Virasoro constraints and polynomial recursion for the linear Hodge integrals

Abstract

The Hodge tau-function is a generating function for the linear Hodge integrals. It is also a tau-function of the KP hierarchy. In this paper, we first present the Virasoro constraints for the Hodge tau-function in the explicit form of the Virasoro equations. The expression of our Virasoro constraints is simply a linear combination of the Virasoro operators, where the coefficients are restored from a power series for the Lambert W function. Then, using this result, we deduce a simple version of the Virasoro constraints for the linear Hodge partition function, where the coefficients are restored from the Gamma function. Finally, we establish the equivalence relation between the Virasoro constraints and polynomial recursion formula for the linear Hodge integrals.

Appendix

1.1 A.1 String and dilaton equations

Here we give a brief argument of the string equation: for $2g-2+n>0$, $1\le j\le g$,

$$\begin{aligned}<\lambda _j \tau _0\prod _{i=1}^n\tau _{d_i}>=\sum _{k=1}^n<\lambda _j \tau _{d_k-1}\prod _{i\ne j}\tau _{d_i}>, \end{aligned}$$

(63)

and dilaton equation: for $2g-2+n>0$, $1\le j\le g$,

$$\begin{aligned}<\lambda _j \tau _1\prod _{i=1}^n\tau _{d_i}>=(2g-2+n)<\lambda _j \prod _{i=1}^n\tau _{d_i}>. \end{aligned}$$

(64)

For the case $m=-1$ in formula (5), we have

$$\begin{aligned}&V_{-1}^{(H)}\cdot \exp (F_H(u,q))\nonumber \\&\quad =\left( L_{-2}+2uL_{-1}+u^2L_0-\frac{u^2}{24}-\sum _{j=1}^{\infty }(-u)^{j-1}j\frac{\partial }{\partial q_j}\right) \cdot \exp (F_H(u,q))=0.\nonumber \\ \end{aligned}$$

(65)

Observe that, for the polynomial $\phi _k(z)$, we have

$$\begin{aligned} \phi _k(-1) = {\left\{ \begin{array}{ll} -1 &{} k=0 \\ 0 &{} k\ge 1 . \end{array}\right. } \end{aligned}$$

Then,

$$\begin{aligned} \sum _{j=1}^{\infty }(-u)^{j-1}j\frac{\partial }{\partial q_j} \widetilde{\phi _{k}}(u,q) = {\left\{ \begin{array}{ll} 1 &{} k=0 \\ 0 &{} k\ge 1 . \end{array}\right. } \end{aligned}$$

Taking into account Eq. (2), we can obtain the following equation from (65): for $2g-2+n>0$,

$$\begin{aligned}&\sum _{\begin{array}{c} 1\le j\le g\\ d_i\ge 0 \end{array}}(-1)^j<\lambda _j \tau _0\prod _{i=1}^n\tau _{d_i}>u^{2j}\prod _{i=1}^n\widetilde{\phi _{d_i}}(u,q) \nonumber \\&\quad =\sum _{\begin{array}{c} 1\le j\le g\\ d_i\ge 0 \end{array}}\sum _{k=1}^n(-1)^j<\lambda _j \tau _{d_k-1}\prod _{i\ne j}\tau _{d_i}>u^{2j}\prod _{i=1}^n\widetilde{\phi _{d_i}}(u,q). \end{aligned}$$

Since $\{\widetilde{\phi _{k}}(u,q)\}$ is a linear independent set for $k\ge 0$, the above equation immediately implies the string Eq. (63).

Next we explain how Eq. (7) gives us the dilaton equation. Let

$$\begin{aligned} c_n=\sum _{k=0}^{\infty }(-1)^k\left( {\begin{array}{c}k+2\\ k\end{array}}\right) u^{k}\frac{\partial }{\partial q_{k+3}}\cdot \widetilde{\phi _n}(u,q). \end{aligned}$$

The cases $n=0$ and $n=1$ are obvious. For $n\ge 2$, we notice that

$$\begin{aligned} c_n=[z^{0}] \left\{ \frac{z^3}{(1+z)^3}\left( \phi _n(z^{-1})\right) \right\} . \end{aligned}$$

Then, from Eq. (42), we have

$$\begin{aligned} c_n= {\left\{ \begin{array}{ll} 1 &{} n=1 \\ 0 &{} n\ne 1 . \end{array}\right. } \end{aligned}$$

