Abstract. In a joint paper with Benjamin Church, a family of 2-generated groups is presented and proved to not be weakly integral. In this informal note I want to explain the strong computational counterpart of this project, and aims to complement the paper. Below you will find SageMath code to reproduce all the computations of the paper and more.
Each section/subsection is collapsible, click on it if you want to read. Please, do not hesitate to contact me with any corrections, questions and/or remarks.
Let $X^{\tt an}$ be the set $X(\mathbb{C})$ of the complex points of $X$ equipped with the analytic topology (see [] for a thorough treatment of the analytification functor). The topological space $X^{\tt an}$ is a connected topological space with the homotopy type of a connected finite CW-complex.
In particular, the (topological) fundamental group $\pi_1(X^{\tt an},x_0)$ (where $x_0 \in X(\mathbb{C})$ is an arbitrary point) is a finitely presented group.
Certainly, any finitely presented group $\pi$ can be realized as the fundamental group of a connected finite CW-complex. Indeed, just consider the CW-complex with just one 0-dim'l cell, one 1-dim'l cell for each generator and a 2-dim'l cell for each relation, gluing it to the 1-dim'l cells by reading the word.
Hence, can we conclude that every finitely presented group can be realized as the (topological) fundamental group of some $X^{\tt an}$?
The answer is NO because we don't know if the CW-complexes used to realize finitely presented groups as fundamental groups have the homotopy type of $X^{\tt an}$ for some $X$. The paper and this note is all about this: to find finitely presented groups that are not the fundamental group of any $X^{\tt an}$.
Let me name the groups that are not fundamental in the previous sense:
It is classical that every free group on finitely many symbols $F_n$ is of geometric origin, since $F_n\simeq\pi_1(\mathbb{P}^1(\mathbb{C})-\{n+1\text{ points}\},x_0)$. We would like to answer the very general question:
Can we decide if a finitely presented group is of geometric origin?
This question has been studied, among others, by de Jong and Esnault in [dJE24] and they found a neccessary condition for a finitely presented group to be of geometric origin. We recall it below.
A few comments on this definition:
Despite we can identify $\mathbb{C}\sim\bar{\mathbb{Q}}_\ell$ and argue as before, we can do better. Since $\pi$ is finitely presented, it is finitely generated in particular. This implies that any representation $\rho:\pi\rightarrow\mathrm{GL}_r(\mathbb{C})$ factors through $\mathrm{GL}_r(A)\hookrightarrow\mathrm{GL}_r(\mathbb{C})$ for some ring $A\subset\mathbb{C}$ of finite type (over $\mathbb{Z}$). Since for almost all prime numbers $\ell$ we have an injection $A\hookrightarrow\bar{\mathbb{Z}}_\ell$ (just take the completion with respect to a prime ideal of $A$ whose residue characteristic is $\ell$) we see that $\rho$ factors through $\mathrm{GL}_r(\bar{\mathbb{Z}}_\ell)$ for almost all prime numbers $\ell$.
Now we remove the parameters $(r,\chi)$ from the definition of weak integrality to say:
The remarkable (and beautiful) theorem of de Jong and Esnault is
This theorem implies a new obstruction for a finitely presented group to be geometric, and de Jong and Esnault were interested in finding examples of groups that are not weakly integral. We present the first example in the next subsection, due to Breuillard.
Since the family of not weakly integrals groups I will tell you about later is a family of 1-relator 2-generated groups indexed by prime powers that are not weakly integral with respect to $(2,\mathbf{1})$ for the corresponding prime, it won't hurt us if we make some remarks about this example.
Let $\pi$ be a finitely generated group $\pi:=\langle s_1,\dots,s_n \vert r_\lambda; \lambda\in\Lambda\rangle$. Let $G/\mathbb{C}$ be a complex linear algebraic group. We think of $G$ as a group of matrices with some properties.
For every $\mathbb{C}$-algebra $R$, any group homomorphism $\rho: \pi\rightarrow G(R)$ is determined by $n$ elements of $S_i\in G(R)$ verifying the conditions $r_\lambda(S_1,\dots,S_n)=I$ for every $\lambda$. If we vary $R$ we obtain an affine subscheme of $G^n$. More explicitly, the functor of points of this affine scheme is the one defined by: \[R\in\mathbb{C}{\tt -alg}\longmapsto \mathrm{Hom}_{\tt Group}(\pi,G(R))\in{\tt Sets}.\]
Observe that, like in the representation theory of finite groups, it is desirable to reduce the amount of information used to describe characters. For example, it would be useful to understand when two elements of $\pi$ are conjugated. For abstract finitely presented groups this may be difficult, but for $t\in\pi$ the representations $\rho$ and $\rho^t$ (precomposition with conjugation by $t$) are conjugate representations. Hence, we can consider the points of the scheme $\mathrm{Hom}_{\tt Group}(\pi,G)$ up to conjugation by $G$, which is a moduli space.
So we define the $G$-character variety of $\pi$, denoted by $M(\pi,G)$ to be the affine scheme $\mathrm{Hom}_{\tt Group}(\pi,G)//G$, where $G$ acts by conjugation. From the point of view of scheme theory this definition is technical (it is a GIT quotient). But, as we will see soon, these problems are not very important for the case of a 2-generated groups $\pi$ and $G=\mathrm{SL}_2$.
To informally motivate the name "character variety", morally the coordinates of the resulting moduli space $M(\pi,G)$ are the functions \[\begin{matrix}x_i:&\mathrm{Hom}_{\tt Group}(\pi,G) &\longrightarrow &\mathbb{C}\\ &\rho &\longmapsto &\mathrm{tr}\ \rho(s_i).\end{matrix}\] These "variables" should be enough to describe every "central" function defined over the scheme $\mathrm{Hom}_{\tt Group}(\pi,G)$. But, as I said, this is just an idea since in general, these functions are not always enough (we need more generators of the group $\pi$, that may be redundant for the group but not for the character variety) and are not neccessarily algebraically independent. For the case of $\mathrm{SL}_2$, we will find the former phenomena for $F_2$. The latter happens for example for $\pi=F_3$ and $G=\mathrm{SL}_2$, but we won't consider this situation here (consult [Gol09] for a good exposition).The polynomials $c_n(t),d_n(t)$ will play an important role later, so we mention a few properties here. By writing $A^{n+1}=A\cdot A^n$ we get the following recursive formulas for them: \[\begin{aligned}c_{n+1}(t)&= tc_n(t)+d_n(t),\\ d_{n+1}(t)&=-c_n(t).\end{aligned}\] If we write the matrix that express the recurrence formulas, we find that \[\begin{pmatrix}c_{n+1}(t) &c_n(t)\\ d_{n+1}(t) &d_n(t)\end{pmatrix} = \begin{pmatrix}t &1\\ -1 &0\end{pmatrix}^n\quad \forall n\in\mathbb{Z}.\] In particular, taking the determinant, it follows that \[c_n(t)^2+d_n(t)^2+tc_n(t)d_n(t)-1=0.\] This implies that $c_n(t)$ and $d_n(t)$ does not vanish simultaneously. For many more properties of these polynomials I refer to the paper, or just observe that $c_n(t)=U_{n-1}(t/2)$ where $U_k$ are the Chebyshev polynomials of the second kind.
