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Let $X=\{p_1,p_2,p_3,p_4\}$ be four points in $\mathbb{R}^2$, not three of them on a line. We use the square bracket notation $$[i,j,k]=\det \begin{pmatrix} p_i & p_j & p_k \\ 1 & 1 & 1 \end{pmatrix},$$ which denotes twice the signed area of the triangle spanned by $p_i,p_j,p_k$. For every pair $i,j$ from $\{1,2,3,4\}$ we set

$$A_{ij}:= \frac{[i,j,a][i,j,b]}{[i,j,k][i,j,l]},\quad \text{with $\{i,j,k,l\}=\{1,2,3,4\}$}.$$

Note that $\{i,j,k,l\}=\{1,2,3,4\}$ implies that $i,j,k,l$ are distinct.

Then for any two points $p_a$ and $p_b$ (not necessarily from $X$) the following holds $$ \sum_{1\le i\lt j\le4} A_{ij}=1. $$ Or if you prefer it to write out the definition of the 6 $A_{ij}$s $$\frac{[1,2,a][1,2,b]}{[1,2,3][1,2,4]}+\frac{[1,3,a][1,3,b]}{[1,3,2][1,3,4]} +\frac{[1,4,a][1,4,b]}{[1,4,3][1,4,2]}+\cdots +\frac{[3,4,a][3,4,b]}{[3,4,1][3,4,2]} =1.$$ This can be shown algebraically, see here on page 717.

I am interested in a geometric interpretation of this invariant. Can this be proven by elementary theorems from (projective?) geometry?

I want to understand the invariant, because I am interested in higher dimensional analogues.

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If $i$ or $j$ equals $k$ or $l$, $A_{ij}$ does not exist. If we fix $k$ and $l$, that leaves only one choice for $i$ and $j$. Something seems amiss. –  robjohn Dec 5 '12 at 11:01
    
@robjohn:how could $i=k$, if $\{i,j,k,l\}=\{1,2,3,4\}$? In the definition of $A_{ij}$, the indices $k,l$ are the ones different from $i,j$. For example $A_{1,2}=\frac{[1,2,a][1,2,b]}{[1,2,3][1,2,4]}$, $A_{1,3}=\frac{[1,3,a][1,3,b]}{[1,3,2][1,3,4]}$, and so on. –  A.Schulz Dec 5 '12 at 11:26
    
Your question appears inconsistent: in the formula with the formal sum the right hand side is $1$, whereas the formula with the explicit summands has a right hand side of $0$. I assume that the latter is correct, but I am not perfectly sure. –  MvG Dec 5 '12 at 12:10
    
@MvG: That was a typo. Fixed. Thanks for pointing it out. –  A.Schulz Dec 5 '12 at 12:28
    
@A.Schulz: nowhere is it stated that $k$ and $l$ are different from each other and from $i$ and $j$. If all are supposed to be distinct, then the sum, which is taken over $i,j\in\{1,2,3,4\},i\neq j$ has only two non-zero terms, swapping $i$ and $j$. Now it appears that you have amended the question and the sum seems to vary all $i,j,k,l$ (the terms with any two indices equal vanish, so there is no need to require uniqueness), even though the sum doesn't reflect this. There are 24 permutations of $1,2,3,4$. You say there are $6$ $A_{ij}$. Which $6$ of the $24$ are supposed to be in the sum? –  robjohn Dec 5 '12 at 14:01
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1 Answer

Observe that the $f$ introduced by your equation $f=0$ is not homogenous in all of the entries, since $a$ and $b$ apear in each of the summands but not in the summand induced by the denominator. On the other hand, each linear combination of Grassmann-Plücker relations will be a homogenous bracket polynomial.

Probably, this is due to the fact, that each of the $A_{ij}$ is not a projective invariant, since the expression is not invariant under the rescaling of the homogenous coordinates of the points. This is due to the way, you defined your bracket which is not the standard way of defining it: in a general setup inside projective geometry, you cannot assume that the coordinates of the points equals 1. To put it differntly, you cant assume a standard embedding of the point in $\mathbb R^2$ into the projective plane.

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Thanks for pointing this out. Maybe saying projective invariant in the title is misleading. Notice that I do not use homogenous coordinates at all (and I never said so). The statement is about points in $\mathbb{R}^2$. In fact, the sum will not give $1$ if you put in homogenized versions of the $p_i$s. The reason I mentioned projective geometry is, because I think tools from projective geometry might be applicable. That's why I had looked at GP-relations. But you are right, the approach to go via GP-relation is questionable. –  A.Schulz Dec 5 '12 at 14:51
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