Constant Q-curvature metrics on conic 4-manifolds

Case 1: Suppose that β ≠ - 1 2 .

Let p ⁢ ( x ) ∈ 𝒫 m . By Euler formula, x i ⁢ D i ⁢ p ⁢ ( x ) = m ⁢ p ⁢ ( x ) . We have

So Δ ⁢ ( p ⁢ ( x ) ⁢ r 4 ⁢ β + 2 ) = f ⁢ ( x ) ⁢ r 4 ⁢ β if

Note if β ∈ ( - 1 , 0 ) and β ≠ - 1 2 , then ( 4 ⁢ β + 2 ) ⁢ ( 4 ⁢ β + 4 + 2 ⁢ m ) cannot be an eigenvalue of r 2 ⁢ Δ . Then, by Lemma 7.1, we see that there exists a p ⁢ ( x ) such that

Now let q ⁢ ( x ) be a homogeneous polynomial with degree m. Then

By applying Lemma 7.1 again, there is a polynomial such that Δ ⁢ ( q ⁢ ( x ) ⁢ r 4 ⁢ β + 4 ) = p ⁢ ( x ) ⁢ r 4 ⁢ β + 2 .

To solve

we first compute

where ϕ ⁢ ( x ) ∈ 𝒫 m - 2 . First, we can solve

as in (A.1). We can also find a polynomial p 2 ⁢ ( x ) such that Δ ⁢ ( p 2 ⁢ ( x ) ⁢ r 4 ⁢ β + 2 ) = ϕ ⁢ ( x ) ⁢ r 4 ⁢ β . Thus,

By a similar argument, there exist q 1 , q 2 such that

hence

This gives the solution g = r 4 ⁢ β + 4 ⁢ ( q 1 ⁢ ( x ) ⁢ log ⁡ r + q 2 ⁢ ( x ) ) for k = 1 in (A.2). We use induction for k ≥ 2 . Suppose that we can find solutions of (7.1) for 0 ≤ l ≤ k - 1 :

where ϕ j ⁢ ( x ) are polynomials. Then we can find p ⁢ ( x ) such that

In the remaining terms ∑ j = 0 k - 1 r 4 ⁢ β ⁢ ϕ j ⁢ ( x ) ⁢ ( log ⁡ r ) j , the degrees of log ⁡ r are strictly smaller than k. Therefore, the remaining terms can be solved by induction and there exist { p l ⁢ ( x ) } such that

Repeat the argument for each f ~ ⁢ ( x ) = p l ⁢ ( x ) ⁢ | x | 2 and we can solve (A.2).

Case 2: β = - 1 2 .

If homogeneous degree m = 0 , we see that Δ 2 ⁢ ( c 16 ⁢ r 2 ⁢ log ⁡ r ) = c ⁢ r - 2 . So this is true for degree 0 polynomial. If m = 1 , direct computation shows that

If m = 2 , we have for i ≠ j ,

For i = j , we compute

Note that Δ 2 ⁢ ( r 4 ⁢ log ⁡ r ) = 7 × 64 + 3 × 128 ⁢ log ⁡ r . Since Δ 2 ⁢ r 4 = 192 , we can still find a solution for Δ 2 ⁢ ( q ⁢ ( x ) ⁢ r 2 ) = x i 2 ⁢ r 2 in the form of (7.2).

For functions in the form of f ⁢ ( x ) ⁢ r - 2 ⁢ ( log ⁡ r ) l in (7.3), we argue by induction with respect to k. Note the above argument is for l = 0 . Suppose that for 0 ≤ l ≤ k - 2 , we have a solution for (7.3). We prove for l = k - 1 . Any quadratic polynomial f ⁢ ( x ) is a linear combination of 1 , x i , x i ⁢ x j , x i 2 . Thus, we only have to consider these four subcases.

Subcase 1: f ⁢ ( x ) = 1 . Take test function c ⁢ r 2 ⁢ ( log ⁡ r ) k . Compute

where c i are polynomials of k and c i = 0 for k = 1 , 2 , … , 4 - i . Let c = 2 - 4 and the first term can be cancelled. The remaining terms f - Δ 2 ⁢ [ c ⁢ r 2 ⁢ ( log ⁡ r ) k ] in (A.3) are lower degrees terms and can be solved by induction.

Subcase 2: f ⁢ ( x ) = x i . We consider a test function of the form c ⁢ x i ⁢ r 2 ⁢ ( log ⁡ r ) k . Direct computation shows

Likewise, take c = ( 48 ⁢ k ) - 1 . Then f ⁢ ( x ) - Δ 2 ⁢ ( c ⁢ x i ⁢ r 2 ⁢ ( log ⁡ r ) k ) has lower degrees in log ⁡ r and we can solve the remaining terms by induction.

Subcase 3: f ⁢ ( x ) = x i ⁢ x j , i ≠ j . By taking out a test function in the form of c ⁢ x i ⁢ x j ⁢ r 2 ⁢ ( log ⁡ r ) k with a proper choice of c, we may reduce the top degree of log ⁡ r . Then the lower degree terms can be solved by induction. The argument is similar to Subcase 2.

Subcase 4: f ⁢ ( x ) = x i 2 . We compute directly

where P and Q are polynomials in one variable and deg ⁡ P ⁢ ( x ) ≤ k - 2 , deg ⁡ Q ⁢ ( x ) ≤ k . Note that the degree of P ⁢ ( x ) is less than k - 1 . Thus, by induction assumption there exists a function P ¯ ⁢ ( x ) in the form of (7.3) such that

For Q ⁢ ( x ) in (A.4), if it has degree k, there is a test function in the form of c ⁢ r 4 ⁢ ( log ⁡ r ) k . We compute

where b i are constants independent of k. Suppose that a k ≠ 0 is the coefficient of the top degree term of Q ⁢ ( x ) . Then

is a polynomial in log ⁡ r with degree less than k. Thus, we can apply the induction assumption to R ⁢ ( log ⁡ r ) to show that there exists a polynomial Q ¯ ⁢ ( x ) with deg ⁡ Q ¯ ⁢ ( x ) ≤ k and

Now let

By (A.4), (A.5) and (A.6),

This completes the last case.