Now consider the "variable" $z:=\mathrm{tr}\ AB$. It is interesting that we can describe the image by a representation $\rho$ of any word $\omega(a,b)\in F_2$ as a "linear" combination of the matrices $I$, $A$, $B$ and $AB$ with polynomial coefficients in the variables $x,y,z$. Indeed, consider the following word:
Morally, every function invariant under conjugation can be expressed as a polynomial in the algebraically independent "coordinates" \[x,y,z:\mathrm{Hom}_{\tt Group}(\pi,\mathrm{SL}_2(\mathbb{C}))\rightarrow \mathbb{C}\] defined respectively by $x:\rho\mapsto \mathrm{tr}\ \rho(a)$, $y:\rho\mapsto \mathrm{tr}\ \rho(b)$ and $z:\rho\mapsto \mathrm{tr}\ \rho(ab)$.
If the group $\pi$ has only one relation (the number of relations does not really matter but this is all we need) $r(a,b)\in F_2$, then $\rho(r)=r(A,B)$ can be written in terms of the matrices $I$, $A$, $B$ and $AB$ with polynomial coefficients. Using Fricke—Klein—Vogt theorem we can present the $\mathrm{SL}_2$-character variety of $\pi$ as a Zariski closed subscheme of $\mathbb{A}^3_{x,y,z}$. In fact, if $r(A,B) = p_I\cdot I + p_A\cdot A + p_B\cdot B + p_{AB}\cdot AB$, then $M(\pi,\mathrm{SL}_2)$ would be
The word $ABA^{-1}B^{-1}$ expressed using the matrices $\{1,A,B,AB\}$ is \[(y^2-1)\cdot I + (x-yz)\cdot A - y\cdot B + z\cdot AB.\] Its trace is $x^2+y^2+z^2-xyz-2$ and condition (b) above is just ${\tt red}:=x^2+y^2+z^2-xyz-4\neq 0$. Hence, we can detect irreducibility using Fricke—Klein—Vogt's coordinates. In particular, using coordinates we just work with the basic open set $\mathrm{D}({\tt red}) \subset M(\pi,\mathrm{SL}_2)$.
Moreover, if we assume $\pi$ is presented with a single relator $r(a,b)$ and write it in terms of $I$, $A$, $B$ and $AB$ as follows: \[r(A,B)=p_I\cdot I + p_A\cdot A + p_B\cdot B + p_{AB}\cdot AB,\] then condition (c) implies that the "variety" of irreps (up to conjugation) in terms of the variables $x$, $y$ and $z$ is simply the following constructible set \[V(p_I,p_A,p_B,p_{AB})\cap \mathrm{D}({\tt red})\subset \mathbb{A}^3_{x,y,z}.\]Now we use the concept of group algebras to give the correct construction. Below we follow closely [BH95]. This subsection is more technical and can be skipped without risk.
We introduced the "variables" $A$ and $B$ in order to make use of Cayley—Hamilton, otherwise it would not make any sense for the symbols $a,b\in\pi$. Nevertheless, we do not need to add new symbols as soon as we can operate linearly with words in $\pi$. In other words, we just need the group algebra $\mathbb{Z}\pi$ to express Cayley—Hamilton in an "universal" way.Consider the two-sided ideal $(h(g+g^{-1})-(g+g^{-1})h;\ g,h\in\pi)\subset\mathbb{Z}\pi$. This ideal "contains" the SL2-Cayley—Hamilton condition for every element $g\in\pi$ (actually for every element of $\mathbb{Z}\pi$) in a "central" form, and every consequence that derives from it. Hence, we consider the non-commutative ring \[H[\pi] := \frac{\mathbb{Z}\pi}{(h(g+g^{-1})-(g+g^{-1})h;\ g,h\in\pi)}.\]
These relations arise form the observation that, if $\rho : \pi \to \mathrm{SL}_2(R)$ is an $R$-valued representation of $\pi$, then by Cayley—Hamilton the matrix $\rho(g)+\rho(g)^{-1}=\mathrm{tr}(\rho(g))\cdot I$ is a central matrix. Therefore, its commutator with any element lies in the kernel of $\tilde\rho : \mathbb{Z}\pi \to \mathrm{Mat}_{2\times2}(R)$ linearly extending $\rho$, i.e. \[\tilde\rho\big(\sum n_g g\big):= \sum n_g\rho(g).\]
With this interpretation in mind, define the $\mathbb{Z}$-linear map $T: H[\pi]\to H[\pi]$ called ''trace'' by the formula $T(x) = x+\iota(x)$ where $\iota$ is the $\mathbb{Z}$-linear extension of $g\mapsto g^{-1}$.
By construction, $T(g)$ is mapped to $\mathrm{tr}\ \rho(g)\cdot I$ under any $\mathbb{Z}$-algebra map $\tilde\rho : H[\pi]\to\mathrm{Mat}_{2\times2}(R)$ extending a representation $\rho:\pi\to\mathrm{SL}_2(R)$. Denote by $TH[\pi]$ the subalgebra of $H[\pi]$ generated by the subset $\mathbb{Z}\cup\{T(x)\vert x\in H[\pi]\}$, which is commutative. Now $H[\pi]$ can be seen as an algebra over the commutative ring $TH[\pi]$.
The key feature of $H[\pi]$ is that, for any commutative ring $R$, representations $\rho:\pi\to\mathrm{SL}_2(R)$ are in bijection with involution preserving algebra homomorphisms $\tilde\rho:H[\pi]\to\mathrm{Mat}_{2\times2}(R)$ (see [BH95, Proposition 1.3]).
The following theorem is a key result for us:
Moreover, if $\pi=F_2$ is the free group on two generators $a,b$, then $TH[F_2]=\mathbb{Z}[x,y,z]$, i.e. the elements $x,y,z$ are algebraically independent [BH95, Proposition 3.2].
Define
In particular, if we take the group $\pi=F_2$ and the homorphism $\varphi=\mathrm{id}$, we find that the algebra $H[F_2]$ is free of rank 4 over $TH[\pi]=\mathbb{Z}[x,y,z]$. These two statements can be understood as another form of Fricke—Klein—Vogt theorem.
Finally, the calculation of $H[F_2]$ allows us to study arbitrary $2$-generated groups via the following general result:
Combining these results for $\pi := \langle a,b\vert w_j(a,b)\rangle$, it follows that
Clearly, absolute irreducibility implies irreducibility. Moreover, over an algebraically closed field both notions coincide (this is Burnside theorem).
The following result makes clear the crucial role played by the polynomial $\mathsf{red}$:
As a corollary, we can understand absolute irreducibility in the following geometric way:
Let $R$ be a commutative ring. Then $\rho:\pi\to\mathrm{SL}_2(R)$ is abs. irred. $\Leftrightarrow$ $\tilde\rho(\mathsf{red})\in R^\times$ $\Leftrightarrow$ $\tilde\rho(\mathsf{red})\in \overline{k}^\times$ for every geom. point $R\to \overline{k}$ [to see this, specialize this property to the morphisms $R\to R/\mathfrak{m} \to (R/\mathfrak{m})^{\texttt{alg}}$ for every maximal ideal $\mathfrak{m}\subset R$] $\Leftrightarrow$ $\rho : \pi \to \mathrm{SL}_2(\overline{k})$ is abs. irred. for every geom. point $R\to\overline{k}$ $\Leftrightarrow$ $\rho:\pi\to\mathrm{SL}_2(\overline{k})$ is irred. for every geom. point $R\to\overline{k}$ [this last equivalence is Burnside's theorem].
