Proof. The morphisms $X_1\times {\mathbb {G}}_m\to X|_{{\mathbb {G}}_m}$ and $X|_{{\mathbb {G}}_m}\to X_1\times {\mathbb {G}}_m$ given at the level of points by $(a,\lambda )\mapsto \lambda \cdot a$ and $a\mapsto ((\pi (a))^{-1}\cdot a,\pi (a))$, respectively, are inverses to each other and are ${\mathbb {G}}_m$-equivariant. This proves the first part of the lemma, and gives us an isomorphism of graded rings $A/(t-1)\otimes {\mathbb {C}}[t,1/t]\simeq A[1/t]$. Taking the degree $0$ parts of both sides, the second assertion follows.

Proof. Since $A$ is connected, if it is a normal ring then it is automatically a normal domain. Since $A$ is Cohen–Macaulay, in order to show that $A$ is normal it suffices by Serre's criterion for normality [Reference Matsumura and ReidMat86, Theorem 23.8, p. 183] to show that $A$ is regular in codimension $1$.

Since $t$ is not nilpotent, the ${\mathbb {G}}_m$-equivariant morphism $\pi : X\to {\mathbb {A}}^1$ is dominant, with isomorphic fibers of dimension $\dim A-1$ over ${\mathbb {G}}_m$. By condition (2), in fact $\pi$ has equidimensional fibers everywhere. By the miracle flatness theorem [Reference Matsumura and ReidMat86, Theorem 23.1, p. 179], since $A$ is Cohen–Macaulay and ${\mathbb {C}}[t]$ is regular it follows that $A$ is flat over ${\mathbb {C}}[t]$. In particular, $t$ is not a zero divisor in $A$.

By Lemma 2.2, we have $X|_{{\mathbb {G}}_m}\simeq X_1\times {\mathbb {G}}_m$, which is regular in codimension $1$ by condition (1). Thus, given any prime ideal ${\mathfrak {p}}\subset A$ of height $\leq 1$ such that $t\notin {\mathfrak {p}}$, the localization $A_{\mathfrak {p}}$ is regular. Next, suppose that ${\mathfrak {p}}\subset A$ is a prime ideal of height $1$ with $t\in {\mathfrak {p}}$. Since $t\in A$ is not a zero divisor, $A/(t)$ is Cohen–Macaulay of dimension $\dim A/(t)=\dim A-1$. Hypothesis (2) implies that $(A/tA)_{\mathfrak {p}}$ is regular, and hence $A_{\mathfrak {p}}$ is regular. This proves the result.

Proof. (1) If $A^{\operatorname{Fil}}$ is an integral domain, then so is $A=A^{\operatorname{Fil}}[1/t]_0$. Conversely, suppose that $A$ is an integral domain. If $ab=0$ in $A^{\operatorname{Fil}}$, then replacing $a$ and $b$ by nonzero terms (if any) of highest degree we may assume that $a$ and $b$ are homogeneous of degree $r$ and $s$, respectively. We then have $ab=0\in A_{r+s}^{\operatorname{Fil}}=\operatorname{Fil}_{r+s}$ within $A$, and hence $a=0$ or $b=0$. This proves that $A^{\operatorname{Fil}}$ is an integral domain.

(2) Given a normal ${\mathbb {Z}}$-graded domain $B$, its degree-$0$ component $B_0$ is also a normal domain. Indeed, it is obvious that $B_0$ is a domain; to see that $B_0$ is also normal, note that if $x\in B$ is integral over $B_0$ then we must have $x\in B_0$ by degree reasons. Applying this to $A=A^{\operatorname{Fil}}[1/t]_0$, we see that if $A^{\operatorname{Fil}}$ is a normal domain then so is $A$.

Fact 2.9 If $\pi$ is a group generated by $a_1,\ldots ,a_m$, then ${\mathbb {Q}}[X(\pi )]$ is generated as a ${\mathbb {Q}}$-algebra by the collection $\{\operatorname{tr}_{a_{i_1}\cdots a_{i_k}}:1 \leq i_1 < \cdots < i_k \leq m\}_{1 \leq k \leq 3}$.

Proof. (1) We have $\operatorname{length}_\sigma (g)=1$ if and only if $g=1$, and there are at most finitely many $\sigma$-words of length $\leq r$ in $\pi$ for any given $r$.

(2) Applying Lemma 2.8 and induction, we see that the class $\operatorname{tr}_{w}\in R_\pi$ of every word $w$ of length $r\geq 4$ in $\pi$ is a linear combination of terms of the form $\operatorname{tr}_{w_1}\cdots \operatorname{tr}_{w_s}$ with $\sum _{i=1}^s\operatorname{length}_\sigma (w_i)\leq r$ and $\operatorname{length}_\sigma (w_i)\leq 3$ for each $i=1,\ldots ,s$. This shows that $R_\pi ^\sigma$ is generated by $(R_\pi ^\sigma )_1\oplus (R_\pi ^\sigma )_2\oplus (R_\pi ^\sigma )_3$. Applying part (1) of this lemma, we obtain the desired result.

Proof. We begin by considering the filtration $\operatorname{Fil}^{\deg }$ on ${\mathbb {C}}[{\mathbb {M}}^m]$. For each $r\geq 0$, a close inspection of the work of Procesi [Reference ProcesiPro76, Theorems 1.1, 1.2, and 1.3] shows that the space of invariants $({\mathbb {C}}[{\mathbb {M}}^m]_r)^{\operatorname{SL}_2}$ is spanned by elements of the form

\[ \operatorname{tr}(w_1)\operatorname{tr}(w_2)\cdots \operatorname{tr}(w_s), \]

where each $w_i$ is a noncommutative monomial in the standard matrix variables $x_1,\ldots ,x_m$ on ${\mathbb {M}}^m$ such that we have $\sum _{i=1}^s \deg (w_i)=r$. Here, the degree $\deg (w_i)$ refers to the total multiplicity of matrices $x_1,\ldots ,x_m$ appearing in the monomial $w_i$. Since $\operatorname{SL}_2$ is linearly reductive, we see that the map

\[ (\operatorname{Fil}_r^{\deg}{\mathbb{C}}[{\mathbb{M}}^m])^{\operatorname{SL}_2}\to(\operatorname{Fil}_r^{\sigma}{\mathbb{C}}[\operatorname{Rep}_\pi])^{\operatorname{SL}_2}, \]

