By Daniel J. Bates, Chris Peterson, Andrew J. Sommese (auth.), Alicia Dickenstein, Frank-Olaf Schreyer, Andrew J. Sommese (eds.)

In the decade, there was a burgeoning of task within the layout and implementation of algorithms for algebraic geometric compuation. a few of these algorithms have been initially designed for summary algebraic geometry, yet now are of curiosity to be used in functions and a few of those algorithms have been initially designed for functions, yet now are of curiosity to be used in summary algebraic geometry.

The workshop on Algorithms in Algebraic Geometry that was once held within the framework of the IMA Annual software yr in functions of Algebraic Geometry via the Institute for arithmetic and Its purposes on September 18-22, 2006 on the collage of Minnesota is one tangible indication of the curiosity. a hundred and ten contributors from 11 international locations and twenty states got here to hear the various talks; speak about arithmetic; and pursue collaborative paintings at the many faceted difficulties and the algorithms, either symbolic and numberic, that remove darkness from them.

This quantity of articles captures many of the spirit of the IMA workshop.

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In particular, the notion of Bruhat order is problematic as already noted in [Eriksson and Linusson, 2000b]. We suspect, for example, that there exist two permutation array schemes X and Y such that Y is reducible, and some but not all components of Y lie in the closure of X . 36 SARA BILLEY AND RAVI VAKIL Although Mnev's theorem is constructive, we have not attempted to explicitly produce a reducible or non-reduced permutation array scheme. 5. Intersecting Schubert varieties. , ... ,E: chosen generally and I:i f(w i ) = (2)' We show there is a unique permutation array P for this problem if X is nonempty, and we identify it .

1, 0 ,0 , OJ [1 ,2,1 , 0 ,0, 0 ) II U 10, 0,0 , 1,0 , OJ (0, 0 , 1, 1, 0 , OJ (0 , 1 ,2, 1,0 ,01 (1 ,3, 3 , 1 ,0, OJ II [0, 0 , 0, 0 , 1 , OJ (0 ,0,0 , I , 1,0] (0,0 ,1,2 ,1, 0 ) [0, I , 3 , 3 , I , OJ (1 , 4 , 6, 4 , 1 , OJ [0, 0 , 0 , 0, 0, 1J [0, 0 , 0 ,0 , I , 1) 10 ,0,0, 1, 2 , 1J 10 , 0 , 1, 3,3, IJ [0 , 1, 4,6,4 , 1J [1, 5 , 10 , 10 , 5 , I ] The vector table associat ed to Ps has one degree of freedom. The vector in position (3,5) is freely chosen to be x + c x 2 • Then for all other points in Ps \ P6 we can solve a rank 2 equation which determines the corresponding vecto r in terms of c.

Notice also how to read u, v, and w from PI, . . , P4 : Pi has one less row than PHI ; listing these excised rows from right to left yields u. Similarly, listing the excised columns from right to left yields v, and listing the excised numbers from right to left yields w (see the example immediately above). , E;, E~. It will be notationally convenient to represent a vector v = (VI, .. + vnx n- l . We choose three flags, or equivalently three "t ransverse" ordered bases, as follows: E! = E~ = E; (1,x,x 2,x3) (x3, x 2 , X , 1) ((x + 1)3, (x + 1)2, (x + 1),1).