Furthermore, after the Structure theorem, the absolute irreducibility of $\rho$ implies $R\otimes_{TH[\pi]}H[\pi]$ is free of rank 4 over $R$, with basis $\{1\otimes1,1\otimes a,1\otimes b,1\otimes ab\}$.
Finally, the next two results allow us two uniquely (over a field up to conjugation) reconstruct a representation from its character:
We combine several results below using schemes.
We already know that, for $F_2=\langle a,b\vert\emptyset\rangle$, $TH[F_2]=\mathbb{Z}[x,y,z]$ is a domain therefore $H[F_2]$ is the free $\mathbb{Z}[x,y,z]$-module with basis $\{1,a,b,ab\}$. In particular, $M(F_2,\mathrm{SL}_2) = \mathbb{A}^3_{x,y,z}$. Furthermore, we know that the scheme of absolutely irreducible representations $M^{\texttt{irr}}(F_2,\mathrm{SL}_2)$ (i.e. irreducible over every geometric point) corresponds to the basic open affine $\mathrm{D}(\mathsf{red}) \subset \mathbb{A}^3_{x,y,z}$ where $\mathsf{red} = x^2+y^2+z^2-xyz-4$.
$H[F_2]$ being free implies we can write $\omega(a,b) = c_1-1 + c_a\cdot a+c_b\cdot b+c_{ab}\cdot ab$ for some polynomials $c_g\in\mathbb{Z}[x,y,z]$ for $g\in\{1,a,b,ab\}$. We immediately obtain \[H[\pi] = \frac{H[F_2]}{(c_1+c_aa+c_bb+c_{ab}ab)}.\] However, this presentation does not lead to a description of $TH[\pi]$. After base change to the ring $\mathbb{Z}[1/2]$, we obtain \[TH[\pi]\otimes\mathbb{Z}[1/2] = \frac{\mathbb{Z}[1/2][x,y,z]}{(T\ g\cdot(\omega(a,b)-1))_{g\in\{1,a,b,ab\}}}.\] Observe that
| $T\ (\omega(a,b)-1)$ | $=$ | $2(c_1-1)+xc_a+yc_b+zc_{ab}$; |
| $T\ a(\omega(a,b)-1)$ | $=$ | $xc_1 + (x^2-2)c_x+zc_b+(xz-y)c_{ab}$; |
| $T\ b(\omega(a,b)-1)$ | $=$ | $yc_1+zc_a+(y^2-2)c_b+(yz-x)c_{ab}$; |
| $T\ ab(\omega(a,b)-1)$ | $=$ | $zc_1+(xz-y)c_a+(yz-x)c_b+(z^2-2)c_{ab}$. |
Moreover, again over $\mathrm{D}(2)\subset\mathrm{Spec}(\mathbb{Z})$, we can identify $M^{\texttt{irr}}(\pi,\mathrm{SL}_2)$ with the intersection of the previous algebraic set with the basic open $\mathrm{D}(\mathsf{red})$ inside $\mathbb{A}^3_{x,y,z}$. In fact, we can do better because for irreducible representations the set $\{1,a,b,ab\}$ is a basis, hence $M^{\texttt{irr}}(\pi,\mathrm{SL}_2)$ is identified with the constructible set $V(c_1-1,c_a,c_b,c_{ab})\cap \mathrm{D}(\mathsf{red})$ inside $\mathbb{A}^3_{x,y,z}$.
Finally, let me mention that we are not interested in absolutely irreducible representations, but on those representations irreducible over $\mathbb{C}$. To study these points, we define $M^{\texttt{gen.irr}}(\pi,\mathrm{SL}_2)$ as the scheme-theoretic image of the morphism $M^{\texttt{irr}}(\pi,\mathrm{SL}_2)_{\mathbb{Q}}\to M(\pi,\mathrm{SL}_2)$. By definition, it follows that $M^{\texttt{gen.irr}}(F_2,\mathrm{SL}_2)=M(F_2,\mathrm{SL}_2)$. In general, it can be shown that [CG24, Proposition 5] $M^{\texttt{gen.irr}}(\pi,\mathrm{SL}_2)$ is cut out by the $w$-elimination ideal of $(c_1-1,c_a,c_b,c_{ab},w\cdot\mathsf{red}-1)\subset \mathbb{Q}[x,y,z,w]$.# The Cayley--Hamilton algebra of F₂ over ℤ
BaseRing.<x,y,z> = PolynomialRing(ZZ,3)
Alg = FreeAlgebra(BaseRing,2)
Mon = Alg.monoid()
a, b = Mon.gens()
mons = [Mon(1), a, b, a*b]
M = MatrixSpace(BaseRing,4) # Matrices for the algebra
mats = [M([0,1,0,0, -1,x,0,0, z-x*y,y,x,-1, -y,z,1,0]),
M([0,0,1,0, 0,0,0,1, -1,0,y,0, 0,-1,0,y])]
H.<a,b> = Alg.quotient(mons, mats)This implementation has a major limitation: $a^{-1}$ cannot be directly computed. But observe that
sage: a*(x-a)
1sage: (x-a)*a
1And we do, for example:
sage: a^2 * b * (x-a)^2 * (y-b)^2 # a²ba⁻²b⁻²
(x*y^2*z-x^2*y-x*z+y) - (x^2*y^2*z-x^3*y-x*y^3-x^2*z+3*x*y)*a - (x*y*z-x^2+1)*b + (x^2*y*z-x^3-x*y^2+2*x)*a*bThe trace map $T:H[F_2]\to TH[F_2]$ is defined as follows:
ts = [2,x,y,z]
def T(w): # T : H[F₂] → TH[F₂]
ws = w.vector() # Coord. wrt. (1,a,b,ab)
t=0
for i in range(4):
t += ts[i]*ws[i]
return tNow we study Breuillard's example. First calculate $T(g \cdot (a^2ba^{-2}b^{-2}-1))$ for $g\in\{1,a,b,ab\}$:
sage: T(a^2 * b * (x-a)^2 * (y-b)^2 - 1)
-x^3*y^2*z + x^4*y + x^2*y^3 + x^2*y*z^2 - 4*x^2*y + y - 2sage: T(a * (a^2 * b * (x-a)^2 * (y-b)^2 - 1))
-x^4*y^2*z + x^5*y + x^3*y^3 + x^3*y*z^2 + x^2*y^2*z - 5*x^3*y - x*y^3 - x*y*z^2 + 5*x*y - x - zsage: T(b * (a^2 * b * (x-a)^2 * (y-b)^2 - 1))
-x^3*y*z + x^4 + x^2*y^2 + x^2*z^2 - 4*x^2 - y + 2sage: T(a*b * (a^2 * b * (x-a)^2 * (y-b)^2 - 1))
-x^3*y^2*z^2 + x^4*y*z + 2*x^2*y^3*z + x^2*y*z^3 - x^3*y^2 - x*y^4 - x*y^2*z^2 - 5*x^2*y*z + x^3 + 5*x*y^2 + x*z^2 - 3*x - zAlso define:
red = T(a * b * (x-a) * (y-b)) - 2
redQ=PolynomialRing(QQ,'x, y, z')(red)
sage: red
-x*y*z + x^2 + y^2 + z^2 - 4Define the SL2-character variety associated to $\langle a,b\vert a^2ba^{-2}b^{-2}\rangle$:
w = a^2 * b * (x-a)^2 * (y-b)^2
I = PolynomialRing(QQ,'x, y, z').ideal(T(w - 1),
T(a * (w - 1)),
T(b * (w - 1)),
T(a*b * (w - 1)))Observe that we are considering the ideal over the ring $\mathbb{Q}[x,y,z]$. Now we can ask SageMath for the primary decomposition of this ideal. Furthermore, we ask if the components are reduced (prime), for their generators and if the points correspond to irreps. (by checking if the reduction of $\mathsf{red}$ in the component is non-zero):
sage: for P in I.primary_decomposition():
print(f'Primary component of dimension {P.dimension()}.')
