Kadomtsev–Petviashvili Turning Points and CKP Hierarchy

Abstract

A characterization of the Kadomtsev–Petviashvili hierarchy of type C (CKP) in terms of the KP tau-function is given. Namely, we prove that the CKP hierarchy can be identified with the restriction of odd times flows of the KP hierarchy on the locus of turning points of the second flow. The notion of CKP tau-function is clarified and connected with the KP tau function. Algebraic–geometrical solutions and in particular elliptic solutions are discussed in detail. A new identity for theta-functions of curves with holomorphic involution having fixed points is obtained.

Appendices

Appendix A: Proof of Lemma 2.1

In this appendix we give a sketch of proof of Lemma 2.1, i.e. we are going to prove that the conditions (2.53) and

$$\begin{aligned} \partial _x\partial _{t_4}\log \tau ^{\mathrm{KP}}\Bigr |_{\mathbf{t}_{\mathrm{e}}=0} =\partial _x\partial _{t_6}\log \tau ^{\mathrm{KP}}\Bigr |_{\mathbf{t}_{\mathrm{e}}=0} =\ldots =0. \end{aligned}$$

(A1)

follow from the constraint

$$\begin{aligned} \partial _{t_2}\log \tau ^{\mathrm{KP}}\Bigr |_{\mathbf{t}_{\mathrm{e}}=0}=0 \quad \text{ for } \text{ all } t_1, t_3, t_5, \ldots \end{aligned}$$

(A2)

(see (2.48)) provided $\tau ^{\mathrm{KP}}$ is a KP tau-function, i.e. satisfies all the equations of the KP hierarchy.

We use the representation of the KP hierarchy in the unfolded form suggested in [36, 37], see also section 3.2 of [38]. Set $F=\log \tau ^{\mathrm{KP}}$ and $F_{k_1,\ldots ,\, k_m}=\partial _{t_{k_1}}\ldots \partial _{t_{k_m}}\! F$. Then the KP hierarchy can be written in the form

$$\begin{aligned} F_{k_1,\ldots , \, k_m}=\sum _{n\ge 1}\sum R_{k_1 ,\ldots ,\, k_m}^{(n)} \! \left( \begin{array}{lll} s_1 &{} \ldots &{} s_n \\ r_1 &{} \ldots &{} r_n \end{array} \right) \partial _x^{r_1}F_{s_1}\ldots \partial _x^{r_n}F_{s_n}, \end{aligned}$$

(A3)

where $m\ge 2$ and $\displaystyle {R_{k_1, \ldots ,\, k_m}^{(n)} \! \left( \begin{array}{lll} s_1 &{} \ldots &{} s_n \\ r_1 &{} \ldots &{} r_n \end{array} \right) }$ are universal rational coefficients. The second sum is taken over all matrices $\displaystyle {\left( \begin{array}{lll} s_1 &{} \ldots &{} s_n \\ r_1 &{} \ldots &{} r_n \end{array} \right) }$ such that $s_i, r_i \ge 1$ with the conditions

$$\begin{aligned} \sum _{i=1}^n (s_i+r_i)=\sum _{i=1}^m k_i, \qquad \sum _{i=1}^n r_i \ge n+m-2. \end{aligned}$$

(A4)

For example [36],

$$\begin{aligned} F_{2,3}=\frac{3}{2}\, \partial _xF_4 -\frac{3}{2}\, \partial _x^3F_2 -3\partial _x F_2 \, \partial _x^2F. \end{aligned}$$

(A5)

From the fact that if $\tau ^{\mathrm{KP}}(x, \mathbf{t})$ is a tau-function, then $\tau ^{\mathrm{KP}}(-x, -\mathbf{t})$ is a tau-function, too (this is a corollary of the Hirota equations), it follows that

$$\begin{aligned} \text{ if } \displaystyle {\sum _{i=1}^n (r_i-1)-m\equiv 1}\hbox { (mod 2), then }\displaystyle {R_{k_1, \ldots ,\, k_m}^{(n)} \! \left( \begin{array}{lll} s_1 &{} \ldots &{} s_n \\ r_1 &{} \ldots &{} r_n \end{array} \right) =0.} \end{aligned}$$

(A6)

First we prove (A1). The proof is by induction. We assume that (A1) is true for $\partial _xF_{2}, \ldots , \partial _xF_{2k}$ (this is certainly true if $k=1$) and will deduce from (A3) that it is true for $k\rightarrow k+1$. From (A2) and (A3) at $m=2$ we have (at $\mathbf{t}_{\mathrm{e}}=0$):

$$\begin{aligned} \begin{array}{c} \displaystyle { 0=F_{2, \, 2k+1}=\sum _{s_1+r_1=2k+3}R^{(1)}_{2, \, 2k+1} \left( \begin{array}{c} s_1\\ r_1\end{array} \right) \partial _x^{r_1}F_{s_1}} \\ \\ \displaystyle {+ \sum _{s_1+s_2+r_1+r_2=2k+3}R^{(2)}_{2, \, 2k+1} \left( \begin{array}{cc} s_1&{}s_2\\ r_1&{}r_2\end{array} \right) \partial _x^{r_1}F_{s_1}\partial _x^{r_2}F_{s_2}+\ldots } \end{array} \end{aligned}$$

