# Representation of $2\times 2$ matrices over perfect fields by a diagonal
quadratic form

Murtuza Nullwala
Ramrao Adik Institute of Technology,
Nerul, 400706.  Anuradha S. Garge
University of Mumbai,
Mumbai, 400098.

###### Abstract

Let $F$ be a perfect field. In this paper, we show that a diagonal quadratic
form $\sum_{i=1}^{m}a_{i}X_{i}^{2}$ over $F$ is universal over $M_{2}(F)$ if
and only if atleast two of $a_{1},a_{2},\ldots,a_{m}$ are non-zero.

Mathematics Subject Classification 2010: 11E04, 15B33. Key words: Quadratic
forms, Perfect fields, Universal.

##  1 Introduction

Let $R$ denote a commutative ring with unity. Let $M_{n}(R)$ denote the set of
all $n\times n$ matrices over $R$. An $n-$ary quadratic form over $R$ is a
homogeneous polynomial $q$ of degree $2$ in $n$ variables. Thus $q$ is of the
form:

$q(X_{1},X_{2},\ldots,X_{n})=\sum_{i,j=1}^{n}a_{ij}X_{i}X_{j}$, where
$a_{ij}\in R$.

The study of representation of all positive integers by a quadratic form is a
classical one. In 1770, Lagrange was the first mathematician to show that any
positive integer can be represented by the quadratic form

$X_{1}^{2}+X_{2}^{2}+X_{3}^{2}+X_{4}^{2}.$ (0.1)

In 1916, S. Ramanujan enlisted $54$ (see [1]) diagonal forms which represent
all positive integers. If
$q(X_{1},X_{2},\ldots,X_{n})=\sum_{i,j=1}^{n}a_{ij}X_{i}X_{j}\in\mathbb{Z}[X_{1},X_{2},\ldots,X_{n}]$
is a quadratic form then we can associate a symmetric matrix $M_{q}$ of order
n, defined as $M_{q}=(m_{ij})$ where $m_{ij}=\dfrac{1}{2}(a_{ij}+a_{ji})$.

######  Definition 1.

(I) A quadratic form $q$ over $\mathbb{Z}$ is said to have integer matrix
$M_{q}$ if all entries of $M_{q}$ are integers.
(II) A quadratic form is said to be positive definite if
$q(x_{1},x_{2},\ldots,x_{n})>0$, $x_{i}\in\mathbb{Z}$ and atleast one
$x_{i}\neq 0$ for some $i,\;1\leq i\leq n$.

######  Definition 2.

A diagonal quadratic form $\sum_{i=1}^{m}a_{i}X_{i}^{2}$, where $a_{i}\in F$
is said to be universal over $F$ if it represents every element of $F.$

In 2000, Manjul Bhargava (see [1]) proved an interesting criterion called the
fifteen theorem, to decide when a positive definite quadratic form with
integer matrix represents all positive integers. We recall this theorem below:

######  Theorem 3.

If a positive definite quadratic form having an associated integer matrix
represents every positive integer up to fifteen, then it represents every
positive integer.

In 1986, Vaserstein (see [7]) proved that for $n\geq 2$, every $n\times n$
integral matrix is a sum of three squares i.e., the quadratic form
$X_{1}^{2}+X_{2}^{2}+X_{3}^{2}$ is universal over $M_{n}(\mathbb{Z}).$ In
2017, Jungin Lee (see [5]) gave a criterion to check when a diagonal quadratic
form $\sum_{i=1}^{n}a_{i}X_{i}^{2}$, $a_{i}\in\mathbb{Z}$ represents all
matrices in $M_{2}(\mathbb{Z})$ i.e. a criterion to check when
$\sum_{i=1}^{n}a_{i}X_{i}^{2}$ is universal over $M_{2}(\mathbb{Z}).$ We
recall below his theorem:

######  Theorem 4.

A quadratic form $\sum_{i=1}^{m}a_{i}X_{i}^{2},$ $a_{i}\in\mathbb{Z}$ is
universal over $M_{2}(\mathbb{Z})$ if and only if there is no prime which
divides $m-1$ numbers of $a_{1},a_{2},\ldots,a_{m}$ and there exist three
numbers of $a_{1},a_{2},\ldots,a_{m}$ which are not multiples of $4$.

We now consider quadratic forms over a field with characteristic not equal to
$2$. Then to every quadratic form $q$ we can still associate a symmetric
matrix $M_{q}$. One has the following result for the universality of the
quadratic form $q$ (see [4]):

######  Theorem 5.

Let $F$ be a field of characteristic not equal to $2$. If for a quadratic form
$q$ over $F,$ $q(x_{1},x_{2},\ldots,x_{n})=0$ for some
$x_{1},x_{2},\ldots,x_{n}\in F$ with atleast one $x_{i}\neq 0$ and the
determinant of the associated matrix $M_{q}$ is non-zero then $q$ is universal
over $F.$

In the light of the paper by Jungin Lee, S. A. Katre posed the following
question: given a diagonal quadratic form
$a_{1}X_{1}^{2}+a_{2}X_{2}^{2}+\cdots+a_{m}X_{m}^{2}$ with $a_{i}\in F,$ when
is it universal over $M_{n}(F),$ $n\geq 2$? In this paper, we answer the above
question for $n=2$ by proving the following result:

Theorem: A diagonal quadratic form $\sum_{i=1}^{m}a_{i}X_{i}^{2}$ over a
perfect field $F$ is universal over $M_{2}(F)$ if and only if atleast two of
$a_{1},a_{2},\ldots,a_{m}$ are nonzero.

The advantage of the above result is that one does not have to go to the
matrix associated with the quadratic form. From the coefficients of the
quadratic form itself, we can easily check the universality.

##  2 Proof of the main result

We prove our main result by proving the following sequence of theorems.

######  Theorem 6.

A quadratic form $aX^{2}$ where $a\in F$ is not universal over $M_{2}(F)$.

###### Proof.

If $a=0$ then clearly it is not universal over $M_{2}(F)$. So we suppose that
$a\neq 0$. Consider $a\left[\begin{array}[]{cc}x&y\\\
z&w\end{array}\right]^{2}=\left[\begin{array}[]{cc}0&1\\\
0&0\end{array}\right]$. Then we get the following system of equations

$\displaystyle a(x^{2}+yz)=0$ (6.1) $\displaystyle ay(x+w)=1$ (6.2)
$\displaystyle az(x+w)=0$ (6.3) $\displaystyle a(yz+w^{2})=0.$ (6.4)

From Equation (6.2) and (6.3), we get $z=0$. Then from Equation (6.1) and
(6.4), we get $x=0$ and $w=0$, a contradiction to Equation (6.2). ∎

######  Theorem 7.

A quadratic form $a_{1}X_{1}^{2}+a_{2}X_{2}^{2}$ where $a_{1},a_{2}\in F$,
char $F\neq 2$ and both $a_{1},a_{2}\neq 0$, is universal over $M_{2}(F)$.

###### Proof.

Consider $a_{1}\left[\begin{array}[]{cc}x_{1}&y_{1}\\\
z_{1}&w_{1}\end{array}\right]^{2}+a_{2}\left[\begin{array}[]{cc}x_{2}&y_{2}\\\
z_{2}&w_{2}\end{array}\right]^{2}=\left[\begin{array}[]{cc}p&q\\\
r&s\end{array}\right]$
which gives us the following system of equations

$\displaystyle
a_{1}x_{1}^{2}+a_{2}x_{2}^{2}+a_{1}y_{1}z_{1}+a_{2}y_{2}z_{2}=p$ (7.1)
$\displaystyle a_{1}y_{1}(x_{1}+w_{1})+a_{2}y_{2}(x_{2}+w_{2})=q$ (7.2)
$\displaystyle a_{1}z_{1}(x_{1}+w_{1})+a_{2}z_{2}(x_{2}+w_{2})=r$ (7.3)
$\displaystyle
a_{1}w_{1}^{2}+a_{2}w_{2}^{2}+a_{1}y_{1}z_{1}+a_{2}y_{2}z_{2}=s.$ (7.4)

From Equation (7.1) and (7.4), we get

$a_{1}y_{1}z_{1}+a_{2}y_{2}z_{2}=p-(a_{1}x_{1}^{2}+a_{2}x_{2}^{2})=s-(a_{1}w_{1}^{2}+a_{2}w_{2}^{2}).$
(7.5)

We will find a solution to the equation

$p-(a_{1}x_{1}^{2}+a_{2}x_{2}^{2})=s-(a_{1}w_{1}^{2}+a_{2}w_{2}^{2})$ (7.6)

such that $x_{1}+w_{1}\neq 0.$ If we do so, as $a_{1}\neq 0$ and $a_{2}\neq
0$, from Equations (7.2) and (7.3), we get the values of $y_{1}$ and $z_{1}$
as