obtained from the surjective morphism

\[ \operatorname{Fil}_r^{\deg}{\mathbb{C}}[{\mathbb{M}}^m]\to \operatorname{Fil}_r^{\sigma}{\mathbb{C}}[\operatorname{Rep}_\pi] \]

of finite-dimensional complex representations of $\operatorname{SL}_2$, remains surjective. In particular, the vector space $(\operatorname{Fil}_r^\sigma {\mathbb {C}}[\operatorname{Rep}_\pi ])^{\operatorname{SL}_2}$ is spanned by elements of the form

\[ \operatorname{tr}_{w_1}\operatorname{tr}_{w_2}\cdots\operatorname{tr}_{w_s}, \]

where each $w_i$ is a product of elements of $\sigma$ (with multiplicities, but without inverses) such that the sum total number of elements appearing is $\leq r$. This shows that $(\operatorname{Fil}_r^\sigma {\mathbb {C}}[\operatorname{Rep}_\pi ])^{\operatorname{SL}_2}\subseteq \operatorname{Fil}_r^{\sigma }R_\pi$ since each element of the above form lies in $\operatorname{Fil}_r^{\sigma }R_\pi$, recalling that the filtration $\operatorname{Fil}^\sigma$ is compatible with multiplication on $R_\pi$. The other containment $\operatorname{Fil}_r^\sigma R_\pi \subseteq (\operatorname{Fil}_r^{\sigma }{\mathbb {C}}[\operatorname{Rep}_\pi ])^{\operatorname{SL}_2}$ follows from the observation that, using the relation ${\operatorname{tr}_{a}}{\operatorname{tr}_{a}}={\operatorname{tr}_{ab}}+{\operatorname{tr}_{ab^{-1}}}$ in $R_\pi$, one can write any ${\operatorname{tr}_{a}}$ with $\operatorname{length}_\sigma (a)\leq r$ as a linear combination of elements of the above form. Finally, since the surjection ${\mathbb {C}}[{\mathbb {M}}^m]\to {\mathbb {C}}[\operatorname{Rep}_\pi ]$ is compatible with the filtrations and the $\operatorname{SL}_2$-actions, we obtain a surjection

\[ ({\mathbb{C}}[{\mathbb{M}}^m]^{\deg})^{\operatorname{SL}_2}\to({\mathbb{C}}[\operatorname{Rep}_\pi]^\sigma)^{\operatorname{SL}_2}=({\mathbb{C}}[\operatorname{Rep}_\pi]^{\operatorname{SL}_2})^\sigma=R_\pi^\sigma, \]

which proves the second part of the proposition.

Proof. Let us write $\operatorname{Rep}_m'=\operatorname{Spec}{\mathbb {C}}[{\mathbb {M}}^m]^{\deg }/(E_1,\ldots ,E_m)$. Our proof of (1) and (2) will proceed by induction on $m$. Both statements are clear when $m=1$ (we note by the Jacobian criterion that the singular locus of $\operatorname{Rep}_1'$ consists of a single point). So let $m\geq 2$. For any $s\in \{1,\ldots ,m\}$, note that we have

\[ {\mathbb{C}}[{\mathbb{M}}^m]^{\deg}/(E_1,\ldots,E_s)={\mathbb{C}}[\operatorname{Rep}_s']\otimes{\mathbb{C}}[(X_{s+1})_{ij},\ldots,(X_m)_{ij}]. \]

In particular, each

\[ {\mathbb{C}}[{\mathbb{M}}^m]^{\deg}/(E_1,\ldots,E_s) \]

is an integral domain, showing that $E_1,\ldots ,E_m$ is a regular sequence in ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$, and ${\mathbb {C}}[\operatorname{Rep}_m']$ is a graded Cohen–Macaulay ring of pure dimension $3m+1$. Consider the ${\mathbb {G}}_m$-equivariant morphism

\[ \pi:\operatorname{Rep}_m'\to{\mathbb{A}}^1 \]

associated to the morphism of graded rings ${\mathbb {C}}[t]\to {\mathbb {C}}[\operatorname{Rep}_m']$ given by $t\mapsto T$. The fiber $\pi ^{-1}(1)=\operatorname{SL}_2^m$ is smooth, and $\pi ^{-1}(0)={\mathbb {M}}_0^m$ is reduced of dimension $3m$. We thus conclude by Lemma 2.3 that ${\mathbb {C}}[\operatorname{Rep}_m']$ is a normal domain. This completes the induction, and we have proven (1) and (2). It remains to prove (3). Note that we have a natural surjective ring homomorphism ${\mathbb {C}}[\operatorname{Rep}_m']\to {\mathbb {C}}[\operatorname{Rep}_m]^\sigma$. Note that ${\mathbb {C}}[\operatorname{Rep}_m]^\sigma$ is integral by Lemma 2.6, and has dimension at least $3m+1$. Thus, we see that ${\mathbb {C}}[\operatorname{Rep}_m']\to {\mathbb {C}}[\operatorname{Rep}_m]^\sigma$ must be an isomorphism, as the former is also integral of dimension $3m+1$. This completes the proof of the lemma.

Proof. First, by a result of Hochster and Roberts [Reference Hochster and RobertsHR74, Main Theorem], we see that $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}$ is Cohen–Macaulay. We next claim that the surjective homomorphism $\Phi _\sigma :({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}\to R_m^\sigma$ constructed in Proposition 2.12 induces an isomorphism

\[ ({\mathbb{C}}[{\mathbb{M}}^m]^{\deg})^{\operatorname{SL}_2}/(E_1,\ldots,E_m)\simeq R_m^\sigma. \]

Indeed, since $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}$ is a pure subring of ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$, the left-hand side is isomorphic to $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg }/(E_1,\ldots ,E_m))^{\operatorname{SL}_2}=({\mathbb {C}}[\operatorname{Rep}_m]^\sigma )^{\operatorname{SL}_2}=R_m^\sigma$ by Lemma 3.1, which is the desired result. Since ${\mathbb {C}}[\operatorname{Rep}_m]^\sigma$ is a normal graded domain by Lemma 3.1, so is $R_m^\sigma$. To show that $R_m^\sigma$ is Cohen–Macaulay, it suffices to show that $E_1,\ldots ,E_m$ is a regular sequence in $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}$. But this follows from the fact that $E_1,\ldots ,E_m$ is a regular sequence in ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$ and that $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}$ is a pure subring of ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$. The desired result follows.

