The Pentagram Integrals on Inscribed Polygons

First of all, we have the cross ratio

$$[t_{1},t_{2},t_{3},t_{4}]=\frac{(t_{1}-t_{2})(t_{3}-t_{4})}{(t_{1}-t_{3})(t_{2}-t_{4})}.$$

Suppose that $\phi$ is a twisted $n$-gon, with monodromy $M$. We let $v_{i}=\phi(i)$. The label $i$ in Figure 2 denotes $v_{i}$, and similarly for the other labels.

We associate to each vertex $v_{i}$ two numbers:

$$x_{i}=[v_{i-2},v_{i-1},((v_{i-2},v_{i-1})\cap(v_{i},v_{i+1}),((v_{i-2},v_{i-1})\cap(v_{i+1},v_{i+2})],$$

$$y_{i}=[v_{i+2},v_{i+1},((v_{i+2},v_{i+1})\cap(v_{i},v_{i-1}),((v_{i+2},v_{i+1})\cap(v_{i-1},v_{i-2})].$$

Here $(a,b)$ denotes the line determined by points $a$ and $b$. For instance, $x_{i}$ is the cross ratio of the $4$ white points in Figure 2. We call the invariants $x_{1},y_{1},x_{2},y_{2},...$ the corner invariants. These invariants form a periodic sequence of length $2n$. That is $x_{k+n}=x_{k}$ and $y_{k+n}=y_{k}$ for all $k$.

We define

$$O_{n}=\prod_{i=1}^{n}x_{i};\hskip 30.0ptE_{n}=\prod_{i=1}^{n}y_{i}.$$

The other monodromy invariants are best defined in an indirect way. Recall that $M$ is the monodromy of our twisted polygon $\phi$. We lift $M$ to an element of $GL_{3}(\mbox{\boldmath{$R$}})$ which we also denote by $M$. We define

$$\Omega_{1}=\frac{{\rm trace\/}^{3}(M)}{{\rm det\/}(M)};\hskip 30.0pt\Omega_{2}=\frac{{\rm trace\/}^{3}(M^{-1})}{{\rm det\/}(M^{-1})}.$$

These quantities are independent of the lift of $M$ and only depend on the conjugacy class of $M$. Finally, these quantities are invariant under the pentagram map.

We define

$$\widetilde{\Omega}_{1}=O_{n}^{2}E_{n}\Omega_{1};\hskip 30.0pt\widetilde{\Omega}_{2}=O_{n}E_{n}^{2}\Omega_{2}.$$

It turns out that these quantities are polynomials in the corner invariants. The remaining monodromy invariants are suitably weighted homogeneous parts of these polynomials.

We have a basic rescaling operation

$$R_{t}(x_{1},y_{1},x_{2},y_{2},...)=(tx_{1},t^{-1}y_{1},tx_{2},t^{-1}y_{2},...).$$

We say that a polynomial in the corner invariants has weight $k$ if

$$R_{t}^{*}(P)=t^{k}P.$$

Here $R_{t}^{*}$ denotes the obvious action of $R_{t}$ on polynomials. In [S3] we show that

$$\widetilde{\Omega}_{1}=\sum_{k=1}^{[n/2]}O_{k};\hskip 30.0pt\widetilde{\Omega}_{2}=\sum_{k=1}^{[n/2]}E_{k}.$$

Here $O_{k}$ and $E_{k}$ are the weight $k$ polynomials in each sum, and $[n/2]$ denotes the floor of $n/2$.

Now we describe the combinatorial formulas for these invariants. Again, the paper [S3] has a proof that the description below coincides with the one in the previous section. In everything we say, the indices are taken cyclically, mod $n$.

We introduce the monomials

$$X_{i}:=x_{i}\,y_{i}\,x_{i+1}.$$

(1)

The monodromy polynomial $O_{k}$ is built out of the monomials $x_{i}$ for $i=1,...,n$ and $X_{j}$ for $j=1,..,n$. We call two monomials consecutive if they involve consecutive or coinciding variables $x_{i}$. More explicitly, we have the following criteria:

1. 1.

   $X_{i}$ and $X_{j}$ are consecutive if $j\in\left\{i-2,\,i-1,\,i,\,i+1,\,i+2\right\};$

2. 2.

   $X_{i}$ and $x_{j}$ are consecutive if $j\in\left\{i-1,\,i,\,i+1,\,i+2\right\};$

3. 3.

   $x_{i}$ and $x_{j}$ are consecutive if $j\in\{i-1,i,i+1\}$.

