Flows on the $\mathbf{PGL(V)}$-Hitchin Component

Abstract

In this article we define new flows on the Hitchin components for $\mathrm {PGL}(V)$. Special examples of these flows are associated to simple closed curves on the surface and give generalized twist flows. Other examples, so called eruption flows, are associated to pair of pants in S and capture new phenomena which are not present in the case when $n=2$. We determine a global coordinate system on the Hitchin component. Using the computation of the Goldman symplectic form on the Hitchin component, that is developed by two of the authors in a companion paper to this article (Sun and Zhang in The Goldman symplectic form on the $\mathrm{PGL} ({V})$-Hitchin component, 2017. arXiv:1709.03589), this gives a global Darboux coordinate system on the Hitchin component.

Appendices

Appendix A. Proof of Proposition 2.23

In this appendix, we give a proof of Proposition 2.23, which we restate here for the convenience of the reader.

Proposition

Let $(F_1,\dots ,F_k)$ be a positive k-tuple of flags in ${\mathcal {F}}(V)$. Then for any positive integers $n_1$, $\dots$, $n_k$ that sum to $d\le n$, we have that

$$\begin{aligned} \dim \left( \sum _{j=1}^kF_j^{(n_j)}\right) =d. \end{aligned}$$

We start with the following two notions.

Definition A.1

 

• (1) A map $\zeta :S^1\rightarrow {\mathbb {P}}(V)$ is convex if $\zeta (x_1)+\dots +\zeta (x_k)$ is a direct sum for all $k\le n:=\dim (V)$ and for all pairwise distinct $x_1$, ..., $x_k$ in $S^1$.

• (2) Let F be a flag in ${\mathcal {F}}(V)$ and let $\zeta :S^1\rightarrow {\mathbb {P}}(V)$ be a convex curve. We say $\zeta$osculatesF if there is some point x in $S^1$ such that $\zeta (x)=F^{(1)}$, and for all $l=1$, ..., $d-1$,

  $$\begin{aligned} \lim _{i\rightarrow \infty }\zeta (x_{1,i})+..+\zeta (x_{l,i})=F^{(l)} \end{aligned}$$

  for all pairwise distinct points l-tuples of points $(x_{1,i},...,x_{l,i})$ in $S^1$ such that $\lim _{i\rightarrow \infty }(x_{1,i},\dots ,x_{l,i})=(x,\dots ,x)$.

For the proof, we use the following facts. The first is due to Fock and Goncharov [FG06, Theorem 1.3 and Section 9.11].

Theorem A.2

([FG06]). Let $(F_1,\dots ,F_k)$ be a positive triple of flags in ${\mathcal {F}}(V)$. Then there is a continuous convex map $\zeta :S^1\rightarrow {\mathcal {F}}(V)$ that osculates $F_l$ for all $l=1$, ..., k.

The second is an immediate consequence of Proposition 2.11.

Proposition A.3

The space ${\mathcal {F}}(V)^3_+$ of positive triples of flags in V is a connected component of the space of ${\mathcal {F}}(V)^{[3]}$ of generic triples of flags in V.

Using these, we prove the following key lemma.

Lemma A.4

Let $k\ge 4$ be an integer, let $(F_1,\dots ,F_k)$ be a positive k-tuple of flags in ${\mathcal {F}}(V)$ and let $m=1$, ..., $n-1$. Let H be the flag defined by

$$\begin{aligned} H^{(i)}:=\left\{ \begin{array}{ll} F_2^{(i)}&{}\quad \text {if } i\le m;\\ F_2^{(m)}+F_3^{(i-m)}&{}\quad \text {if }i>m. \end{array}\right. \end{aligned}$$

Then $(F_1,H,F_4,\dots ,F_k)$ is a positive $(k-1)$-tuple of flags.

Proof

Note that $(F_1,F_2,F_4,\dots ,F_k)$ is a positive k-tuple of flags and $H^{(1)}=F_2^{(1)}$. Thus, by Proposition 2.19, it is sufficient to prove that $(F_1,H,F_4)$ is a positive triple of flags.

