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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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    @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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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.2012-12-05