$y_{1}=\dfrac{q-a_{2}y_{2}(x_{2}+w_{2})}{a_{1}(x_{1}+w_{1})}$ (7.7)
$z_{1}=\dfrac{r-a_{2}z_{2}(x_{2}+w_{2})}{a_{1}(x_{1}+w_{1})}.$ (7.8)

Now consider Equation (7.6) in the form
$a_{1}x_{1}^{2}+a_{2}x_{2}^{2}-a_{1}w_{1}^{2}-a_{2}w_{2}^{2}=p-s$ and choose
$x_{2}=w_{2}=0.$ Thus, our equation now reads
$a_{1}x_{1}^{2}-a_{1}w_{1}^{2}=p-s.$ Case(I): If $p-s=0$, as $a_{1}\neq 0$
take $x_{1}=w_{1}$. Since char $F\neq 2$ by selecting any nonzero value for
$x_{1}$ we have $x_{1}+w_{1}\neq 0.$

Case(II): If $p-s\neq 0$, we get $x_{1}^{2}-w_{1}^{2}=\dfrac{p-s}{a_{1}}$ i.e.
$(x_{1}-w_{1})(x_{1}+w_{1})=\dfrac{p-s}{a_{1}}$. We shall choose
$x_{1}$,$w_{1}$ such that $x_{1}-w_{1}=\dfrac{p-s}{a_{1}}$ and
$x_{1}+w_{1}=1.$ To see this, adding the two equations we get
$2x_{1}=\dfrac{p-s+a_{1}}{a_{1}}$ i.e. $x_{1}=\dfrac{p-s+a_{1}}{2a_{1}}$ and
$w_{1}=1-x_{1}=\dfrac{a_{1}-(p-s)}{2a_{1}}$.
From each of the cases above, we have $x_{1}+w_{1}\neq 0$. Now from Equation
(7.7) and (7.8), since we have chosen $x_{2}=w_{2}=0,$ we have the following
values of $y_{1}$ and $z_{1}:$

$\displaystyle y_{1}=\dfrac{q}{a_{1}(x_{1}+w_{1})}$ (7.9) $\displaystyle
z_{1}=\dfrac{r}{a_{1}(x_{1}+w_{1})}.$ (7.10)

Now from Equation (7.5), we get

$a_{2}y_{2}z_{2}=-a_{1}y_{1}z_{1}+s-a_{1}w_{1}^{2}$ (7.11)
$\text{i.e.}\;y_{2}z_{2}=\dfrac{1}{a_{2}}(-a_{1}y_{1}z_{1}+s-a_{1}w_{1}^{2}).$
(7.12)

Then by fixing any non-zero value of $z_{2}$ we get the value of $y_{2}$.
Hence, we get a solution to the system of equations. ∎

######  Definition 8.

A field $F$ is said to be perfect if it either has characteristic $0$ or when
characteristic $p>0$ then every element of $F$ is a $p^{th}$ power i.e.
$F^{p}=F$ where $F^{p}=\\{x^{p}:x\in F\\}$.

######  Lemma 9.

Let $F$ be a perfect field of characteristic $2.$ Then the map $f:F\rightarrow
F$ defined by $f(x)=x^{2}$ is an isomorphism.

###### Proof.

We have $f(x+y)=(x+y)^{2}=x^{2}+2xy+y^{2}=x^{2}+y^{2}=f(x)+f(y)$ and
$f(xy)=x^{2}y^{2}=f(x)f(y)$. As $f$ is a non-zero homomorphism it implies that
$f$ is injective (see [2]). Also $F$ is a perfect field of char $F=2$ which
implies $F=F^{2}$ and hence $f$ is onto. Therefore, $f$ is an isomorphism. ∎

######  Theorem 10.

A quadratic form $a_{1}X_{1}^{2}+a_{2}X_{2}^{2}$ where $a_{1},a_{2}\in F$, $F$
is a perfect field of char $F=2$ and both $a_{1},a_{2}\neq 0$ is universal
over $M_{2}(F)$.

###### Proof.

Consider the following system of equations

$\displaystyle
a_{1}x_{1}^{2}+a_{2}x_{2}^{2}+a_{1}y_{1}z_{1}+a_{2}y_{2}z_{2}=p$ (10.1)
$\displaystyle a_{1}y_{1}(x_{1}+w_{1})+a_{2}y_{2}(x_{2}+w_{2})=q$ (10.2)
$\displaystyle a_{1}z_{1}(x_{1}+w_{1})+a_{2}z_{2}(x_{2}+w_{2})=r$ (10.3)
$\displaystyle
a_{1}w_{1}^{2}+a_{2}w_{2}^{2}+a_{1}y_{1}z_{1}+a_{2}y_{2}z_{2}=s.$ (10.4)

From Equations (10.1) and (10.4), we will find solution to the equation

$a_{1}x_{1}^{2}+a_{2}x_{2}^{2}-a_{1}w_{1}^{2}-a_{2}w_{2}^{2}=p-s.$ (10.5)

Since char $F=2$, we may consider the following equation

$a_{1}x_{1}^{2}+a_{2}x_{2}^{2}+a_{1}w_{1}^{2}+a_{2}w_{2}^{2}=p+s.$ (10.6)

Case (I): If $p+s=0$ then take $x_{2}=w_{1}=0$, $w_{2}=1$ and then from Lemma
9 as $a_{1}\neq 0$ and $a_{2}\neq 0$, $a_{1}x_{1}^{2}=a_{2}$ has a non-zero
solution.
Case (II): If $p+s\neq 0$ then take $x_{2}=w_{1}=w_{2}=0$ then from Lemma 9,
$a_{1}x_{1}^{2}=p+s$ has a non-zero solution in $F$ as $a_{1}\neq 0$ and
$p+s\neq 0$.
From each of the above cases, we have $x_{1}+w_{1}\neq 0$ and further
following the proof of Theorem 7, we have a solution to the required system of
equations. ∎

######  Theorem 11.

A diagonal quadratic form $\sum_{i=1}^{m}a_{i}X_{i}^{2}$ over a perfect field
$F$ is universal over $M_{2}(F)$ if and only if atleast two elements of
$a_{1},a_{2},\ldots,a_{m}$ are nonzero.

###### Proof.

Without loss of generality let $a_{1},a_{2}\neq 0$ and $X_{i}=0$ for $3\leq
i\leq m$. Then from Theorem 7 and 10, $\sum_{i=1}^{m}a_{i}X_{i}^{2}$ is
universal. Conversely, suppose $\sum_{i=1}^{m}a_{i}X_{i}^{2}$ is universal
then from Theorem 6, we have the result. ∎

######  Remark 12.

We demonstrate the importance of a perfect field (see [6]) by showing that the
theorem may fail, if this condition does not hold. Consider the field
$F=\mathbb{Z}_{2}(X).$ This field is not perfect since $X$ is not a square in
$\mathbb{Z}_{2}(X);$ for suppose
$X=\dfrac{(a_{0}+a_{1}X+a_{2}X^{2}+\cdots+X^{k})^{2}}{(b_{0}+b_{1}X+b_{2}X^{2}+\cdots+X^{t})^{2}}$
which implies that
$X(b_{0}+b_{1}X+b_{2}X^{2}+\cdots+X^{t})^{2}=(a_{0}+a_{1}X+a_{2}X^{2}+\cdots+X^{k})^{2}$.
By comparing degrees, we get $2t+1=2k$, a contradiction. We now prove that the
quadratic form $X_{1}^{2}+X_{2}^{2}$ is not universal over
$M_{2}(\mathbb{Z}_{2}(X)).$ To see this, consider

$\left[\begin{array}[]{cc}x_{1}&y_{1}\\\
z_{1}&w_{1}\end{array}\right]^{2}+\left[\begin{array}[]{cc}x_{2}&y_{2}\\\
z_{2}&w_{2}\end{array}\right]^{2}=\left[\begin{array}[]{cc}p&q\\\
r&s\end{array}\right]$

which gives us the following system of equations:

$\displaystyle x_{1}^{2}+x_{2}^{2}+y_{1}z_{1}+y_{2}z_{2}=p$ (12.1)
$\displaystyle y_{1}(x_{1}+w_{1})+y_{2}(x_{2}+w_{2})=q$ (12.2) $\displaystyle
z_{1}(x_{1}+w_{1})+z_{2}(x_{2}+w_{2})=r$ (12.3) $\displaystyle
w_{1}^{2}+w_{2}^{2}+y_{1}z_{1}+y_{2}z_{2}=s.$ (12.4)

Existence of solution to the equation

$x_{1}^{2}+x_{2}^{2}+w_{1}^{2}+w_{2}^{2}=p+s$ (12.5)

is necessary for the existence of a solution to the original system of
equations. Since Char $F=2$, Equation (12.5) can be written as

$(x_{1}+x_{2}+w_{1}+w_{2})^{2}=p+s.$ (12.6)

Since $\mathbb{Z}_{2}(X)$ is not a perfect field the equation $x^{2}=a$ need
not have a solution for every $a\in\mathbb{Z}_{2}(X).$ (For example, $a=X$
works as shown above.) This proves that $X_{1}^{2}+X_{2}^{2}$ is not universal
over $M_{2}(\mathbb{Z}_{2}(X)).$

##  3 Algorithm of the proof

For a given matrix $A$ over $\mathbb{Q}$ and $a_{1},a_{2}\in\mathbb{Q}$,
following is the algorithm to find entries of the matrices $X_{1}$ and $X_{2}$
such that $a_{1}X_{1}^{2}+a_{2}X_{2}^{2}=A$ and the program can be implemented
in Python language.

