TASK 2: Computing splitting types

On this page we collect ideas, algorithms, scripts, etc related to finding splitting types of free graded modules over R = K[S, T], especially the modules Ω(φ) for morphisms as in Exercise 37.

We will upload to this page

splittingcalculator.gp Fully functional program for computing the splitting type of Ω(φ) or E_x(φ) for a given morphism into P^5. Will work on editing for clarity and efficiency.

splittingtype_calculator.rtf Small update on Joe's program, doesn't ask for the degree and computes the splitting type for the extended tangent bundle via the unverified formula for it in terms of the cotangent bundle.

A few rational curves I've found along with their splitting types - I searched for ones that had no two components equal:

G = (S^5, S^4*T + T^4*S, T^5, S^5 + T*S^4 + T^5, S^5 + T^4*S + T^5, S^5 + T*S^4 + T^4*S + T^5) has splitting type -5,-5,-6,-6,-8. If the answer to Exercise 37 is correct, this means the splitting type of E_X(φ) is 5, 5, 1, 1, -7.

G = (S^5, S^4*T +T^4*S, S^5 + T^5, T*S^4 + T^5, T^4*S + T^5, T*S^4 + T^4*S + T^5) has splitting type -5,-5,-6,-6,-8. If the answer to Exercise 37 is correct, this means the splitting type of E_X(φ) is 5, 5, 1, 1, -7.

G = (T*S^4, T^2*S^3, S^5 + T^2*S^3, T^4*S, S^3*T^2 + T^5, S^5 + T^2*S^3 + T^4) has splitting type -5,-6,-6,-6,-7. If the answer to Exercise 37 is correct, this means the splitting type of E_X(φ) is 5, 1, 1, 1, -3.

G = (T*S^4, S^4*T + T^2*S^3, S^5 + T*S^4 + T^2*S^3, T^4*S, T*S^4 + T^2*S^3 + T^5, S^5 + T*S^4 + T^2*S^3 + T^5) has splitting type -5,-6,-6,-6,-7. If the answer to Exercise 37 is correct, this means the splitting type of E_X(φ) is 5, 1, 1, 1, -3.

I'm getting more…I'll figure out a systematic way to display these tomorrow…

Musings If our results for the final few exercises are correct, then the splitting types in question must squeeze between two bounds, namely each e_i must be greater than or equal to -(5/4)d and the sum of the e_i must be equal to -6d where in both cases d is the degree of the morphism (for e_i generating the cotangent bundle). This restricts the types of splitting types we might expect, though knowing this ahead of time might not be helpful.

More usefully, this bound gives us that free curves can exist in degrees 4,8,12, and 16 and very free curves can exist in degrees 5,9,10,13,14, and 15. Moreover, very free curves can exist in every degree greater than 16. So at least we have some idea of where to start looking.

In general it seems that degrees divisible by 4 are best for searching for free rational curves, and degrees congruent to 1 (mod 4) are best for searching for very free rational curves, insofar as they generally allow for a higher number of nonnegative respectively positive possible splitting types.

Some More Thoughts Given that (G_0,…,G_5) is a morphism of degree d on the fermat X, it seems that G_0^(2^n),…,G_5^(2^n) is a morphism of degree d * 2^n on X. This is because ^(2^n) is a ring homomorphism and for a in F_2, a^(2^n) = a, so G_i^(2^n) is just G_i with degree of S and T multiplied by (2^n) in each term. Consider the following examples:

find_splittingTypeE(S^10,T^2*S^8 + T^8*S^2,T^10,S^10 + T^2*S^8 + T^10,S^10 + T^8 * S^2 + T^10,S^10 + T^2*S^8 + T^8*S^2 + T^10,10) %6 = [10, 10, 2, 2, -14] ? find_splittingTypeE(T^2*S^8,T^4*S^6,S^10 + T^4*S^6,T^8*S^2,T^4*S^6 + T^10, S^10 + T^4*S^6 + T^10,10) %7 = [10, 2, 2, 2, -6]

which correspond to these degree 5 results: [S^5, T*S^4 + T^4*S, S^5 + T^5, T*S^4 + T^5, T^4*S + T^5, T*S^4 + T^4*S + T^5] [5, 5, 1, 1, -7] [T*S^4, T^2*S^3, S^5 + T^2*S^3, T^4*S, T^2*S^3 + T^5, S^5 + T^2*S^3 + T^5] [5, 1, 1, 1, -3]

