Renormalization Fixed Point of the KPZ Universality Class

Abstract

The one dimensional Kardar–Parisi–Zhang universality class is believed to describe many types of evolving interfaces which have the same characteristic scaling exponents. These exponents lead to a natural renormalization/rescaling on the space of such evolving interfaces. We introduce and describe the renormalization fixed point of the Kardar–Parisi–Zhang universality class in terms of a random nonlinear semigroup with stationary independent increments, and via a variational formula. Furthermore, we compute a plausible formula the exact transition probabilities using replica Bethe ansatz. The semigroup is constructed from the Airy sheet, a four parameter space-time field which is the Airy\(_2\) process in each of its two spatial coordinates. Minimizing paths through this field describe the renormalization group fixed point of directed polymers in a random potential. At present, the results we provide do not have mathematically rigorous proofs, and they should at most be considered proposals.

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Acknowledgments

The authors would like to thank H. Spohn, J. Baik, E. Cator, E. Corwin, K. Khanin, B.Valko, B.Virag, A. Borodin, the special semesters on Random Matrix Theory and its Applications at MSRI, and Dynamics and Transport in Disordered Media at the Fields Institute. IC was partially supported by the NSF PIRE Grant OISE-07-30136, the Grant DMS-1208998 as well as by Microsoft Research and MIT through the Schramm Memorial Fellowship, by the Clay Mathematics Institute through the Clay Research Fellowship, by the Institute Henri Poincare through the Poincare Chair, and by the Packard Foundation through a Packard Fellowships for Science and Engineering. JQ is supported by the NSERC. DR was partially supported by NSERC, by a Fields-Ontario Postdoctoral Fellowship, by Fondecyt Grant 1120309, by Conicyt Basal-CMM, and by Programa Iniciativa Científica Milenio Grant number NC130062 through Nucleus Millenium Stochastic Models of Complex and Disordered Systems.

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Correspondence to Ivan Corwin.

Appendix: Two-Point Distribution Function for the KPZ Fixed Point with Flat Initial Condition

Appendix: Two-Point Distribution Function for the KPZ Fixed Point with Flat Initial Condition

The goal of this appendix is to obtain a formula for the two-point distribution function of the KPZ fixed point with flat initial condition based on the formulas proposed in Sect. 3, and compare it with the two-point distribution function for the Airy\(_1\) process. In (13) we need to take

$$\begin{aligned} f=f_L=0\cdot \mathbf {1}_{[-L,L]}+\infty \cdot \mathbf {1}_{[-L,L]^\mathrm{c}}\qquad \text {and}\qquad s(x)= {\left\{ \begin{array}{ll} s_1 &{} \text {if }\, x=0,\\ s_2 &{} \text {if } \, x=u,\\ \infty &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Here, \(L\) is a positive real number introduced because, we recall, in (13) \(f\) is supposed to be infinite outside some compact set. Our interest is the case \(L\rightarrow \infty \).

Using these choices, what we want to compute is

$$\begin{aligned} \lim _{L\rightarrow \infty }\mathbb {P}\!\left( T_2f_L(0)\le s_0,\,T_2f_L(r)\le s_1\right) . \end{aligned}$$

We have chosen here to take \(T_t\) with \(t=2\) to simplify our computations. Observe that in order to use the KPZ fixed point formula to compute this probability we need to use the discrete version for the part involving \(s(x)\) and the continuum version for the flat initial condition, but one can check that this does not introduce any difficulty.

Going back to the formula, note that in our case \(L_s=0\), \(R_s=r\) and \(-L_c=R_c=L\). Observe also that, since the function \(c\) in (13) is defined from \(f\) via \(f(y)=-(t/2)^{1/3}(c(y)-(y-L_s)^2)\) and we are taking \(t=2\), we have \(c(y)=y^2\). Therefore the KPZ fixed point formula reads, in our case,

$$\begin{aligned} \mathbb {P}\!\left( T_2f_L(0)\le s_0,\,T_2f_L(r)\le s_1\right) =\det (I-K_{\mathrm{Ai}}+G_LK_{\mathrm{Ai}}), \end{aligned}$$
(28)