Let $\zeta$ be a continuous convex map that osculates the flags $F_1$, ..., $F_4$. For all $l=1$, $\dots$, 4, let $x_l$ be the point in $S^1$ such that $\zeta (x_l)=F_l^{(1)}$. It is a consequence of Proposition 2.19 that ${\mathcal {F}}(V)^4_+$ is open in the space of pairwise transverse quadruple of flags. Thus, there are pairwise disjoint open intervals $I_1$, ..., $I_4\subset S^1$ such that $I_l$ contains $x_l$, and satisfy the following property: For all cyclically ordered $(n-1)$-tuple of points ${\mathbf {y}}_l:=(y_{l,1},\dots ,y_{l,n-1})$ in $I_l$, let $F_{{\mathbf {y}}_l}$ be the flag in ${\mathcal {F}}(V)$ given by $F_{\mathbf{y}_l}^{(i)}:=\sum ^i_{j=1}\zeta (y_{l,j})$. Then $(F_{\mathbf{y}_1},\dots ,F_{{\mathbf {y}}_4})$ is a positive quadruple of flags.

Now, for each $i=1$, ..., $n-1$, let $f_i:[0,1]\rightarrow S^1$ be continuous paths such that

• $f_i(0)=y_{2,i}$.

• $f_i(1)=\left\{ \begin{array}{ll} y_{2,i}&{}\quad \text {if } i\le m;\\ y_{3,i-m}&{}\quad \text {if }i>m. \end{array}\right.$

• $f_i(0)\le f_i(t)\le f_i(1)\le f_i(0)$ for all t in [0, 1].

• $\big (f_1(t),\dots ,f_{n-1}(t)\big )$ is a cyclically ordered $(n-1)$-tuple of points in $\partial \Gamma$ for all t in [0, 1].

Such paths $f_i$ exist because $({\mathbf {y}}_2,{\mathbf {y}}_3)$ is a cyclically ordered $(2n-2)$-tuple of points.

Since $\zeta$ is convex, we may define $G(t)=G({\mathbf {y}}_2,\mathbf{y}_3,t)$ to be the flag in ${\mathcal {F}}(V)$ given by $G(t)^{(j)}:=\sum ^j_{i=1}\zeta (f_i(t))$. Since $G(0)=F_{\mathbf{y}_2}$, we see that $(F_{{\mathbf {y}}_1},G(0),F_{{\mathbf {y}}_4})$ is a positive triple of flags. The convexity of $\zeta$ also implies that $(F_{{\mathbf {y}}_1},G(t),F_{{\mathbf {y}}_4})$ is a generic for all t. Since ${\mathcal {F}}(V)^3_+$ is a connected component of ${\mathcal {F}}(V)^{[3]}$ and $t\mapsto G(t)$ is continuous, it follows that $(F_{\mathbf{y}_1},G(1),F_{{\mathbf {y}}_4})$ is also a positive triple of flags.

Observe that

• for $l=1$, 4, the limit of $F_{{\mathbf {y}}_l}$ as $\mathbf{y}_l$ converges to $(x_l,\dots ,x_l)$ is $F_l$,

• the limit of $G({\mathbf {y}}_2,{\mathbf {y}}_3,1)$ as $(\mathbf{y}_2,{\mathbf {y}}_3)$ converges to $(x_2,\dots ,x_2,x_3,\dots ,x_3)$ is H.

Since ${\mathcal {F}}(V)^3_+$ is closed in the space of pairwise transverse triple of flags, this implies that $(F_1,H,F_4)$ is a positive triple of flags. $\square$

Proof of Proposition 2.23

We prove this by induction on k. For the base case $k=3$, this is an immediate consequence of Proposition 2.11.

