Criterion for being a simple vector

Question

1) I was wondering if there exists a criterion (of (a) combinatorial or (b) geometric nature) for a sum of simple vectors $V\in\wedge^k(\mathbb R^n)$,

$V=e_{a_{11}}\wedge\cdots\wedge e_{a_{1k}} + \cdots + e_{a_{m1}}\wedge\cdots\wedge e_{a_{mk}}$, where $e_1,\cdots, e_n$ is a basis of $\mathbb R^n$

to be a simple vector: in other words, I want to know what should one verify in order to be sure that $V$ can be written as $V=v_1\wedge\cdots\wedge v_k$ for some vectors $v_i\in\mathbb R^n$.

2) What happens if we allow noninteger coefficients? In other words, how does the answer change if we consider question 1) for

$V=c_1e_{a_{11}}\wedge\cdots\wedge e_{a_{1k}} + \cdots + c_me_{a_{m1}}\wedge\cdots\wedge e_{a_{mk}}$, where now $c_i\in\mathbb R$.

3) any reference or hint about some (name of) branch of mathematics dealing with this is also welcome!

Answer 1

You are looking for the Plucker relations, a collection of quadratic polynomials in the $c_m$'s which vanish if and only if $V$ can be written as $v_1 \wedge v_2 \wedge \cdots \wedge v_k$. You should be able to read about them in most books on algebraic geometry. For example, Griffiths-Harris or Miller-Sturmfels will definitely discuss this.

If you are comfortable with representation theory of $GL_n$, they can be described very briefly: Let $\omega_k$ be the weight of the highest weight vectors in $\bigwedge^k(\mathbb{R}^n)$. There is (up to scalar multiple) a unique $GL_n$ equivariant map from $\bigwedge^k(\mathbb{R}^n)^{\otimes 2}$ to the representation of $GL_n$ with highest weight $2 \omega_k$. Let the kernel of this map be $K$. The plucker relations say that $V \otimes V$ is in $K$.

It is not difficult to write them down explicitly, but there are a lot of signs to get right, so I'll just refer you to the sources cited above.

There is no significant simplification in assuming that the $c$'s are $0$ or $1$.

Since you tagged this combinatorics and hyperplane arrangements, you may want to know the following: If $V$ is simple, then the set of $k$-tuples $(a_{m1}, \ldots, a_{mk})$ for which $c_m$ is nonzero form the bases of a matroid. There is no useful converse to this statement. Essentially by definition, a matroid $M$ is realizable if and only if you can find vectors $v_1$, $v_2$, ..., $v_k$ such that the bases of $M$ correspond to the nonzero terms in $v_1 \wedge v_2 \wedge \cdots \wedge v_k$. But there is no good criterion for a matroid to be realizable.