Proof. For $m=2$ and $F_2$ a free group on generators $\sigma =\{a_1,a_2\}$, we may verify the functional equation by direct computation as follows. Recall the isomorphism $(\operatorname{tr}_{a_1},\operatorname{tr}_{a_2},\operatorname{tr}_{a_1a_2}):X_2\simeq {\mathbb {A}}^3$ from Example 2.7. It follows $R_2^\sigma \simeq {\mathbb {C}}[X_1,X_2,X_3,T]$ with each $X_1,X_2,T$ given degree 1 and $X_3$ given degree $2$, so that the Krull dimension of $R_2^\sigma$ is $4$ and

\[ H_2(t)=\frac{1}{(1-t)^3(1-t^2)} \]

from which the desired symmetry follows. For $m\geq 3$, we have the following result of Le Bruyn [Reference Le BruynLeB85] (based on rational expressions due to Weyl and Schur, cf. [Reference Le BruynLeB85]): the Hilbert series $h_m(t)$ of the graded algebra ${\mathbb {C}}[{\mathbb {M}}^m]^{\operatorname{SL}_2}$ has the functional equation $h_m(1/t)=-t^{4m}h_m(t)$. Our Hilbert series $H_m(t)$ is related to the series $h_m(t)$ by

\[ H_m(t)=\frac{(1-t^2)^m}{1-t}h_m(t), \]

where the factor $(1-t^2)^m/(1-t)$ arises as we take the $\operatorname{SL}_2$-invariants of

\[ {\mathbb{C}}[{\mathbb{M}}^m]^{\deg}/(E_1,\ldots,E_m) \]

instead of ${\mathbb {C}}[{\mathbb {M}}^m]$, noting that $E_1,\ldots ,E_m$ is a regular sequence of homogeneous elements of degree $2$ in ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$. Thus, the functional equation of $H_m(t)$ follows from that of $h_m(t)$. Now, the Krull dimension of $R_m^\sigma$ is $(3m-3)+1$. Hence, our last claim follows from the functional equation and Proposition 3.2, by the result of Stanley [Reference StanleySta78, Theorem 4.4] (see also [Reference Bruns and HerzogBH93, Corollary 4.4.6, p. 177]).

Proof. We have an identification $R_{g,n}^\sigma =R_m^\sigma$. We claim that $F_1,\ldots ,F_n$ is a regular sequence in $R_m^\sigma =({\mathbb {C}}[\operatorname{Rep}_m]^\sigma )^{\operatorname{SL}_2}$. Indeed, using the fact that $({\mathbb {C}}[{\mathbb {M}}^m]^{\deg })^{\operatorname{SL}_2}$ is a pure subring of ${\mathbb {C}}[{\mathbb {M}}^m]^{\deg }$, it suffices to note that ${\mathbb {C}}[\operatorname{Rep}_m]^\sigma /(F_1,\ldots ,F_s)$ is an integral domain (of successively lower dimension) for $1\leq s\leq n-1$. This follows from an easy application of Lemma 2.3 (as in the proof of Lemma 3.1). Now, by Theorem 3.3, $R_m^\sigma$ is a graded Gorenstein ring of dimension $3m-2$ with canonical module $R_m^\sigma (-2m-1)$. Hence, it follows that

\[ R_{g,n,k}^\sigma=R_{g,n}^\sigma/(F_1,\ldots,F_n) \]

is a graded Gorenstein ring of dimension $3m-2-n=6g+2n-5$ with canonical module

\[ R_{g,n,k}^\sigma(-2m-1+(n-1+4g+n-1))=R_{g,n,k}^\sigma(-1), \]

provided that we know $R_{g,n,k}^\sigma$ is a domain. Indeed, $R_{g,n,k}^\sigma$ is a graded Cohen–Macaulay ring, being a quotient of $R_{g,n}^\sigma$ by a regular sequence $F_1,\ldots ,F_n$, and its Hilbert series is $(1-t)^{n-1}(1-t^{4g+n-1})$ times the Hilbert series of $R_{g,n}^\sigma$; by applying parts (c) and (b) of [Reference Bruns and HerzogBH93, Corollary 4.4.6, p. 177] and [Reference Bruns and HerzogBH93, Theorem 4.4.5, p. 176], we would then deduce the above.

It thus remains to show that $R_{g,n,k}^\sigma$ is a normal domain. By Lemma 2.3, it suffices to show the following:

1. (1) $R_{g,n,k}$ is a normal ring and

2. (2) $R_{g,n,k}^\sigma /(t)$ is reduced of dimension $6g+2n-6$.

If $\chi (\Sigma _{g,n})\leq -3$, then (1) and (2) will follow from Propositions 4.7 and 4.9 proved in § 4. Let us now consider the cases where $|\chi (\Sigma _{g,n})|\leq 2$. If $\chi (\Sigma _{g,n})=-2$, then (1) and (2) follow from Propositions 4.8 and 4.9 for $(g,n)=(1,2)$, and they follow from Example 3.4(3) and Proposition 4.9 for $(g,n)=(0,4)$. The case $\chi (\Sigma _{g,n})=-1$ remains. The case $(g,n)=(0,3)$ is trivial, and for $(g,n)=(1,1)$ Example 3.4(2) verifies (1). It thus remains to verify (2) for $(g,n)=(1,1)$. But we find that

\[ R_{1,1,k}^\sigma/(t)\simeq{\mathbb{C}}[x,y,z]/(z(z-xy)), \]

whose spectrum is the union of the plane $\{z=0\}$ with the surface $\{z=xy\}$ in ${\mathbb {A}}_{(x,y,z)}^3$, and hence is reduced and of dimension $2$. This completes the proof.

Proof. The claim $p_F\in \{0,1\}$ follows from the fact that

\[ \operatorname{Spec}(R_{g,n,k}^\sigma/t)=({\mathbb{C}}[\operatorname{Rep}_{g,n,k}']/(t))^{\operatorname{SL}_2} \]

is reduced by Proposition 4.9. To see that $q_F\in \{0,1,2\}$, note that $B_F$ is generated in degrees $1$, $2$, and $3$ by Lemma 2.11 so we must have $q_F\in \{0,1,2,3\}$. Suppose toward contradiction that $q_F=3$, so that $(B_F)_1=(B_F)_2=0$. Given any $a_1,a_2,a_3\in \pi _1(\Sigma _{g,n})$, on $R_{g,n,k}$ we have the relations

\[ \operatorname{tr}_{a_1a_2a_3}+\operatorname{tr}_{a_1a_3a_2}=\operatorname{tr}_{a_1a_2}\operatorname{tr}_{a_3}+\operatorname{tr}_{a_1a_3}\operatorname{tr}_{a_2} +\operatorname{tr}_{a_2a_3}\operatorname{tr}_{a_1}-\operatorname{tr}_{a_1}\operatorname{tr}_{a_2}\operatorname{tr}_{a_3} \]

