Correction to: Representation growth and rational singularities of the moduli space of local systems

Abstract

We explain and correct a mistake in Section 2.6 and Appendix C of the first and second author’s paper “Representation Growth and Rational Singularities of the Moduli Space of Local Systems” [1].

1 Correction to: Invent. math. (2016) 204:245–316 https://doi.org/10.1007/s00222-015-0614-8

We use throughout the notation and conventions of [1]. The source of the mistake is the description of the set $S_2$ in page 272. The elements

$$\begin{aligned} (\{ (d,d-1),(d-1,d-1)\},(d,d-1)),\; ( \{(d-1,d-1),(d-1,d)\},(d-1,d) ) \end{aligned}$$

(1)

are not considered. This spoils the proof of Lemma 2.40 (the blue and green subgraphs are no longer trees), which is used to prove Theorem 2.1.

2 A straightforward correction

The right description of $S_2$ is

$$\begin{aligned} S_2= \left\{ (\{(i,j),(j,{l})\},(i,{l})) \in I^{(2)} \times J\, |\, \{ i, l\}=\{d-1,d\} \text { and } j= \left\lfloor \frac{i+{l}}{2} \right\rfloor ,\right. \\ \left. \text { or } \{ i, l\}\ne \{d-1,d\} \text { and } j= \left\lceil \frac{i+{l}}{2}\right\rceil +\delta _{i,{l}} \right\} . \end{aligned}$$

Then, $\varGamma _2$ is the polygraph attached to the graph $\varGamma _3=(I,E)$, with

$$\begin{aligned} E= \left\{ \{(i,j),(j,{l})\} \in I^{(2)} \, | \, \{ i, l\}=\{d-1,d\} \text { and } j= \left\lfloor \frac{i+{l}}{2} \right\rfloor , \right. \\ \left. \text { or } \{ i, l\}\ne \{d-1,d\} \text { and } j= \left\lceil \frac{i+{l}}{2}\right\rceil +\delta _{i,{l}} \, \right\} . \end{aligned}$$

As for Fig. 1, 2 and 3 in [1], the two edges corresponding to (1) are missing.

Additionally, there are some mismatches concerning Fig. 2 and $\omega _3$ in [1]. The simplest way to make the labels of Fig. 2 match is fist to multiply $\omega _3$ in page 273 by 5, obtaining

$$\begin{aligned} w_3((i,j))(m)= {\left\{ \begin{array}{ll}5^{|i-j|+1} &{}\text {if }m\equiv i-j\text { (mod 3)}\\ 3 \cdot 5^{|i-j|} &{}\text {if }m\equiv (i-j-sign(i-j+1/2))\text { (mod 3)}\\ 0 &{}\text {if }m\equiv (i-j+sign(i-j+1/2))\text { (mod 3)},\\ \end{array}\right. } \end{aligned}$$

(2)

and then consider the colouring [red is $m=0$, green is $m=2$, blue is $m=1$], so that we just have to swap the values of the blue and green labels along the diagonal of Fig. 2 in page 314.

Image 1

Graph $\varGamma _3$ with the weights $\omega _3$ for the case $d = 6$

Finally, the key to prove now Lemma 2.40 in the closest way to that of [1] is redefining $\omega _3$ at the nodes $(d-1,d-1)$, $(d-1,d)$ and $(d,d-1)$:

$$\begin{aligned} \omega _3(d-1,d-1)(0)&=3\cdot 5^1,&\omega _3(d-1,d)(0)&=5^2 ,&\omega _3(d,d-1)(0)&=3\cdot 5^0 , \\ \omega _3(d-1,d-1)(2)&=3\cdot 5^1,&\omega _3(d-1,d)(2)&=4\cdot 5^1 ,&\omega _3(d,d-1)(2)&= 0,\\ \omega _3(d-1,d-1)(1)&=0 ,&\omega _3(d-1,d)(1)&=0 ,&\omega _3(d,d-1)(1)&=5^2, \end{aligned}$$

so that the resulting diagram for $d=6$ (Fig. 2 in [1]) is given by Image 1.

With the redefinition of $\omega _3$ above, we only need to add to the proof of Lemma 2.40 in [1] an analysis of the edges around $(d-1,d-1)$. They look, for $d\ge 3$, like the ones in Image 1. We thus have forests with maximal degree $\le 3$, as we need. Finally, the cases $d=1,2$ are straigthforward.

Image 2

Graph $\varGamma _3$ with $I\ne J$ for the case $d = 6$

3 An alternative solution

We indicate here how to get an alternative solution that, although requires more changes, would keep better the original intuition for the proof.

At the beginning of Section 2.6 in [1], recall that $L=\{1,\ldots ,d\}$. Stay with $J=L\times L{\setminus } \{(d,d)\}$ and replace I by $I=L\times L{\setminus } \{(1,1)\}$. This entails changes in $S_j$ and $\varGamma _j$, which we omit here for the sake of brevity. With the definition of $\omega _3$ as in (2) (without any redefinition) and the same conventions for the colours as described in Sect. 1, the corresponding $\varGamma _3$ for $d=6$ is given by Image 2.

The general proof then follows along the same lines as the original one.