An identity with vector products

Question

Consider $a$, $b$, $c$, $a'$, $b'$, $c'$ points in $\mathbb{R}^3$. Denote by $[ab]$ the cross product of the vectors $a$, $b$, $(a b)$ their dot product, and $[abc] = ([ab]c)$ their mixed product. Let $x=[bc]$, $x'=[b'c']$, $y =[ca]$, $y'=[c'a']$, $z=[ab]$, $z'=[a'b']$. Show that $$[[xx'][yy'][zz']]=[abc][a'b'c'][[aa'][bb'][cc']]$$

Notes: This is from Gurevich, Foundations of the Theory of Algebraic Invariants, p. 138.

Why is the identity important? If $a$, $b$, $c$, $a'$, $b'$, $c'$ are viewed as points in $\mathbb{P}^2(\mathbb{R})$ then $[[bc][b'c']]$ is the point of intersection of the lines $bc$, $b'c'$. So the LHS gives the condition that the points $bc\cap b'c'$, $ca\cap c'a'$, $ab\cap a'b'$ are collinear, while the third factor on RHS is the condition that the lines $aa'$, $bb'$, $cc'$ is concurrent. So this would give a proof of Desargues' theorem, and its converse.

Here are identities that are listed as helping in the proof:

$$[[ab]x]= (ax)b-(bx)a \\ [[ab][cd]] = [acd]b-[bcd]a = [abc]d-[abd]c$$

(another identity $([ab][cd]) = (ac)(bd)-(ad)(bc)$ might be useful here)

From the second identity we get $$[[ab][cd][ef]]=[acd][bef]-[bcd][aef]=[abc][def]-[abd][cef]$$

This would allow us to tackle the LHS of the required identity, but I am a bit stuck at the moment. Maybe both sides can be expressed in terms of mixed products, and then we can deduce all from the Plucker relations.

Any ideas/hints would be very helpful!

Answer 1

The identity can be derived using a special case of Cauchy-Binet formula $$ (p\times q)\cdot(u\times v) =\det\begin{bmatrix} p\cdot u & p\cdot v\\ q\cdot u & q\cdot v\\ \end{bmatrix} $$ or, in your symbols, $$ [pq[uv]]=(pu)(qv)-(pv)(qu).\tag{0} $$ Put $p=[xx'],\,q=[yy'],\,u=z$ and $q=z'$, we get $$ \left[[xx'][yy'][zz']\right] =[xx'z][yy'z']-[xx'z'][yy'z]\tag{1} $$ Apply $(0)$ again, we further obtain \begin{align} [xx'z] &=[xx'[ab]] =(xa)(x'b)-(xb)(x'b) =(xa)(x'b),\\ [yy'z']&=\cdots=(ya')(y'b'),\\ [xx'z']&=\cdots=-(xb')(x'a'),\\ [yy'z]&=\cdots=-(yb)(y'a).\\ \end{align} Since $(xa)=(yb)=[abc]$ and $(y'b')=(x'a')=[a'b'c']$, the RHS of $(1)$ can be rewritten as $$ \left[[xx'][yy'][zz']\right] =[abc][a'b'c']\left\{(x'b)(ya')-(xb')(y'a)\right\}. $$ Now the result follows because \begin{aligned} &(x'b)(ya')-(xb')(y'a)\\ &=([b'c']b)([ca]a')-([bc]b')([c'a']a)\\ &=[b'c'b][caa']-[bcb'][c'a'a]\\ &=[bb'c'][aa'c]-[bb'c][aa'c']\quad\text{by permutations}\\ &=[[bb'][aa'][c'c]]\quad\text{by a parallel to } (1)\\ &=[[aa'][bb'][cc']].\\ \end{aligned}

Answer 2

We'll use the equality for the triple product of cross products:

$$[[ab][cd][ef]]= [acd][bef]-[bcd][aef]$$

We get with the above $$[[xx'][yy'][zz']]=[xyy'][x'zz']-[xzz'][x'yy']$$

Now substitute the values of $x$, $x'$, $\ldots$ in the triple products on the RHS. We get $$[xyy']=[[bc][ca][c'a']]= [bca][cc'a']-[cca][bc'a']=[bca][cc'a']\\ [x'zz']=[[b'c'][ab][a'b']]=[b'ab][c'a'b']-[c'ab][b'a'b']=[b'ab][c'a'b']\\ [xzz']=[[bc][ab][a'b']]=[bab][ca'b']-[cab][ba'b']=-[cab][ba'b']\\ [x'yy']=[[b'c'][ca][c'a']]=[b'ca][c'c'a']-[c'ca][b'c'a']=-[c'ca][b'c'a']$$

Putting it all together we get $$[[xx'][yy'][zz']]=[abc][a'b'c']([cc'a'][b'ab]- [ba'b'][c'ca])$$

Now, in the bracket on RHS we have (using the skew-symmetry of the triple product): $$[abb'][a'cc']-[a'bb'][acc']=[[aa'][bb'][cc']]$$