1\. Import the packages numpy and fraction for declaring and displaying the
array values.
2\. Declare a two dimensional array A
3\. p=A[0][0]
4\. q=A[0][1]
5\. r=A[1][0]
6\. s=A[1][1]
7\. Initialize the values a1, a2.
8\. Initialize x2=0, w2=0 and z2=1.
9\. if p-s==0:
10\. x1=w1=1
11\. y1=q/(a1*(x1+w1))
12\. z1=r/(a1*(x1+w1))
13\. y2=(s-a1*y1*z1-a1*pow(w1,2))/(a2)
14\. Display y1,y2 and z1
15\. else:
16\. x1=(p-s+a1)/(2*a1)
17\. w1=(a1-p+s)/(2*a1)
18\. y1=q/(a1*(x1+w1))
19\. z1=r/(a1*(x1+w1))
20\. y2=(s-a1*y1*z1-a1*pow(w1,2))/(a2)
21\. Display x1, w1, y1, y2, z1

######  Example 13.

Consider $F=\mathbb{Q}$ and a diagonal quadratic form
$q(X_{1},X_{2})=2X_{1}^{2}+X_{2}^{2}$. We have $\text{char}\;F=0$. Let
$A=\left[\begin{array}[]{cc}\frac{1}{5}&2\\\ 0&-1\end{array}\right]\in
M_{2}(\mathbb{Q})$. By following the steps of Theorem 7, we will find a
solution to equation

$2\left[\begin{array}[]{cc}x_{1}&y_{1}\\\
z_{1}&w_{1}\end{array}\right]^{2}+\left[\begin{array}[]{cc}x_{2}&y_{2}\\\
z_{2}&w_{2}\end{array}\right]^{2}=\left[\begin{array}[]{cc}\frac{1}{5}&2\\\
0&-1\end{array}\right].$ (13.1)

Let $a_{1}=2,a_{2}=1,p=\dfrac{1}{5},q=2,r=0,s=-1$. Take $x_{2}=w_{2}=0$ and
$z_{2}=1$. We have $p-s=\dfrac{6}{5}\neq 0$. We get
$x_{1}=\dfrac{p-s+a_{1}}{2a_{1}}=\dfrac{4}{5}$ and
$w_{1}=\dfrac{a_{1}-(p-s)}{2a_{1}}=\dfrac{1}{5}$. Then we get
$y_{1}=\dfrac{q}{a_{1}(x_{1}+w_{1})}=1$,
$z_{1}=\dfrac{r}{a_{1}(x_{1}+w_{1})}=0$ and
$y_{2}=\dfrac{1}{a_{2}}(-a_{1}y_{1}z_{1}+s-a_{1}w_{1}^{2})=\dfrac{-27}{25}$.

Acknowledgement: We sincerely thank Professor S. Kotyada for useful
discussions. We also thank Professor S. A. Katre for all his support and
encouragement and Ms. Nivedita Elanshekar for helping in computational part.

## References

  * [1] Manjul Bhargava, On the Conway-Schneeberger Fifteen Theorem, Quadratic forms and their applications (Dublin, 1999), Contemporary Math; 272, American Mathematical Society, 2000, Providence, RI, 27-37.
  * [2] David S. Dummit, Richard M. Foote, Abstract Algebra, Wiley.
  * [3] Malcolm Griffin, Mark Krusemeyer, Matrices as sums of squares, Linear and Multilinear Algebra, 1977, Volume 5, 33-44.
  * [4] T. Y. Lam, Introduction to Quadratic forms over fields, Graduate studies in Mathematics, American Mathematical Society, Volume 67.
  * [5] Jungin Lee, Integral matrices as diagonal quadratic forms, Linear and Multilinear Algebra, Volume 66, 2018, Issue 4.
  * [6] Lawrence Ervin Gosnell, Quadratic forms over fields of characteristic 2, Blacksburg, Virginia, June 1973.
  * [7] L. N. Vaserstein, Every integral matrix is the sum of three squares, Linear and Multilinear Algebra, 1986, Volume 20, Issue 1, 1-4.
  * [8] L. N. Vaserstein, On the sums of powers of matrices, 1987, Volume 21, Issue 3.

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}\gamma=-1,\\\ \mathcal{C}_{4}&\text{ if }\alpha=-1\text{ and
}\gamma\neq-1,\\\ \mathcal{C}_{5}&\text{ if }\alpha,\gamma=-1\text{ and
}\beta\neq 0.\\\ \end{array}\right.$

This gives

$F_{i,2,k}={\footnotesize\begin{pmatrix}0&1&0&0&0\\\
q\left(q-1\right)^{2}&q\left(q-2\right)^{2}&q\left(q-2\right)\left(q-1\right)&q\left(q-2\right)\left(q-1\right)&q\left(q-1\right)^{2}\\\
0&q\left(q-2\right)&0&q\left(q-1\right)&0\\\
0&q\left(q-2\right)&q\left(q-1\right)&0&0\\\ 0&q-1&0&0&0\end{pmatrix}},$

with $i$ the row index and $k$ the column index. By completely similar
arguments, one can show that

$\begin{gathered}F_{i,3,k}={\tiny\begin{pmatrix}0&0&1&0&0\\\
0&q\left(q-2\right)&0&q\left(q-1\right)&0\\\
q\left(q-1\right)&0&q\left(q-2\right)&0&q\left(q-1\right)\\\ 0&q&0&0&0\\\
0&0&q-1&0&0\end{pmatrix}},\qquad F_{i,4,k}={\tiny\begin{pmatrix}0&0&0&1&0\\\
0&q\left(q-2\right)&q\left(q-1\right)&0&0\\\ 0&q&0&0&0\\\
q\left(q-1\right)&0&0&q\left(q-2\right)&q\left(q-1\right)\\\
0&0&0&q-1&0\end{pmatrix}}\\\ \\\ \text{and
}F_{i,5,k}={\tiny\begin{pmatrix}0&0&0&0&1\\\ 0&q-1&0&0&0\\\ 0&0&q-1&0&0\\\
0&0&0&q-1&0\\\ q-1&0&0&0&q-2\end{pmatrix}}.\end{gathered}$

Applying (3.5) now yields

$Z_{\pi}(L)=q^{6}(q-1)^{7}{\tiny\begin{pmatrix}q^{2}+q-1&q^{3}\left(q-2\right)^{2}&q^{3}\left(q-2\right)&q^{3}\left(q-2\right)&\left(q-1\right)^{2}\left(q+1\right)\\\
q^{3}\left(q-2\right)^{2}&q^{4}\left(q^{2}-3q+3\right)^{2}&q^{4}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{4}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)^{2}\left(q-1\right)\\\
q^{3}\left(q-2\right)&q^{4}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{4}\left(q^{2}-3q+3\right)&q^{4}\left(q-2\right)^{2}&q^{3}\left(q-2\right)\left(q-1\right)\\\
q^{3}\left(q-2\right)&q^{4}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{4}\left(q-2\right)^{2}&q^{4}\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)\left(q-1\right)\\\
\left(q-1\right)^{2}\left(q+1\right)&q^{3}\left(q-2\right)^{2}\left(q-1\right)&q^{3}\left(q-2\right)\left(q-1\right)&q^{3}\left(q-2\right)\left(q-1\right)&\left(q-1\right)\left(q^{3}-q^{2}+1\right)\end{pmatrix}}$

and precomposing with $\eta^{-1}$ gives

$\tilde{Z}(L)=q^{6}(q-1)^{5}\\\
{\tiny\begin{pmatrix}\left(q-1\right)^{2}\left(q^{2}+q-1\right)&q^{2}\left(q-2\right)^{2}&q^{2}\left(q-2\right)\left(q-1\right)&q^{2}\left(q-2\right)\left(q-1\right)&\left(q-1\right)^{3}\left(q+1\right)\\\
q^{3}\left(q-2\right)^{2}\left(q-1\right)^{2}&q^{3}\left(q^{2}-3q+3\right)^{2}&q^{3}\left(q-2\right)\left(q-1\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)\left(q-1\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)^{2}\left(q-1\right)^{2}\\\
q^{3}\left(q-2\right)\left(q-1\right)^{2}&q^{3}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-1\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)^{2}\left(q-1\right)&q^{3}\left(q-2\right)\left(q-1\right)^{2}\\\
q^{3}\left(q-2\right)\left(q-1\right)^{2}&q^{3}\left(q-2\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)^{2}\left(q-1\right)&q^{3}\left(q-1\right)\left(q^{2}-3q+3\right)&q^{3}\left(q-2\right)\left(q-1\right)^{2}\\\
\left(q-1\right)^{4}\left(q+1\right)&q^{2}\left(q-2\right)^{2}\left(q-1\right)&q^{2}\left(q-2\right)\left(q-1\right)^{2}&q^{2}\left(q-2\right)\left(q-1\right)^{2}&\left(q-1\right)^{2}\left(q^{3}-q^{2}+1\right)\end{pmatrix}}.$

