# A form $F$ is simple if and only if $F\wedge F=0$?

Gravitation by Charles W. Misner, Kip Throne and John Wheeler page 93 Box 4.1 point 5 b. Applications: a.

"In four dimensions, all 0 forms, 1- forms, 3-forms, and 4-forms are simple. A 2-form $$F$$ is generally a sum of two simple forms, e.g., $$F=-edt\wedge dx+hdy\wedge dz$$; it is simple if and only if $$F\wedge F=0$$" where simple form is defined to be the "one that can be written as a wedge product of $$p$$ 1-forms: $$\sigma_p=\alpha \wedge...(p~factors)\wedge \gamma$$."

Question:

1. Does the conclusion here $$F\wedge F=0$$ specifically applied to $$F=-edt\wedge dx+hdy\wedge dz$$? or does it apply to any 2 form in 4 dimension? or does it apply to any 2 form in any dimesion?

2. The statement "A form $$F$$ is simple if and only if $$F\wedge F=0$$": I have a hard time to prove if $$F\wedge F=0$$ then $$F$$ is simple for general cases. Could you demonstrate a general proof, please?

Let $$F\in \Omega^p(\mathbb{R}^d)$$ be an arbitrary $$p$$-form in $$d$$ dimensions, where $$1\leq p\leq d$$.

1. Clearly any form $$F$$ squares to zero: $$F\wedge F=0$$ if $$p$$ is odd or if $$2p>d$$.

2. Clearly any simple/decomposable $$F$$ squares to zero: $$F\wedge F=0$$.

3. Proposition 1: The opposite implication $$\forall F\in \Omega^p(\mathbb{R}^d):~~F\wedge F=0~\Rightarrow~ F\text{ decomposable}\tag{1}$$ holds if and only if $$p=2$$ and $$d\geq 4$$.

4. Tomas Brauner's answer shows the "if" part of Proposition 1 by using the fact that any 2-form/antisymmetrix matrix can be brought on 2-block diagonal form.

5. The "only if" part of Proposition 1 can be seen by finding suitable counterexamples. E.g. in the case $$p=4$$ and $$d\geq 8$$, one may check that $$F~=~(\mathrm{d}x^1 \wedge \mathrm{d}x^2+ \mathrm{d}x^3 \wedge \mathrm{d}x^4)\wedge \mathrm{d}x^5 \wedge \mathrm{d}x^6 \tag{2}$$ squares to zero but is not$$^{\dagger}$$ decomposable.

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$$^{\dagger}$$ Sketched indirect proof of non-decomposability of (2): Assume that the 4-form (2) $$F~=~A_1\wedge A_2 \wedge A_3 \wedge A_4\tag{3}$$ is decomposable with four 1-forms $$A_k~=~\sum_{j=1}^d a_{kj} \mathrm{d}x^j, \qquad k~\in~\{1,2,3,4\}.\tag{4}$$ Next explore that various contractions yield $$\forall j\geq 7:~~ i_j F~=~0,\tag{5}$$ and $$i_1i_3F~=~0,\quad i_1i_4F~=~0,\quad i_2i_3F~=~0, \quad i_2i_4F~=~0,\tag{6}$$ to derive that $$\forall k\in\{1,2,3,4\}~\forall j\geq 7:~~ a_{kj}~=~0,\tag{7}$$ and that $$\forall k,\ell\in\{1,2,3,4\}:$$ $$a_{k1}a_{\ell 3}~=~a_{k3}a_{\ell 1}~\wedge~ a_{k1}a_{\ell 4}~=~a_{k4}a_{\ell 1}~\wedge~ a_{k2}a_{\ell 3}~=~a_{k3}a_{\ell 2}~\wedge~ a_{k2}a_{\ell 4}~=~a_{k4}a_{\ell 2}, \tag{8}$$ and therefore $$\forall k,\ell\in\{1,2,3,4\}:~~ (a_{k 1}, a_{k 2}, a_{k 3}, a_{k 4}) ~~\parallel~~(a_{\ell 1}, a_{\ell 2}, a_{\ell 3}, a_{\ell 4}).\tag{9}$$

Deduce that the four 1-forms (4) are spanned by $$\mathrm{d}x^5$$, $$\mathrm{d}x^6$$ and only one more 1-form, and hence the four 1-forms (4) are not linearly independent. Contradiction. $$\Box$$

Let me add an argument that is much less general than the answer by Mike Stone, but somewhat more elementary. It is valid for 2-forms regardless of the dimension of the base space the forms act on.

So, a general 2-form $$F$$ can be written in an arbitrarily chosen basis of 1-forms $$e^i$$ as $$F=\frac12f_{ij}e^i\wedge e^j$$. The matrix of coefficients $$f_{ij}$$ is antisymmetric, and thus can be made block-diagonal by a suitable change of basis. In other words, the form $$F$$ can always be expanded as $$F=f_{12}e^1\wedge e^2+f_{34}e^3\wedge e^4+\dotsb.$$ It follows immediately that $$F\wedge F=0$$ if and only if the above expansion contains at most one term (that is, exactly one term as long as the form $$F$$ is nonzero). This answers your questions.

The general condition for an $$n$$-form with components $$T^{\mu_1,\mu_2,\ldots \mu_n}$$ to be writable as a single wedge product $$T=U_1\wedge U_2,\wedge\ldots \wedge U_n$$ is $$T^{\mu_1\mu_2\ldots\mu_{n-1}[ \nu_1}T^{\nu_2\nu_3\ldots \nu_{n+1}]}=0$$ for all possible choices of the indices. where $$[\ldots]$$ means antisymmetrize. Thus a three index antisymmetric tensor is decomposable if $$T^{i_1i_2j_1}T^{j_2j_3j_4}-T^{i_1i_2 j_2}T^{j_1j_3j_4}+T^{i_1i_2j_3}T^{j_1j_2j_4}-T^{i_1i_2 j_4}T^{j_1j_2j_3}=0$$ These are the Plücker conditions. It is fairly easy to show that they are necessary, but I do not know an easy way to show that they are sufficient.