where

$$\begin{aligned} G_L(x,y)=\iint du\,dm\left[ \bar{P}_{s_1+m+u}e^{-rH}\bar{P}_{s_2+m+u}e^{rH}\right] \!(u,x)\frac{\partial }{\partial m}\left[ e^{LH}\Theta ^{s^2-m}_{[-L,L]}e^{LH}\right] \!(y,u), \end{aligned}$$

where \(\Theta ^{s^2-m}_{[-L,L]}\) is exactly the same operator that gives GOE in [16] (the exact formula will not be relevant). Note that

$$\begin{aligned} \bar{P}_{a_1+b}e^{-rH}\bar{P}_{a_2+b}e^{rH}(z_1,z_2)=\bar{P}_{a_1}e^{-rH}\bar{P}_{a_2}e^{rH}(z_1-b,z_2-b), \end{aligned}$$

so integrating by parts we obtain

$$\begin{aligned} G_L(x,y)= & {} \iint dm\,du\left[ \delta _{s_1+m}e^{-rH}\bar{P}_{s_2+m}e^{rH}+\bar{P}_{s_1+m}e^{-rH}\delta _{s_2+m}e^{rH}\right] \!(0,x-u)\\&\quad \cdot \left[ e^{LH}\Theta ^{s^2-m}_{[-L,L]}e^{LH}\right] \!(y,u) \end{aligned}$$

The arguments in [16] show that \(K_{\mathrm{Ai}}e^{LH}\Theta ^{s^2-m}_{[-L,L]}e^{LH}K_{\mathrm{Ai}}\) converges to \(K_{\mathrm{Ai}}(I-\varrho _{-m})K_{\mathrm{Ai}}\) in trace class norm as \(L\rightarrow \infty \), where

$$\begin{aligned} \varrho _a(x,y)=\delta _{x+y=2a}. \end{aligned}$$

Hence one expects

$$\begin{aligned} \lim _{L\rightarrow \infty }\mathbb {P}\!\left( T_2f_L(0)\le s_0,\,T_2f_L(r)\le s_1\right) =\det (I-K_{\mathrm{Ai}}-GK_{\mathrm{Ai}}), \end{aligned}$$

with

$$\begin{aligned}&G(x,y)\\&\quad =\iint du\,dm\left[ \delta _{s_1+m}e^{-rH}\bar{P}_{s_2+m}e^{rH}+\bar{P}_{s_1+m}e^{-rH}\delta _{s_2+m}e^{rH}\right] (0,x-u) [I-\varrho _{-m}](y,u). \end{aligned}$$

Write \(G=\bar{G}-\Gamma \), where the two terms come from \(I\) and \(\varrho _{-m}\) in the last expression. We will first look at the term involving \(I\). It is given by

The last integral equals \(e^{-rH}P_{s_2-s_1}e^{rH}(0,x-y)\), and thus \(\bar{G}=I\). Then

$$\begin{aligned} \det \!\big (I-K_{\mathrm{Ai}}+GK_{\mathrm{Ai}})=\det \!\big (I-\Gamma K_{\mathrm{Ai}}\big ). \end{aligned}$$
(29)

Next we look at \(\Gamma \). We have

$$\begin{aligned} \Gamma (x,y)= & {} \int du\,e^{-rH}\bar{P}_{s_2-s_1}e^{rH}(0,x-u)\varrho _{s_1}(y,u)\\&\quad +\iint dm\,du\,\bar{P}_{s_1+m}e^{-rH}(0,s_2+m)e^{rH}(s_2+m,x-u)\varrho _{-m}(y,u). \end{aligned}$$

Write \(\Gamma \) as \(G_1+G_2\). Then \(G_1(x,y)=e^{-rH}\bar{P}_{s_2-s_1}e^{rH}(0,x+y-2s_1)\). By the Baker-Campbell-Hausdorff formula (BCH) we have

$$\begin{aligned} e^{-r\Delta }e^{-rH}&=e^{r^3/3}e^{r^2\nabla }e^{-r\xi }\\ e^{rH}e^{r\Delta }&=e^{-r^3/3}e^{r\xi }e^{-r^2\nabla }. \end{aligned}$$