This matrix can be diagonalized as follows:

$\tilde{Z}(L)=q^{6}(q-1)^{5}A{\footnotesize\begin{pmatrix}q^{3}&&&&\\\
&q(q-1)^{2}&&&\\\ &&q^{3}(q-1)^{2}&&\\\ &&&q^{3}(q-1)^{2}&\\\
&&&&q^{3}(q-1)^{4}\end{pmatrix}}A^{-1}\\\ \\\ \text{with
}A={\footnotesize\begin{pmatrix}1&1&0&1&1\\\ q&0&0&-q(q-1)&q(q-1)^{2}\\\
-q&0&1&0&q(q-1)\\\ -q&0&-1&q(q-2)&q(q-1)\\\ q-1&-1&0&q-1&q-1\end{pmatrix}}.$

######  Remark 3.9.

We see that $Z_{\pi}(L)$ is symmetric. This can be explained by the fact that
for each $i$ and $j$ we have an isomorphism:

$\begin{array}[]{ccc}\\{(g,g_{1},g_{2})\in\mathcal{C}_{i}\times
G^{2}:g[g_{1},g_{2}]\in\mathcal{C}_{j}\\}&\overset{\sim}{\longleftrightarrow}&\\{(g,g_{1},g_{2})\in\mathcal{C}_{j}\times
G^{2}:g[g_{1},g_{2}]\in\mathcal{C}_{i}\\}\\\
(g,g_{1},g_{2})&\mapsto&(g[g_{1},g_{2}],g_{2},g_{1})\\\
(g[g_{1},g_{2}],g_{2},g_{1})&\mapsfrom&(g,g_{1},g_{2})\end{array}$

Hence the classes of both sides are equal in
$\text{K}(\textbf{Var}_{\mathbb{C}})$, and so
$Z_{\pi}(L)(T_{i})|_{T_{j}}=Z_{\pi}(L)(T_{j})|_{T_{i}}$. This does not hold
for any TQFT, but it relies on the fact that the $C_{i}$ are points.

######  Theorem 3.10.

The virtual class of the $\mathbb{U}_{3}$-representation variety
$\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{g})$ is

$[\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{g})]=q^{3g-3}(q-1)^{2g}\left(q^{2}(q-1)^{2g+1}+q^{3g}(q-1)^{2}+q^{3g}(q-1)^{4g}+2q^{3g}(q-1)^{2g+1}\right).$

###### Proof.

One can check that

$A^{-1}=\frac{1}{q^{3}}{\footnotesize\begin{pmatrix}(q-1)^{2}&1&1-q&1-q&(q-1)^{2}\\\
q^{2}(q-1)&0&0&0&-q^{2}\\\
q(q-2)(q-1)&-q(q-2)&q^{3}-2q^{2}+2q&-2q(q-1)&q^{3}-3q^{2}+2q\\\
2q-2&-2&q-2&q-2&2q-2\\\ 1&1&1&1&1\end{pmatrix}}.$

By matrix multiplication, we find that

$\displaystyle\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{g})$
$\displaystyle=\frac{1}{[\mathbb{U}_{3}]^{g}}\tilde{Z}(L)^{g}(T_{1})|_{T_{1}}$
$\displaystyle=q^{3g-3}(q-1)^{2g}\left(q^{2}(q-1)^{2g+1}+q^{3g}(q-1)^{2}+q^{3g}(q-1)^{4g}+2q^{3g}(q-1)^{2g+1}\right).$

∎

######  Remark 3.11.

In particular, for small values of $g$, we find

$\displaystyle[\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{1})]$
$\displaystyle=q^{3}(q-1)^{4}(q^{2}+q-1),$
$\displaystyle[\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{2})]$
$\displaystyle=q^{7}\left(q-1\right)^{6}\left(q^{8}-6q^{7}+15q^{6}-18q^{5}+9q^{4}+q^{3}-3q^{2}+3q-1\right),$
$\displaystyle[\mathfrak{X}_{\mathbb{U}_{3}}(\Sigma_{3})]$
$\displaystyle=q^{11}\left(q-1\right)^{8}\left(q^{14}-10q^{13}+45q^{12}-120q^{11}+210q^{10}-250q^{9}+200q^{8}\right.$
$\displaystyle\qquad\left.-100q^{7}+25q^{6}+q^{5}-5q^{4}+10q^{3}-10q^{2}+5q-1\right).$

As in Remark 3.4, the factor $(q-1)^{2g+2}$ can be explained from the actions
of $(\mathbb{C}^{*})^{2g}$ (given by scaling the $A_{i},B_{i}$) and
$(\mathbb{C}^{*})^{2}$ (given by conjugating with
$\left(\begin{smallmatrix}1&&\\\ &x&\\\
&&y\end{smallmatrix}\right),x,y\in\mathbb{C}^{*}$).

###  3.3 Upper triangular $4\times 4$ matrices

The last case we will treat is the group $\mathbb{U}_{4}$ of upper triangular
$4\times 4$ matrices. We can use the same strategies as in the previous case
of $\mathbb{U}_{3}$, but all computations are done using Algorithm 4.1. Source
code for these computations is given in [27]. The group $\mathbb{U}_{4}$
contains sixteen unipotent conjugacy classes [3]. We consider the following
representatives of these classes

$\begin{gathered}{\tiny\begin{pmatrix}1&&&\\\ &1&&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&&\\\ &1&&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&1&\\\ &1&&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&&1\\\ &1&&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&&\\\ &1&1&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&&\\\ &1&&1\\\ &&1&\\\
&&&1\end{pmatrix},}\\\ {\tiny\begin{pmatrix}1&&&\\\ &1&&\\\ &&1&1\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&&\\\ &1&1&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&&\\\ &1&&1\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&&\\\ &1&&\\\ &&1&1\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&1&\\\ &1&&1\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&1&\\\ &1&&\\\ &&1&1\\\
&&&1\end{pmatrix},}\\\ {\tiny\begin{pmatrix}1&&&\\\ &1&1&\\\ &&1&1\\\
&&&1\end{pmatrix},\begin{pmatrix}1&&&1\\\ &1&1&\\\ &&1&\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&&\\\ &1&1&\\\ &&1&1\\\
&&&1\end{pmatrix},\begin{pmatrix}1&1&1&\\\ &1&&\\\ &&1&1\\\
&&&1\end{pmatrix}},\end{gathered}$

which we denote in order by $\xi_{1},\ldots,\xi_{16}$. Explicitly, the
conjugacy classes are given by

$\displaystyle\mathcal{C}_{1}$
$\displaystyle=\\{a_{0,1}=a_{0,2}=a_{0,3}=a_{1,2}=a_{1,3}=a_{2,3}=0\\},$
$\displaystyle\mathcal{C}_{2}$
$\displaystyle=\\{a_{1,2}=a_{1,3}=a_{2,3}=0,a_{0,1}\neq 0\\},$
$\displaystyle\mathcal{C}_{3}$
$\displaystyle=\\{a_{0,1}=a_{1,2}=a_{1,3}=a_{2,3}=0,a_{0,2}\neq 0\\},$
$\displaystyle\mathcal{C}_{4}$
$\displaystyle=\\{a_{0,1}=a_{0,2}=a_{1,2}=a_{1,3}=a_{2,3}=0,a_{0,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{5}$
$\displaystyle=\\{a_{0,1}=a_{2,3}=a_{0,3}a_{1,2}-a_{0,2}a_{1,3}=0,a_{1,2}\neq
0\\},$ $\displaystyle\mathcal{C}_{6}$
$\displaystyle=\\{a_{0,1}=a_{0,2}=a_{1,2}=a_{2,3}=0,a_{1,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{7}$
$\displaystyle=\\{a_{0,1}=a_{0,2}=a_{1,2}=0,a_{2,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{8}$ $\displaystyle=\\{a_{2,3}=0,a_{0,1}\neq
0,a_{1,2}\neq 0\\},$ $\displaystyle\mathcal{C}_{9}$
$\displaystyle=\\{a_{1,2}=a_{2,3}=0,a_{0,1}\neq 0,a_{1,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{10}$
$\displaystyle=\\{a_{1,2}=a_{0,2}a_{2,3}+a_{0,1}a_{1,3}=0,a_{0,1}\neq
0,a_{2,3}\neq 0\\},$ $\displaystyle\mathcal{C}_{11}$
$\displaystyle=\\{a_{0,1}=a_{1,2}=a_{2,3}=0,a_{0,2}\neq 0,a_{1,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{12}$
$\displaystyle=\\{a_{0,1}=a_{1,2}=0,a_{0,2}\neq 0,a_{2,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{13}$ $\displaystyle=\\{a_{0,1}=0,a_{1,2}\neq
0,a_{2,3}\neq 0\\},$ $\displaystyle\mathcal{C}_{14}$
$\displaystyle=\\{a_{0,1}=a_{2,3}=0,a_{1,2}\neq
0,a_{0,3}a_{1,2}-a_{0,2}a_{1,3}\neq 0\\},$ $\displaystyle\mathcal{C}_{15}$
$\displaystyle=\\{a_{0,1}\neq 0,a_{1,2}\neq 0,a_{2,3}\neq 0\\},$
$\displaystyle\mathcal{C}_{16}$ $\displaystyle=\\{a_{1,2}=0,a_{0,1}\neq
0,a_{2,3}\neq 0,a_{0,2}a_{2,3}+a_{0,1}a_{1,3}\neq 0\\}.$

