Suppose we are given a $k$-dimensional complex vector space. Consider the second exterior power of $V$, that is $\Lambda^2V$. Denote $X=\{v_1\wedge v_2\colon v_1,v_2\in V\}$.

Now I have two questions:

1) Is it true that if $k=4$, then $x\in X$ if and only if $x\wedge x=0\in \Lambda^4 V$?

2) Let $k>4$. Does there exist $x\in \Lambda^2 V\setminus X$ such that $x\wedge x\neq 0$?

Thank you, C.

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    $\begingroup$ You defined $X$ as a subset of $\Lambda^4V$ (the squares of the elements of $\Lambda^2V$) so (1) and (2) do not make sense as written. $\endgroup$ – Mariano Suárez-Álvarez May 27 '12 at 19:48
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    $\begingroup$ Fine work with Caterina Sforza. If you had been leading the siege at Forli, I know you would have stayed there to the end. Oh, what was Machiavelli really like? $\endgroup$ – Will Jagy May 27 '12 at 19:59
  • $\begingroup$ Mariano Suárez-Alvarez, of course, it was a misprint. Thanks for spotting it. $\endgroup$ – Cesare Borgia May 27 '12 at 20:06
  • $\begingroup$ Dear Cesare, I think there still is a misprint in your definition of $X$: don't you mean $v\in \Lambda ^2 V$ ? $\endgroup$ – Georges Elencwajg May 27 '12 at 20:15
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    $\begingroup$ Is there still a problem with the definition of $X$? If it is supposed to consist of those elements of $\bigwedge^2 V$ that are "squares", then $X = \{0\}$ because $v \wedge v = 0$ for any $v \in V = \bigwedge^1 V$ for degree reasons. $\endgroup$ – Gunnar Þór Magnússon May 27 '12 at 20:29

I think that the following argument answers your question #1 in the affirmative. We need a bit more detailed description of those elements of $\Lambda^2V$ that are not in the set $X$.

Step 1. If the four vectors $v_1,v_2,v_3,v_4$ are linearly dependent, then the element $$ z=v_1\wedge v_2+v_3\wedge v_4\in X. $$ Proof. Assume that, for example, $v_4=av_1+bv_2+cv_3$ for some constants $a,b,c.$ By plugging this in we can easily rewrite $z$ in the form $$ z=(v_1+bv_3)\wedge v_2+cv_3\wedge v_1. $$ If here $b=0$, then we get $z=v_1\wedge(v_2-cv_3)\in X$. If $b\neq0$, then we can rewrite $z$ as $$ z=(v_1+bv_3)\wedge (v_2+\frac{c}{b}v_1) $$ again verifying the claim.

Step 2. If $\dim V=4$, then any element of $\Lambda^2V$ of the form $$ z=v_1\wedge v_2+v_3\wedge v_4+v_5\wedge v_6 $$ with $\{v_1,v_2,v_3,v_4\}$ linearly independent can be written as a sum of two elements of $X$.

Proof. The first four vectors form a basis of $V$, so we can write $$ v_5=u_1+u_2,\ \text{where}\ u_1=av_1+b v_2,\quad u_2=cv_3+dv_4 $$ for some constants $a,b,c,d$. If $u_1=0$, then the vectors $\{v_3,v_4,v_5,v_6\}$ are linearly dependent, and the claim follows from the result of Step1. Similarly we see that the case $u_2=0$ is easy, so we concentrate on the case $u_1\neq0\neq u_2$. Next we rewrite the wedge products $v_1\wedge v_2$ and $v_3\wedge v_4$ using $u_1$ and $u_2$ as the other factor. This is, again, easy. Say, if $a\neq0$, then $$ v_1\wedge v_2=(\frac1a u_1-\frac bav_2)\wedge v_2=u_1\wedge(\frac1a v_2), $$ and if $a=0, b\neq0$ we proceed analogously. So we can write $v_1\wedge v_2=u_1\wedge v_1'$ and $v_3\wedge v_4=u_2\wedge v_2'$. Putting all this together we can write $$ z=u_1\wedge(v_1'+v_6)+u_2\wedge(v_2'+v_6) $$ as a sum of two elements of $X$.

The following is now obvious by induction on the number of summands from $X$.

Lemma. If $k=4$, then the elements $z=\Lambda^2V\setminus X$ can be written in the form $$z=v_1\wedge v_2+ v_3\wedge v_4$$ with $v_1,v_2,v_3,v_4$ linearly independent.

We can then wrap up question #1. If $z\in \Lambda^2V\setminus X$, then $z=v_1\wedge v_2+ v_3\wedge v_4$, where the vectors $v_i,i=1,2,3,4,$ form a basis of $V$. Then $$ z\wedge z=2 v_1\wedge v_2\wedge v_3\wedge v_4\neq0, $$ as an even number of swaps are needed to bring $v_3\wedge v_4\wedge v_1\wedge v_2$ into the form $v_1\wedge v_2\wedge v_3\wedge v_4$.

  • $\begingroup$ It seems to me that everything in $\Lambda^2$ can be written as a sum of $\lfloor(k/2)\rfloor$ elements of $X$. Surely this is known? I'm afraid I don't have a reference, though. $\endgroup$ – Jyrki Lahtonen May 30 '12 at 11:01

As a general remark, you are asking how to recognize perfect wedges in $\bigwedge^2 V$ via equations in higher exerior powers. This is related to the Plucker relations for Grassmanians.

E.g. a non-zero perfect wedge $v_1\wedge v_2$ in $\bigwedge^2 V$ corresponds to a $2$-plane in $V$ (the span of $v_1$ and $v_2$), are you are asking whether or not the single equation $x \wedge x = 0$ cuts out the locus of $2$-planes as a subvariety of $\bigwedge^2 V$.

The answer is yes when $\dim V = 4$, but no in general.


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