# What are the necessary and sufficient conditions for a two-form to be an exterior product of two one-forms?

Consider a two-form $$\gamma \in \Lambda^2(V)$$ where $$V$$ is a real vector space. Now I would like to know the necessary and sufficient conditions for $$\gamma$$ to be expressible as an exterior product of two one-forms, $$\gamma=\alpha \wedge \beta,\, \alpha,\beta \in \Lambda(V)$$.

Obviously, a necessary condition for the decomposition to hold is that any exterior power of $$\gamma$$ vanishes, $$\gamma\wedge...\wedge\gamma = 0$$. But it is not clear to me whether this condition is sufficient.

• I recommend that you search for "Plucker quadratic relations". – Jason Starr Dec 17 '18 at 11:09

## 2 Answers

While Plücker relations give the general theory, a direct answer to your question is the following: a 2-form $$\gamma$$ is decomposable (is a product of two 1-forms) iff $$(\iota_v \gamma) \wedge \gamma = 0$$ for any vector $$v$$, where $$\iota_v \gamma$$ is the usual contraction of a vector with a 2-form, that is, $$(\iota_v \gamma)(u) = \gamma(v,u)$$ for any other vector $$u$$.

The proof is straightforward. Pick some vector $$v$$ such that $$\alpha = \iota_v \gamma \ne 0$$. If it doesn't exist, then $$\gamma = 0$$, which is the trivial case. Then $$\alpha \wedge \gamma = 0$$ implies* that there exists $$\beta$$ such that $$\gamma = \alpha \wedge \beta$$ and you are done.

* If the implication is not obvious, extend $$e_1 = \alpha$$ to a basis $$e_i$$ in $$\Lambda(V)$$ and expand $$\gamma$$ in the induced basis $$e_i\wedge e_j$$. Clearly, $$e_1\wedge \gamma = 0$$ iff the only non-zero coefficients in the expansion are in front of the terms $$e_1 \wedge e_j$$.

Let $$e_1$$, ..., $$e_n$$ be a basis of $$V$$. Let $$\gamma = \sum c_{ij} e_i \otimes e_j$$, so $$C = (c_{ij})$$ is a skew symmetric matrix. Then $$\gamma$$ factors as $$\alpha \wedge \beta$$ if and only if $$C$$ has rank $$\leq 2$$.

Proof: Let $$C$$ be a skew symmetric matrix, we must show that $$C$$ can be written as $$a b^T - b a^T$$ for vectors $$a$$ and $$b$$ if and only if $$\mathrm{rank}(C) \leq 2$$. The condition is clearly necessary, since $$\mathrm{rank}(a b^T) = \mathrm{rank}(b a^T) \leq 1$$.

Choose a $$2$$-dimensional subspace $$\mathrm{Span}(v_1, v_2)$$ containing the image of $$C$$. Since $$C^T = -C$$, this space also contains the image of $$C^T$$. Replacing $$C$$ by $$SCS^T$$, we may assume that $$v_1$$ and $$v_2$$ are the first two basis vectors.

Then the conditions on the images of $$C$$ and $$C^T$$ say that $$C$$ is $$0$$ outside the intersection of the first two rows and the first two columns, and skew symmetry further says that the upper left $$2 \times 2$$ block is of the form $$\left( \begin{smallmatrix} 0&c \\ -c&0 \end{smallmatrix} \right)$$. For such matrices, the claim is obvious. $$\square$$

The OP asks whether it is enough to ask that $$\gamma \wedge \gamma$$ vanish (well, the OP asks about all higher wedge powers vanishing as well, but that clearly follows from $$\gamma \wedge \gamma=0$$.) The answer is yes. Any $$2$$-form can be written as $$u_1 \wedge v_1 + u_2 \wedge v_2 + \cdots + u_r \wedge v_{r}$$ where $$2r$$ is the rank of the matrix $$C$$ and $$u_1$$, ..., $$u_r$$, $$v_1$$, ..., $$v_{r}$$ are linearly independent; this is more or less the classification of skew symmetric bilinear forms. Then $$\gamma^{\wedge k} = k! \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq r} u_{i_1} \wedge v_{i_1} \wedge u_{i_2} \wedge v_{i_2} \wedge \cdots \wedge u_{i_k} \wedge v_{i_k}.$$ This is clearly nonzero if and only if $$k \leq r$$. In particular, $$r \leq 1$$ if and only if $$\gamma \wedge \gamma=0$$.

The condition that $$\gamma \wedge \gamma =0$$, expanded in coordinates, states that the $$4 \times 4$$ principal Pfaffians of $$C$$ vanish. It is a nice high school algebra challenge to show that this implies the $$3 \times 3$$ minors of $$C$$ vanish, which is the more standard algebraic test for a matrix to have rank $$\leq 2$$.