As before, we write $T_{i}=[C_{i}]_{C_{i}}\in\text{K}(\textbf{Var}/C_{i})$ and
consider $V=\langle T_{1},\ldots,T_{16}\rangle$. We wish to compute
$Z_{\pi}(L)(T_{1})$ first and deduce the other columns from this column as we
did before. We have

$Z_{\pi}(L)(T_{1})|_{T_{i}}=[\\{(g_{1},g_{2})\in\mathbb{U}_{4}:[g_{1},g_{2}]\in\mathcal{C}_{i}\\}]$

In terms of coordinates, this will yield systems of equations in 20 variables.
Using Algorithm 4.1 we obtain

$Z_{\pi}(L)(T_{1})=q^{12}(q-1)^{9}{\footnotesize\begin{pmatrix}q^{3}+4q^{2}-6q+4\\\
q^{3}\left(q-2\right)\left(q^{2}+q-1\right)\\\
q\left(q^{2}-q+1\right)\left(q^{2}+q-4\right)\\\
\left(q-1\right)\left(q^{3}+3q^{2}-6q+4\right)\\\
q^{2}\left(q-2\right)\left(q^{3}+q-1\right)\\\
q\left(q^{4}+q^{3}-8q^{2}+9q-4\right)\\\
q^{2}\left(q-2\right)\left(q^{3}+2q^{2}-4q+2\right)\\\
q^{6}\left(q-2\right)^{2}\\\
q^{3}\left(q-2\right)\left(q-1\right)^{2}\left(q+1\right)\\\
q^{3}\left(q-2\right)\left(q-1\right)\left(q^{2}-q-1\right)\\\
q\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)\\\
q^{2}\left(q-2\right)\left(q-1\right)\left(q^{3}-2q+2\right)\\\
q^{6}\left(q-2\right)^{2}\\\
q^{2}\left(q-2\right)\left(q-1\right)^{2}\left(q^{2}+q+1\right)\\\
q^{6}\left(q-2\right)^{3}\\\
q^{3}\left(q-2\right)\left(q^{4}-3q^{3}+2q^{2}-1\right)\end{pmatrix}}.$

Completely analogous to the previous section, the other columns can be
computed from this result via

$Z_{\pi}(L)(T_{j})|_{T_{i}}=\sum_{k=1}^{16}F_{ijk}\cdot
Z_{\pi}(L)(T_{1}|_{T_{k}}).$

where

$F_{ijk}=[\\{g\in\mathcal{C}_{j}:g\xi_{k}\in\mathcal{C}_{i}\\}].$

Again, see [27] for the actual computations. As usual, the map
$\eta=\pi_{!}\pi^{*}$ is diagonal, with $\eta(T_{i})=[\mathcal{C}_{i}]T_{i}$.
Composing $Z_{\pi}(L)$ with $\eta^{-1}$ gives us $\tilde{Z}(L)$, which can be
found in Appendix B. After diagonalizing the reduced TQFT
$\tilde{Z}(L)=Z_{\pi}(L)\circ\eta^{-1}$, we obtain the following result.

######  Theorem 3.12.

The virtual class of the $\mathbb{U}_{4}$-representation variety
$\mathfrak{X}_{\mathbb{U}_{4}}(\Sigma_{g})$ is

$[\mathfrak{X}_{\mathbb{U}_{4}}(\Sigma_{g})]=q^{8g-2}\left(q-1\right)^{4g+2}+q^{8g-2}\left(q-1\right)^{6g+1}+q^{10g-4}\left(q-1\right)^{2g+3}\\\
+q^{10g-4}\left(q-1\right)^{4g+1}\left(2q^{2}-6q+5\right)^{g}+3q^{10g-4}\left(q-1\right)^{4g+2}+q^{10g-4}\left(q-1\right)^{6g+1}\\\
+q^{12g-6}\left(q-1\right)^{8g}+q^{12g-6}\left(q-1\right)^{2g+3}+3q^{12g-6}\left(q-1\right)^{4g+2}+3q^{12g-6}\left(q-1\right)^{6g+1}.$

######  Remark 3.13.

For small values of $g$, we have

$\displaystyle[\mathfrak{X}_{\mathbb{U}_{4}}(\Sigma_{1})]$
$\displaystyle=q^{6}\left(q-1\right)^{5}\left(q^{3}+4q^{2}-6q+4\right),$
$\displaystyle[\mathfrak{X}_{\mathbb{U}_{4}}(\Sigma_{2})]$
$\displaystyle=q^{15}\left(q-1\right)^{7}\left(q^{12}-9q^{11}+36q^{10}-81q^{9}+108q^{8}-76q^{7}-11q^{6}+124q^{5}-219q^{4}\right.$
$\displaystyle\qquad\left.+222q^{3}-126q^{2}+36q-3\right),$
$\displaystyle[\mathfrak{X}_{\mathbb{U}_{4}}(\Sigma_{3})]$
$\displaystyle=q^{23}\left(q-1\right)^{9}\left(q^{22}-15q^{21}+105q^{20}-455q^{19}+1365q^{18}-3000q^{17}+4975q^{16}\right.$
$\displaystyle\qquad\left.-6300q^{15}+6075q^{14}-4366q^{13}+2136q^{12}-93q^{11}-2139q^{10}+5157q^{9}\right.$
$\displaystyle\qquad\left.-8101q^{8}+8885q^{7}-6746q^{6}+3465q^{5}-1196q^{4}+329q^{3}-110q^{2}+35q-5\right).$

###  3.4 Moduli space of representations and character variety

Let $X$ be a path-connected topological space with finitely generated
fundamental group and $G$ a linear algebraic group over an algebraically
closed field $k$. There is an action of $G$ on the affine representation
variety $\mathfrak{X}_{G}(X)$ given by conjugation, and one can look at the
geometric invariant theory (GIT) quotient

$\mathcal{M}_{G}(X)=\mathfrak{X}_{G}(X)\sslash G,$

defined as the spectrum of the ring of invariants,
$\operatorname{Spec}(\mathcal{O}_{\mathfrak{X}_{G}(X)})^{G}$. This scheme is
known as the moduli space of $G$-representations.

For reductive groups $G$, the GIT quotient is finitely generated over $k$ by a
theorem of Nagata [22]. We specialize to the case $G=\mathbb{U}_{n}$, which
are non-reductive groups, so the GIT quotient is not guaranteed to be finitely
generated over $k$. However, the GIT quotient is a categorical quotient [23].
For this reason, we will find the GIT quotient by constructing the categorical
quotient of $\mathfrak{X}_{G}(X)$ by $G$. We begin with the following lemma.

######  Lemma 3.14.

Let $X$ be a variety over an algebraically closed field $k$ equipped with a
$G$-action. Let $\pi:X\to Y$ be a $G$-equivariant morphism of varieties over
$k$, such that the action of $G$ on $Y$ is trivial. Assume that there exists a
$G$-equivariant morphism $\sigma:Y\to X$ a such that
$\pi\circ\sigma=\text{id}_{Y}$. If for any $x\in X$ the Zariski-closure of the
$G$-orbit of $x$ contains $\sigma(\pi(x))$, then $\pi$ is a categorical
quotient.

###### Proof.

Let $f:X\to Z$ be any $G$-invariant morphism. We need to show there exists a
unique $G$-invariant morphism $g:Y\to Z$ such that $f=g\circ\pi$. If such a
morphism exists, it must be given by $g=f\circ\sigma$ since
$f\circ\sigma=(g\circ\pi)\circ\sigma=g$. This already shows uniqueness.

We show now that for $g=f\circ\sigma$ we have $f=g\circ\pi$. Take $x\in X$ and
note that as $f$ is $G$-invariant, so $f(\tilde{x})=f(x)$ for any $\tilde{x}$
in the orbit of $x$. By continuity we find that $f(\sigma(\pi(x)))=f(x)$
finishing the proof. ∎

We apply this lemma to the case of $G=\mathbb{U}_{2}$.

