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.

  • 4
    $\begingroup$ I recommend that you search for "Plucker quadratic relations". $\endgroup$ – Jason Starr Dec 17 '18 at 11:09

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 $\leq 2$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.