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?

 

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7 thoughts on “Trace test

  1. Jason Starr says:

    I think that I misunderstood something. Begin with F(x,y) = (x^3+x^2-y^2)x. The extra x factor is just there to insure that there is a nontrivial factorization, which I believe was part of the setup. There are power series roots r_1 = y + s_1 and r_2 = -y + s_2. These are roots of the first factor for F(x,y). The sum of these roots is not linear in y.

  2. JSE says:

    Crap, I messed this up somehow, I started writing the post thinking about homogeneous forms and then changed it. Give me a sec and I’ll work this out.

  3. JSE says:

    Wait now I’m confused again about whether I was confused. The first factor x^3 + x^2 – y^2, considered as a polynomial in x, has three roots, whose sum is -1, right? I don’t know if I get which two roots you mean.

  4. Jason Starr says:

    @JSE. Maybe I misunderstood your post. I thought that we are to sum up roots in pairs. For the polynomial $x^3+x^2-y^2$, the “pair sums” of roots are, to order $y^3$, $y+s_1 + y^2$, $-y+s_2+y^2$, and $s_1+s_2$. If we sum up all three roots, we get the coefficient of $x^2$, which is $1$. In general, if we assume that the polynomial has total degree $n$ and that the coefficient of $x^n$ is nonzero (so equal to $1$ after scaling), then certainly the coefficient of $x^{n-1}$ has degree zero or one in $y$, or else the degree is strictly larger than $n$. Maybe that is the point about root sums being linear.

  5. Jason Starr says:

    Typo correction: Those first two pair sums should have constant term $-1$.

  6. Jason Starr says:

    Now I believe that I understand. The example in your post is for an irreducible quadratic factor. The analogous result holds for a proper irreducible factor of any degree, but you have to consider a partial sum of roots with more than two summands. Let $F(x,y)$ be an (inhomogeneous) polynomial of total degree $n$ and such that the coefficient of $x^n$ is nonzero. Via Puiseux series, we can factor $F(x,y)$ as $(x-r_1(y))…(x-r_n(y))$. If $F(x,y)$ has a nontrivial factorization in $k[x,y]$, then there exists a nonempty proper subset of these roots, say $r_1(y),…,r_m(y)$ after reordering, whose partial sum $r_1(y)+…+r_m(y)$, as a Puiseux series, is actually an (affine) linear polynomial in $y$. Thus, if every nonempty, proper partial sum is not (affine) linear in $y$, then $F(x,y)$ is irreducible. This is straightforward to prove using elementary algebra via the observation I made in my previous comment: for both $F(x,y)$ and every irreducible factor of $F(x,y)$, since the $x$-degree equals the total degree, the “next-to-leading” coefficient in $x$ is an (affine) linear polynomial in $y$. Note, this sufficient criterion for irreducibility is not necessary: the polynomial $(x-y)^4-y^3$ is irreducible, but there are pair sums that are affine linear in $y$.

  7. JSE says:

    Exactly! It’s about all partial sums, not just pairwise sums. It’s a test for whether a subset of the roots could be the set of roots of an irreducible algebraic factor.

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