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While reading this paper, I have seen the following claim stated without a proof:

Let $V$ be an $n$-dimensional vector space over a field, and let $\alpha,\beta \in \bigwedge^k V$ be decomposable and non-zero.

Suppose that $$\alpha=(u_1 \wedge \dots \wedge u_r) \wedge v_1 \wedge \dots \wedge v_{k-r},\beta=(u_1 \wedge \dots \wedge u_r) \wedge w_1 \wedge \dots \wedge w_{k-r},$$

where $$\text{span}(u_1 ,\dots,u_r)=\text{span}(u_1 ,\dots,u_r,v_1\dots v_{k-r}) \cap \text{span}(u_1 ,\dots,u_r,w_1\dots w_{k-r}) $$

(i.e. $\text{span}(u_1 ,\dots,u_r)$ is the intersection of the subspaces corresponding to the decomposable tensors $\alpha,\beta$.)

Then, if $\alpha+\beta$ is decomposable and non-zero, then so is $ v_1 \wedge \dots \wedge v_{k-r}+ w_1 \wedge \dots \wedge w_{k-r} $.

In other words, we can "mod out" the "common intersection" of $\alpha$ and $\beta$.

How to prove this statement?

Here is my failed attempt:

We can write $$\tilde u_1 \wedge \dots \wedge \tilde u_k=\alpha+\beta=(u_1 \wedge \dots \wedge u_r) \wedge \gamma, \tag{1}$$

where $\gamma=v_1 \wedge \dots \wedge v_{k-r}+ w_1 \wedge \dots \wedge w_{k-r} $. By wedging this equality with $u_i$, we see that $\text{span}(u_1,\dots,u_r) \subseteq \text{span}(\tilde u_1,\dots,\tilde u_k)$. Thus, we can assume W.L.O.G that $\tilde u_i=u_i$ for $1 \le i \le r$.

Rewriting, we have
$$ u_1 \wedge \dots \wedge u_k=\alpha+\beta=(u_1 \wedge \dots \wedge u_r) \wedge \gamma, \tag{2}$$

Now, it suffices to prove that $\gamma \wedge u_j=0$ for $k < j \le r$, since this would imply that the dimension of the subspace of $V$ annihilating $\gamma$ is at least $r-k$. Since it is also not greater than $r-k$, it must be $k$. This implies that $\gamma$ is decomposable.

So, we now prove that $\gamma \wedge u_j=0$ for $k < j \le r$: We can complete $(u_1,\dots,u_k)$ into a basis $(u_1,\dots,u_n)$ of $V$. Now we write $\gamma=\sum a^{i_1,\dots,i_{r-k}}u_{i_1} \wedge \dots \wedge u_{i_{r-k}}$. Since $\alpha+\beta \neq 0$, there is at least one summand in $\gamma$ that is not composed entirely from wedge of $u_1,\dots,u_r$....

(I don't see how to continue).

In fact, it seems that equation $(2)$ should imply that $\gamma$ is not necessarily decomposable, since it is not uniquely determined by it:

Indeed, if we modify $\gamma$ by adding decomposable elements which involve any of the $u_1,\dots,u_r$, the RHS does not change, so the equation still holds. It seems to me then, that we should be able to convert $\gamma$ to be a non-decomposable element.

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I will just try and explain my interpretation of the reduction used in the paper. (i.e. I haven't tried to explain the second paragraph of their proof showing that a $p$-vector of the form $u_1\wedge\dots\wedge u_p+w_1\wedge\dots w_p$ is indecomposable whenever $u_1,\dots,u_p,w_1,\dots,w_p$ are linearly independent and $p>1.$ If you find any problem in that part then ask.)

With your notation let $U=\mathrm{span}(u_1,\dots,u_r).$ There is a well=defined map defined by

\begin{align} \mu&:&&\bigwedge^{k-r}(V/U)&&\to &&\bigwedge^k V\\ &&&[x_1]\wedge\dots\wedge [x_{k-r}]&&\mapsto &&u_1\wedge\dots\wedge u_r\wedge x_1\wedge\dots\wedge x_{k-r} \end{align} and extending linearly.

I believe the fact being used is that decomposable $k$-vectors in the image of this map are the image of a decomposable $(k-r)$-vector. (In fact the preimage is unique since $\mu$ is an inclusion.)

Consider $\xi=y_1\wedge \dots\wedge y_k$ in the image of $\mu.$ Then $\xi$ is annihilated by wedging with any $u_i.$ This implies each $u_i$ is in the span of $y_1,\dots,y_k.$ So by extending $\{u_1,\dots,u_r\}$ to a basis of $\{y_1,\dots,y_k\}$ (ultimately, the Steinitz exchange lemma) and scaling, we can write $\xi=u_1\wedge\dots\wedge u_r\wedge x_1\wedge\dots\wedge x_{k-r}$ as required.

In the proof in the paper, the second paragraph is effectively taking place in $\bigwedge^{k-r}(V/U)$ (with different notation). So it assumes $[v_1]\wedge\dots\wedge[v_{k-r}]+[w_1]\wedge\dots\wedge[w_{k-r}]$ is decomposable and derives a contradition if $k-r>1.$ It should also be noted that the quotients $[v_1],\dots,[w_{k-r}]$ are still linearly independent.

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  • $\begingroup$ Thank you! This is a truly wonderful answer. Just one question to make sure I did not miss anything: We really need the fact that $\mu$ is injective, right? Because in our setting, we know that $\alpha+\beta=\mu([v_1]\wedge\dots\wedge[v_{k-r}]+[w_1]\wedge\dots\wedge[w_{k-r}])$ is decomposable, and we want to deduce that the sum $[v_1]\wedge\dots\wedge[v_{k-r}]+[w_1]\wedge\dots\wedge[w_{k-r}]$ is decomposable... $\endgroup$ – Asaf Shachar Feb 8 at 6:41
  • $\begingroup$ Thus, it does not suffice to know that there exists a decomposable preimage, but we need the uniqueness to ensure that it is "the" preimage $[v_1]\wedge\dots\wedge[v_{k-r}]+[w_1]\wedge\dots\wedge[w_{k-r}]$ itself that is really decomposable, right? I must add that I truly appreciate your insightful and elegant answers, I have noticed that you have answered many of my questions lately. Thanks again. $\endgroup$ – Asaf Shachar Feb 8 at 6:41
  • $\begingroup$ @AsafShachar: you're right, the injectivity is the important bit here. $\endgroup$ – Dap Feb 8 at 13:32

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