######  Lemma 3.15.

There exists an isomorphism of varieties
$\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})\simeq(\mathbb{G}_{m}^{*})^{4g}$ over
$\mathbb{C}$.

###### Proof.

Let

$M=\Big{\\{}(A_{1},B_{1},\ldots,A_{g},B_{g})\in\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g}):\text{all
$A_{i},B_{i}$ are diagonal}\Big{\\}},$

be the variety $(\mathbb{G}_{m})^{4g}$ and consider the morphism
$\pi:\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g})\to M$ that sends every
component $\left(\begin{smallmatrix}a_{i}&b_{i}\\\
0&c_{i}\end{smallmatrix}\right)$ to the component
$\left(\begin{smallmatrix}a_{i}&0\\\ 0&c_{i}\end{smallmatrix}\right)$. Take
$\sigma:M\to\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g})$ to be the inclusion.
Now indeed $\pi$ is $\mathbb{U}_{2}$-invariant and for any

$A={\footnotesize\left(\begin{pmatrix}a_{1}&b_{1}\\\
0&c_{1}\end{pmatrix},\ldots,\begin{pmatrix}a_{2g}&b_{2g}\\\
0&c_{2g}\end{pmatrix}\right)}\in\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g})$

we find that

$\lim_{x\to 0}{\footnotesize\begin{pmatrix}x&0\\\
0&1\end{pmatrix}}\;A\;{\footnotesize\begin{pmatrix}x&0\\\
0&1\end{pmatrix}^{-1}}={\footnotesize\left(\begin{pmatrix}a_{1}&0\\\
0&c_{1}\end{pmatrix},\ldots,\begin{pmatrix}a_{2g}&0\\\
0&c_{2g}\end{pmatrix}\right)}=\sigma(\pi(A)),$

hence $\sigma(\pi(A))$ lies in the analytic-closure (so also Zariski-closure)
of the orbit of $A$. By Lemma 3.14 we conclude that
$\pi:\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g})\to M$ is the categorical
quotient of $\mathfrak{X}_{\mathbb{U}_{2}}(\Sigma_{g})$ by the action of
$\mathbb{U}_{2}$ providing the required isomorphism
$\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})\simeq(\mathbb{C}^{*})^{4g}$. ∎

We remark that the moduli space of $G$-representations
$\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})$ is a variety even though
$\mathbb{U}_{2}$ is a non-reductive linear group.

We turn our attention to the $G$-character variety of $X$. We write
$\Gamma=\pi_{1}(X)$, which is assumed to be finitely generated. The character
of a representation $\rho\in\mathfrak{X}_{G}(X)$ is defined as the map

$\chi_{\rho}:\Gamma\to k:\gamma\mapsto\text{tr}(\rho(\gamma)),$

and the character map as

$\chi:\mathfrak{X}_{G}(X)\to k^{\Gamma}:\rho\mapsto\chi_{\rho}.$

The image of $\chi$ is called the $G$-character variety, denoted
$\chi_{G}(X)$. By the results from [6], there exists a finite set of elements
$\gamma_{1},\ldots,\gamma_{a}\in\pi_{1}(X)$ such that $\chi_{\rho}$ is
determined by the characters
$\chi_{\rho}(\gamma_{1}),\ldots\chi_{\rho}(\gamma_{a})$ for any $\rho$. This
way $\chi_{G}(X)$ can be identified with the image of the map
$\mathfrak{X}_{G}(X)\to
k^{a}:\rho\mapsto(\chi_{\rho}(\gamma_{1}),\ldots,\chi_{\rho}(\gamma_{a}))$,
which gives the $G$-character variety the structure of a variety. This
structure is independent of the chosen $\gamma_{i}$.

Note that the character map $\chi$ is a $G$-invariant morphism: indeed the
trace map is invariant under conjugation. By the universal property of the
categorical quotient, we obtain an induced map

$\overline{\chi}:\mathcal{M}_{G}(X)\to\chi_{G}(X).$

In the case of $G=\text{SL}_{n}(\mathbb{C})$, $\text{Sp}_{2n}(\mathbb{C})$ or
$\text{SO}_{2n+1}(\mathbb{C})$ this map is an isomorphism ([6, 7, 16]).
However, $\overline{\chi}$ fails to be an isomorphism for the surfaces
$\Sigma_{g}$ and the non-reductive groups $G=\mathbb{U}_{2}$ with $g\geq 1$.

######  Theorem 3.16.

The map

$\overline{\chi}:\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})\to\chi_{\mathbb{U}_{2}}(\Sigma_{g})$

is not an isomorphism.

###### Proof.

Arguing as in Lemma 3.14, the map $\overline{\chi}$ must be given by
$\overline{\chi}(A)=\chi_{A}$. But this cannot be an isomorphism: for general
$A\in\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})$ one can consider
$B=\left(\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right)A\left(\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right)^{-1}$ (where the diagonal entries of $A$ are
interchanged), and we have $\chi_{A}=\chi_{B}$, even though in general $A\neq
B$. Therefore, the moduli space $\mathcal{M}_{\mathbb{U}_{2}}(\Sigma_{g})$ is
not isomorphic to the character variety $\chi_{\mathbb{U}_{2}}(\Sigma_{g})$
through the natural map. ∎

This argument can easily be generalized to the case $G=\mathbb{U}_{n}$ for any
$n\geq 2$. Namely, similar to before take

$M=\Big{\\{}\left(A_{1},B_{1},\ldots,A_{2g},B_{2g}\right)\in\mathfrak{X}_{\mathbb{U}_{n}}(\Sigma_{g}):\text{all
$A_{i},B_{i}$ are diagonal}\Big{\\}}$

and $\pi:\mathfrak{X}_{\mathbb{U}_{n}}(\Sigma_{g})\to M$ the map that sets all
off-diagonal entries to zero, and
$\sigma:M\to\mathfrak{X}_{\mathbb{U}_{n}}(\Sigma_{g})$ the inclusion. Indeed
$\sigma\circ\pi=\text{id}$ and one easily checks that $\sigma(\pi(A))$ lies in
the closure of the orbit of $A$ for any
$A\in\mathfrak{X}_{\mathbb{U}_{n}}(\Sigma_{g})$ (e.g. conjugate $A$ with
${\footnotesize\begin{pmatrix}x^{n-1}&&&\\\ &x^{n-2}&&\\\ &&\ddots&\\\
&&&1\end{pmatrix}}$ and take the limit $x\to 0$). This way we find that
$\mathcal{M}_{\mathbb{U}_{n}}(\Sigma_{g})\simeq(\mathbb{G}_{m})^{2ng}$. Again
note that the natural map
$\overline{\chi}:\mathcal{M}_{\mathbb{U}_{n}}(\Sigma_{g})\to\chi_{\mathbb{U}_{n}}(\Sigma_{g})$
cannot be an isomorphism since there are symmetries (permuting diagonal
entries) that are invariant under the character map proving Theorem 1.2.

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##  4 Appendix A

In this section, we describe our algorthim for computing classes of affine
varieties over $\mathbb{C}$ in the Grothendieck ring of varieties
$\text{K}(\textbf{Var}_{\mathbb{C}})$ in terms of
$q=[\mathbb{A}^{1}_{\mathbb{C}}]$.

Let $S=\\{x_{1},\ldots,x_{n}\\}$ be a finite set (of variables), and $F,G$ be
finite subsets of $\mathbb{C}[S]$. Then we write $X(S,F,G)$ for the (reduced)
subvariety of $\mathbb{A}_{\mathbb{C}}^{n}$ given by $f=0$ for all $f\in F$
and $g\neq 0$ for all $g\in G$. For example,

$\mathbb{A}_{\mathbb{C}}^{n}=X(\\{x_{1},\ldots,x_{n}\\},\varnothing,\varnothing)\quad\text{
and }\quad\text{GL}_{2}(\mathbb{C})=X(\\{a,b,c,d\\},\varnothing,\\{ad-bc\\}).$

For convenience we will write $\text{ev}_{x}(f,u)$ with $f\in\mathbb{C}[S]$
for the polynomial where $x\in S$ in $f$ is substituted for
$u\in\mathbb{C}[S]$. Then we write

$\text{ev}_{x}(F,u)=\\{\text{ev}_{x}(f,u):f\in F\\}$

and

$\text{ev}_{x}(F,u,v)=\\{v^{\deg_{x}(f)}\cdot\text{ev}_{x}(f,u/v):f\in F\\}$

for a set of polynomials $F$ and $x\in S$ and $u,v\in\mathbb{C}[S]$. Note that
for $\text{ev}_{x}(F,u,v)$ the substituted polynomials are multiplied by a
suitable number of factors $v$, in order to clear denominators.

We describe our recursive algorithm which computes the class $[X(S,F,G)]$ in
terms of $q=[\mathbb{A}^{1}_{\mathbb{C}}]$.

