TASK 3: Theoretical thoughts

On this page we collect observations about our problem. Here is a list of things to look at (not necessarily in this order):

• What does the preprint of Mingmin Shen (on the reading page say in our language?
• Can we reprove Mingmin's result using simple techniques with graded modules?
• Given a morphism φ as in Exercise 37 define an R-bilinar pairing E_X(φ) × Ω(φ) —> R
• Use the pairing and Exercise 37 to say something intelligent about splitting types
• Formulate conjectures about splitting types and prove them
• Add your own ideas here

Using the Hilbert polynomial, it's not difficult to see that the sum of the component e_i for the splitting type of the extended tangent bundle must equal the degree of the morphism. Also, as we proved in class, the sum of the component e_i for the splitting type of the cotangent bundle must equal -6d for a degree d morphism.

We will upload to this page clearly written short notes addressing various issues:

• Solution to Exercise 37: (Not completely proven) ∀ e_i ∈ Ω(φ), ∃ (e_i)' ∈ E_X(φ) given by (e_i)' = 4e_i + 5d, where d is the degree of the morphism.

Here is an R- bilinear pairing H: E_X(φ) × Ω(φ) —> R:

• Let [(A_0,…,A_5), (B_0,…,B_5)] be an element of E_X(φ) × Ω(φ). Then define H[(A_0,…,A_5), (B_0,…,B_5)] = A_0B_0 + … + A_5B_5. Let's check that this is R-bilinear:
• If (C_0,…,C_5) ∈ E_X(φ), then H[(A_0,…,A_5)+(C_0,…,C_5),(B_0,…,B_5)] = H[(A_0 + C_0,…,A_5+C_5),(B_0,…,B_5)] = (A_0 + C_0)B_0 + … +(A_5 + C_5)B_5 = (A_0B_0 + … + A_5B_5) + (C_0B_0 + … + C_5B_5) = H[(A_0,…,A_5),(B_0,…,B_5)] + H[(C_0,…,C_5),(B_0,…,B_5)].
• If f ∈ R, then H[f(A_0,…,A_5),(B_0,…,B_5)] = H[(fA_0,…,fA_0),(B_0,…,B_5)] = fA_0B_0 + … + fA_5B_5 = f(A_0B_0 + … + A_5B_5) = fH[(A_0,…,A_5), (B_0,…,B_5)].
• Similarly one can check linearity in the other coordinate.
• As Prof. de Jong mentioned, this pairing is a graded pairing.
• Non-degeneracy: In general this bilinear pairing will not be degenerate, in the sense that H does not map everything to 0. Consider the morphism φ = (S^2, S^2, ST, ST, T^2, T^2). Then (T^2, 0, ST, 0, 0, 0) ∈ Ω(φ), and as we verified in class, (T^8, 0, S^4T^4, 0, 0, 0) ∈ E_X(φ). However, H[(T^8, 0, S^4T^4, 0, 0, 0),(T^2, 0, ST, 0, 0, 0)] = T^10 + S^5T^5 which is not the zero polynomial in R.
• I typed up some notes about Bilinear Pairings, just to recall some results and definitions. The last result in this note (in bold) could help us in proving that the bilinear pairing H is nondegenerate. The notes use localization, and I apologize for that because we have not done localization in class. In particular the rank of the matrix of the bilinear pairing is the same as the rank of the matrix over the field of fractions of R. Bilinear Pairings
• The Matrix of the bilinear pairing H. There might be a clever way to see why this matrix has nonzero determinant. Matrix of H
• Alas, the bilinear form H is degenerate, because ker(Ω(φ) —–> Hom_R(E_X(φ),R)) = <(G_0^4,…,G_5^4)>.