print(f' Is it prime? {P.is_prime()}.')
print(f' It is generated by {P.gens()}.')
print(f' Is red a unit in the quotient? {redQ.reduce(P)}')
print()
Primary component of dimension 0.
Is it prime? True.
It is generated by [y + 1, x + 2*z, 2*z^2 - 1].
Is red a unit in the quotient? -3/2
Primary component of dimension 1.
Is it prime? True.
It is generated by [y - 2, x - z].
Is red a unit in the quotient? 0
Since these components are defined over the field $\mathbb{Q}$, we know that the coordinates $(x,y,z)$ define a unique representation up to conjugation. Hence, after this calculation we conclude that $M(\langle a,b\vert a^2ba^{-2}b^{-2}\rangle,\mathrm{SL}_2)$ over $\mathbb{Q}$ has a 1-dim'l irreducible component of reducible representations, corresponding to the curve $(t,2,t)$; and a 0-dim'l irreducible component with just two points corresponding to two irreps. with traces $y=-1$, $x = \mp \sqrt{2}$ and $z=\pm\frac{\sqrt{2}}{2}$.
This proves Breuillard's claim that this group is not of geometric origin: the group does not have integral irreps. so it is not weakly integral.
Some remarks:
l, k = 2, 2
n = l^k
w = a^(n*(n-1)) * b * (x-a)^n * (y-b)^2
I = PolynomialRing(QQ,'x, y, z',order='lex').ideal(T(w - 1),
T(a * (w - 1)),
T(b * (w - 1)),
T(a*b * (w - 1)))sage: for P in I.primary_decomposition():
print(f'Primary component of dimension {P.dimension()}.')
print(f' Is it prime? {P.is_prime()}.')
print(f' It is generated by {P.gens()}.')
print(f' Is red a unit in the quotient? {redQ.reduce(P)}')
print()
Primary component of dimension 0.
Is it prime? True.
It is generated by [8*z^8 - 32*z^6 + 40*z^4 - 16*z^2 + 1, y + 1, x - 8*z^7 + 28*z^5 - 28*z^3 + 8*z].
Is red a unit in the quotient? x^2 + x*z + z^2 - 3
Primary component of dimension 0.
Is it prime? True.
It is generated by [8*z^4 - 8*z^2 + 1, y - 1, x - 2*z].
Is red a unit in the quotient? 3*z^2 - 3
Primary component of dimension 1.
Is it prime? True.
It is generated by [y^9 - 9*y^7 + 27*y^5 - 30*y^3 + 9*y - z^8 + 8*z^6 - 20*z^4 + 16*z^2 - 2, x*z^7 - 6*x*z^5 + 10*x*z^3 - 4*x*z - y^8 + 8*y^6 - 20*y^4 + 16*y^2 - y*z^6 + 5*y*z^4 - 6*y*z^2 + y - 2, x*y^4 - x*y^3*z^6 + 5*x*y^3*z^4 - 6*x*y^3*z^2 + x*y^3 - 3*x*y^2 + 2*x*y*z^6 - 10*x*y*z^4 + 12*x*y*z^2 - 2*x*y + x + y^4*z^5 - 4*y^4*z^3 + 3*y^4*z - y^3*z + y^2*z^7 - 9*y^2*z^5 + 22*y^2*z^3 - 13*y^2*z + 2*y*z - z^7 + 7*z^5 - 14*z^3 + 7*z, x^2 - x*y*z + y^2 + z^2 - 4].
Is red a unit in the quotient? 0Observe that Newton polygon of the polynomial $8z^8 - 32z^6 + 40z^4 - 16z^2 + 1$ with respect to 2-valuation has a single slope $3/8$ with multiplicity $8$. This implies that all the traces $z$ are not integral. It can be checked (repeat the previous computation but use the lexicographic order with $y>z>x$) that $x$ is integral.
Similarly, the Newton polygon of the polynomial $8z^4 - 8z^2 + 1$ with respect to the 2-valuation has a single slope $3/4$. As before, it can be checked that $x$ is integral.
Finally, the points in the 1-dim'l component correspond to reducible representations. This proves that this group is not of geometric origin using de Jong—Esnault obstruction.
For $\ell^k=3$, use again lines 1-19 and do:
l, k = 3, 1
n = l^k
w = a^(n*(n-1)) * b * (x-a)^n * (y-b)^2
I = PolynomialRing(QQ,'x, y, z',order='lex').ideal(T(w - 1),
T(a * (w - 1)),
T(b * (w - 1)),
T(a*b * (w - 1)))sage: for P in I.primary_decomposition():
print(f'Primary component of dimension {P.dimension()}.')
print(f' Is it prime? {P.is_prime()}.')
print(f' It is generated by {P.gens()}.')
print(f' Is red a unit in the quotient? {redQ.reduce(P)}')
print()
Primary component of dimension 0.
Is it prime? True.
It is generated by [3*z^3 - 3*z + 1, y + 1, x - 3*z^2 + 2].
Is red a unit in the quotient? x^2 + x*z + 1/3*x - 7/3
Primary component of dimension 0.
Is it prime? True.
It is generated by [3*z^3 - 3*z + 1, y - 1, x + 3*z^2 - 2].
Is red a unit in the quotient? x^2 - x*z - 1/3*x - 7/3
Primary component of dimension 1.
Is it prime? True.
It is generated by [y^4 - 4*y^2 - z^3 + 3*z + 2, x*z^2 - x - y^3 - y*z + 3*y, x*y*z - x*y - y^2 - z^2 + z + 2, x*y^2 - x*z - x - y*z + y, x^2 - x*y + z - 2].
Is red a unit in the quotient? 0As before, it follows that $\Gamma_3$ is not weakly integral.