Next, we prove the inductive step. Suppose that $k\ge 4$. Let H be the flag in ${\mathcal {F}}(V)$ defined by

$$\begin{aligned} H^{(i)}:=\left\{ \begin{array}{ll} F_2^{(i)}&{}\quad \text {if } i\le n_2;\\ F_2^{(n_2)}+F_3^{(i-n_2)}&{}\quad \text {if }i> n_2. \end{array}\right. \end{aligned}$$

By Lemma A.4, we see that $(F_1,H,F_4,\dots ,F_k)$ is a positive $(k-1)$-tuple of flags. We further observe that

$$\begin{aligned} \sum _{j=1}^kF_j^{(n_j)}=F_1^{(n_1)}+H^{(n_2+n_3)}+\sum _{j=4}^kF_j^{(n_j)}. \end{aligned}$$

The statement now follows by applying the inductive hypothesis to the right hand side. $\square$

Appendix B. Proof of Lemma 4.1

In this appendix, we prove Lemma 4.1, which we restate here for the reader’s convenience.

Lemma

Let ${\mathcal {V}}$ denote the set of vertices of ${\widetilde{{\mathcal {T}}}}$, and for non-negative integers j, let $\xi _j$ be a Frenet curve in $\widetilde{\mathrm {Fre}}(V)$. If $\lim _{j\rightarrow \infty }\xi _j(p)=\xi _0(p)$ for all vertices p in ${\widetilde{{\mathcal {V}}}}$, then $\lim _{j\rightarrow \infty }\xi _j(p)=\xi _0(p)$ for all points p in $\partial \Gamma$.

Let x, y, z be the vertices of a triangle in ${\widetilde{\Theta }}$, let w be any point in $\partial \Gamma \setminus {\mathcal {V}}$, and assume without loss of generality that $y<z<x<w<y$ in this cyclic order in $S^1$. By Proposition 2.19, to conclude that $\lim _{j\rightarrow \infty }\xi _j(w)=\xi _0(w)$, it is sufficient to show that

• (1) For any pair ${\mathbf {i}}$ of positive integers that sum to n,

  $$\begin{aligned} C^{\mathbf{i}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(w))=\lim _{j\rightarrow \infty }C^{\mathbf{i}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(w)). \end{aligned}$$

• (2) For any triple of positive integers ${\mathbf {i}}$ that sum to n,

  $$\begin{aligned} T^{\mathbf{i}}(\xi _0(x),\xi _0(w),\xi _0(y))=\lim _{j\rightarrow \infty }T^{\mathbf{i}}(\xi _j(x),\xi _j(w),\xi _j(y)). \end{aligned}$$

Since ${\mathcal {V}}$ is dense in $\partial \Gamma$, there are sequences $\{a_k\}_{k=1}^\infty$ and $\{b_k\}_{k=1}^\infty$ in ${\mathcal {V}}$ such that $\lim _{k\rightarrow \infty }a_k=\lim _{k\rightarrow \infty }b_k=w$ and $y<z<x<a_k<w<b_k<y$ for all k.

Proof of (1)

Since $\xi _0$ is continuous, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }C^{\mathbf{i}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(a_k))= & {} \lim _{k\rightarrow \infty }C^{\mathbf{i}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(b_k))\\= & {} C^{\mathbf {i}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(w)) \end{aligned}$$

for all pairs of positive integers ${\mathbf {i}}$ that sum to n. Furthermore, it is also well-known (see for example Proposition 2.12 of [Zha15a]) that

$$\begin{aligned} C^{\mathbf {i}}(\xi _j(x),\xi _j(a_k),\xi _j(y),\xi _j(w)),\,\,\,\,C^\mathbf{i}(\xi _j(x),\xi _j(w),\xi _j(y),\xi _j(b_k))>1 \end{aligned}$$

for all non-negative integers j and all positive integers k. In particular,

$$\begin{aligned} \begin{array}{l} C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(a_k))>C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(w))\\ >C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(b_k)). \end{array} \end{aligned}$$

