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Let $X$ be a finite-dimensional vector space and let $\Lambda^p(X)$ be the $p$th exterior power of $X$. My picture of an elementary element $x_1 \wedge \ldots \wedge x_p$ in the exterior power is something like: $x_1 \wedge \ldots \wedge x_p=0$, if the vectors are dependent, and otherwise $x_1 \wedge \ldots \wedge x_p$ is like a $p$-dimensional subspace of $X$ (the one spanned by the vectors) together with a volume form on that subspace. Of course, this picture does not really say how to think of higher rank elements in $\Lambda^p(X)$ which generally are linear combinations of elementary elements. Sometimes this isn't so bad. If you have an element of the form $\alpha = \sum_{i=1}^k \alpha_i$ where each $\alpha_i$ is elementary and where the subspaces corresponding to each $\alpha_i$ are independent, then you can sort of think of having a "direct sum" of elementary elements.

Question: It seems plausible that every element of $\Lambda^p(X)$ can be expressed in this way. Is this actually true? That is can any element of $\Lambda^p(X)$ be written as $$(x_1 \wedge \ldots \wedge x_p) + (x_{p+1} \wedge \ldots \wedge x_{2p}) + \ldots + (x_{pk -p + 1} \wedge \ldots \wedge x_{pk})$$ where $x_1,x_2,\ldots,x_{pk}$ are linearly independent? In this situation, is $k$ the rank of this element (in the sense that the element cannot be expressed as a sum of fewer than $k$ elementary elements).

Note that, if the answer to this question is "yes", then it follows that the rank $r$ of an element of $\Lambda^p(X)$ satsfies $pr \leq \dim(X)$.

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I have managed to show that the answer to this question is "no". If $\dim(X) \geq 5$, then not every element $\alpha \in \Lambda^3(X)$ can be written in the form I inquired about. You can see this by thinking about the kernel of the map \begin{align*} T_\alpha : X \to \Lambda^4(X) && T_\alpha(x) = \alpha \wedge x. \end{align*} Two observations:

  1. If $\alpha$ is "elementary", i.e. $\alpha = x \wedge y \wedge z$ for linearly independent vectors $x,y,z \in X$, then the kernel of $T_\alpha$ can be shown to equal $\mathrm{span}\{x,y,z\}$, so the dimension of the kernel is 3.

  2. If $\alpha$ is a sum of $k \geq 2$ elementary elements which are "independent", i.e. $\alpha = (x_1 \wedge y_1 \wedge z_1) + (x_2 \wedge y_2 \wedge z_2) + \ldots + (x_k \wedge y_k \wedge z_k)$ for linearly independent vectors $x_1,y_1,z_1,\ldots,x_k,y_k,z_k \in X$, then the kernel of $T_\alpha$ can be shown to be trivial.

Now, let $x,y_1,y_2,z_1,z_2 \in X$ be linearly independent and consider $$\alpha = (x \wedge y_1 \wedge z_1) + (x \wedge y_2 \wedge z_2) \in \Lambda^3(X).$$ One can check that, for this element, the kernel of $T_\alpha$ is $\mathrm{span}(x)$. Since the kernel of $T_\alpha$ is neither 3-dimensional, nor 0-dimensional, this $\alpha$ cannot be expressed as the sum of one or more "independent" elementary elements.

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  • $\begingroup$ Cool observation! NB: the right term for "elementary" is decomposable. $\endgroup$ – lisyarus Sep 22 '19 at 11:03

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