Trace test

Jose Rodriguez gave a great seminar here yesterday about his work on the trace test, a numerical way of identifying irreducible components of varieties.  In Jose’s world, you do a lot of work with homotopy; if a variety X intersects a linear subspace V in points p1, p2, .. pk, you can move V a little bit and numerically follow those k points around.  If you move V more than a little bit — say in a nice long path in the Grassmannian that loops around and returns to its starting point — you’ll come back to p1, p2, .. pk, but maybe in a different order.  In this way you can compute the monodromy of those points; if it’s transitive, and if you’re careful about avoiding some kind of discriminant locus, you’ve proven that p1,p2…pk are all on the same component of V.

But the trace test is another thing; it’s about short paths, not long paths.  For somebody like me, who never thinks about numerical methods, this means “oh we should work in the local ring.”  And then it says something interesting!  It comes down to this.  Suppose F(x,y) is a form (not necessarily homogenous) of degree at most d over a field k.  Hitting it with a linear transformation if need be, we can assume the x^d term is nonzero.  Now think of F as an element of k((y))[x]:  namely $F = x^d + a_1(y) x^{d-1} + \ldots + a_d(y).$

Letting K be the algebraic closure of k((y)), we can then factor F as (x-r_1) … (x-r_d).  Each of these roots can be computed as explicitly to any desired precision by Hensel’s lemma.  While the r_i may be power series in k((y)) (or in some algebraic extension), the sum of the r_i is -a_1(y), which is a linear function A+by.

Suppose you are wondering whether F factors in k[x,y], and whether, for instance, r_1 and r_2 are the roots of an irreducible factor of F.  For that to be true, r_1 + r_2 must be a linear function of y!  (In Jose’s world, you grab a set of points, you homotopy them around, and observe that if they lie on an irreducible component, their centroid moves linearly as you translate the plane V.)

Anyway, you can falsify this easily; it’s enough for e.g. the quadratic term of r_1 + r_2 to be nonzero.  If you want to prove F is irreducible, you just check that every proper subset of the r_i sums to something nonlinear.

1.  Is this something I already know in another guise?
2.  Is there a nice way to express the condition (which implies irreducibility) that no proper subset of the r_i sums to something with zero quadratic term?

New SIAM journal on applied algebra

As you know I’m a fan of “applied pure math.”  You can be an applied mathematician and not do PDEs or numerical analysis now!  And to reflect this new reality, there’s a new journal, SIAM Journal on Applied Algebra and Geometry (SIAGA).  They’re accepting submissions starting March 23; I actually have a little something that might be finished by then that would be perfect!

There are a bunch of cool pictures on the SIAGA flyer:  UC-Berkeley grad student Anna Seigal is explaining them on her blog, one by one.  The first entry depicts the algebraic variety you encounter when you try to find a point minimizing the summed distances to k other points in the plane.  Handsome!

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Applied Algebra Days this weekend in Madison

One of the exciting aspects of math at Wisconsin is the new emphasis on what I call “applied pure math” — that is, applied math that doesn’t involve PDEs or numerical analysis.  If you’re in town and want to see what this looks like, you can come to the first Applied Algebra Days conference, featuring Ronny Hadani, Pablo Parrilo, Olga Holtz, and lots of other interesting people.  And as an added bonus it’s in the shiny new Wisconsin Institutes for DiscoverySchedule hereAbstracts here.

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