Here \(\xi \) denotes the independent variable, so that \((e^{r\xi }f)(x)=e^{rx}f(x)\). Using this we have

$$\begin{aligned} G_1(x,y)&=e^{r\Delta }e^{-r\Delta }e^{-rH}\bar{P}_{s_2-s_1}e^{rH}e^{r\Delta }e^{-r\Delta }(0,x+y-2s_1)\\&=e^{r\Delta }e^{r^3/3}e^{r^2\nabla }e^{-r\xi }\bar{P}_{s_2-s_1}e^{-r^3/3}e^{r\xi }e^{-r^2\nabla }e^{-r\Delta }(0,x+y-2s_1)\\&=e^{r\Delta }\bar{P}_{s_2-s_1-r^2}e^{-r\Delta }(0,x+y-2s_1), \end{aligned}$$

where in the last equality we have used the identities \(e^{-r\xi }\bar{P}_{a}e^{r\xi }=\bar{P}_a\) and \(e^{r^2\nabla }\bar{P}_ae^{-r^2\nabla }=\bar{P}_{a-r^2}\). Here, and below, we are writing expressions involving \(e^{-r\Delta }\) with \(r>0\). This is justified as in [31] because this operator is always applied after \(B_0\) (or \(K_{\mathrm{Ai}}=B_0P_0B_0\)), which is given by

$$\begin{aligned} B_0(x,y)=\mathrm{Ai}(x+y). \end{aligned}$$

The kernel \(G_2\) is a bit more complicated. By BCH we have

$$\begin{aligned} e^{a\nabla }e^{tH}=e^{at}e^{tH}e^{a\nabla }. \end{aligned}$$

Using this we may write

$$\begin{aligned} e^{-rH}(0,s_2+m)=e^{-rH}e^{-s_2\nabla }(0,m)=e^{-s_2r}e^{-s_2\nabla }e^{-rH}(0,m) \end{aligned}$$

and, similarly,

$$\begin{aligned} e^{rH}(s_2+m,x+y+2m)&=e^{s_2\nabla }e^{rH}e^{-2m\nabla }(m,x+y)=e^{2mr}e^{(-2m+s_2)\nabla }e^{rH}(m,x+y)\\&=e^{2mr}e^{s_2\nabla }e^{rH}(-m,x+y)=e^{(2m+s_2)r}e^{rH}(-m,x+y-s_2), \end{aligned}$$

so that

$$\begin{aligned} G_2(x,y)=\int _{-s_1}^\infty dm\,e^{-rH}(-s_2,m)e^{rm}e^{rH}(-m,x+y-s_2)e^{rm}. \end{aligned}$$

Observe that the first factor equals \(e^{-rH}e^{r\xi }(-s_2,m)\) while the second one equals \(e^{-r\xi }e^{rH}(-m,x+y-s_2)\). By BCH again one has \(e^{-r\xi }e^{rH}=e^{-r\Delta }e^{-r^2\nabla }e^{-r^3/3}\) and \(e^{-rH}e^{r\xi }=e^{r\Delta }e^{r^2\nabla }e^{r^3/3}\), so using this and the symmetry of the heat kernel above gives

$$\begin{aligned} G_2(x,y)&=\int _{-s_1}^\infty dm\,e^{r\Delta }(-s_2+r^2,m)e^{-r\Delta }(-m,x+y-s_2+r^2)\\&=\int _{-\infty }^{s_1} dm\,e^{r\Delta }(s_2-r^2,m)e^{-r\Delta }(m,x+y-s_2+r^2)\\&=e^{r\Delta }\bar{P}_{s_1-s_2+r^2}e^{-r\Delta }(0,x+y-2s_2+2r^2). \end{aligned}$$

Putting the formulas for \(G_1\) and \(G_2\) together with (28) and (29), after taking \(L\rightarrow \infty \), the conclusion is that

$$\begin{aligned} \mathbb {P}\!\left( T_2f(0)\le s_0,\,T_2f(r)\le s_1\right) =\det \!\big (I-\Gamma K_{\mathrm{Ai}}\big ), \end{aligned}$$

where

$$\begin{aligned} \Gamma (x,y)=e^{r\Delta }\bar{P}_{s_2-r^2-s_1}e^{-r\Delta }(0,x+y-2s_1) +e^{r\Delta }\bar{P}_{s_1-s_2+r^2}e^{-r\Delta }(0,x+y-2s_2+2r^2). \end{aligned}$$

Observe that the \(r^2\) corresponds just to a parabolic shift, so writing \(\tilde{s}_1=s_1=s_1+0^2\) and \(\tilde{s}_2=s_2+r^2\) we get

$$\begin{aligned}&\Gamma (x,y)=\Gamma _1(x,y)+\Gamma _2(x,y)=e^{r\Delta }\bar{P}_{\tilde{s}_2-\tilde{s}_1}e^{-r\Delta }(0,x+y-2\tilde{s}_1)\\&\quad \quad \quad \quad \quad +\,e^{r\Delta }\bar{P}_{\tilde{s}_1-\tilde{s}_2}e^{-r\Delta }(0,x+y-2\tilde{s}_2). \end{aligned}$$

This could already be considered a working formula.

