# Self-dual condition implies dimension 2 modulo 4

Reading the article of Henneaux and Teitelboim Dynamics of Chiral (Self-Dual) p-Forms they state that "In order for self-dual fields to exist it is necessary that $$F$$ and $$^*F$$ should have the same number of components. It is also necessary that the square of the operation of taking the dual should give $$+1$$. These two demands restrict the spacetime dimension to be equal to $$2$$ modulo $$4$$".

My attempt was the following. If $$A$$ is a $$p$$-form, then $$F=dA$$ is a $$(p+1)$$-form of the form \begin{align} F=\frac{1}{(p+1)!}F_{\mu_1\mu_2\dots\mu_{p+1}}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_{p+1}}\,. \end{align} Its Hodge dual is \begin{align} ^*F=\frac{\sqrt{|g|}}{(p+1)!(n-(p+1))!}F_{\mu_1\dots\mu_{p+1}}\epsilon^{\mu_1\dots\mu_{p+1}}{}_{\mu_{p+2}\dots \mu_n}dx^{\mu_{p+2}}\wedge \dots\wedge dx^{\mu_n}\,. \end{align} Equating both sides, one realize that the same number of components implies equation $$p=n-(p+2)$$, which yields $$n=2(p+2)$$. On the other side, on a Lorentzian manifold, the square of the Hodge operator is $${^*}{^*}F=(-1)^{1+(p+1)(n-(p+1))}F$$. Taking this operation equals to $$+1$$, we arrive to the condition \begin{align} 1+(p+1)(n-(p+1))=2k\,, \end{align} with $$k=0,1,2,3\dots$$. Solving this condition implies that $$n=\sqrt{8k}$$?

How can I get to the condition that the authors say?

It's somewhat simpler to frame this in terms of $$q=p+1$$, in which case your conditions become that $$q=n-q$$ and that $$q(n-q)$$ is odd. The former condition implies that $$n=2q$$ must be even, while the latter implies that $$q$$ must be odd; the desired result follows immediately.