Consider the point configuration

and the numbers q1=1, q2=0, q3=1, q4=0, q5=1, q6=0. The corresponding matroid Gromov-Witten enumerative problem is to find the number of triangles whose vertices are on three generic lines, and whose sides pass through 3 generic points. The ``high school strategy'' to solve this problem is as follows: guess one vertex 1 of the triangle on the corresponding line, then project that through one of the given points to another line, then do this again, and again, eventually arriving back at a point 1' on the original straight line. Your guess was correct if the point you arrive at is exactly 1. Of course, there is 0 chance it happens. Click here, and you can play with your initial choices of 1. Just use the cursor to move point 1 on its line, and find how many times you can align 1 with 1'.

(1) You probably found that the number of good guesses is 2. One proof (over C) of this statement is that the trasformation 1 -> 1' is a projective transformation, hence it is of the form x -> (ax+b)/(cx+d) [in affine coordinates]. The number of fixed points of this transformation is clearly 2.

(2) Another, geometric, proof of this fact, is a nice treat in projective geometry: the famous Steiner construction.

(3) A third proof (overkill) is computing the equivariant class of the above matroid, (c1-d1-d2-d3)(c1-d3-d4-d5)(c1-d5-d6-d1), and finding the coefficient of d1.d3.d5. This must be (-1) times the number of solutions.

Now consider the configuration

and the numbers q1=1, q2=1, q3=1, q4=1, q5=0, q6=0. You may experience with this problem here by using your cursor to move point 2 on its line, and find how many times you can align it with 2'. Note that this is almost the same as above with the only difference that the centers of projections are collinear. Probably you found that there is 1 correct solutions and there is another `fake' one. The `solution' when all points are collinear is called fake, because although here our high shool strategy makes 2 to coincide with 2', but the resulting configuration is not in the relevant matrix matroid variety. That is, the resulting configuration is not a limit of Menelaus configurations. An even further way of saying this is that the fake solution satisfies the naive equations of the matrix matroid variety, but not the other ones.

Remark that the x -> (ax+b)/(cx+d) argument also gives the incorrect solution 2. The correct solution is 1, and the equivariant cohomology evidence is that the coefficient of d1.d2.d3.d4 in the relevant equivariant class [Y] is 1. Observe that this variety is not a product of matrix Schubert varieties, so Giambelli type formulas will not tell us [Y].

Consider the configuration

and the numbers q1=1, q2=0, q3=0, q4=0, q5=1, q6=1, q7=1, q8=0, q9=0, q10=0, q11=2. You can experience with the interactive diagram here. Use the cursor to move point 1 on its (yellow) line and find how many times you can make the three black lines concurrent.

The solution is 3, which can be obtained with elementary calculation (degree of a 3 by 3 determinant of linear forms is 3), or by checking the coefficient of d1.d5.d6.d7.(d12)^2 in the relevant equivariant class. This equivariant class is again an easy one, it is a product of Giambelli formulas: (c1-d1-d2-d5)(c1-d1-d3-d6)(c1-d1-d4-d7)(c1-d5-d8-d11)(c1-d6-d9-d11)(c1-d7-d10-d11).

Here is a more challenging one

with the numbers q1=1, q2=1, q3=0, q4=1, q5=1, q6=1, q7=1. Click here and use the cursor to move points 1 and 2 on their lines and find how many times you can make the yellow lines e and f go through the points 4 and 2, respectively.

The relevant equivariant class has a term d1.d2.d4.d5.d6.d7 with coefficient 1, hence the number of solutions is 1.

This problem can not be set up as a one-variable problem---ie. you have to move two points e.g. as presented: 1 and 2. Note the infinitely many fake solutions when points 1, 2 and 3 are on one line (try it!). These correspond to a component of variety corresponding to the naive ideal, whose dimension is LARGER than expected. You may also want to detect the singularity when both 1 and 2 approach point A.

Consider the Pappus configuration P1,...,P9. Suppose e1,...,e8 are generic lines on the plane, and Q is a generic point. How many Pappus configurations exist with Pi \in ei (i=1...8), and P9=Q ? The solution is 5 (!!!). Challenge: create a Cinderella2 illustration... (I failed).