Similarly, for $\ell^k=5$:
l, k = 5, 1
n = l^k
w = a^(n*(n-1)) * b * (x-a)^n * (y-b)^2
I=PolynomialRing(QQ,'x, y, z',order='degrevlex(2),lex(1)').ideal(T(w - 1),
T(a * (w - 1)),
T(b * (w - 1)),
T(a*b * (w - 1)))sage: for P in I.primary_decomposition():
print(f'Primary component of dimension {P.dimension()}.')
print(f' Is it prime? {P.is_prime()}.')
print(f' It is generated by {P.gens()}.')
print(f' Is red a unit in the quotient? {redQ.reduce(P)}')
print()
Primary component of dimension 0.
Is it prime? True.
It is generated by [25*z^10 - 125*z^8 + 200*z^6 - 15*z^5 - 125*z^4 + 25*z^3 + 25*z^2 - 10*z + 1, y + 1, x - 25*z^9 + 120*z^7 - 180*z^5 + 15*z^4 + 105*z^3 - 21*z^2 - 20*z + 6].
Is red a unit in the quotient? x^2 + x*z + z^2 - 3
Primary component of dimension 0.
Is it prime? True.
It is generated by [25*z^10 - 125*z^8 + 200*z^6 - 15*z^5 - 125*z^4 + 25*z^3 + 25*z^2 - 10*z + 1, y - 1, x + 25*z^9 - 120*z^7 + 180*z^5 - 15*z^4 - 105*z^3 + 21*z^2 + 20*z - 6].
Is red a unit in the quotient? x^2 - x*z + z^2 - 3
Primary component of dimension 1.
Is it prime? True.
It is generated by [x^2 - x*y*z + y^2 + z^2 - 4, x*y^7*z^7 - x*y^7*z^6 - 6*x*y^7*z^5 + 5*x*y^7*z^4 + 10*x*y^7*z^3 - 6*x*y^7*z^2 - 4*x*y^7*z + x*y^7 - y^8*z^6 + y^8*z^5 + 5*y^8*z^4 - 4*y^8*z^3 - 6*y^8*z^2 + 3*y^8*z + y^8 - 6*x*y^5*z^7 + 6*x*y^5*z^6 + 36*x*y^5*z^5 - 30*x*y^5*z^4 - 60*x*y^5*z^3 + 36*x*y^5*z^2 + 24*x*y^5*z - 6*x*y^5 - y^6*z^8 + y^6*z^7 + 14*y^6*z^6 - 13*y^6*z^5 - 50*y^6*z^4 + 38*y^6*z^3 + 52*y^6*z^2 - 25*y^6*z - 8*y^6 + 10*x*y^3*z^7 - 10*x*y^3*z^6 - 60*x*y^3*z^5 + 50*x*y^3*z^4 + 100*x*y^3*z^3 - 60*x*y^3*z^2 - 40*x*y^3*z + 10*x*y^3 + 5*y^4*z^8 - 5*y^4*z^7 - 50*y^4*z^6 + 45*y^4*z^5 + 150*y^4*z^4 - 110*y^4*z^3 - 140*y^4*z^2 + 65*y^4*z + 20*y^4 - 4*x*y*z^7 + 4*x*y*z^6 + 24*x*y*z^5 - 20*x*y*z^4 - 40*x*y*z^3 + 24*x*y*z^2 + 16*x*y*z - 4*x*y - 6*y^2*z^8 + 6*y^2*z^7 + 52*y^2*z^6 - 46*y^2*z^5 - 140*y^2*z^4 + 100*y^2*z^3 + 120*y^2*z^2 - 54*y^2*z - 16*y^2 + z^8 - z^7 - 8*z^6 + 7*z^5 + 20*z^4 - 14*z^3 - 16*z^2 + 7*z + 2, y^9 - x*y^6*z^14 + 13*x*y^6*z^12 - 66*x*y^6*z^10 + 165*x*y^6*z^8 - 210*x*y^6*z^6 + 126*x*y^6*z^4 - 28*x*y^6*z^2 + x*y^6 + y^7*z^13 - 12*y^7*z^11 + 55*y^7*z^9 - 120*y^7*z^7 + 126*y^7*z^5 - 56*y^7*z^3 + 7*y^7*z - 9*y^7 + 5*x*y^4*z^14 - 65*x*y^4*z^12 + 330*x*y^4*z^10 - 825*x*y^4*z^8 + 1050*x*y^4*z^6 - 630*x*y^4*z^4 + 140*x*y^4*z^2 - 5*x*y^4 + y^5*z^15 - 20*y^5*z^13 + 150*y^5*z^11 - 550*y^5*z^9 + 1050*y^5*z^7 - 1008*y^5*z^5 + 420*y^5*z^3 - 50*y^5*z + 27*y^5 - 6*x*y^2*z^14 + 78*x*y^2*z^12 - 396*x*y^2*z^10 + 990*x*y^2*z^8 - 1260*x*y^2*z^6 + 756*x*y^2*z^4 - 168*x*y^2*z^2 + 6*x*y^2 - 4*y^3*z^15 + 66*y^3*z^13 - 432*y^3*z^11 + 1430*y^3*z^9 - 2520*y^3*z^7 + 2268*y^3*z^5 - 896*y^3*z^3 + 102*y^3*z - 30*y^3 + x*z^14 - 13*x*z^12 + 66*x*z^10 - 165*x*z^8 + 210*x*z^6 - 126*x*z^4 + 28*x*z^2 - x + 3*y*z^15 - 46*y*z^13 + 282*y*z^11 - 880*y*z^9 + 1470*y*z^7 - 1260*y*z^5 + 476*y*z^3 - 52*y*z + 9*y, x*y^8 - x*y^6*z^13 + 12*x*y^6*z^11 - 55*x*y^6*z^9 + 120*x*y^6*z^7 - 126*x*y^6*z^5 + 56*x*y^6*z^3 - 7*x*y^6*z - 7*x*y^6 + y^7*z^12 - 11*y^7*z^10 + 45*y^7*z^8 - 84*y^7*z^6 + 70*y^7*z^4 - 21*y^7*z^2 - y^7*z + y^7 + 5*x*y^4*z^13 - 60*x*y^4*z^11 + 275*x*y^4*z^9 - 600*x*y^4*z^7 + 630*x*y^4*z^5 - 280*x*y^4*z^3 + 35*x*y^4*z + 15*x*y^4 + y^5*z^14 - 19*y^5*z^12 + 132*y^5*z^10 - 435*y^5*z^8 + 714*y^5*z^6 - 546*y^5*z^4 + 154*y^5*z^2 + 6*y^5*z - 7*y^5 - 6*x*y^2*z^13 + 72*x*y^2*z^11 - 330*x*y^2*z^9 + 720*x*y^2*z^7 - 756*x*y^2*z^5 + 336*x*y^2*z^3 - 42*x*y^2*z - 10*x*y^2 - 4*y^3*z^14 + 62*y^3*z^12 - 374*y^3*z^10 + 1110*y^3*z^8 - 1680*y^3*z^6 + 1204*y^3*z^4 - 322*y^3*z^2 - 10*y^3*z + 14*y^3 + x*z^13 - 12*x*z^11 + 55*x*z^9 - 120*x*z^7 + 126*x*z^5 - 56*x*z^3 + 7*x*z + x + 3*y*z^14 - 43*y*z^12 + 242*y*z^10 - 675*y*z^8 + 966*y*z^6 - 658*y*z^4 + 168*y*z^2 + 4*y*z - 7*y].