I conjecture that for a given morphism (G_0,…,G_5) with splitting type of E_x given by e_i, the splitting type of the morphism (G_0^(2^n),…,G_5^(2^n)) is given by (2^n)e_i (this holds in the examples). In particular, if this holds then given a (very) free rational curve of degree d, there must exist (very) free rational curves of degree d * 2^n for all positive n. We could use this to eliminate potential degrees: for example, since there are no very free rational curves of degree 12 = 6 * 2, there cannot be very free rational curves of degree 6 (which we already knew). I do not think that working in the opposite direction is possible, but it would be very useful if it is. -Joe

Note that by induction we can reduce the proof of Joe's claim to the case where n = 1. The proof outlined in TASK III for Exercise 37 can be modified to prove Joe's statement for n = 1. So, his conjecture is true. - Rankeya

Splitting Type Data

degree5_someresults.pdf generated by this program:degree5_splittingtypes.txt that uses John's rational curve generator with Joe's splitting type computer

degree4_splittingtypes_.txt: I ran this for 3 1/2 hours straight and found not a single interesting (without duplicate polynomial entries) curve. Maybe we can try to prove that there aren't any? Update: This is worth attempting, because if a non-duplicate curve exists then it is free–its splitting type would start with one 4 and then necessarily have 0s follow. But it seems that all degree 4 curves are composed of 3 duplicate polynomials.

* Adding to Joe's observation, this could a problem that the computer can handle. We just look at all 6*5 matrices over F_2 of maximal rank, and see which of morphisms given by these matrices satisfy the Fermat hypersurface and have gcd 1. What is great is that if one such morphism is found, it will definitely be free due to the observation made in TASK III - Rankeya

degree8_results.pdf: A single non-duplicate curve found after 30,000,000 iterations of factor.gp

a_free_curve_of_degree_8.pdf: A free curve of degree 8!!!

A bogus idea for degree d = 5 (Rankeya)

A polynomial of degree d in F_2[S,T] is just a d+1 tuple of elements of F_2. For d = 5, we have so far gotten splitting types where at least one of the f_i = 5. This means that G_0, …, G_5 are linearly dependent over F_2. If one the f_i = 5, the morphism is neither free nor very free (see discussion on TASK III page regarding degree d not being divisible by 4). Now, the data of a morphism of degree 5 is the same as a 6 by 6 matrix with entries in F_2. So, to prove that very free/ free morphisms of degree 5 exist or don't exist, we first need to come up with G_0, …, G_5 which are linearly independent over k. This is equivalent to listing all the elements of GL_6(F_2), then deciding which of those give us morphisms from P^1 to our Fermat hypersurface. I don't know if PARI can just list all elements of GL_6(F_2), or if listing elements of GL_6(F_2) is computationally feasible. But, this is an idea. The cardinality of GL_6(F_2) is (2^6 - 1)(2^6 - 2)(2^6 - 2^2) (2^6 - 2^3)(2^6 - 2^4)(2^6 - 2^5). Also, the order of the rows do not matter, so we can further divide by a factor of 6! = 720.

Later Edit: Earlier I thought that just finding an element of GL_6(F_2) satisfying the Fermat hypsersurface, might not be enough. But this really is sufficient.

Generalizing bogus idea for degrees d such that 4 does not divide d, where d ≥ 5

In TASK III, I showed that for such d, if one of the f_i = d, then the morphism cannot be free or very free. So, we need to look at those morphisms φ = (G_0,…,G_5) such that G_0, …, G_5 are linearly independent over F_2, as otherwise we will get at least one f_i = d. This first amounts to finding 6*(d+1) matrices over F_2 with maximal rank, and then choosing those polynomial G_0,…,G_5 given by these matrices which satisfy the Fermat hypersurface, and having gcd 1. This might narrow down our search space. Note that this does not guarantee that we will find a free or very free morphism, but I think we can be more selective about which morphisms we need to look at. The cardinality of the set of such matrices is (2^{d+1} - 1)(2^{d+1} - 2)(2^{d+1} - 2^2)(2^{d+1} - 2^3)(2^{d+1} - 2^4)(2^{d+1} - 2^5).

This might have a shot because PARI seems to be quick at handling matrices over F_2, and we can write an efficient algorithm for PARI to display 6*(d+1) matrices of maximal rank, without having to go through all 6*(d+1) matrices over F_2.

task_ii.txt · Last modified: 2012/07/02 12:27 by rankeya
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