Hence, using Eq. (49) and the above result, we can obtain the following equation from Eq. (7): for $2g-2+n>0$,

$$\begin{aligned}&\sum _{\begin{array}{c} 1\le j\le g\\ d_i\ge 0 \end{array}}(-1)^j<\lambda _j\tau _1\prod _{i=1}^n\tau _{d_i}>u^{2j}\prod _{i=1}^n\widetilde{\phi _{d_i}}(u,q)\\&\quad =(2g-2+n)\sum _{\begin{array}{c} 1\le j\le g\\ d_i\ge 0 \end{array}}(-1)^j<\lambda _j\prod _{i=1}^n\tau _{d_i}>u^{2j}\prod _{i=1}^n\widetilde{\phi _{d_i}}(u,q). \end{aligned}$$

This proves the dilaton equation Eq. (64).

1.2 A.2 Some technical results

We recall a method introduced in [18]. For $k=1,2,\dots$, let

$$\begin{aligned} \mathcal {D}_{-k}=z^{1+k}\frac{\mathrm {d}}{\mathrm {d}z}-kz^k, \quad A(u)=-\sum _{k=1}^{\infty }a_ku^k\mathcal {D}_{-k}. \end{aligned}$$

Then

$$\begin{aligned} \exp (A(u))=\exp \left( -\sum a_ku^kz^{1+k}\frac{\partial }{\partial z}\right) \exp (g(uz)). \end{aligned}$$

It is easy to verify that, for $m,n>0$, $[\mathcal {D}_{-m},\mathcal {D}_{-n}]=(n-m)\mathcal {D}_{-m-n}.$ Then there exists a unique sequence of numbers $\{d_n\}$, such that

$$\begin{aligned} \frac{\partial }{\partial u}e^{A(u)}=\left( \sum _{n=1}^{\infty }d_nu^{n-1}\mathcal {D}_{-n}\right) e^{A(u)}. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \frac{\partial }{\partial u}e^{A(u)}= \left( \sum _{n=1}^{\infty }d_nu^{n-1}z^{1+k}\frac{\partial }{\partial z}+ \exp \left( -\sum _{k=1}^{\infty }a_ku^kz^{1+k}\frac{\partial }{\partial z}\right) \cdot \frac{\partial }{\partial u} g(uz) \right) e^{A(u)}. \end{aligned}$$

This gives us

$$\begin{aligned} \frac{\partial }{\partial u} g(uz)=\exp \left( \sum _{k=1}^{\infty }a_ku^kz^{1+k}\frac{\partial }{\partial z}\right) \cdot \left( -\sum _{n=1}^{\infty } d_nu^{n-1}nz^n\right) . \end{aligned}$$

And

$$\begin{aligned} z\frac{\mathrm {d}}{\mathrm {d}z}g(z)=\exp \left( \sum a_kz^{1+k}\frac{\mathrm {d}}{\mathrm {d}z}\right) \cdot \left( -\sum _{n=1}^{\infty }d_{n}nz^{n}\right) . \end{aligned}$$

(66)

We refer the readers to [18] for more details about the above argument.

Lemma 15

$$\begin{aligned} \phi _{n}(z) =\sum _{i=0}^n(-1)^i(2n-2i-1)!!C_i\left( f^{2n-2i+1}\right) _{+}. \end{aligned}$$

Proof

First, we know that $z=(f)_{+}$. By induction, assume that, for $n=k$,

$$\begin{aligned}\phi _{k}(z)=\sum _{i=1}^k(-1)^i(2k-2i-1)!!C_i(f^{2k-2i+1})_{+}.\end{aligned}$$

When $n=k+1$, we have, by Eq. (40),

$$\begin{aligned} \phi _{k+1}(z)&= (2k+1)!!\left( f^{2k+3}\right) _{+}-\sum _{i=1}^{k+1} C_i\phi _{k+1-i}(z)\\&=(2k+1)!!\left( f^{2k+3}\right) _{+}-\sum _{n=1}^{k+1}(2k-2n+1)!!\left( f^{2k-2n+3}\right) _{+}\\&\quad \times \sum _{j=0}^{n-1}(-1)^jC_jC_{n-j}\\&=\sum _{n=0}^{k+1}(-1)^n(2k-2n+1)!!C_n\left( f^{2k-2n+3}\right) _{+}. \end{aligned}$$