What comes next is trying to put the formula we got in a form which makes the comparison with the Airy\(_1\) formula easier. Writing \(K_{\mathrm{Ai}}=B_0P_0B_0\) and using the cyclic property of the determinant we have

$$\begin{aligned} \det (I-\Gamma K)=\det (I-P_0B_0\Gamma B_0). \end{aligned}$$

Now

$$\begin{aligned} B_0\Gamma _1B_0(x,y)=\iint dz_1\,dz_2\,\mathrm{Ai}(x+z_1)e^{r\Delta }\bar{P}_{\tilde{s}_2-\tilde{s}_1}e^{-r\Delta }(0,z_1+z_2-2\tilde{s}_1)\mathrm{Ai}(z_2+y). \end{aligned}$$

Shifting \(z_1\) to \(z_1-x\) and \(z_2\) to \(z_2-z_1+x\) gives

$$\begin{aligned} B_0\Gamma _1B_0(x,y)=\iint dz_1\,dz_2\,\mathrm{Ai}(z_1)e^{r\Delta }\bar{P}_{\tilde{s}_2-\tilde{s}_1}e^{-r\Delta }(0,z_2-2\tilde{s}_1)\mathrm{Ai}(z_2-z_1+x+y). \end{aligned}$$

But \(\int \! dz\mathrm{Ai}(z)\mathrm{Ai}(a-z)=2^{-1/3}\mathrm{Ai}(2^{-1/3}a)\), so letting

$$\begin{aligned} \widetilde{B}_0(x,y)=2^{-1/3}\mathrm{Ai}(2^{-1/3}(x+y)) \end{aligned}$$

we have deduced that

$$\begin{aligned} B_0\Gamma _1B_0(x,y)=\Gamma _1\widetilde{B}_0(0,x+y), \end{aligned}$$

and of course the same holds with \(\Gamma _2\) instead. Letting

$$\begin{aligned} \Lambda (x,y)=\Gamma \widetilde{B}_0(0,x+y) \end{aligned}$$

we have

$$\begin{aligned} \mathbb {P}\!\left( T_2f(0)\le \tilde{s}_0,\,T_2f(r)\le \tilde{s}_1\right) =\det \!\big (I-P_0\Lambda \big ). \end{aligned}$$

Finally we change variables \(x\mapsto 2^{1/3}x\), \(y\mapsto 2^{1/3}y\) in the determinant. This changes the kernel \(P_0\Lambda (x,y)\) to \(2^{1/3}P_0\Lambda (2^{1/3}x,2^{1/3}y)\). Writing this explicitly for the term involving \(\Gamma _1\) (and dropping the \(P_0\) for a moment) gives

$$\begin{aligned}&2^{1/3}\int dw\,\Gamma _1(0,w)\widetilde{B}_0(w,2^{1/3}x+2^{1/3}y)\\&\quad =2^{1/3}\int dz\, e^{r\Delta }(0,z)P_{\tilde{s}_2-\tilde{s}_1}e^{-r\Delta }\widetilde{B}_0(z,2^{1/3}x+2^{1/3}y)\\&\quad =2^{2/3}\int dz\,e^{r\Delta }(0,2^{1/3}z)P_{\tilde{s}_2-\tilde{s}_1}e^{-r\Delta }\widetilde{B}_0(2^{1/3}z,2^{1/3}x+2^{1/3}y). \end{aligned}$$