Since $\lim _{j\rightarrow \infty }\xi _j(p)=\xi _0(p)$ for any vertex p in ${\mathcal {V}}$, we see that

$$\begin{aligned} C^{{\mathbf {i}}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(w))= & {} \lim _{k\rightarrow \infty }C^{{\mathbf {i}}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(a_k))\\= & {} \lim _{k\rightarrow \infty }\lim _{j\rightarrow \infty }C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(a_k))\\\ge & {} \lim _{j\rightarrow \infty }C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(w))\\\ge & {} \lim _{k\rightarrow \infty }\lim _{j\rightarrow \infty }C^{{\mathbf {i}}}(\xi _j(x),\xi _j(z),\xi _j(y),\xi _j(b_k))\\= & {} \lim _{k\rightarrow \infty }C^{{\mathbf {i}}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(b_k))\\= & {} C^{{\mathbf {i}}}(\xi _0(x),\xi _0(z),\xi _0(y),\xi _0(w)). \end{aligned}$$

This proves (1). $\square$

To prove (2), we need to use the following lemma. Let $\mathbf{i}:=(i_1,i_2,i_3)$ be a triple of positive integers that sum to n. For any generic quadruple of flags $F_1,F_2,F_3,F_4$ in ${\mathcal {F}}(V)$, define

$$\begin{aligned} \begin{array}{l}T^{(i_1,i_2,F_4,i_3)}(F_1,F_2,F_3)\\ := \displaystyle \frac{F_1^{(i_1+1)}\wedge F_2^{(i_2)}\wedge F_3^{(i_3-1)}\cdot F_1^{(i_1-1)}\wedge F_2^{(i_2)}\wedge F_4^{(1)}\wedge F_3^{(i_3)}\cdot F_1^{(i_1)}\wedge F_2^{(i_2-1)}\wedge F_3^{(i_3+1)}}{F_1^{(i_1+1)}\wedge F_2^{(i_2-1)}\wedge F_3^{(i_3)}\cdot F_1^{(i_1)}\wedge F_2^{(i_2)}\wedge F_4^{(1)}\wedge F_3^{(i_3-1)}\cdot F_1^{(i_1-1)}\wedge F_2^{(i_2)}\wedge F_3^{(i_3+1)}} \end{array} \end{aligned}$$

Lemma B.1

Let $\xi :S^1\rightarrow {\mathcal {F}}(V)$ be a Frenet curve and let $x_1<x_4<x_2<x_5<x_3<x_1$ lie in $S^1$ in this cyclic order. For each $m=1$, $\dots$, 5, let $F_m:=\xi (x_m)$, then

$$\begin{aligned} T^{(i_1,i_2,F_4,i_3)}(F_1,F_2,F_3)>T^{\mathbf{i}}(F_1,F_2,F_3)>T^{(i_1,i_2,F_5,i_3)}(F_1,F_2,F_3). \end{aligned}$$

(Recall that we assume $\dim (V)\ge 3$.)

Proof

Let $K:=F_1^{(i_1-1)}+F_2^{(i_2-1)}+F_3^{(i_3-1)}$. For $m=1$, 2, 3, let $L_{m}\subset V$ be a line such that $F_m^{(i_m-1)}+L_{m}=F_m^{(i_m)}$, and let $P_{m}\subset V$ be a plane such that $F_m^{(i_m-1)}+P_{m}=F_m^{(i_m+1)}$. For any point x in $S^1$, let

$$\begin{aligned} L_x:=\left\{ \begin{array}{ll} \xi ^{(1)}(x)&{}\quad \text {if }x\ne x_1,\,x_2,\,x_3;\\ L_{m}&{}\quad \text {if }x=x_m; m=1,\,2,\,3, \end{array}\right. \,\,\,P_x:=\left\{ \begin{array}{ll} \xi ^{(2)}(x)&{}\quad \text {if }x\ne x_1,\,x_2,\,x_3;\\ P_{m}&{}\quad \text {if }x=x_m; m=1,\,2,\,3, \end{array}\right. \end{aligned}$$

and let $H:=L_{x_1}+L_{x_2}+L_{x_3}(=L_1+L_2+L_3)$. Then define $\xi ':S^1\rightarrow {\mathcal {F}}(H)$ by

$$\begin{aligned} \xi '^{(1)}(x):=(K+L_x)\cap H, \,\,\,\,\,\xi '^{(2)}(x):=(K+P_x)\cap H. \end{aligned}$$

One can verify that $\xi '$ does not depend on the choices of $L_{m}$ and $P_{m}$, and is Frenet.