This completes the proof. $\square$

1.3 A.3 More proof of Proposition 14

We consider the correspondence $L_n\rightarrow l_n, M_k\rightarrow m_k, n\partial q_n\rightarrow z^n$, where

$$\begin{aligned} l_n&= -z^{1+n}\frac{\mathrm {d}}{\mathrm {d}z}-\frac{1}{2}nz^n-z^n\\ m_k&=\frac{1}{2}z^k \left( z\frac{\mathrm {d}}{\mathrm {d}z}+\frac{1}{2}\right) ^2+\frac{(k+1)}{2}z^k\left( z\frac{\mathrm {d}}{\mathrm {d}z}+\frac{1}{2}\right) + \frac{1}{12}(k+1)(k+2)z^k. \end{aligned}$$

Then

$$\begin{aligned}&\left[ \frac{1}{n}z^n,l_k\right] =z^{n+k} \quad \text{ and }\quad \left[ \frac{1}{n}z^n,m_k\right] =l_{n+k},\\&[l_m,l_n]=(m-n)l_{m+n}+\frac{m^3-m}{12}\delta _{m+n,0},\\&[l_n,m_k]=(2n-k)m_{n+k}+\frac{n^3-n}{12}z^{n+k}. \end{aligned}$$

These formulas agree with the formula presented in Sect. 4.2. Let

$$\begin{aligned} \Phi _z^{+} =\exp \left( \sum _{m=1}^{\infty }a_mz^{1+m}\frac{\mathrm {d}}{\mathrm {d}z}\right) . \end{aligned}$$

Then,

$$\begin{aligned} \exp \left( \sum _{m=1}^{\infty }a_m \left( -z^{1+m}\frac{\mathrm {d}}{\mathrm {d}z}-\frac{1}{2}mz^m-z^m\right) \right)&= e^{-\Phi _z^{+}}\frac{1}{(1+h)}=\frac{1}{1+z}e^{-\Phi _z^{+}}. \end{aligned}$$

Now, by the definition of $m_k$, we have

$$\begin{aligned} m_{-4}=\frac{1}{2}z^{-4}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) ^2-z^{-4}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) -\frac{1}{8}z^{-4}. \end{aligned}$$

From Eq. (6), we can see that

$$\begin{aligned} \frac{1}{1+z}e^{-\Phi _z^{+}}m_{-4}e^{\Phi _z^{+}}(1+z)=m_0+4m_{-1}+6m_{-2}+4m_{-3}+m_{-4}+Q(z), \end{aligned}$$

where Q(z) is a polynomial. To compute Q(z), we first compute

$$\begin{aligned}&\frac{1}{1+z}\frac{1}{2}\frac{(1+z)^4}{z^4} \left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) ^2 (1+z)\\&\quad = \frac{1}{2}\frac{(1+z)^4}{z^4}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) ^2+ \frac{(1+z)^3}{z^3}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) +\frac{(1+z)^3}{2z^3}. \end{aligned}$$

Then we have

$$\begin{aligned}&e^{-\Phi _z^{+}}m_{-4} e^{\Phi _z^{+}} =\frac{1}{2} \frac{(1+z)^4}{z^4}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) ^2-\frac{(1+z)^3}{z^4}\left( z\frac{\mathrm {d}}{\mathrm {d}z}\right) -\frac{1}{8}\eta ^{-4}. \end{aligned}$$

Hence

$$\begin{aligned}&\frac{1}{1+z}e^{-\Phi _z^{+}}m_{-4} e^{\Phi _z^{+}}(1+z) \\&\quad =m_0+4m_{-1}+6m_{-2}+4m_{-3}+m_{-4}+\frac{1}{8}z^{-4}+\frac{1}{3}z^{-3}+\frac{1}{4}z^{-2}-\frac{1}{24}-\frac{1}{8}\eta ^{-4}. \end{aligned}$$

This shows that

$$\begin{aligned} Q(z)=\frac{1}{8}z^{-4}+\frac{1}{3}z^{-3}+\frac{1}{4}z^{-2}-\frac{1}{24}-\frac{1}{8}\eta ^{-4}. \end{aligned}$$

And therefore,

$$\begin{aligned} e^U\left( \frac{1}{8}q_4\right) e^{-U}=\frac{1}{8}\sum _{i=-4}^{\infty } [z^{i}](\eta ^{-4})u^{i+4}\alpha _{i}. \end{aligned}$$

Remark

It is definitely possible to prove Proposition 14 directly using the operators $l_n$ and $m_k$. But we think the use of power series like (66) and (6) simplifies a lot of computations, while the direct computation still needs to take care of the nested commutators of differential operators involving $l_n$ and $m_k$.