Dual field strength equals to itself

[ismaili]

$$\epsilon^{\mu_1\mu_2\cdots\mu_k\ \mu_{k+1}\cdots\mu_D} \epsilon_{\mu_1\cdots\mu_k\ \nu_{k+1}\cdots\nu_{D}} = (-1)k!\delta^{\mu_{k+1}}_{[\nu_{k+1}}\cdots\delta^{\mu_D}_{\nu_D]}\quad\cdots(*)$$
where we are working in the Minkowski space.
And the definition of square bracket on indices is like this,
$$A_{[\mu_1\mu_2\mu_3]} =\frac{1}{3!}(A_{\mu_1\mu_2\mu_3}-A_{\mu_1\mu_3\mu_2}+\cdots)$$

I'm wondering the validity of this identity, while I'm trying to check the possibility that the duality of a field strength is equal to itself.
In 4D, this seems to be impossible, because, if
$$\tilde{F}_{\mu\nu} = \frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma} = F_{\mu\nu}$$
, we multiply the both sides of the above equation by $$\epsilon^{\mu\nu\alpha\beta}/2$$ and sum over $$\mu,\nu$$, we will see
$$\quad\Rightarrow\quad F_{\alpha\beta} = -F_{\alpha\beta}$$
where eq(*) is used.
But if eq(*) is true, then in any dimension, the dual of a field strength is never equal to itself!
Since the main problem of the above calculation is that minus sign due to the signature of Minkowski space.

However, I'm sure that in 2D, the self dual field strength exists.
So, is my identity eq(*) correct? If it's correct, where is my argument wrong?
Thank you so much.

Sincerely

 

[Valerie Park]

,TaeYoungA:In $D$ dimensions the equation $(*)$ is true. You can easily check it using Young symmetrizers, for example. For a rank $2k$ antisymmetric tensor we have:$$A_{\mu_1 \cdots \mu_{2k}} = \frac{1}{(2k)!} \sum_{\sigma \in S_{2k}} (-1)^{\sigma} \sigma (A_{\mu_1 \cdots \mu_{2k}})$$where $S_{2k}$ is the permutation group of $2k$ elements and $(-1)^{\sigma}$ is the signature of the permutation $\sigma$.Now, if we contract the two Levi-Civita tensors with the antisymmetric tensor we obtain:$$A^{\mu_1 \cdots \mu_{2k}} \epsilon_{\mu_1 \cdots \mu_{2k} \nu_{2k+1} \cdots \nu_D} = \frac{1}{(2k)!} \sum_{\sigma \in S_{2k}} (-1)^{\sigma} \sigma (A^{\mu_1 \cdots \mu_{2k}}) \epsilon_{\mu_1 \cdots \mu_{2k} \nu_{2k+1} \cdots \nu_D}$$On the other hand, the identity $(*)$ implies:$$A^{\mu_1 \cdots \mu_{2k}} \epsilon_{\mu_1 \cdots \mu_{2k} \nu_{2k+1} \cdots \nu_D} = (-1)^{k} k! \delta^{\mu_{2k+1}}_{[\nu_{2k+1}} \cdots \delta^{\mu_D}_{\nu_D]} A^{\mu_1 \cdots \mu_{2k}}$$Comparing the two expressions we see that they are equal if and only if $$\sum_{\sigma \in S_{2k}} (-1)^