Let $O$ be a monomial obtained by the product of the monomials $X_{i}$ and $x_{j}$, that is,

$$O=X_{i_{1}}\cdots{}X_{i_{s}}\,x_{j_{1}}\cdots{}x_{j_{t}}.$$

Such a monomial is called admissible if no two of the monomials are consecutive. For every admissible monomial, define the weight $|O|=s+t$ and the sign $\mathrm{sign}(O)=(-1)^{s}$. One then has

$$O_{k}=\sum_{|O|=k}\mathrm{sign}(O)\,O;\hskip 30.0ptk\in\left\{1,2,\ldots,\left[\frac{n}{2}\right]\right\}.$$

For example, if $n\geq 5$ we obtain the following two polynomials:

$$O_{1}=\sum_{i=1}^{n}\left(x_{i}-x_{i}\,y_{i}\,x_{i+1}\right),\qquad O_{2}=\sum_{i=1}^{n}\left(x_{i}\,x_{i+2}-x_{i}\,y_{i}\,x_{i+1}\,x_{i+3}\right),$$

The same formula works for $E_{k}$, if we make all the same definitions with $x$ and $y$ interchanged. More precisely, one builds the polynomials $E_{k}$ from the monomials $y_{i}$ and $Y_{j}:=y_{j-1}x_{j}y_{j}$ with the same restriction that no consecutive monomials are allowed and the same definitions of the weight and sign.

We note that the dihedral symmetry

$$\sigma(x_{i})=y_{-i},\ \sigma(y_{i})=x_{-i}$$

(2)

interchanges the polynomials $O_{k}$ and $E_{k}$.

Now we describe determinantal formulas for the monodromy invariants; these formulas did not appear in our previous papers on the subject.

For positive integers $k>l$, we define the four-diagonal determinant

$$F_{l}^{k}=\left|\begin{array}[]{ccccccc}1&x_{k}&X_{k-1}&0&0&\dots&0\\ 1&1&x_{k-1}&X_{k-2}&0&\dots&0\\ 0&1&1&x_{k-2}&X_{k-2}&\dots&0\\ \dots&\dots&\dots&\dots&\dots&\dots&\dots\\ 0&\dots&\dots&\dots&1&1&x_{l+1}\\ 0&\dots&\dots&\dots&0&1&1\end{array}\right|,$$

where $x_{i}$ and $y_{i}$ are the corner invariants and $X_{i}$ is as in (1). By convention,

$$F_{k+2}^{k}=0,\ F_{k+1}^{k}=1,\ F_{k}^{k}=1.$$

Then one has the following formula for the monodromy invariants $O_{k}$:

$$\sum_{i=0}^{[n/2]}O_{i}=F_{1}^{n}+F_{0}^{n-1}-F_{1}^{n-1}+x_{n}y_{n}x_{1}F_{2}^{n-1}.$$

(3)

Similarly, for $E_{k}$, define

$$G_{p}^{q}=\left|\begin{array}[]{ccccccc}1&y_{p+1}&Y_{p+2}&0&0&\dots&0\\ 1&1&y_{p+2}&Y_{p+3}&0&\dots&0\\ 0&1&1&y_{p+3}&Y_{p+4}&\dots&0\\ \dots&\dots&\dots&\dots&\dots&\dots&\dots\\ 0&\dots&\dots&\dots&1&1&y_{q}\\ 0&\dots&\dots&\dots&0&1&1\end{array}\right|.$$

Then one has:

$$\sum_{i=0}^{[n/2]}E_{i}=G_{0}^{n-1}+G_{1}^{n}-G_{1}^{n-1}+y_{n}x_{1}y_{1}G_{1}^{n-2}.$$

(4)

Formulas (3) and (4) simplify if one considers an open version of the monodromy invariants: instead of having a periodic “boundary condition” $x_{i+n}=x_{i},y_{i+n}=y_{i}$, assume that $x_{i}=0$ for $i\leq 0$ and $i\geq n+1$, and $y_{i}=0$ for $i\leq-1$ and $i\geq n$. With this “vanishing at infinity” boundary conditions, the monodromy invariants are given by a single determinant:

$$\sum_{i=0}^{[n/2]}O_{i}=F_{0}^{n},\quad\sum_{i=0}^{[n/2]}E_{i}=G_{-1}^{n-1}.$$

We shall encounter an open version of monodromy invariants in Section 4.3.

Now we are going to extract the main combinatorial information from the discussion at the end of the last section.

We fix some integer $N>0$ and consider the set of length $N$ lists in the letters $\{A,X,B\}$. We consider two lists the same if they are cyclic permutations of each other. We say that a cyclic sentence is an equivalence class of such strings. To illustrate our notation by way of example, $(AXABA)$ denotes the equivalence class of $AXABA$. Here $N=5$. We let $\cal S$ denote the set of such sentences, with the dependence on $N$ understood.