Is red a unit in the quotient? 0As we saw, after writing $\omega_\ell:= a^{\ell(\ell-1)}ba^{-\ell}b^{-2}$ with respect to the basis $\{1,a,b,ab\}$ we directly obtain equations cutting out irreps. by comparing with the element $1$. This is equivalent to compare the coordinates of the words $a^{\ell(\ell-1)}b$ and $b^2a^\ell$. We have the following:
From the last pair of equations we get the following: The equality between polynomials is equivalent to
To determine the exact order of $\rho(a)$ we proceed as follows: First observe that the only SL2-matrices of order $2$ are $\pm I$. Write $A:=\rho(a)$. We know $A\neq\pm I$ because $\rho$ is assumed to be irreducible. This implies $A^2 \neq I$. Moreover, $A^\ell = d_\ell(x) + c_\ell(x)A$ and $c_\ell(x)\neq0$. This implies that $A^\ell\neq\pm I$. In particular, $A^{2\ell}\neq I$. Finally, if $y+1=0$, then $A^{\ell^2}=-I$ so $A$ has order exactly $2\ell^2$. If $y-1=0$ then already $A^{\ell^2}=1$ and $A$ has order $\ell^2$.
This implies that $A$ is diagonalizable, with eigenvalues $\ell^2$-th root of unity if $y-1=0$ or eigenvalues $2\ell^2$-th roots of unity if $y+1=0$. From now on we assume $y-1=0$, the other case being identical. Therefore \[x = \zeta_{\ell^2}^j + \zeta_{\ell^2}^{-j}\] for some $j\in\{0,\dots,\ell^2-1\}$ prime to $\ell$ and $\zeta_{\ell^2}=\mathrm{exp}(2\pi i/\ell^2)$.
Then, $z = c_{\ell+1}(x)$ can be expressed as \[z=\frac{1}{\zeta_{\ell^2}^j}\frac{1-\zeta_{\ell^2}^{2j(\ell^2+1)}}{1-\zeta_{\ell^2}^{2j}}.\] Using well known formulas for the valuation of cyclotomic numbers, it follows that $z$ is $\ell'$-integral if and only if $\ell'\neq\ell$, and its $\ell$-adic valuation is $-\frac{1}{\ell}$.
Moreover, we saw that both $x$ and $y$ are integral. So non-integrality occurs only for $z$. This fact is used to prove that any of the points $(x,y,z)$ we found satisfy the equation $\mathsf{red}=0$, otherwise $z$ would be an algebraic integer. This justifies why we didn't have to struggle with $\mathsf{red}$, and just use that $\{1,a,b,ab\}$ is a basis.
Finally observe that $\Gamma_\ell$ has $\ell(\ell-1)$ non-conjugated irreps. Half of them with $\mathrm{tr}\ \rho(b)=1$ and the other half with $\mathrm{tr}\ \rho(b)=-1$.
The proof for arbitrary $\ell$, $k$ is along these lines. A feature of the proof is that it treats all the cases uniformly.The idea is the following: Write $a^{n_1}b^{m_1}a^{n_2}b^{m_2}$ as $c_1 + c_a a + c_b b + c_{ab} ab$, for some $c_g \in \mathbb{Z}[x,y,z]$. Note that the powers $a^{n_i}$ (resp. $b^{m_i}$) can be written in terms of $\{1,a\}$ (resp. $\{1,b\}$) using the polynomials $c_{n_i}(x),d_{n_i}(x)$ (resp. $c_{m_i}(y),d_{m_i}(y)$).
Therefore, we can write $c_g$ as a polynomial in terms of $x,y,z$ and $c_{n_i},d_{n_i},c_{m_i},d_{m_i}$. After finding these expressions, we study the ideal $V(c_1-1,c_a,c_b,c_{ab})$ which contains the SL2-irreps. as the open subset $\mathrm{D}(\mathsf{red})$. The surprise is that, if we treat the polynomials $c_{\ast_i},d_{\ast_i}$ for $\ast\in\{n,m\}$ as independent polynomial variables $\gamma_{\ast_i},\delta_{\ast_i}$, then these equations impose certain universal conditions on $\gamma$'s and $\delta$'s.
More precisely, consider the following ring
With these new variables, we think of $a^{n_i}$ (resp. $b^{m_i}$) as the expression $\delta_{n_i}+\gamma_{n_i}a$ (resp. $\delta_{m_i}+\gamma_{m_i}b$). In fact, since $c_{n_i}(x)^2+xc_{n_1}(x)d_{n_1}(x)+d_{n_1}(x)^2-1=0$, these variables are not algebraically independent and we will impose the determinant conditions $\gamma_{n_i}^2 + x\gamma_{n_i}\delta_{n_i}+\delta_{n_i}^2-1$ and $\gamma_{m_i}^2 + x\gamma_{m_i}\delta_{m_i}+\delta_{m_i}^2-1$.
In order to simplify the defining equations, we present the relation $a^{n_1}b^{m_1}a^{n_2}b^{m_2}=1$ in a more symmetrical way: \[a^{n_2}b^{m_2}=b^{-m_1}a^{-n_1}.\] This requires us to express $a^{-n_1}$ (resp. $b^{-m_1}$) in terms of the variables $\gamma_{n_1},\delta_{n_1}$ (resp. $\gamma_{m_1},\delta_{m_1}$). But note that $c_{-n}(t)=-c_n(t)$ and $d_{-n}(t)=c_{n+1}(t)=tc_{n}(t)+d_{n}(t)$; hence we think of $a^{-n_1}$ as the algebra element $(x\gamma_{n_1}+\delta_{n_1})\cdot 1 - \gamma_{n_1}\cdot a$ and similarly for $b^{-m_1}$.
We compare the following expressions component-wise
# The Cayley--Hamilton algebra of F₂ over ℚ
BaseRing.<z,cn1,dn1,cm1,dm1,cn2,dn2,cm2,dm2,x,y> = PolynomialRing(QQ,11,order='lex')
Alg = FreeAlgebra(BaseRing,2)
Mon = Alg.monoid()
a, b = Mon.gens()
mons = [Mon(1), a, b, a*b]
M = MatrixSpace(BaseRing,4) # Matrices for the algebra
mats = [M([0,1,0,0, -1,x,0,0, z-x*y,y,x,-1, -y,z,1,0]),
M([0,0,1,0, 0,0,0,1, -1,0,y,0, 0,-1,0,y])]
H.<a,b> = Alg.quotient(mons, mats)
wr = (dn2+cn2*a)*(dm2+cm2*b)
wl = (y*cm1+dm1-cm1*b)*(x*cn1+dn1-cn1*a)Observe that we defined the Cayley—Hamilton algebra of $F_2$ over $\mathbb{Q}$ instead of $\mathbb{Z}$. This is in order to be able to use SageMath methods to study the ideal $I$. Moreover, we use a very specific ordering of the variables to improve performance.