Furthermore, from the definition of the triple ratio, we see that

$$\begin{aligned} T^{(i_1,i_2,F_4,i_3)}(F_1,F_2,F_3)= & {} T^{(1,1,\xi '(x_4),1)}(\xi '(x_1),\xi '(x_2),\xi '(x_3)),\\ T^{{\mathbf {i}}}(F_1,F_2,F_3)= & {} T^{(1,1,1)}(\xi '(x_1),\xi '(x_2),\xi '(x_3)),\\ T^{(i_1,i_2,F_5,i_3)}(F_1,F_2,F_3)= & {} T^{(1,1,\xi '(x_5),1)}(\xi '(x_1),\xi '(x_2),\xi '(x_3)). \end{aligned}$$

Thus, it is sufficient to prove this lemma in the case when $\dim (V)=3$. That is a straightforward computation (see Proposition 2.3.4 of [Zha15b]). $\square$

Proof of (2)

The Frenet property of $\xi _0$ implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }T^{(i_1,i_2,\xi _0(a_k),i_3)}(\xi _0(x),\xi _0(w),\xi _0(y))= & {} T^{{\mathbf {i}}}(\xi _0(x),\xi _0(w),\xi _0(y))\\= & {} \lim _{k\rightarrow \infty }T^{(i_1,i_2,\xi _0(b_k),i_3)}(\xi _0(x),\xi _0(w),\xi _0(y)) \end{aligned}$$

for all ordered triples of positive integers $\mathbf{i}:=(i_1,i_2,i_3)$ that sum to n. Also, by Lemma B.1, we have

$$\begin{aligned} \begin{array}{l} T^{(i_1,i_2,\xi _j(a_k),i_3)}(\xi _j(x),\xi _j(w),\xi _j(y))>T^{{\mathbf {i}}}(\xi _j(x),\xi _j(w),\xi _j(y))\\ >T^{(i_1,i_2,\xi _j(b_k),i_3)}(\xi _j(x),\xi _j(w),\xi _j(y)) \end{array} \end{aligned}$$

for all non-negative integers j and all positive integers k.

Since $\lim _{j\rightarrow \infty }\xi _j(p)=\xi _0(p)$ for all vertices p in ${\mathcal {V}}$, this implies that

$$\begin{aligned} T^{{\mathbf {i}}}(\xi _0(x),\xi _0(w),\xi _0(y))= & {} \lim _{k\rightarrow \infty }T^{(i_1,i_2,\xi _0(a_k),i_3)}(\xi _0(x),\xi _0(w),\xi _0(y))\\= & {} \lim _{k\rightarrow \infty }\lim _{j\rightarrow \infty }T^{(i_1,i_2,\xi _j(a_k),i_3)}(\xi _j(x),\xi _j(w),\xi _j(y))\\\ge & {} \lim _{j\rightarrow \infty }T^{{\mathbf {i}}}(\xi _j(x),\xi _j(w),\xi _j(y))\\\ge & {} \lim _{k\rightarrow \infty }\lim _{j\rightarrow \infty }T^{(i_1,i_2,\xi _j(b_k),i_3)}(\xi _j(x),\xi _j(w),\xi _j(y))\\= & {} \lim _{k\rightarrow \infty }T^{(i_1,i_2,\xi _0(b_k),i_3)}(\xi _0(x),\xi _0(w),\xi _0(y))\\= & {} T^{{\mathbf {i}}}(\xi _0(x),\xi _0(w),\xi _0(y)). \end{aligned}$$

This proves (2). $\square$