We single out certain strings of letters, and to each of these special words we assign a coefficient and a weight. Here is the list.

$$\matrix{{\rm word\/}&{\rm coefficient\/}&{\rm weight\/}\cr X&1&0\cr XA&1&1\cr XAA&1&1\cr XBA&1&1\cr AXA&2&1\cr AAA&-1&1\cr BA&-1&1\cr ABA&-1&1\cr XXA&-2&1\cr}$$

We say that a parsing of a cyclic sentence is a description of the cyclic sentence as a concatenation of words. For example

$$(ABA/XA);\hskip 30.0pt(XAA/BA)$$

are the only two parsings of $(ABAXA)$. We define the coefficient of a parsing to be the product of the coefficients of the word. We define the weight of the parsing to be the sum of the weights of the words. Both parsings above have coefficient $-1$ and weight $2$.

For each cyclic sentence $S$, we define $c(S,w)$ to be the total sum of the coefficients of the weight $w$ parsings of $S$. For instance, when $S=(ABAXA)$, we have $c(S,2)=-2$ and otherwise $c(S,w)=0$. For a more streamlined equation, we define

$$|S|=\sum_{w=0}^{\infty}c(S,w)t^{w}.$$

Here $|S|$ is a polynomial in $t$ that encodes all the coefficients at once. For example

$$|(ABAXA)|=-2t^{2}.$$

Let $\overline{S}$ denote the cyclic sentence obtained by reversing $S$. In view of Section 3, Theorem 1.1 is equivalent to the following result.

In light of the work in the previous section, Theorem 4.1 implies Theorem 1.1. The rest of this chapter is devoted to proving Theorem 4.1.

Before we tackle Theorem 4.1, we slightly modify our puzzle for the sake of convenience.

By Lemma 4.2, we can simply throw out any strings that contain $XB$, and we may drop the word $XBA$ from our list of words. There is a similar cancellation involving $XXA$. Within a parsing, the occurence of $.../X/XA/...$ has weight $1$ and coefficient $1$ whereas the occurence of $.../XXA/...$ has weight $1$ and coefficient $-2$. If we have a parsing that involves $.../XXA/...$ we can create a new parsing by replacing the last $.../XXA/...$ with $.../X/XA/...$. The new parsing cancels out “half” of the original. Thus, we may consider an alternate puzzle where the parsing $/X/XA/$ is forbidden and the word list is

$$\matrix{{\rm word\/}&{\rm coefficient\/}&{\rm weight\/}\cr X&1&0\cr XA&1&1\cr XAA&1&1\cr AXA&2&1\cr AAA&-1&1\cr BA&-1&1\cr ABA&-1&1\cr XXA&-1&1\cr}$$

All we have done is dropped $XBA$ from the list and changed the weight of $XXA$ from $-2$ to $-1$. We call this last puzzle the tight puzzle. Establishing Theorem 4.1 for the tight puzzle is the same as establishing these results for the original one.

As an intermediate step to proving Theorem 4.1, we state a variant of the result. We consider bi-infinite strings in the letters $\{A,B,X\}$, where there are only finitely many $A$s and $B$s. We say that two such strings are equivalent if one of them is a shift of the other one. We say that an open sentence is an equivalence class of such strings. We use finite strings to denote sentences, with the understanding that the left and right of the finite string is to be padded with an infinite number of $X$s. Thus, $ABAXA$ refers to the bi-infinite sentence $...XXABAXAXX...$. We define parsings just as in the cyclic case. For instance, here are all the parsings of this sentence

• •

  $/ABA/XA/(-1)$

• •

  $/XXA/BA/XA/(1)$

(Recall that we have forbidden $/X/XA$.) The first of these have weight $2$ and the last one has weight $3$. We have put the coefficients next to the parsings in each case. Our notation is such that the left and right sides of each expression are padded with $.../X/X/...$. Based on the list above, we have

$$|ABAXA|=-t^{2}+t^{3}.$$

Here is the variant of Theorem 4.1 for open sentences.

Theorem 4.1 implies Theorem 4.3 in a fairly direct way. For instance, suppose we are interested in proving Theorem 4.3 for an open sentence $S$. We say that the span of $S$ is the combinatorial distance between the first and last non-$X$ letter of $S$. For instance, the span of $ABAXA$ is $4$. Supposing that $S$ has span $s$, we simply create a cyclic sentence of length (say) $s+10$ by padding the nontrivial part of $S$ with $X$s and then taking the cyclic equivalence class. Call this cyclic sentence $S^{\prime}$. We clearly have

$$|S|=|S^{\prime}|=|\overline{S}^{\prime}|=|\overline{S}|.$$

The middle equality is Theorem 4.1. The end inequalities are obvious.

Now we turn to the proof of Theorem 4.3. In the next result, $W$ stands for a finite string in the letters $A,B,X$.