(A7)

Separating the term with $r_1=1$ in the first sum in the right hand side of (A7), we write it as

$$\begin{aligned} 0=F_{2, \, 2k+1}=R^{(1)}_{2, \, 2k+1} \left( \begin{array}{c} 2k+2\\ 1\end{array} \right) \partial _x F_{2k+2} \,\, +\; \text{ all } \text{ the } \text{ rest }. \end{aligned}$$

(A8)

Now, recalling the condition (A6), we see that the non-zero coefficients at the different terms in the right hand side are when $\displaystyle {\sum _{i=1}^n s_i =n-1 \; \text{(mod } \text{2) }}$. From this it follows that for both odd and even n at least one of the $s_i$’s must be even (and less then $2k+2$). Therefore, “all the rest” terms vanish by the induction assumption. Since the coefficient $\displaystyle {R^{(1)}_{2, \, 2k+1} \left( \begin{array}{c} 2k+2\\ 1\end{array} \right) }$ is not equal to zero (see [36]), we conclude from (A8) that $\partial _x F_{2k+2}=0$.

Next we are going to prove that if $\partial _xF_{2k}=0$ for all $k\ge 1$ and all $t_1, t_3, \ldots$, then $F_{k_1,\ldots ,\, k_m}=0$ for all even $k_1, \ldots , k_m$ and odd $m\ge 3$. As soon as $m+1$ and all $k_i$’s are even, we can, using (A4), rewrite the condition (A6) in the form

$$\begin{aligned} \sum _{i=1}^n s_i \equiv n \,\,\, \text{(mod } \text{2) }. \end{aligned}$$

(A9)

But if at least one of $s_i$ in (A3) is even, then the corresponding term vanishes because $F_{2k}=0$ for all $k\ge 1$. Therefore, all the $s_i$’s must be odd, i.e., $s_i=2l_i+1$ and so the condition (A9) is satisfied which means that the coefficient $\displaystyle {R_{k_1, \ldots ,\, k_m}^{(n)} \! \left( \begin{array}{lll} s_1 &{} \ldots &{} s_n \\ r_1 &{} \ldots &{} r_n \end{array} \right) }$ vanishes. This proves that $F_{k_1, \ldots ,\, k_m}=0$.

Appendix B: Proof of Eq. (4.21)

Here we prove the matrix identity (4.21).

First of all we note that $\dot{L}_{ik}=-(\dot{x}_i-\dot{x}_k)\Phi '(x_i-x_k)$, and, therefore, we have $\dot{L} =-[\dot{X}, B]$. To transform the commutators $[L,B]+[L,D]$, we use the identity

$$\begin{aligned} \Phi (x )\Phi '(y)-\Phi (y)\Phi '(x)=\Phi (x+y)(\wp (x) -\wp (y)). \end{aligned}$$

(B1)

With the help of it we get for $i\ne k$

$$\begin{aligned}&-\Bigl ([L,B]+[L,D]\Bigr )_{ik} \\&\quad =\, \sum _{j\ne i,k}\Phi (x_i-x_j)\Phi '(x_j-x_k)- \sum _{j\ne ,k}\Phi ' (x_i-x_j)\Phi (x_j-x_k) \\&\qquad +\, \Phi (x_i-x_k)\Bigl (\sum _{j\ne k}\wp (x_j-x_k)-\sum _{j\ne i}\wp (x_i-x_j)\Bigr )=0, \end{aligned}$$

so we see that $[L,B]+[L,D]$ is a diagonal matrix. To find its matrix elements, we use the limit of (B1) at $y=-x$:

$$\begin{aligned} \Phi (x)\Phi '(-x)-\Phi (-x)\Phi '(x)=\wp '(x) \end{aligned}$$

which leads to

$$\begin{aligned}&-\Bigl ([L,B]+[L,D]\Bigr )_{ii} \\&\quad =\, \sum _{j\ne i}\Bigl (\Phi (x_i-x_j)\Phi ' (x_j-x_i)-\Phi ' (x_i-x_j)\Phi (x_j-x_i)\Bigr ) =\sum _{j\ne i}\wp '(x_i-x_j)=D'_{ii}, \end{aligned}$$

so we finally obtain the matrix identity

$$\begin{aligned}{}[L,B]+[L,D]=-D'. \end{aligned}$$

(B2)

Combining the derivatives of (B1) w.r.t. x and y, we obtain the identity

$$\begin{aligned} \Phi (x)\Phi ''(y)-\Phi (y)\Phi ''(x)=2\Phi '(x+y)(\wp (x)-\wp (y)) +\Phi (x+y)(\wp '(x)-\wp '(y))\nonumber \\ \end{aligned}$$

(B3)

which allows us to prove the matrix identity

$$\begin{aligned}{}[L,C]=-2[D, B]+D'L+LD', \end{aligned}$$

(B4)

which is used, together with (B2), to transform $\dot{L}+[L,M]$ to the form (4.21).