People usually say that the quantum cohomology of $\mathbb{P}^1$ Frobenius manifold $QH^*(\mathbb{P}^1)$, corresponds to dispersionless extended Toda hierarchy (e.g. page 6 of https://arxiv.org/pdf/math/0308152.pdf). I'm trying to understand this correspondence in more detail but I'm some having troubles. More specifically how do I recover the $\tau$-function? Should I be able to get the Frobenius potential $F$ of $QH^*(\mathbb{P}^1)$ from the $\tau$-function by \begin{equation} F = \frac{1}{2}(t^1)^2t^2 + e^{t^2} = \lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau |_{t^{\alpha,p > 0} = 0, t^\alpha := t^{\alpha,0}} ? \end{equation} (Where $\epsilon \rightarrow 0$ is the dispersionless limit. I'm sort of getting this from Theorem 6.1 of https://arxiv.org/pdf/hep-th/9407018.pdf, but the result is for KdV). If I didn't set $t^{\alpha,p > 0} = 0$, should I get the Gromov-Witten potential of $\mathbb{P}^1$ with descendences from $\lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau$?

Here is what I have gathered so far (mainly from https://arxiv.org/pdf/nlin/0306060.pdf), please correct me if I misunderstood something. The extended Toda hierarchy is given by the bihamiltonian structure on the loop space $\mathcal{L}(\mathbb{R}^2)$: \begin{align*} \{u(x),u(y)\}_1 = \{v(x), v(y)\}_1 &= 0, \\ \{u(x), v(y)\}_1 &= \frac{1}{\epsilon}(e^{\epsilon \partial_x} - 1)\delta(x - y),\\ \{u(x), u(y)\}_2 &= \frac{1}{\epsilon}(e^{\epsilon \partial_x}e^{v(x)} - e^{v(x)}e^{-\epsilon\partial_x})\delta(x-y)\\ \{u(x),v(y)\}_2 &= \frac{1}{\epsilon}u(x)(e^{\epsilon\partial_x} - 1)\delta(x-y)\\ \{v(x),v(y)\}_2 &= \frac{1}{\epsilon}(e^{\epsilon\partial_x} - e^{-\epsilon\partial_x})\delta(x-y). \end{align*} From this, I'm guessing some sort of Miura transformation can be used to put $\{.,.\}_2 - \lambda\{.,.\}_1$ into the normal form and from that we get a set of Hamiltonian densitites $\{h_{\alpha,p} = h_{\alpha,p}(x,u,v,u_x,v_x,...)\}$. The Hamiltonians are given by $H_{\alpha, p} := \int_0^{2\pi}h_{\alpha,p}dx$. The $t^{\alpha,p}$-evolution is \begin{equation} \frac{\partial w^\alpha}{\partial t^{\beta,p}} = \{w^\alpha,H_{\beta,p}\}_1, \qquad w = u,v;\ \alpha,\beta = 1,2;\ p \geq 0. \end{equation} All of these came from Theorem 3.1 of https://arxiv.org/pdf/nlin/0306060.pdf. From here we can use the standard result of biharmiltonian theory that $\partial h_{\alpha,p-1}/\partial t^{\beta,q} = \partial h_{\beta,q-1}/\partial t^{\alpha,p} =: \frac{1}{\epsilon}(e^{\epsilon\partial_x} - 1)\Omega_{\alpha,p;\beta,q}$, therefore there is a function $\log \tau$ such that \begin{equation} \Omega_{\alpha,p;\beta,q} = \epsilon^2\frac{\partial^2\log \tau}{\partial t^{\alpha,p}\partial t^{\beta,q}}. \end{equation}

Ideally, I would like to calculate $\log\tau$ from here and directly and answer my question above. However, the explicit form of $h_{\alpha,p}$ as given in Theorem 3.1 is complicated and I'm not sure how to proceed. Also, I have never seen anyone written down the equation $F = \lim_{\epsilon \rightarrow 0}\epsilon^2\log \tau |_{t^{\alpha,p > 0} = 0, t^\alpha := t^{\alpha,0}}$ explicitely anywhere, so maybe it could be wrong. Could anyone give me some advices or correct my understand or provide me with a suitable reference?


1 Answer 1


I would say that basically everything you wrote is correct, and in particular the equation $F=\lim_{\epsilon\to 0} \epsilon^2 \log \tau|_{t^{\alpha,p>0}=0,t^{\alpha,0}=t^\alpha}$. It is true that, more in general, $\epsilon^2 \log \tau$ gives you the total descendant potential $\mathcal F(t^{*,*};\epsilon)$ at all genera.

Notice that you don't even need to specify which coordinates you are using (the Miura transformation you write about), because no dynamical variable appears here, only the time variables $t^{\alpha,p}$.

However the correct equation for the tau structure $h_{\alpha,p}$ is $$\frac{\partial h_{\alpha,p-1}}{\partial t^{\beta,q}} = \frac{\partial h_{\beta,q-1}}{\partial t^{\alpha,p}} = \partial_x \Omega_{\alpha,p;\beta,q}$$ (the dispersive Poisson structure only enters this equation through the time derivatives $\partial_{t^{\alpha,p}}$). This equation relates to the tau function / total descendant potential via $$\left. \frac{\partial h_{\alpha,p-1}}{\partial t^{\beta,q}}\right|_{w^\gamma_k=(w^{\mathrm{top}})^\gamma_k(x=0,t^{*,*};\epsilon)} = \frac{\partial^3 \mathcal F}{\partial t^{1,0} \partial t^{\alpha,p} \partial t^{\beta,q}},$$ where $w^\mathrm{top}$ is the topological solution, with initial datum $(w^\mathrm{top})^\gamma(x,t^{*,*}=0;\epsilon) = x \delta^\gamma_1$ (here $w^\gamma$ are the normal coordinates $w^\gamma=\eta^{\gamma \mu}h_{\mu,-1}$ of the tau structure).

About your question on the explicit computation of the tau function starting from the hierarchy and its tau structure, you are right to say it is not a trivial exercise. The best way I can think of is to use the explicit relation between the Dubrovin-Zhang tau structure and the Lax representation of the extended Toda system. You can find this information scattered in the literature, including the paper of Carlet, Dubrovin and Zhang you already cited.

However we also gathered this information in section 6.1 of our paper with A. Buryak, https://arxiv.org/abs/1411.6797. It is about the double ramification hierarchy, but since it also deals with its relation to the Dubrovin-Zhang construction, you find the change of coordinates between each of the tree representations: Lax representation, DR hierarchy and DZ hierarchy. A word of warning: the Dubrovin-Zhang hierarchy comes in two version, with ancestors or descendants psi-classes. You need to use Givental's $S$-matrices as explained in the paper to pass from one to the other.

Finally, I gathered this example and several others (including ILW and Gelfand-Dickey) at the end of my review https://arxiv.org/abs/1703.00232. That too is about the double ramification hierarchy, but again you will find the change of coordinates between Lax representation, DR hierarchy and DZ hierarchy. All those formulae come from our papers with Buryak, Dubrovin and Guéré, so please look into the references therein to find out more.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.