Proof. Let $n=|V(G)|$ and $m=|E(G)|$ . Since $(G,\ast )$ is $\Gamma $ -rigid, there is a $\Gamma $ -symmetric infinitesimally rigid realisation $(G, \ast ,p)$ of $(G,\ast )$ . Since $|V(G)|\geq 2d$ , we can take such a realisation so that the affine span of $p(V(G))$ is at least $d-1$ . In particular, the space of trivial motions of $(G,*,p)$ has dimension $\binom {d+1}{2}$ .

Recall that the rigidity matrix $R(G,p)$ represents a linear map from $\mathbb {R}^{dn}$ to $\mathbb {R}^m$ . We shall decompose $\mathbb {R}^{dn}$ and $\mathbb {R}^m$ into two subspaces whose elements are either symmetric or anti-symmetric with respect to $\Gamma $ and $*$ . Specifically, let

$$ \begin{align*} M_{\mathrm{sym}}&:=\{ \dot{p}:V(G)\rightarrow \mathbb{R}^d \mid \dot{p}(u^*)=I_{t,d} \dot{p}(u)\text{ for all } u\in V(G)\}, \\ M_{\mathrm{ant}}&:=\{ \dot{p}:V(G)\rightarrow \mathbb{R}^d \mid \dot{p}(u^*)=-I_{t,d} \dot{p}(u)\text{ for all } u\in V(G)\}, \\ S_{\mathrm{sym}}&:=\{ w:E(G)\rightarrow \mathbb{R} \mid w(e^*)=w(e) \text{ for all } e\in E(G)\} \text{ and} \\ S_{\mathrm{ant}}&:=\{ w:E(G)\rightarrow \mathbb{R} \mid w(e^*)=-w(e) \text{ for all } e\in E(G)\}. \end{align*} $$

Then $\mathbb {R}^{dn}=M_{\mathrm {sym}}\oplus M_{\mathrm {ant}}$ and $\mathbb {R}^m=S_{\mathrm {sym}}\oplus S_{\mathrm {ant}}$ . Observe further that the rigidity matrix maps $M_{\mathrm {sym}}$ to $S_{\mathrm {sym}}$ and $M_{\mathrm {ant}}$ to $S_{\mathrm {ant}}$ . Indeed, if $\dot {p}\in M_{\mathrm {sym}}$ , then for any edge $e=uv$ , we have

$$ \begin{align*} (p(u^*)-p(v^*))\cdot (\dot{p}(u^*)-\dot{p}(v^*))&= (I_{t,d}(p(u^*)-p(v^*)))\cdot I_{t,d}(\dot{p}(u^*)-\dot{p}(v^*)) \\ &=(p(u)-p(v))\cdot (\dot{p}(u)-\dot{p}(v)), \end{align*} $$

which implies that the image of $\dot {p}$ belongs to $S_{\mathrm {sym}}$ . A similar calculation shows the corresponding property for $M_{\mathrm {ant}}$ .

Let T be the space of trivial infinitesimal motions of $(G,p)$ . A canonical basis of T consists of $\binom {d}{2}$ infinitesimal rotations about the subspaces spanned by each set of $(d-2)$ axes and d translations along each axis. Due to the structure of $I_{t,d}$ , it follows that the $\binom {t+1}{2}$ -dimensional space of isometries in the subspace spanned by the first t axes and the $\binom {d-t}{2}$ -dimensional space of rotations rotating in the subspace spanned by the last $d-t$ axes are contained in $M_{\mathrm {sym}}$ . One can also directly check that the remaining $\binom {d+1}{2} - \binom {t+1}{2}-\binom {d-t}{2}$ elements of the canonical basis belong to $M_{\mathrm {ant}}$ . Hence,

$$ \begin{align*} \dim T\cap M_{\mathrm{sym}}= \binom{t+1}{2}+\binom{d-t}{2} \text{ and } \dim T\cap M_{\mathrm{ant}} = \binom{d+1}{2} - \binom{t+1}{2}-\binom{d-t}{2}. \end{align*} $$

The infinitesimal rigidity of $(G,p)$ implies that $\ker R(G,p)=T$ . Since the rigidity matrix maps $M_{\mathrm {sym}}$ to $S_{\mathrm {sym}}$ and $M_{\mathrm {ant}}$ to $S_{\mathrm {ant}}$ , this gives

$$ \begin{align*} \frac{m}{2}&=\dim S_{\mathrm{sym}}\geq \dim M_{\mathrm{sym}}-\dim T\cap M_{\mathrm{sym}}=\frac{dn}{2}- \left(\binom{t+1}{2}+\binom{d-t}{2}\right) \text{ and}\\ \frac{m}{2}&=\dim S_{\mathrm{ant}}\geq \dim M_{\mathrm{ant}}-\dim T\cap M_{\mathrm{ant}}=\frac{dn}{2}- \left( \binom{d+1}{2} - \binom{t+1}{2}-\binom{d-t}{2}\right), \end{align*} $$

as required.

Proof. Suppose that $\dot {p}$ is an infinitesimal flex of $(G,p)$ . By assumption, there exist skew symmetric matrices $A_i$ of size d and $t_i \in \mathbb R^d$ for $ i =1,2$ such that $\dot {p}(v) = A_ip(v) +t_i$ for $v \in V(G_i)$ . We can choose coordinates so that $p(w) = 0$ for some $w \in V(G_1) \cap V(G_2)$ . It follows that $A_1 0 +t_1 = \dot {p}(w) = A_2 0 + t_2$ . Therefore, $t_1 = t_2$ , and so $A_1p(v) = A_2p(v)$ for all $v \in V(G_1) \cap V(G_2)$ . Therefore, the skew symmetric matrix $A_1 - A_2$ vanishes on a space of dimension $d-1$ , and so must be $0$ , as required.