Consider Identity 4. There are two kinds of parsings of $WAXA$. One kind has the form $W/AXA$ and the other kind has the form $/WA/XA$. Note that $AXA$ has weight $1$ and coefficient $2$ and $XA$ has weight $1$ and coefficient $1$. From this, we see that $c(WAXA,w)=2c(W,w-1)+c(WA,w-1)$ for all $w$. Identity 4 follows immediately. Identity 5 has the same proof. $\spadesuit$

Suppose now that $W$ is a shortest word for which we don’t know the result of this lemma. We know that $W$ has length at least $3$, so we can write $W=VR$, where $R$ has length $3$ and ends in $A$. Consider the case when $R=AXA$.

By induction, we have

$$|ABAV|+t|AV|+t|V|=0.$$

(8)

$$|ABAVA|+t|AVA|+t|VA|=0.$$

(9)

Using Identity 4 of Lemma 4.4 (three times) we have

$$t|VAXA|=t^{2}|VA|+2t^{2}|V|.$$

(10)

$$t|AVAXA|=t^{2}|AVA|+2t^{2}|AV|.$$

(11)

$$|ABAVAXA|=t|ABAVA|+2t|ABAV|.$$

(12)

When we add together the right hand sides of Equations 10, 11, 12, we get $0$, thanks to Equations 8 and 9. Hence, when we add the left hand sides of Equations 10, 11, 12, we also get $0$. But this last sum is exactly the identity we wanted to prove.

A similar argument works when $R$ is any of the other $3$-letter strings that appear in Lemma 4.4. The only case we haven’t considered is the case when $R=XBA$, but these strings are forbidden. $\spadesuit$

Now that we have Lemma 4.4 and Lemma 4.5, our proof of Theorem 4.3 goes by induction. First of all, we check Theorem 4.3 for all strings having span at most $3$. Suppose then that $W$ is the shortest open sentence for which we do not know Theorem 4.3. We can write $W=VR$ where $R$ is some string of length $3$ that ends in $A$.

Let’s consider the case when $R=XAA$. Then we have

$$|W|=|VXAA|=t|V|=t|\overline{V}|=|AAX\overline{V}|=|\overline{W}|.$$

The second equality is Identity 2 of Lemma 4.4. The third equality is the induction assumption. The fourth equality is Identity 2 of Lemma 4.5. A similar argument works when $R$ is any of the $3$ letter strings in Lemma 4.4. The final case, $R=XBA$, is forbidden.

This completes the proof of Theorem 4.3.

We need to mention another convention before we launch into the cyclic case. Besides cyclic and open sentences, there is one more case we can consider. We introduce the notation $[W]$ to denote an open word whose parsings cannot be created by padding $X$s onto the left and right of $W$. We will illustrate what we have in mind by way of example. Setting $W=ABAXA$, the parsings of the open string $W$ are

$$/AXA/BA/;\hskip 30.0pt/XXA/X/ABA.$$

However, the second parsing involves two $X$s that have been padded onto the left of $W$. Only the first parsing of $W$ is also a parsing of the locked string $[W]$. We let $|[W]|$ be the polynomial that encodes the weights and coefficients of all the parsings of $[W]$.

Below we prove the following result.

Lemma 4.7 allows us to finish the proof of Theorem 4.1. Our proof goes by induction on the number of $Xs$ in the word $W$. Let $W$ be a word having the smallest number of $X$s, for which we do not know Theorem 4.1.

After cyclically permuting the letters in $W$, we can write $W=VX$. By Lemma 4.7, we have

$$|(W)|=|(VX)|=-2|(VB)|+|V|.$$

Applying Lemma 4.7 to $\overline{W}$, we have

$$|(\overline{W})=|(\overline{V}X)|=-2|(\overline{V}B)|+|\overline{V}|=-2|\overline{VB}|+|\overline{V}|.$$

Setting $Y=VB$, we have

$$|(W)|=-2|(Y)|+|V|=-2|(\overline{Y})|+|\overline{V}|=|(\overline{W})|.$$

The middle equality comes from Theorem 4.3 (to handle $V$) and the induction assumption (to handle $Y$.)

It remains only to prove Lemma 4.7. We will establish some auxilliary relations in this section, and then use them in the next section to prove Lemma 4.7.

To illustrate the idea, we assume that $W$ ends with $AXA$, that is, $W=VAXA$. Then, by Identity 4 of Lemma 4.4,

$$|W|=|VAXA|=t|VA|+2t|V|,$$

$$|[XW|=|[XVAXA]|=t|[XVA]|+2t|[XV]|,$$

and

$$|[AXW]|=|[AXVAXA]|=t|AXVA|+2t|[AXV]|.$$

By the induction assumption,

$$|VA|=|[XVA]|-|[AXVA]|,\ \ |V|=|[XV]|-|[AXV]|,$$

and the result follows for $W$. $\spadesuit$