Self-adjoint commuting ordinary differential operators

Abstract

In this paper we study self-adjoint commuting ordinary differential operators of rank two. We find sufficient conditions when an operator of fourth order commuting with an operator of order 4g+2 is self-adjoint. We introduce an equation on potentials V(x),W(x) of the self-adjoint operator $L=(\partial_{x}^{2}+V)^{2}+W$ and some additional data. With the help of this equation we find the first example of commuting differential operators of rank two corresponding to a spectral curve of higher genus. These operators have polynomial coefficients and define commutative subalgebras of the first Weyl algebra.

Appendix

In this Appendix we study (23).

Lemma 5

There exists a Zariski open set U⊂C ⁴ such that for every (α ₀,…,α ₃)∈U polynomial F _g (z) has no multiple roots of order greater than two.

Proof

To prove the lemma we represent F _g in the form

$$F_g(z)=F_g^0(z)+\alpha_3F_g^1(z)+O\bigl(\alpha_3^2\bigr), $$

and prove that $F_{g}^{0}(z)$ has g double roots and one simple root.

Let us consider (20)–(22). We put $\delta_{g}=\alpha_{3}^{g}$, then

$$\delta_{g-1}=\alpha_3^{g-1}\frac{\alpha_2g^2+z}{2g-1}. $$

Moreover, from (20) it follows that Q has the form

$$\begin{aligned} Q =&\alpha_3^gx^g+\dots+ \alpha_3^sx^s\bigl(p_{s}+\alpha_3q_{s}+O\bigl(\alpha_3^2\bigr)\bigr)+\cdots \\ &{}+\bigl(p_{0}+\alpha_3q_{0}+O\bigl(\alpha_3^2\bigr)\bigr) \end{aligned}$$

(24)

for some p _i =p _i (z),q _i =q _i (z). Let us note that from (22) it follows that

$$p_g=1,\qquad p_{g-1}=\frac{\alpha_2g^2+z}{2g-1},\qquad q_g=0,\qquad q_{g-1}=0. $$

Let us substitute (24) into (15). We get

$$F_g(z)=p_0^2(z)z+\alpha_3p_0(z)\bigl(\alpha_1p_1(z)+2q_0(z)z\bigr)+O\bigl(\alpha_3^2\bigr), $$

so,

$$F_g^0(z)=p_0^2(z)z,\qquad F_g^1(z)=p_0(z)\bigl(\alpha_1p_1(z)+2q_0(z)z\bigr). $$

Let us find p _i . For this we again substitute (24) into (19) and find the coefficients at $\alpha_{3}^{i+1}x^{i}$. These coefficients must be equal to zero. This gives us

$$p_i=\frac{2(i+1)(\alpha_2(i+1)^2+z)}{(2i+1)(g^2+g-i^2-i)}p_{i+1},\quad 0\leq i\leq g-1. $$

Hence

$$p_i(z)=\bigl(\alpha_2(i+1)^2+z\bigr)\cdots\bigl(\alpha_2g^2+z\bigr)A_i,\quad 0\leq i\leq g-1, $$

where A _i is a constant. Lemma 5 is proved. □