and

\begin{align*} \operatorname{tr}_{a_1a_2a_3}\operatorname{tr}_{a_1a_3a_2}&=(\operatorname{tr}_{a_1}^2+\operatorname{tr}_{a_2}^2+\operatorname{tr}_{a_3}^2)+(\operatorname{tr}_{a_1a_2}^2+\operatorname{tr}_{a_2a_3}^2+\operatorname{tr}_{a_1a_3}^2)\\ &\quad -(\operatorname{tr}_{a_1}\operatorname{tr}_{a_2}\operatorname{tr}_{a_1a_2}+\operatorname{tr}_{a_2}\operatorname{tr}_{a_3}\operatorname{tr}_{a_2a_3}+\operatorname{tr}_{a_1}\operatorname{tr}_{a_3}\operatorname{tr}_{a_1a_3})\\ &\quad +\operatorname{tr}_{a_1a_2}\operatorname{tr}_{a_2a_3}\operatorname{tr}_{a_1a_3}-4. \end{align*}

Since the right-hand sides of both equations are zero on $B_F$, which is an integral domain, this shows that $\operatorname{tr}_{a_1a_2a_3}=\operatorname{tr}_{a_1a_3a_2}=0$ on $B_F$. This shows that we must have $(B_F)_3=0$ as well, a contradiction. Hence, we must have $q_F\in \{0,1,2\}$.

Proof. Combining Corollary 3.5 with the result of Watanabe [Reference WatanabeWat81, Corollary 2.9], we find that $K+E'+E=\operatorname{Div}(f)$ for some $f\in {\mathbb {C}}(Z)$ (the function field of $Z$) and

\[ E'=\sum_{\substack{q_F\neq0}}\frac{q_F'-1}{q'_F}\operatorname{Proj} B_F, \]

where we have set $q_F'=q_F/\gcd (p_F,q_F)$ for $q_F\neq 0$, so that in particular $q_F'=1$ if $p_F=0$. But in light of Lemma 3.6, we have

\begin{align*} 0&\sim K+E'+E\\ &=K+\sum_{\substack{p_F=1\\q_F=2}}\frac{1}{2}\operatorname{Proj} B_F+\left\{\sum_{\substack{p_F=1\\q_F=2}}\frac{1}{2}\operatorname{Proj} B_F+\sum_{\substack{p_F=1\\q_F=1}}\operatorname{Proj} B_F\right\}\\ &=K+D \end{align*}

from which we conclude the result.

Proof. (1) For any $a,v\in {\mathbb {M}}({\mathbb {C}})$, we have $\det (a+\varepsilon v)=\det (a)+\varepsilon \operatorname{tr}(va^*)$ within ${\mathbb {C}}[\varepsilon ]/(\varepsilon ^2)$, from which the first statement follows. In particular, $\dim T_{a}{\mathbb {M}}_{\det a}=3$ provided that $a\neq 0$. Since $\operatorname{tr}(uaa^*)=\operatorname{tr}(aua^*)=\det (a)\operatorname{tr}(u)=0$ for every $u\in \mathfrak {sl}_2({\mathbb {C}})$, we have $\mathfrak {sl}_2({\mathbb {C}})\cdot a\subseteq T_{a}{\mathbb {M}}_{\det a}$ and $a\cdot \mathfrak {sl}_2({\mathbb {C}})\subseteq T_a{\mathbb {M}}_{\det a}$. Lastly, if $\det (a)\neq 0$ then $\dim \mathfrak {sl}_2({\mathbb {C}})\cdot a=\dim a\cdot \mathfrak {sl}_2({\mathbb {C}})=3$ from which the equalities follow by dimension reasons.

(2) For any $a,v\in {\mathbb {M}}({\mathbb {C}})$ we have $\operatorname{tr}(a+\varepsilon v)=\operatorname{tr}(a)+\varepsilon \operatorname{tr}(v)$ within ${\mathbb {C}}[\varepsilon ]/(\varepsilon ^2)$, from which the first statement follows. We have $\operatorname{tr}([v,a]a^*)=\operatorname{tr}(vaa^*-ava^*)=\det (a)\operatorname{tr}(v)-\det (a)\operatorname{tr}(v)=0$ and $\operatorname{tr}([v,a])=0$ for any $v\in \mathfrak {sl}_2({\mathbb {C}})$, and hence $[\mathfrak {sl}_2({\mathbb {C}}),a]\subseteq T_{a,{\rm Char}}$. Finally, if $a$ is nonscalar, then we have $\dim _{\mathbb {C}} T_{a}{\mathbb {M}}_{\det a,\operatorname{tr} a}=2$ since $v\mapsto \operatorname{tr}(va^*)$ and $w\mapsto \operatorname{tr}(v)$ are linearly independent on ${\mathbb {M}}({\mathbb {C}})$. On the other hand, $\dim _{\mathbb {C}}[\mathfrak {sl}_2({\mathbb {C}}),a]=2$ if $a$ is nonscalar, and hence $T_{a}{\mathbb {M}}_{\det a,\operatorname{tr} a}=[\mathfrak {sl}_2({\mathbb {C}}),a]$ by dimension reasons.

Proof. Assume that $a\in {\mathbb {M}}_t^m({\mathbb {C}})$ satisfies neither of the conditions (1) and (2). We must show that $d_aF:T_a{\mathbb {M}}_t^m\to {\mathbb {C}}^n$ is surjective. Now, the first $n-1$ coordinates of $d_aF$ are given by

\[ (u_1,\ldots,u_m)\mapsto(\operatorname{tr}(u_{2g+1}),\ldots,\operatorname{tr}(u_{2g+n-1})), \]

for each $u=(u_1,\ldots ,u_m)\in T_a{\mathbb {M}}_{t}^m$. Since $a_{2g+j}$ is nonscalar for $j\in \{1,\ldots ,n-1\}$ by hypothesis, each of the functionals $u\mapsto \operatorname{tr}(u_{2g+j})$ is nonzero. Thus, to show surjectivity of $d_aF$ it remains to show that $(0,\ldots ,0,1)\in \operatorname{Im} d_aF$.

