# Can we “mod out” a common subspace in the Grassmannian inside the exterior algebra?

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.

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.

• 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... – Asaf Shachar Feb 8 at 6:41
• 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. – Asaf Shachar Feb 8 at 6:41
• @AsafShachar: you're right, the injectivity is the important bit here. – Dap Feb 8 at 13:32