Let $V$ be a vector space of $\dim n$ over $K$. Let $P$ be the set of all pure products of the form $v_1 \bigwedge v_2$. How to prove that there is a one-one correspondence between the one dimensional subspaces in $P$ and the two dimensional subspaces of $V$.

Try: the correspondence is: subspace spanned by $v_1 \bigwedge v_2$ <-------> subspace spanned by $(v_1,v_2)$. I showed that, subspace spanned by $(v_1,v_2)$ ------> subspace spanned by $v_1 \bigwedge v_2$ is well defined. I cannot show that the reverse arrow is well defined. Please help.

  • 1
    $\begingroup$ Hint: If $(v_1,v_2)$ is a basis for a subspace of $V$, then you can get to any other basis by the operation $(v_1,v_2)\mapsto (v_1,rv_1+v_2)$, scalar multiplication of $v_i$, and switching the order. What effect do these operations have on the wedge $v_1\wedge v_2$? $\endgroup$ – Cheerful Parsnip Mar 27 '13 at 19:52
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    $\begingroup$ @JimConant I strongly suspect you are proving the wrong implication. You show $sp(v_1,v_2) = sp(v_3,v_4) \implies v_1\wedge v_2 = v_3\wedge v_4$. OP is interested in the reverse. $\endgroup$ – Lord_Farin Mar 27 '13 at 20:00
  • $\begingroup$ Welcome user1 and +1 , since your question is well-formulated and interesting. You are asking about a particular case of the Plücker embedding: if you google this two-word expression you will find many interesting online documents about the subject. $\endgroup$ – Georges Elencwajg Mar 27 '13 at 20:31
  • $\begingroup$ @Lord_Farin: Ah, you are right. I read it too quickly. Thanks. $\endgroup$ – Cheerful Parsnip Mar 27 '13 at 20:46

$P$ is stable by multiplication by elements of the multiplicative group $K^\times$, and so it is enough to produce a bijection between $K^\times$-orbits of non-zero elements of $P$ and 2-d subspaces of $V$. In what follows I assume all the basics of multilinear algebra.

You have already defined a candidate function, sending $K\{v_1,v_2 \}$ to the $K^\times$ orbit of $v_1 \wedge v_2$. This map is surjective since any non-zero wedge $v_1 \wedge v_2$ has $v_1$ and $v_2$ linearly independent. It remains to show that it is injective. So let's suppose that $v_1 \wedge v_2=x(w_1 \wedge w_2)$ for some non-zero $x \in K^\times$. It follows that $v_1 \wedge w_1 \wedge w_2=0$, and hence $v_1 \in K\{w_1,w_2\}$ (I consider this assertion part of the "basics" of multilinear algebra, but understand I am unusual here, so am going to include a proof sketch below anyway). By symmetry $v_2 \in K\{w_1,w_2\}$, and now it follows that $K\{v_1,v_2\}=K\{w_1,w_2\}$.

Proof that $v_1 \wedge w_1 \wedge w_2=0$ implies $v_1$ in the span of $w_1$ and $w_2$: We may complete $w_1,w_2$ to a basis of $V$. Write $v_1=\sum a_i w_i$ and compute $$0=v_1 \wedge w_1 \wedge w_2=\sum_{i=3}^n a_i w_i \wedge w_1 \wedge w_2.$$ Now the vectors on the RHS are linearly independent, implying the assertion.

  • $\begingroup$ Very nice! ${}{}$ $\endgroup$ – Georges Elencwajg Mar 27 '13 at 22:31

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