Let $u\in \mathfrak {sl}_2({\mathbb {C}})$ be given. We then have $[u,b_{g+j}]\in T_{b_{g+j}}{\mathbb {M}}_t$ for $j\in \{1,\ldots ,n-1\}$, and the composition $T_{b_{g+j}}{\mathbb {M}}_t\hookrightarrow T_a{\mathbb {M}}_t^m\to {\mathbb {C}}^n$ with $d_aF$ sends

\[ [u,b_{g+j}]\mapsto(0,\ldots,0,\operatorname{tr}([u,b_{g+j}]b^{g+j}))=(0,\ldots,0,\operatorname{tr}(u[b_{g+j},b^{g+j}])). \]

Similarly, after a short computation (using the fact $u^*=-u$ for $u\in \mathfrak {sl}_2({\mathbb {C}})$) we also find that for $i\in \{1,\ldots ,g\}$ the composition $T_{(a_{2i-1},a_{2i})}{\mathbb {M}}_t^2\hookrightarrow T_a{\mathbb {M}}_t^m\to {\mathbb {C}}^n$ with $d_aF$ sends

\begin{gather*} (ua_{2i-1}-a_{2i-1}u,-a_{2i}u)\mapsto(0,\ldots,0,\operatorname{tr}(u[a_{2i-1},a_{2i}a_{2i-1}^*a_{2i}^*b^i])),\\ (a_{2i-1}u,0)\mapsto(0,\ldots,0,\operatorname{tr}(u[a_{2i},a_{2i-1}^*a_{2i}^*b^ia_{2i-1}])),\\ (0,a_{2i}u)\mapsto(0,\ldots,0,\operatorname{tr}(u[a_{2i-1}^*,a_{2i}^*b^ia_{2i-1}a_{2i}])),\\ (-a_{2i-1}u,ua_{2i}-a_{2i}u)\mapsto(0,\ldots,0,\operatorname{tr}(u[a_{2i}^*,b^ia_{2i-1}a_{2i}a_{2i-1}^*])). \end{gather*}

In particular, summing up the above tangent vectors in $T_a{\mathbb {M}}_t^m$ we find that

\[ (ua_{2i-1}-a_{2i-1}u,ua_{2i}-a_{2i}u)\mapsto(0,\ldots,0,\operatorname{tr}(u[b_i,b^i])). \]

Thus, if $a$ does not satisfy condition (2), at least one of the expressions above must be nonzero for some $u\in \mathfrak {sl}_2({\mathbb {C}})$. This implies that $(0,\ldots ,0,1)\in \operatorname{Im} d_aF$, and hence $d_aF$ is surjective, as desired.

Proof. First, $\Phi$ is dominant by Lemma A.3. Note that $\operatorname{SL}_2^{2g}\times \prod _{i=1}^n\operatorname{SL}_{2,k_i}$ is a complete intersection scheme of dimension $6g+2n$, and in particular Cohen–Macaulay. The target $\operatorname{SL}_2$ of $\Phi$ is regular of dimension $3$. By the miracle flatness theorem, to prove that $\Phi$ is flat it suffices to show that the fibers of $\Phi$ are equidimensional, of dimension $6g+2n-3$. Consider the composition $\operatorname{tr}\circ \Phi :\operatorname{SL}_2^{2g}\times \prod _{i=1}^n\operatorname{SL}_{2,k_i}\to {\mathbb {A}}^1$. Since $\operatorname{tr}\circ \Phi$ is dominant, it is flat. In particular, the fibers of $\operatorname{tr}\circ \Phi$ are all of pure dimension $6g+2n-1$.

Next, note that $\operatorname{SL}_2$ acts by conjugation on the domain and target of $\Phi$, and $\Phi$ is equivariant with respect to this action. Hence, given any $l\in {\mathbb {C}}$ the fibers of the restriction $\Phi :\Phi ^{-1}(\operatorname{SL}_{2,l})\to \operatorname{SL}_{2,l}$, away from the fibers over $\pm \mathbf 1$, are isomorphic hence equidimensional of dimension $6g+2n-3$.

Proof. Let us denote by $F_+$ (respectively $F_-$) the fiber of $\Phi$ above $\mathbf 1$ (respectively $-\mathbf 1$). We shall proceed by induction on $(g,n)$. Consider first the case where $n=0$. We must then have $g\geq 2$. Consider the projection

\[ \pi:F_{\pm}\subset\operatorname{SL}_2^{2g}\to\operatorname{SL}_2^2 \]

onto the last two copies of $\operatorname{SL}_2$. Given $(b_1,b_2)\in \operatorname{SL}_2^2({\mathbb {C}})$, the dimension of the fiber $\pi ^{-1}(b_1,b_2)$ is estimated as follows.

1. (1) If $\langle b_1,b_2\rangle \neq \pm \mathbf 1$, then $\dim \pi ^{-1}(b_1,b_2)=6(g-1)-3$ by Lemma 4.3.

2. (2) If $\langle b_1,b_2\rangle =\pm \mathbf 1$, then $\dim \pi ^{-1}(b_1,b_2)\leq 6(g-1)-2$.

Here, part (2) follows from the inductive hypothesis and the fact that, when $g=2$ so that $g-1=1$, the fiber of $\langle -,-\rangle :\operatorname{SL}_2^2\to \operatorname{SL}_2$ above $\pm \mathbf 1$ has dimension at most $4$ by Lemma A.4. Since the locus of $(b_1,b_2)\in \operatorname{SL}_2^2$ defined by part (1) is 6-dimensional, and the locus defined by condition (2) is at most 4-dimensional by Lemma A.4, we see that $\dim F_\pm \leq 6g-3$ as desired. Consider next the case where $n\geq 1$. If $g=0$, then we must have $n\geq 3$. Consider the projection

\[ \pi:F_{\pm}\subset\operatorname{SL}_2^{2g}\times\prod_{i=1}^n\operatorname{SL}_{2,k_i}\to\operatorname{SL}_{2,k_n} \]

onto the last factor of the product. Given $b\in \operatorname{SL}_{2,k_n}({\mathbb {C}})$, the dimension of the fiber $\pi ^{-1}(b)$ is estimated as follows.

1. (1) If $b\neq \pm \mathbf 1$, then $\dim \pi ^{-1}(b)=6g+2(n-1)-3$ by the Lemma 4.3.

2. (2) If $b=\pm \mathbf 1$, then $\dim \pi ^{-1}(b)\leq 6g+2(n-1)- 2$.