Relationship between splitting types and (very) freeness: Let f_1, …, f_5 be the splitting type of E_X(φ). Then we have:

• If f_1 > 0, …, f_5 > 0, then φ is very free
• If f_1 ≥ 0, …, f_5 ≥ 0, then φ is free
• If φ is free, then f_1 ≥ 0, …, f_5 ≥ 0
• If φ is very free, then either (1) f_1 > 0, …, f_5 > 0, or (2) exactly one f_i = 0 and the others are all > 0. I am not convinced that the second case can really happen, but I do not know how to prove that it doesn't.
• Define an E-free curve as one all of whose f_i ≥ 0 and E-very free curve as one whose f_i are all > 0. Then Mike's arguments apply to E-freeness and E-very freeness, which we see from the equivalences above is almost the same thing as free and very free.

Complete solution of Exercise 37 (Updated). Exercise 37 (-Rankeya)

A minor observation (This is saying nothing new, but I just want to keep it on record): Let d be the degree of the morphism φ. If one of the f_i = d, say without loss of generality f_1, then we must have f_2 + f_3 + f_4 + f_5 = 0. So, if φ is free/ very free, then f_2 = f_3 = f_4 = f_5 = 0. This means that 4e_2 = 4e_3 = 4e_4 = 4e_5 = -5d. The latter equalities are only possible if 4|d. So, if d is not divisible by 4, and one of the f_i = d (which seems to be happening a lot in our case), then φ cannot be free.

Also, this observation shows that there cannot be a very free morphism satisfying condition (2) above, whose degree is not divisible by 4. Moreover, every free morphism of degree not divisible by 4 must be E-very free.

Morphisms of degree d ≤ 4 Let φ be one such morphism. Then the R linear map of graded modules R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d) —–> R induces a k linear map (R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d)⊕ R(-d))_d ——> R_d. This is a k linear map from a vector space of dimension 6 to a vector space of dimension d+1 ≤ 5. Hence, it must have a nonzero kernel. In particular for d ≤ 3, the kernel must have dimension at least 2 (By Rank- Nullity). Hence, there exists e_i, e_j (i =\= j) such that e_i = -d = e_j. Then, f_i = d = f_j. But then the sum of the f_k's cannot equal d without at least one of the f_k's being less than 0. So, for d ≤ 3, there are no free or very free morphisms.

The case d = 4 is more interesting. If the kernel of the k linear map (R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_4 ——> R_4 has dimension more than 2, then by an argument similar to the case for d ≤ 3, the morphism is neither free nor very free. But, R_4 has dimension 5 over k. If the kernel of (R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_4 ——> R_4 has dimension 1 (note the kernel will always have positive dimension), then (R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_4 ——> R_4 is surjective. So, (R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_{4+1} ——> R_{4+1} is a surjective k linear map by Lemma in class. Now, R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_{4+1} has dimension 12, and R_{4+1} has dimension 6. So, the kernel has dimension 6. The generator of the previous kernel gives 2 generators of this kernel. So, we get 4 more generators in this kernel. This gives us a total of 5 generators, and so we have e_1 = e_2 = e_3 = e_4 = -4-1 and e_5 = -4. So, f_1 = f_2 = f_3 = f_4 = 0, and f_5 = 4. So, the morphism is free, but not very free.

What this shows us is that for d = 4, the morphism is never very free.

The morphism is not free if ker(R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_4 ——> R_4) has dim ≥ 2.

If ker(R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4)⊕ R(-4))_4 ——> R_4) has dim 1, then the morphism is free, and has splitting type [0,0,0,0,4].

Observation about the e_i's and f_i's This is more sort of an observation for me to keep in mind which I am just recording here. If other people can make good use of it, great!

We know that e_1 + e_2 + e_3 + e_4 + e_5 = -6d. We also know that e_i ≤ -d, because there is nothing to the left of (R(-d)^6)_d. So,

- There cannot be an e_i such that e_i < -2d. For if there is, say wlog that e_1 < -2d. Then, e_2 + … + e_5 = -6d - e_1 > -4d. But, this is impossible since e_i ≤ -d. So, we always have that -2d ≤ e_i ≤ -d. This implies that -3d ≤ f_i ≤ d.

Possible proof that there are no free degree 5 morphismsThere are no free degree 5 morphisms