mu_1\ldots \mu_n]} = \sum_{i_r = r}^{n} \sum_{i_{r-1}}^{i_r-1} \cdots \sum_{i_1 = 1}^{i_2 - 1} \text{sgn}\binom{1\cdots n}{i_1 \cdots i_n} P_{r,i_r} P_{r-1,i_{r-1}}\cdots P_{2,i_2} P_{1,i_1} G^{\{\mu_1,\ldots\mu_r\}} T^{\{\mu_{r+1},\ldots\mu_n\}}$$ where $i_{r+1},\ldots,i_n$ is the ordered list of all indices occurring in $T$. SO for instance if we had the case $$G^{\{\mu_3,\mu_4, \mu_6, \mu_7\}} T^{\{\mu_2,\mu_5,\mu_1\}},$$ then $(i_1,i_2,i_3,i_4,i_5,i_6, i_7) = (3,4,6,7, 1,2,5)$, since we first order the sublists. It turns out that under this swapping process the first $(i_1,\ldots, i_r)$ will in fact remain ordered. Ideally I would still like to simplify this somewhat into a recursive form such as that of the anti-symmetrizer.


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