Here, part (2) follows from the inductive hypothesis, from Lemma A.4 in the case $(g,n)=(1,1)$, and from the observation that the fiber of $\operatorname{SL}_{2,l}\times \operatorname{SL}_{2,\pm l}\to \operatorname{SL}_2$, $(a_1,a_2)\mapsto a_1a_2$ over $\pm \mathbf 1$ has dimension $2$ in the case $(g,n)=(0,3)$. Since the locus of $b$ satisfying (1) has dimension $2$, we find that $\dim F_\pm \leq 6g+2n-3$ as desired.

Proof. First, $\Psi$ is dominant by Lemma A.8. Note that ${\mathbb {M}}_{0}^{2g}\times {\mathbb {M}}_{0,0}^n$ is a complete intersection scheme of dimension $6g+2n$, and in particular Cohen–Macaulay. The target ${\mathbb {M}}_0$ of $\Psi$ is regular away from $\{0\}$, of dimension $3$. By the miracle flatness theorem, to prove that $\Psi$ is flat over ${\mathbb {M}}_0\setminus \{0\}$ it suffices to show that the fibers of $\Psi$ are equidimensional, of dimension $6g+2n-3$. Consider the composition $\operatorname{tr}\circ \Psi _{g,n}:{\mathbb {M}}_{0}^{2g}\times {\mathbb {M}}_{0,0}^n\to {\mathbb {A}}^1$. Since $\operatorname{tr}\circ \Psi$ is dominant, it is flat. In particular, the fibers of $\operatorname{tr}\circ \Psi$ are all of pure dimension $6g+2n-1$.

Next, note that $\operatorname{SL}_2$ acts by conjugation on the domain and target of $\Psi$, and $\Psi$ is equivariant with respect to this action. Hence, given any $l\in {\mathbb {C}}$ the fibers of the restriction $\Psi :\Psi ^{-1}({\mathbb {M}}_{0,l})\to {\mathbb {M}}_{0,l}$ are isomorphic, hence equidimensional of dimension $6g+2n-3$, away from the fiber over $0$.

Proof. Since the domain ${\mathbb {M}}_0^{2g}\times {\mathbb {M}}_{0,0}^n$ of $\Psi$ has dimension $6g+2n$ and is integral, if $\dim \Psi ^{-1}(0)\geq 6g+2n$ then $\Psi$ must be identically zero, contradicting the fact that $\Psi$ is dominant by Lemma A.8.

Proof. Since $\operatorname{Rep}_{g,n,k}$ is a complete intersection of dimension $6g+2n-3$ in the regular scheme ${\mathbb {M}}^{m}$, it is Cohen–Macaulay. As a consequence of Serre's criterion for normality [Reference Matsumura and ReidMat86, Theorem 23.8, p. 183], a Cohen–Macaulay ring is normal if and only if it is regular in codimension $1$. It thus suffices to demonstrate that $\operatorname{Rep}_{g,n,k}$ is regular in codimension $1$. Consider the restriction of $F=(F_1,\ldots ,F_n)$ above to $\operatorname{SL}_2^m\subset {\mathbb {M}}^m\times {\mathbb {A}}^1$. Since $\operatorname{SL}_2^m$ is regular, the singular locus of $\operatorname{Rep}_{g,n,k}=F^{-1}(0)$ is the intersection of $\operatorname{Rep}_{g,n,k}$ with the critical locus of $F$. It suffices to show that the locus in $\operatorname{Rep}_{g,n,k}$ defined by each of the conditions (1) and (2) of Lemma 4.2 has dimension at most $(6g+2n-3)-2$. We shall represent points of $\operatorname{Rep}_{g,n,k}$ by sequences $(a_1,\ldots ,a_{2g+n-1})$ of matrices, and adopt the same notations $b_I$ and $b^I$ for the various products of matrices as used in Lemma 4.2.

First, consider the locus $W\subset \operatorname{Rep}_{g,n,k}$ defined by condition (1) of Lemma 4.2, which plays a role only when $n\geq 2$. For each $i=1,\ldots ,n-1$ and $s\in \{\pm 1\}$, the locus $\{b_{g+i}=s\mathbf 1\}\subset \operatorname{Rep}_{g,n,k}$ is isomorphic to $\operatorname{Rep}_{g,n-1,k'}$ where $k'$ is the $(n-1)$-tuple obtained from $k$ by omitting $k_i$ and replacing $k_n$ by $sk_n$. By our assumption on $(g,n)$, we have $2g+(n-1)-1\geq 2$, and we find by Lemma 4.4

\[ \dim\{b_{g+i}=s\mathbf1\}=\dim\operatorname{Rep}_{g,n-1,k'}=6g+2(n-1)-3=(6g+2n-3)-2. \]

As $W=\bigcup _{i=1}^{n-1}\{b_{g+i}=\pm \mathbf 1\}$, it has codimension $2$ in $\operatorname{Rep}_{g,n,k}$ as desired.

Next, let $Z\subset \operatorname{Rep}_{g,n,k}$ be the locus defined by condition (2) of Lemma 4.2. We stratify $Z$ further into three subloci $Z_1$, $Z_2$, and $Z_3$, and estimate their dimensions as follows.

(1) Let $Z_1\subset Z$ be the sublocus consisting of $(a_1,\ldots ,a_m)$ such that $b_i=\pm \mathbf 1$ for some $i\in \{1,\ldots ,g\}$ or $b_{g+j}b_{g+j+1}=\pm \mathbf 1$ for some $j\in \{1,\ldots ,n-2\}$. We claim that $\dim Z_1\leq (6g+2n-3)-2$.

Given $i\in \{1,\ldots ,g\}$ and $s\in \{\pm 1\}$, consider the locus where $b_i=s\mathbf 1$. By Lemma A.4, $(a_{2i-1},a_{2i})$ must vary over a locus of dimension at most $4$. For fixed $(a_{2i-1},a_{2i})$ the remaining matrices in the sequence $(a_1,\ldots ,a_m)$ vary over a locus isomorphic to $\operatorname{Rep}_{g-1,n,k'}$ for $k=(k_1,\ldots ,k_{n-1},sk_n)$, which has dimension $\dim \operatorname{Rep}_{g-1,n,k'}=6(g-1)+2n-3$ by Lemma 4.4 since $2(g-1)+n-1\geq 2$ by our assumption. Thus, the locus in $Z_1$ where $b_i=s\mathbf 1$ for $i\in \{1,\ldots ,g\}$ has dimension bounded by

\[ 4+(6(g-1)+2n-3)=(6g+2n-3)-2. \]

Given $j\in \{1,\ldots ,n-2\}$ and $s\in \{\pm 1\}$, consider the locus where $b_{g+j}b_{g+j+1}=s\mathbf 1$. The pair $(b_{g+j},b_{g+j+1})$ then varies over a locus of dimension at most $2$. For fixed $(b_{g+j},b_{g+j+1})$ the remaining matrices in the sequence $(a_1,\ldots ,a_m)$ vary over a locus isomorphic to $\operatorname{Rep}_{g,n-2,k'}$ where $k'$ is obtained from $k$ by removing $k_{j}$ and $k_{j+1}$, and replacing $k_n$ by $sk_n$. We have $\dim \operatorname{Rep}_{g,n-2,k'}=6g+2(n-2)-3$ by Lemma 4.4 since $2g+(n-2)-1\geq 2$ by our assumption. Thus, the given locus has dimension bounded by

\[ 2+(6g+2(n-2)-3)=(6g+2n-3)-2. \]

This shows that $\dim Z_1\leq (6g+2n-3)-2$.