######  Algorithm 4.1.

Let $X=X(S,F,G)$ for some $S,F$ and $G$ as above.

  1. 1.

If $F$ contains a non-zero constant or if $0\in G$, then $X=\varnothing$, so
$[X]=0$.

  2. 2.

If $F=\varnothing$ and $G=\varnothing$, then
$X=\mathbb{A}_{\mathbb{C}}^{\\#S}$, so $[X]=q^{\\#S}$.

  3. 3.

If some $x\in S$ ‘does not appear’ in any $f\in F$ and any $g\in G$, then we
can factor $X\simeq\mathbb{A}^{1}_{\mathbb{C}}\times X^{\prime}$ with
$X^{\prime}=X(S\setminus\\{x\\},F,G)$. We have $[X]=q[X^{\prime}]$.

  4. 4.

If $f=u^{n}$ (with $n>1$) for some $f\in F$ and $u\in\mathbb{C}[S]$, then we
can replace $f$ with $u$, not changing $X$. That is,
$X=X(S,F\setminus\\{f\\}\cup\\{u\\},G)$. Similarly, if $g=u^{n}$ (with $n>1$)
for some $g\in G$ and $u\in\mathbb{C}[S]$, then
$X=X(S,F,G\setminus\\{g\\}\cup\\{u\\})$.

  5. 5.

If some $f\in F$ is univariate in $x\in S$, we write
$f=(x-\alpha_{1})\cdots(x-\alpha_{m})$, and we have
$[X]=\sum_{i=1}^{m}[X_{i}]$ with
$X_{i}=X(S\setminus\\{x\\},\text{ev}_{x}(F\setminus\\{f\\},\alpha_{i}),\text{ev}_{x}(G,\alpha_{i}))$.

  6. 6.

If $f=uv$ for some $f\in F$ and $u,v\in\mathbb{C}[S]$ (both not constant),
then $X_{1}=X(S,F\setminus\\{f\\}\cup\\{u\\},G)=X\cap\\{u=0\\}$ and
$X_{2}=X(S,F\setminus\\{f\\}\cup\\{v\\},G\cup\\{u\\})=X\cap\\{u\neq 0,v=0\\}$
define a stratification for $X$, and thus $[X]=[X_{1}]+[X_{2}]$.

  7. 7.

If $f=xu+v$ for some $f\in F$, $x\in S$ and $u,v\in\mathbb{C}[S]$ with $x$ not
appearing in $u$ and $v$, then we consider the following cases. For any point
$p$ of $X$, either $u(p)=0$, implying $v(p)=0$ as well, or $u(p)\neq 0$,
implying $x(p)=-v(p)/u(p)$. Therefore $[X]=[X_{1}]+[X_{2}]$ with
$X_{1}=X(S,F\setminus\\{f\\}\cup\\{u,v\\},G)$ and
$X_{2}=X(S,\text{ev}_{x}(F\setminus\\{f\\},-v,u)),\text{ev}_{x}(G,-v,u)\cup\\{u\\})$.

  8. 8.

Suppose $f=x^{2}u+xv+w$ for some $f\in F$, $x\in S$ and
$u,v,w\in\mathbb{C}[S]$ with $x$ not appearing in $u,v$ and $w$. Moreover,
suppose that the discriminant $D=v^{2}-4uw$ is a square, i.e. we can write
$D=h^{2}$ for some $h\in\mathbb{C}[S]$. Then for any point $p$ of $X$, we
consider the following cases. Either $u(p)=0$, in which case $(xv+w)(p)=0$. If
$u(p)\neq 0$, we distinguish between $D(p)=0$ and $D(p)\neq 0$. In the first
case we find that $x(p)=\left(\frac{-v}{2u}\right)(p)$, and in the latter case
we have the two possibilities $x(p)=\left(\frac{-v\pm h}{2u}\right)(p)$. Hence
$[X]=[X_{1}]+[X_{2}]+[X_{3}]+[X_{4}]$, with

$\displaystyle X_{1}$
$\displaystyle=X(S,F\setminus\\{f\\}\cup\\{u,xv+w\\},G),$ $\displaystyle
X_{2}$
$\displaystyle=X(S,\text{ev}_{x}(F\setminus\\{f\\},-v,2u)\cup\\{D\\},\text{ev}_{x}(G,-v,2u)\cup\\{u\\}),$
$\displaystyle X_{3}$
$\displaystyle=X(S,\text{ev}_{x}(F\setminus\\{f\\},-v-h,2u),\text{ev}_{x}(G,-v-h,2u)\cup\\{u,D\\}),$
$\displaystyle X_{4}$
$\displaystyle=X(S,\text{ev}_{x}(F\setminus\\{f\\},-v+h,2u),\text{ev}_{x}(G,-v+h,2u)\cup\\{u,D\\}).$

  9. 9.

If $G\neq\varnothing$, pick any $g\in G$. We have $[X]=[X_{1}]-[X_{2}]$ where
$X_{1}=X(S,F,G\setminus\\{g\\})$ and $X_{2}=X(S,F\cup\\{g\\},G)$.

An implementation of this algorithm in Python can be found at [27], together
with the code for the computations done Sections 3.2 and 3.3.

##  5 Appendix B

The map $\tilde{Z}(L)$ for $G=\mathbb{U}_{4}$ is given by the following
matrix.

$\left[\begin{array}[]{c|cccc}&T_{1}&T_{2}&T_{3}&T_{4}\\\ \hline\cr
T_{1}&q^{6}\left(q-1\right)^{5}\left(q^{3}+4q^{2}-6q+4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}+q-1\right)&q^{6}\left(q-1\right)^{4}\left(q^{2}-q+1\right)\left(q^{2}+q-4\right)&q^{6}\left(q-1\right)^{5}\left(q^{3}+3q^{2}-6q+4\right)\\\
T_{2}&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q-1\right)&q^{9}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\left(q^{2}+q-1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q-1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q-1\right)\\\
T_{3}&q^{7}\left(q-1\right)^{5}\left(q^{2}-q+1\right)\left(q^{2}+q-4\right)&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q-1\right)&q^{7}\left(q-1\right)^{4}\left(q^{5}-q^{4}-2q^{3}+5q^{2}-5q+4\right)&q^{7}\left(q-1\right)^{5}\left(q^{2}-q+1\right)\left(q^{2}+q-4\right)\\\
T_{4}&q^{6}\left(q-1\right)^{6}\left(q^{3}+3q^{2}-6q+4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q-1\right)&q^{6}\left(q-1\right)^{5}\left(q^{2}-q+1\right)\left(q^{2}+q-4\right)&q^{6}\left(q-1\right)^{5}\left(q^{4}+2q^{3}-8q^{2}+10q-4\right)\\\
T_{5}&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+q-1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{5}&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{2}+q+1\right)\\\
T_{6}&q^{7}\left(q-1\right)^{5}\left(q^{4}+q^{3}-8q^{2}+9q-4\right)&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q+1\right)&q^{7}\left(q-1\right)^{4}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)&q^{7}\left(q-1\right)^{5}\left(q^{4}+q^{3}-8q^{2}+9q-4\right)\\\
T_{7}&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2q^{2}-4q+2\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-2q+2\right)&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2q^{2}-4q+2\right)\\\
T_{8}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}\\\
T_{9}&q^{9}\left(q-2\right)\left(q-1\right)^{7}\left(q+1\right)&q^{9}\left(q-1\right)^{6}\left(q+1\right)\left(q^{2}-3q+3\right)&q^{9}\left(q-2\right)\left(q-1\right)^{7}\left(q+1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{7}\left(q+1\right)\\\
T_{10}&q^{9}\left(q-2\right)\left(q-1\right)^{6}\left(q^{2}-q-1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{9}\left(q-2\right)\left(q-1\right)^{4}\left(q^{4}-3q^{3}+2q^{2}-1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{6}\left(q^{2}-q-1\right)\\\
T_{11}&q^{7}\left(q-1\right)^{5}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)&q^{8}\left(q-2\right)\left(q-1\right)^{7}\left(q+1\right)&q^{7}\left(q-1\right)^{4}\left(q^{2}-q+1\right)\left(q^{4}-2q^{3}-q^{2}+5q-4\right)&q^{7}\left(q-1\right)^{5}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)\\\
T_{12}&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}-2q+2\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}+2\right)&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}-2q+2\right)\\\
T_{13}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}\\\
T_{14}&q^{8}\left(q-2\right)\left(q-1\right)^{7}\left(q^{2}+q+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)\left(q-1\right)^{6}&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{4}-q^{3}+1\right)\\\
T_{15}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{5}\\\
T_{16}&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{4}-3q^{3}+2q^{2}-1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}-3q+3\right)&q^{9}\left(q-2\right)\left(q-1\right)^{4}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{4}-3q^{3}+2q^{2}-1\right)\end{array}\right.$