As we already know, we obtain:
sage: wr
dn2*dm2 + cn2*dm2*a + dn2*cm2*b + cn2*cm2*a*bsage: wl
(cn1*cm1*z+cn1*dm1*x+dn1*cm1*y+dn1*dm1) - cn1*dm1*a - dn1*cm1*b - cn1*cm1*a*bNow define $I$, check that the ideal is radical and compute its primary components, checking they are actually prime ideals, and henceforth the decomposition corresponds to the minimal decomposition into irreducibles of the corresponding scheme:
I = BaseRing.ideal(wr.vector()[0]-wl.vector()[0],
wr.vector()[1]-wl.vector()[1],
wr.vector()[2]-wl.vector()[2],
wr.vector()[3]-wl.vector()[3],
cn1^2+x*cn1*dn1+dn1^2-1,
cn2^2+x*cn2*dn2+dn2^2-1,
cm1^2+y*cm1*dm1+dm1^2-1,
cm2^2+y*cm2*dm2+dm2^2-1)sage: I.radical()==I
Truesage: for P in I.primary_decomposition():
print('Primary component generated by:')
for g in P.gens():
print(f' {g}')
print(f' Is it prime? {P.is_prime()}')
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 - dm2
cm1 - cm2
dn1 + dn2
cn1 + cn2
z*dn2^2*dm2^2 - z*dn2^2 - z*dm2^2 + z + 2*cn2*dn2*cm2*dm2 + cn2*dn2*dm2^2*y + cn2*dn2*y + dn2^2*cm2*dm2*x + dn2^2*x*y + cm2*dm2*x + dm2^2*x*y
z*dn2^2*cm2 - z*cm2 - cn2*dn2*cm2*y - 2*cn2*dn2*dm2 - dn2^2*cm2*x*y - dn2^2*dm2*x - dm2*x
z*cn2*dm2^2 - z*cn2 - cn2*cm2*dm2*x - cn2*dm2^2*x*y - 2*dn2*cm2*dm2 - dn2*dm2^2*y - dn2*y
z*cn2*cm2 + cn2*dm2*x + dn2*cm2*y + 2*dn2*dm2
Is it prime? True
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 + dm2
cm1 + cm2
dn1 - dn2
cn1 - cn2
z*dn2^2*dm2^2 - z*dn2^2 - z*dm2^2 + z + 2*cn2*dn2*cm2*dm2 + cn2*dn2*dm2^2*y + cn2*dn2*y + dn2^2*cm2*dm2*x + dn2^2*x*y + cm2*dm2*x + dm2^2*x*y
z*dn2^2*cm2 - z*cm2 - cn2*dn2*cm2*y - 2*cn2*dn2*dm2 - dn2^2*cm2*x*y - dn2^2*dm2*x - dm2*x
z*cn2*dm2^2 - z*cn2 - cn2*cm2*dm2*x - cn2*dm2^2*x*y - 2*dn2*cm2*dm2 - dn2*dm2^2*y - dn2*y
z*cn2*cm2 + cn2*dm2*x + dn2*cm2*y + 2*dn2*dm2
Is it prime? True
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
dn2 + 1
cn2
dm1 - cm2*y - dm2
cm1 + cm2
dn1 + 1
cn1
Is it prime? True
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
dn2 - 1
cn2
dm1 + cm2*y + dm2
cm1 - cm2
dn1 + 1
cn1
Is it prime? True
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
dn2 + 1
cn2
dm1 + cm2*y + dm2
cm1 - cm2
dn1 - 1
cn1
Is it prime? True
Primary component generated by:
cm2^2 + cm2*dm2*y + dm2^2 - 1
dn2 - 1
cn2
dm1 - cm2*y - dm2
cm1 + cm2
dn1 - 1
cn1
Is it prime? True
Primary component generated by:
dm2 + 1
cm2
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 + 1
cm1
dn1 - cn2*x - dn2
cn1 + cn2
Is it prime? True
Primary component generated by:
dm2 - 1
cm2
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 + 1
cm1
dn1 + cn2*x + dn2
cn1 - cn2
Is it prime? True
Primary component generated by:
dm2 - 1
cm2
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 - 1
cm1
dn1 - cn2*x - dn2
cn1 + cn2
Is it prime? True
Primary component generated by:
dm2 + 1
cm2
cn2^2 + cn2*dn2*x + dn2^2 - 1
dm1 - 1
cm1
dn1 + cn2*x + dn2
cn1 - cn2
Is it prime? TrueThe sets of generators for the first two printed irreducible components are not minimal. For example:
sage: P = BaseRing.ideal(cm2^2 + cm2*dm2*y + dm2^2 - 1,
cn2^2 + cn2*dn2*x + dn2^2 - 1,
dm1 - dm2,
cm1 - cm2,
dn1 + dn2,
cn1 + cn2,
z*cn2*cm2 + cn2*dm2*x + dn2*cm2*y + 2*dn2*dm2)
Ideal (cm2^2 + cm2*dm2*y + dm2^2 - 1, cn2^2 + cn2*dn2*x + dn2^2 - 1, dm1 - dm2, cm1 - cm2, dn1 + dn2, cn1 + cn2, z*cn2*cm2 + cn2*dm2*x + dn2*cm2*y + 2*dn2*dm2) of Multivariate Polynomial Ring in z, cn1, dn1, cm1, dm1, cn2, dn2, cm2, dm2, x, y over Rational FieldNow check that the three generators we left out are in this ideal $P$:
sage: z*dn2^2*dm2^2 - z*dn2^2 - z*dm2^2 + z + 2*cn2*dn2*cm2*dm2 + cn2*dn2*dm2^2*y + cn2*dn2*y + dn2^2*cm2*dm2*x + dn2^2*x*y + cm2*dm2*x + dm2^2*x*y in P
Truesage: z*dn2^2*cm2 - z*cm2 - cn2*dn2*cm2*y - 2*cn2*dn2*dm2 - dn2^2*cm2*x*y - dn2^2*dm2*x - dm2*x in P
Truesage: z*cn2*dm2^2 - z*cn2 - cn2*cm2*dm2*x - cn2*dm2^2*x*y - 2*dn2*cm2*dm2 - dn2*dm2^2*y - dn2*y in P
TrueThe following graph summarizes all this information and the relative position of the irreducible components: each node correspond to one of the prime ideals we just found, and an edge indicates the corresponding sum is proper. The graph is interactive, just place the mouse on top of a node or an edge to see the generators of the corresponding ideal:
Observe that this ring factors through the quotient $R/J$, where $J$ is the ideal generated by the 'determinant' relations $\gamma_{n_i}^2+x\gamma_{n_i}\delta_{n_i}+\delta_{n_i}^2-1$, $\gamma_{m_i}^2+y\gamma_{m_i}\delta_{m_i}+\delta_{m_i}^2-1$ for $i=1,2$.