Now \(e^{r\Delta }(0,2^{1/3}z)=(4\pi r)^{-1/2}e^{-(2^{1/3}z)^2/4r}=2^{-1/3}e^{2^{-2/3}r\Delta }(0,z)\). Likewise one can check that \(e^{-r\Delta }\widetilde{B}_0(2^{1/3}z,2^{1/3}x+2^{1/3}y)=2^{-1/3}e^{-2^{-2/3}r\Delta }B_0(z,x+y)\). Hence the last integral can be rewritten as \(e^{2^{-2/3}r\Delta }P_{2^{-1/3}(\tilde{s}_2-\tilde{s}_1)}e^{-2^{-2/3}r\Delta }B_0\). The same of course holds for the term with \(\Gamma _2\). Hence the final formula becomes

$$\begin{aligned} \mathbb {P}\!\left( T_2f(0)\le s_1,\,T_2f(r)\le s_2+r^2\right) =\det \!\big (I-P_0\widetilde{\Lambda }\big ), \end{aligned}$$
(30)

where \(\widetilde{\Lambda }(x,y)=\widetilde{\Gamma }B_0(0,x+y)\) and

$$\begin{aligned} \widetilde{\Gamma }(x,y)= & {} e^{2^{-2/3}r\Delta }\bar{P}_{2^{-1/3}s_2-2^{-1/3}s_1}e^{-2^{-2/3}r\Delta }(0,x+y-2^{2/3}s_1)\\&\quad +\,e^{2^{-2/3}r\Delta }\bar{P}_{2^{-1/3}s_1-2^{-1/3}s_2}e^{-2^{-2/3}r\Delta }(0,x+y-2^{2/3}s_2). \end{aligned}$$

In light of the version of the Airy\(_1\) formula proved in [31]

$$\begin{aligned}&\mathbb {P}\!\left( \mathcal {A}_1(0)\le 2^{-1/3}s_1,\mathcal {A}_1(2^{-2/3}r)\le 2^{-1/3}s_2\right) \nonumber \\&\quad =\det \!\big (I-B_0+\bar{P}_{2^{-1/3}s_1}e^{2^{-2/3}r\Delta }\bar{P}_{2^{-1/3}s_2}e^{-2^{-2/3}r\Delta }B_0\big ) \end{aligned}$$
(31)

this suggests the conjecture

$$\begin{aligned} \mathbb {P}\!\left( T_2f(0)\le s_1,\,T_2f(r)\le s_2+r^2\right) =\mathbb {P}\!\left( \mathcal {A}_1(0)\le 2^{-1/3}s_1,\mathcal {A}_1(2^{-2/3}r)\le 2^{-1/3}s_2\right) . \end{aligned}$$
(32)

Unfortunately, we have not been able to check the equality of the determinants in (30) and (31), and in fact it is not at all clear whether the equality is true. The kernels in the two formulas have many similarities, but observe in particular how the variables \(x,y\) appear in an odd position in \(\widetilde{\Lambda }(x,y)\).

The formula (30) does satisfy some basic reality checks. The kernel \(\widetilde{\Lambda }\) is symmetric in \(s_1,s_2\), which implies the same symmetry for the two-point function. Taking \(s_1\rightarrow \infty \) yields \(F_\mathrm{GOE}(4^{1/3}s_2)\), which is the one-point marginal of \(2^{1/3}\mathcal {A}_1(\cdot )\). Similarly, setting \(r=0\) yields \(F_\mathrm{GOE}(4^{1/3}(s_1\wedge s_2))\). These three facts can be checked more or less directly from (30). An additional, more complicated, reality check which can be performed is the following (we will omit the argument, which is not hard but involves a relatively long computation). Fix some \(s\in \mathbb {R}\) and let \(g(r)=\mathbb {P}\big (\mathcal {A}_1(0)\le 2^{-1/3}s,\mathcal {A}_1(2^{-2/3}r)\le 2^{-1/3}s\big )\) and \(\tilde{g}(r)= \mathbb {P}\big (T_2f(0)\le s,\,T_2f(r)\le s+r^2\big )\). Then \(g'(0)=\tilde{g}'(0)\).

We have also performed some limited numerics—yet they do not give a definitive answer as to the validity (or lack thereof) of this equality.

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Corwin, I., Quastel, J. & Remenik, D. Renormalization Fixed Point of the KPZ Universality Class. J Stat Phys 160, 815–834 (2015). https://doi.org/10.1007/s10955-015-1243-8

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Keywords

  • KPZ equation
  • KPZ universality class
  • KPZ fixed point
  • Airy sheet