$\begin{array}[]{cccc}T_{5}&T_{6}&T_{7}&T_{8}\\\ \hline\cr
q^{6}\left(q-2\right)\left(q-1\right)^{4}\left(q^{3}+q-1\right)&q^{6}\left(q-1\right)^{4}\left(q^{4}+q^{3}-8q^{2}+9q-4\right)&q^{6}\left(q-2\right)\left(q-1\right)^{4}\left(q^{3}+2q^{2}-4q+2\right)&q^{9}\left(q-2\right)^{2}\left(q-1\right)^{3}\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{9}\left(q-2\right)\left(q-1\right)^{6}\left(q+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{10}\left(q-2\right)\left(q-1\right)^{5}&q^{7}\left(q-1\right)^{4}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-2q+2\right)&q^{10}\left(q-2\right)^{2}\left(q-1\right)^{4}\\\
q^{6}\left(q-2\right)\left(q-1\right)^{6}\left(q^{2}+q+1\right)&q^{6}\left(q-1\right)^{5}\left(q^{4}+q^{3}-8q^{2}+9q-4\right)&q^{6}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2q^{2}-4q+2\right)&q^{9}\left(q-2\right)^{2}\left(q-1\right)^{4}\\\
q^{7}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\left(q^{4}+q-1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{5}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{10}\left(q-2\right)\left(q-1\right)^{5}&q^{7}\left(q-1\right)^{4}\left(q^{5}-8q^{3}+17q^{2}-13q+4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2q^{2}-4q+2\right)&q^{10}\left(q-2\right)^{2}\left(q-1\right)^{4}\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2q^{2}-4q+2\right)&q^{8}\left(q-1\right)^{4}\left(q^{5}-q^{4}-7q^{3}+19q^{2}-15q+4\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{3}\\\
q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{12}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-q^{2}+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{9}\left(q-2\right)\left(q-1\right)^{4}\left(q^{4}-3q^{3}+2q^{2}-1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{10}\left(q-2\right)\left(q-1\right)^{6}&q^{7}\left(q-1\right)^{4}\left(q^{6}-3q^{5}+q^{4}+9q^{3}-18q^{2}+13q-4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}-2q+2\right)&q^{10}\left(q-2\right)^{2}\left(q-1\right)^{5}\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}-2q+2\right)&q^{8}\left(q-1\right)^{6}\left(q^{4}-2q^{3}-q^{2}+7q-4\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}\\\
q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{7}\left(q-1\right)^{6}\left(q+1\right)\left(q^{2}+1\right)\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{6}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}\\\
q^{11}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{9}\left(q-2\right)\left(q-1\right)^{4}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\end{array}$

$\begin{array}[]{cccc}T_{9}&T_{10}&T_{11}&T_{12}\\\ \hline\cr
q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q+1\right)&q^{7}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-q-1\right)&q^{6}\left(q-1\right)^{3}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)&q^{6}\left(q-2\right)\left(q-1\right)^{4}\left(q^{3}-2q+2\right)\\\
q^{9}\left(q-1\right)^{5}\left(q+1\right)\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{9}\left(q-2\right)\left(q-1\right)^{6}\left(q+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}\\\
q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q+1\right)&q^{8}\left(q-2\right)\left(q-1\right)^{3}\left(q^{4}-3q^{3}+2q^{2}-1\right)&q^{7}\left(q-1\right)^{3}\left(q^{2}-q+1\right)\left(q^{4}-2q^{3}-q^{2}+5q-4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}+2\right)\\\
q^{7}\left(q-2\right)\left(q-1\right)^{6}\left(q+1\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}-q-1\right)&q^{6}\left(q-1\right)^{4}\left(q^{5}-2q^{4}-2q^{3}+9q^{2}-9q+4\right)&q^{6}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-2q+2\right)\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{3}&q^{11}\left(q-2\right)\left(q-1\right)^{5}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}\\\
q^{8}\left(q-2\right)\left(q-1\right)^{4}\left(q^{3}-q^{2}+1\right)&q^{8}\left(q-2\right)\left(q-1\right)^{3}\left(q^{4}-3q^{3}+2q^{2}-1\right)&q^{7}\left(q-1\right)^{3}\left(q^{6}-3q^{5}+q^{4}+9q^{3}-18q^{2}+13q-4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-2q+2\right)\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-2q+2\right)&q^{8}\left(q-1\right)^{5}\left(q^{4}-2q^{3}-q^{2}+7q-4\right)\\\
q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{4}\\\
q^{9}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\left(q^{3}-q^{2}+1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{9}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-q^{2}+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}\\\
q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{9}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\left(q^{4}-3q^{3}+3q^{2}+1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{3}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{8}\left(q-2\right)\left(q-1\right)^{5}\left(q^{3}-q^{2}+1\right)&q^{8}\left(q-2\right)\left(q-1\right)^{3}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)&q^{7}\left(q-1\right)^{3}\left(q^{7}-4q^{6}+6q^{5}-q^{4}-11q^{3}+19q^{2}-13q+4\right)&q^{7}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}+2\right)\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{8}\left(q-2\right)\left(q-1\right)^{6}\left(q^{3}+2\right)&q^{8}\left(q-1\right)^{4}\left(q^{6}-4q^{5}+6q^{4}-q^{3}-8q^{2}+11q-4\right)\\\
q^{12}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{6}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}\\\
q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}&q^{12}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}-3q+3\right)&q^{9}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\left(q^{5}-4q^{4}+6q^{3}-3q^{2}-1\right)&q^{9}\left(q-2\right)\left(q-1\right)^{3}\left(q^{6}-5q^{5}+9q^{4}-7q^{3}+2q^{2}-1\right)&q^{11}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}-3q+3\right)\end{array}$

$\left.\begin{array}[]{cccc}T_{13}&T_{14}&T_{15}&T_{16}\\\ \hline\cr
q^{9}\left(q-2\right)^{2}\left(q-1\right)^{3}&q^{6}\left(q-2\right)\left(q-1\right)^{5}\left(q^{2}+q+1\right)&q^{9}\left(q-2\right)^{3}\left(q-1\right)^{2}&q^{7}\left(q-2\right)\left(q-1\right)^{2}\left(q^{4}-3q^{3}+2q^{2}-1\right)\\\
q^{11}\left(q-2\right)^{3}\left(q-1\right)^{3}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{10}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{10}\left(q-2\right)\left(q-1\right)^{5}&q^{10}\left(q-2\right)^{3}\left(q-1\right)^{3}&q^{8}\left(q-2\right)\left(q-1\right)^{2}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)\\\
q^{9}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{6}\left(q-2\right)\left(q-1\right)^{4}\left(q^{4}-q^{3}+1\right)&q^{9}\left(q-2\right)^{3}\left(q-1\right)^{3}&q^{7}\left(q-2\right)\left(q-1\right)^{3}\left(q^{4}-3q^{3}+2q^{2}-1\right)\\\
q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{7}\left(q-1\right)^{5}\left(q+1\right)\left(q^{2}+1\right)\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{3}\\\
q^{10}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{10}\left(q-2\right)\left(q-1\right)^{5}&q^{10}\left(q-2\right)^{3}\left(q-1\right)^{3}&q^{8}\left(q-2\right)\left(q-1\right)^{2}\left(q^{5}-4q^{4}+5q^{3}-2q^{2}+1\right)\\\
q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)\left(q-1\right)^{2}\left(q^{2}-3q+3\right)^{2}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{11}\left(q-2\right)\left(q-1\right)^{2}\left(q^{2}-3q+3\right)^{2}&q^{9}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)\left(q^{5}-4q^{4}+6q^{3}-3q^{2}-1\right)\\\
q^{10}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{10}\left(q-2\right)\left(q-1\right)^{6}&q^{10}\left(q-2\right)^{3}\left(q-1\right)^{4}&q^{8}\left(q-2\right)\left(q-1\right)^{2}\left(q^{6}-5q^{5}+9q^{4}-7q^{3}+2q^{2}-1\right)\\\
q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{5}&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\\\
q^{12}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}&q^{12}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-2\right)\left(q-1\right)^{2}\left(q^{2}-3q+3\right)^{2}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)\\\
q^{11}\left(q-2\right)\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{7}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)\left(q^{5}-q^{4}+1\right)&q^{11}\left(q-2\right)^{2}\left(q-1\right)^{3}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{4}\\\
q^{12}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}&q^{12}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{12}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)^{3}&q^{12}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}\\\
q^{11}\left(q-2\right)^{2}\left(q-1\right)^{4}\left(q^{2}-3q+3\right)&q^{11}\left(q-2\right)^{3}\left(q-1\right)^{5}&q^{11}\left(q-2\right)\left(q-1\right)^{3}\left(q^{2}-3q+3\right)^{2}&q^{9}\left(q-1\right)^{2}\left(q^{2}-3q+3\right)\left(q^{6}-5q^{5}+10q^{4}-9q^{3}+3q^{2}+1\right)\end{array}\right]$