Let $I^{\texttt{e}}$ be the extension of $I$ by the morphism $R/J \to \mathbb{Z}[x,y,z]$. Observe that $I^{\texttt{e}}$ is the variety of generically irreducible representations, up to a bunch of representations that may be reducible. We obtain a canonical morphism between algebraic varieties \[\mathsf{Spec}\ \mathbb{Z}[x,y,z]/I^{\texttt{e}} \to \mathsf{Spec}\ R/I,\] and we obtain from the decomposition of the target into irreducibles $\mathsf{Spec}\ R/I = \mathsf{V}(\wp_1)\cup\cdots\cup\mathsf{V}(\wp_{10})$ a decomposition of the domain into closed subschemes $M_i := \mathsf{V}(\wp_i^{\texttt{e}})$. Likewise, a decomposition of $M^{\texttt{gen.irr}}(\pi,\mathrm{SL}_2)$ (which is an open subscheme of $\mathsf{Spec}\ \mathbb{Z}[x,y,z]/I^{\texttt{e}}$) into closed subschemes $M^\circ_i := M_i \cap M^{\texttt{gen.irr}}(\pi,\mathrm{SL}_2)$ follows.
Going back to matrices, let $\rho:\pi\to\mathrm{SL}_2(K)$ be any representation and write as usual $A:=\rho\ a$, $B:=\rho\ b$. As we did in the paper, there is a potential confusion while comparing two sets of conditions: for $n\in\mathbb{Z}$ and $s\in\{\pm1\}$, the equality $A^n=s\cdot I$ is not equivalent to $c_n(\mathrm{tr}\ A)=0$ and $d_n(\mathrm{tr}\ A)=s$.
First, if $c_n(\mathrm{tr}\ A)=0$, then $d_n(\mathrm{tr}\ A)^2=1$; and $A^n = d_n(\mathrm{tr}\ A)\cdot I$. But, if $A^n = s\cdot I$, then $c_n(\mathrm{tr}\ A)=0$ if and only if $\mathrm{tr}\ A\neq 2$ or $n=0$.
As a result, the two equivalent conditions are, for $s\in\{\pm\}$ and $n\in\mathbb{Z}$: $c_n(\mathrm{tr}\ A)=0$ and $d_n(\mathrm{tr}\ A)=s$ if and only if ($n=0$ and $s=1$) or ($A\neq\pm I$ and $A^n=s\cdot I$). So, we say for $n\in\mathbb{Z}$ and $s\in\{\pm1\}$ that $A^n =_{\texttt{nt}} s\cdot I$ (nontrivial identity) if either $n=0$ or $A\neq\pm I$.
With this notation we can study the points of the closed subschemes $M_i$ using matrices $A$, $B$. We have the following description (further details in Section 3.1 of the paper):
| $\begin{aligned}\rho\in M_1: &A^{n_1}=_{\texttt{nt}} -I, A^{n_2}=_{\texttt{nt}} -I,\\ &B^{m_1+m_2}=_{\texttt{nt}}I.\end{aligned}$ | $\begin{aligned}\rho\in M_2: &A^{n_1}=_{\texttt{nt}} -I, A^{n_2}=_{\texttt{nt}} I,\\ &B^{m_1+m_2}=_{\texttt{nt}}-I.\end{aligned}$ |
| $\begin{aligned}\rho\in M_3: &A^{n_1}=_{\texttt{nt}} I, A^{n_2}=_{\texttt{nt}} I,\\ &B^{m_1+m_2}=_{\texttt{nt}}I.\end{aligned}$ | $\begin{aligned}\rho\in M_4: &A^{n_1}=_{\texttt{nt}} I, A^{n_2}=_{\texttt{nt}} -I,\\ &B^{m_1+m_2}=_{\texttt{nt}}-I.\end{aligned}$ |
| $\begin{aligned}\rho\in M_5: &A^{n_1+n_2}=_{\texttt{nt}} I,\\ &B^{m_1}=_{\texttt{nt}} -I, B^{m_2}=_{\texttt{nt}}-I.\end{aligned}$ | $\begin{aligned}\rho\in M_6: &A^{n_1+n_2}=_{\texttt{nt}} -I,\\ &B^{m_1}=_{\texttt{nt}} -I, B^{m_2}=_{\texttt{nt}}I.\end{aligned}$ |
| $\begin{aligned}\rho\in M_7: &A^{n_1+n_2}=_{\texttt{nt}} I,\\ &B^{m_1}=_{\texttt{nt}} I, B^{m_2}=_{\texttt{nt}}I.\end{aligned}$ | $\begin{aligned}\rho\in M_8: &A^{n_1+n_2}=_{\texttt{nt}}-I,\\ &B^{m_1}=_{\texttt{nt}} I, B^{m_2}=_{\texttt{nt}}-I.\end{aligned}$ |
| $\begin{aligned}\rho\in M_9: &A^{n_1-n_2}=_{\texttt{nt}}-I,\\ &B^{m_1-m_2}=_{\texttt{nt}}I,\\ &\mathrm{tr}\ A^{n_1}B^{m_1}=0.\end{aligned}$ | $\begin{aligned}\rho\in M_{10}: &A^{n_1-n_2}=_{\texttt{nt}}I,\\ &B^{m_1-m_2}=_{\texttt{nt}}-I,\\ &\mathrm{tr}\ A^{n_1}B^{m_1}=0.\end{aligned}$ |
Working out these equations and studying the solutions, it is possible to describe the dimension of each closed $M_i$ imposing diophantine conditions on the exponents $n_i$ and $m_i$. Consider the following condition: given two non-zero integers $N_1,N_2$ and signs $s_1,s_2\in\{\pm1\}$; we say $\mathsf{C}(N_1,N_2,s_1,s_2)$ holds if the system \[\begin{matrix}&c_{N_1}(t)=0, &d_{N_1}(t)=s_1,\\ &c_{N_2}(t)=0, &d_{N_2}(t)=s_2;\end{matrix}\] has a solution in $\mathbb{C}$ (or any other char. 0 algebraically closed field). Writing $g:=\mathrm{gcd}(N_1,N_2)$, it is not hard to see that $\mathsf{C}(N_1,N_2,s_1,s_2)$ is equivalent to
| $\mathsf{C}(N_1,N_2,+1,+1)$: | $g>2$. |
| $\mathsf{C}(N_1,N_2,-1,+1)$: | $g>1 \wedge \nu_2(N_1)<\nu_2(N_2)$. |
| $\mathsf{C}(N_1,N_2,+1,-1)$: | $g>1 \wedge \nu_2(N_1)>\nu_2(N_2)$. |
| $\mathsf{C}(N_1,N_2,-1,-1)$: | $g>1 \wedge \nu_2(N_1)=\nu_2(N_2)$. |
Observe that the conditions on the $2$-adic valuations may imply the condition on the $\mathrm{gcd}$.
With this notation we can write down the formula for the dimension of the closed subschemes $M_i$. Below the conditions are hierarchized from top to bottom, that is, if any condition is not verified then the one inmediately below should be prioritized.
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[BH95]
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Brumfiel G. W.; Hilden, H. M. SL2-representations of finitely presented groups. MR1339764. Zbl 0838.20006. doi.org/10.1090/conm/187. |
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Church, B.; García-Cortés, F. SL2-character varieties of 2-generated groups and failure of weak integrality. arXiv:2410.23233. |
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[dJE24]
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de Jong, J.; Esnault. H. Integrality of the Betti moduli space. MR4684597. Zbl 1540.14027. doi.org/10.1090/tran/8928. |
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[Gol09]
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Goldman, W. M. Trace coordinates on Fricke spaces of some simple hyperbolic surfaces. MR2497777. Zbl 1175.30043. arXiv:0901.1404. |