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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.2012-12-05
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    @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.2012-12-05
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    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.2012-12-05
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    @MvG: That was a typo. Fixed. Thanks for pointing it out.2012-12-05
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    @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?2012-12-05
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    @robjohn: As you noticed there was a typo. It has to be $i in the summation (not $i\neq j$). On the other hand, it is very clear that all the $i,j,k,l$ have to be different, since we said in the definition $\{i,j,k,l\}=\{1,2,3,4\}$.2012-12-05
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    @A.Schulz: The fact that $i,j,k,l$ have to be distinct because $\{i,j,k,l\}=\{1,2,3,4\}$ is subtle and easily missed. Perhaps a "note that this implies $i,j,k,l$ are distinct" would be helpful.2012-12-05

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