(2) Let $Z_2\subset Z\setminus Z_1$ be the sublocus consisting of $(a_1,\ldots ,a_m)$ with $\operatorname{tr}(b_i)=\pm 2$ for some $i\in \{1,\ldots ,g\}$ or $\operatorname{tr}(b_{g+j}b_{g+j+1})=\pm 2$ for some $j\in \{1,\ldots ,n-2\}$. We claim that $\dim Z_2\leq (6g+2n-3)-2$.

Given $i\in \{1,\ldots ,g\}$, consider the locus $\operatorname{tr}(b_i)=\pm 2$. By Lemma 4.3, the pair $(a_{2i-1},a_{2i})$ varies over a locus of dimension at most $5$. For fixed $(a_{2i-1},a_{2i})$, by Lemma A.2 the product $b^i$ must vary over a locus of dimension at most $1$. We have two possibilities.

1. (a) We have $b^i\in \{\pm \mathbf 1\}$. We must have $2(g-1)+(n-1)-1\geq 2$ since $Z_2$ lies in the complement of $Z_1$. By Lemma 4.4, for fixed $(a_{2i-1},a_{2i})$ the remaining matrices in $(a_1,\ldots ,a_m)$ vary over a locus of dimension at most $6(g-1)+2(n-1)-3$.

2. (b) We have $b^i\neq \pm \mathbf 1$. By Lemma 4.3, for fixed $(a_{2i-1},a_{2i})$ and $b^i$ the remaining matrices in $(a_1,\ldots ,a_m)$ vary over a locus of dimension at most $6(g-1)-2(n-1)-3$.

Thus, we see that the locus of $(a_1,\ldots ,a_m)$ in $Z$ satisfying $\operatorname{tr}(b_i)=\pm 2$ for some $i\in \{1,\ldots ,g\}$ has dimension bounded by

\[ 5+1+(6(g-1)+2(n-1)-3)=(6g+2n-3)-2. \]

Given $j\in \{1,\ldots ,n-2\}$, consider the locus $\operatorname{tr}(b_{g+j}b_{g+j+1})=\pm 2$. By Lemma 4.3, we see that the pair $(b_{g+j},b_{g+j+1})$ varies over a locus of dimension $3$. For fixed $(b_{g+j},b_{g+j+1})$, by Lemma A.2 the product $b^{\{g+j,g+j+1\}}$ must vary over a locus of dimension at most $1$. We have two possibilities.

1. (a) We have $b^{\{g+j,g+j+1\}}\in \{\pm \mathbf 1\}$. We must have $2g+(n-3)-1\geq 2$ since $Z_2$ lies in the complement of $Z_1$. By Lemma 4.4, for fixed $(b_{g+j},b_{g+j+1})$ the remaining matrices in $(a_1,\ldots ,a_m)$ vary over a locus of dimension at most $6g+2(n-3)-3$.

2. (b) We have $b^{\{g+j,g+j+1\}}\neq \pm \mathbf 1$. By Lemma 4.4, for fixed $(b_{g+j},b_{g+j+1})$ and $b^{\{g+j,g+j+1\}}$ the remaining matrices in $(a_1,\ldots ,a_m)$ vary over a locus of dimension at most $6g-2(n-3)-3$.

Thus, we see that the locus of $(a_1,\ldots ,a_m)$ in $Z$ with $\operatorname{tr}(b_{g+j}b_{g+j+1})=\pm 2$ for some $j\in \{1,\ldots , n-2\}$ has dimension bounded by

\[ 3+1+(6g+2(n-3)-3)=(6g+2n-3)-2. \]

This shows that $\dim Z_2\leq (6g+2n-3)-2$.

(3) We claim that $Z_3=Z\setminus (Z_1\cup Z_2)$ must have $\dim Z_3\leq (6g+2n-3)-2$. Suppose first that $g\geq 1$. The pair $(a_1,a_2)$ varies over a locus of dimension $6$. For fixed $(a_1,a_2)$, since $\operatorname{tr}(b_1)\neq \pm 2$ we see by Lemma A.2 that there are only finitely many possible values of $b^i$. For fixed $(a_1,a_2)$ and value of $b^i$, the remaining matrices in $(a_1,\ldots ,a_m)$ vary over a locus of dimension $6(g-1)+2(n-1)-3$ by Lemma 4.3 (noting that $2(g-1)+(n-1)-1\geq 2$ if $b^i\in \{\pm \mathbf 1\}$ by assumption that $Z_3\cap Z_1=\emptyset$). Hence, for $g\geq 1$, $Z_3$ has dimension bounded by

\[ 6+(6(g-1)+2(n-1)-3)=(6g+2n-3)-2. \]

Arguing similarly, if $g=0$ so that $n\geq 5$, $Z_3$ has dimension bounded by

\[ 4+(6g+2(n-3)-3)=(6g+2n-3)-2. \]

Therefore, $\dim Z_3\leq (6g+2n-3)-2$.

From the above computations, we conclude that $\dim Z\leq (6g+2n-3)-2$. Hence, $\operatorname{Rep}_{g,n,k}$ is normal, as desired.

Proof. Since $\operatorname{Rep}_{1,2,k}$ is a complete intersection of dimension $7$ in the regular scheme ${\mathbb {M}}^m$, it is in particular Cohen–Macaulay. By Serre's criterion, it suffices to show that $\operatorname{Rep}_{1,2,k}$ is regular in dimension $1$. It suffices to show that locus defined by each of the conditions (1) and (2) of Proposition 4.2 is of dimension at most $5$. We shall represent a point of $\operatorname{Rep}_{1,2,k}$ by a triple of matrices $(a_1,a_2,a_3)\in \operatorname{SL}_2^3$ satisfying $\operatorname{tr}(a_3)=k_1$ and $\operatorname{tr}(\langle a_1,a_2\rangle a_3)=k_2$, and adopt the notations of § 4.1.

Consider first the locus of $\operatorname{Rep}_{1,2,k}$ where $a_i=\pm \mathbf 1$ for some $i\in \{1,2,3\}$. By Lemmas 4.3 and A.4, we see that this locus has dimension at most $3+2-1=4$ when $i\in \{1,2\}$, and at most $3+3-1=5$ when $i=3$. The union $Y=\bigcup _{i=1}^3\{a_i=\pm \mathbf 1\}$ includes the locus of $\operatorname{Rep}_{1,2,k}$ defined by condition (1) of Proposition 4.2, and is at most 5-dimensional.

It remains to estimate the dimension of the locus $Z\subset \operatorname{Rep}_{1,2,k}\setminus Y$ defined by condition (2) of Proposition 4.2. In what follows, we stratify $Z$ into subloci and estimate their respective dimensions.

(1) Let $Z_1\subset Z$ be the locus where $\operatorname{tr}(a_1)\neq \pm 2$. From the condition

\[ [a_1^{-1},a_2^{-1}a_3a_1a_2]=0, \]

and Lemma A.6, we see that for given $a_1$ there are only finitely many possible values of $a_2^{-1}a_3a_1a_2$. The sublocus where $a_3a_1=\pm \mathbf 1$ has dimension at most $2+3=5$, so we may restrict our attention to the sublocus of $Z_1$ where $a_3a_1\neq \pm \mathbf 1$. For fixed $a_1$, there are only finitely many possible values of $\operatorname{tr}(a_3a_1)$, and hence $a_3$ varies over a locus of dimension $1$, and for fixed $a_1$ and $a_3$ since there are only finitely many possible values of $a_2^{-1}a_3a_1a_2$ showing that $a_2$ varies over a locus of dimension $1$ (the centralizer of $a_1a_3\neq \pm \mathbf 1$ being 1-dimensional). Thus, the sublocus of $Z_1$ where $a_3a_1\neq \pm \mathbf 1$ has dimension at most $3+1+1=5$. Therefore, we conclude that $\dim Z_1\leq 5$.

(2) Let $Z_2\subset Z\setminus Z_1$ be the locus where $\operatorname{tr}(a_2)\neq \pm 2$. We then consider

\[ [a_2,a_1^{-1}a_2^{-1}a_3a_1]=0, \]

and by the same argument as in part (1) we conclude that $\dim Z_2\leq 5$.

(3) Let $Z_3\subset Z\setminus (Z_1\cup Z_2)$ be the locus where $\operatorname{tr}(a_1a_2)\neq \pm 2$. By the condition

\[ [a_1a_2,a_1^{-1}a_2^{-1}a_3]=0, \]

and Lemma A.6, for fixed $a_1$ and $a_2$ there are at most finitely many possible values of $a_1^{-1}a_2^{-1}a_3$ and hence of $a_3$. Thus, $\dim Z_4\leq 2+2=4$.

(4) Let $Z_4=Z\setminus (Z_1\cup Z_2\cup Z_3)$. We must have $\operatorname{tr}(a_1),\operatorname{tr}(a_2),\operatorname{tr}(a_1a_2)\in \{\pm 2\}$, and hence $(a_1,a_2)$ varies over a locus of dimension at most $3$. Hence, $\dim Z_4\leq 3+2=5$.

This completes the proof that $\dim Z\leq 5$, and hence $\operatorname{Rep}_{1,2,k}$ is normal.

Proof. First, the locus in $H_{g,n}$ determined by the condition $a_i=0$ has dimension

\begin{gather*} 3+6(g-1)+2(n-1)=(6g+2n-3)-2\quad \text{if } i\in\{1,\ldots,2g\},\\ 6g+2(n-2)=(6g+2n-3)-1\quad \text{if } i\in\{2g+1,\ldots,2g+n-1\}. \end{gather*}

Therefore, the union $Y=\bigcup _{i=1}^m\{a_i=0\}$ has dimension at most $(6g+2n-3)-1$. Since ${\rm Crit}(E)\cap H_{g,n}$ and the locus defined by condition (1) of Lemma 4.2 both lie in $Y$, their dimensions also cannot be more than $(6g+2n-3)-1$.

Next, let $Z\subset H_{g,n}\setminus Y$ be the locus defined by condition (2) of Lemma 4.2. We must show that $\dim Z\leq (6g+2n-3)-1$. Let us first treat the cases $(g,n)=(0,4)$ and $(1,2)$ separately.

(1) Let $(g,n)=(0,4)$. The locus $Z$ consists of $(a_1,a_2,a_3)\in {\mathbb {M}}_{0,0}^3({\mathbb {C}})$ with $a_i$ nonzero and $[a_1,a_2a_3]=[a_2,a_3a_1]=[a_3,a_1a_2]=0$. Suppose that one of $a_1a_2$, $a_2a_3$, and $a_3a_1$ is nonzero; say $a_1a_2\neq 0$. Then we must have $\operatorname{tr}(a_1a_2)\neq 0$ by Lemma A.8(1). But the condition $[a_3,a_1a_2]=0$ implies that $a_1a_2$ is a scalar multiple of $a_3$ and hence $\operatorname{tr}(a_1a_2)=0$, a contradiction. Arguing similarly for the other cases, we conclude that

\[ a_1a_2=a_2a_3=a_3a_1=0. \]

The condition $\operatorname{tr} a_2=\operatorname{tr} a_3=0$ then implies that $a_2$ and $a_3$ are both scalar multiples of $a_1$. Thus, $Z$ is 4-dimensional.

(2) Let $(g,n)=(1,2)$. The locus $Z$ consists of $(a_1,a_2,a_3)\in {\mathbb {M}}_0^2\times {\mathbb {M}}_{0,0}({\mathbb {C}})$ with $a_i$ nonzero satisfying, among other things, $[\langle a_1,a_2\rangle ,a_3]=0$. By Lemma A.2(1), this implies that $\langle a_1,a_2\rangle$ is a scalar multiple of $a_3$ and in particular $\operatorname{tr}\langle a_1,a_2\rangle =0$. The condition $\operatorname{tr}(\langle a_1,a_2\rangle a_3)=0$ then shows that in fact $\langle a_1,a_2\rangle a_3=0$ by Lemma A.8(1). The sublocus of $A$ where $\langle a_1,a_2\rangle \neq 0$ has dimension