Proof. Let V, respectively W, be the eigenspace of $\gamma $ corresponding to the eigenvalue $1$ , respectively $-1$ , and let $\pi _V$ , respectively $\pi _W$ , be the orthogonal projection from $\mathbb R^d$ onto V, respectively W. Since P is generic in $\mathbb R^d$ and $\pi _V, \pi _W$ are projections onto coordinate subspaces, it follows that $\pi _V(P)$ is generic in V and $\pi _W(P)$ is generic in W. Therefore, $\dim (\mathrm {aff}(\pi _V(P))) = \min \{n-1,t\}$ and $\dim (\mathrm {lin}(\pi _W(P))) = \min \{n,d-t\}$ .

The facts that $\mathrm {aff}(\pi _V(P)) \subseteq V$ , $\mathrm {lin}(\pi _W(P))\subseteq W$ and $V\cap W=\{{\mathbf {0}}\}$ (since $V,W$ are eigenspaces for distinct eigenvalues of $\gamma $ ) now give

$$ \begin{align*} \dim(\mathrm{aff}(P \cup \gamma(P))) = \dim (\mathrm{aff}(\pi_V(P)) + \mathrm{lin}(\pi_W(P)))&=\dim\mathrm{aff}(\pi_V(P)) + \dim\mathrm{lin}(\pi_W(P))\\ &= \min\{n,d-t\} +\min\{n-1,t\}, \end{align*} $$

as required.

Proof. Let $(G,\ast ,p)$ be a generic $\Gamma $ -framework. Then, for $i=1,2$ , $(H_i, \ast , p|_{V_i})$ is generic and therefore infinitesimally rigid. Now we observe that, by Lemma 2.6, the dimension of $\mathrm {aff}(p(V_1 \cap V_2))$ is $\min \{|V_1\cap V_2| - m +\min \{m/2,d-t\} +\min \{m/2-1,t\},d\}$ . The theorem now follows by applying Theorem 2.4 to the frameworks $(H_i,p|_{V_i})$ , $i=1,2$ .

Proof. Since $|C| \leq d-1$ , we can choose $y \in \mathbb R^d$ so that $y \not \in \mathrm {lin}\{p(w)-p(u):w \in C\}$ . Now, for each $t \in \mathbb R$ , define $q_t:V(G') \to \mathbb R^d$ by

$$ \begin{align*}q_t(v) = \left\{ \begin{array}{ll} p(v) &\text{if }v \in V(G), \\ p(u)+ty &\text{if } v = u'.\\ \end{array}\right. \end{align*} $$

It will suffice to show that for sufficiently small nonzero t, $R(G',q_t)$ is row-independent.

Let L be the matrix obtained by replacing the $uu'$ row of $R(G,q_0)$ by the row vector that has y is the u column-set, $-y$ is the $u'$ column-set and zeroes in all other entries.

Proof of claim.

Suppose that $\lambda $ is an element of the left kernel of L. We can think of $\lambda $ as a real valued function on $E(G')$ . Define $\pi : E(G) \to \mathbb R$ by

$$ \begin{align*}\pi(e) = \left\{ \begin{array}{ll} \lambda(e) &\text{if }e \not\in \{uw:w \in C\}, \\ \lambda(uw)+\lambda(u'w) &\text{if } e = uw, w \in C.\\ \end{array}\right. \end{align*} $$

Now, using the facts that $q_0(u) = q_0(u') = p(u)$ , $q_0(w) = p(w)$ for $w \in V(G)$ and $\lambda L = 0$ , we can readily show that $\pi $ , considered as row vector with entries indexed by $E(G)$ , is an element of the left kernel of $ R(G,p)$ . Since $R(G,p)$ is row-independent, it follows that $\pi =0$ . Therefore, $\lambda (e) = 0$ for all $e \not \in \{uw: w \in C\} \cup \{u'w:w \in C\}$ and $\lambda (uw)+\lambda (u'w) = 0$ for $w \in C$ . Now, using the fact that $\lambda L = 0$ , and considering the uth column-set of L, we have

$$ \begin{align*}\lambda(uu')y + \sum_{w \in C} \lambda(uw) (p(u)-p(w)) = 0.\end{align*} $$

Our choice of y and the fact that $\{p(u)-p(w):w \in C\}$ is linearly independent implies that $\lambda (uw) = 0$ for $w \in C$ and $\lambda (uu') =0$ . So $\lambda = 0$ , and the claim is proved.

Finally, we consider, for $t\neq 0$ , the matrix $M_t$ which is obtained by multiplying the $uu'$ row of $R(G',q_t)$ by $\frac 1t$ . It is clear that $\operatorname {\mathrm {rank}}(M_t) = \operatorname {\mathrm {rank}}(R(G',q_t))$ . It is also clear that $\lim _{t \to 0}(M_t) = L$ . Using the lower semicontinuity of the rank function and Claim 2.10, we see that, for sufficiently small nonzero t, $\operatorname {\mathrm {rank}}(R(G,q_t)) = \operatorname {\mathrm {rank}}(M_t) \geq \operatorname {\mathrm {rank}}(L) = |E(G')|$ . This completes the proof of the theorem.

Proof. Observe that $X_{C \cup \{u\}} = X_C$ since $u \not \in X$ . If $|X_C| = 0$ , then for any $\Gamma $ -generic framework $(G/uv,\ast ,p)$ in $\mathbb R^d$ , $p(C\cup \{u\})$ is a generic set of d points in $\mathbb R^d$ . If $|X_C| = 2$ , then $p(C \cup \{u\})$ is an affinely independent set of d-points in $\mathbb R^d$ . In both cases, it follows that $\{p(u) - p(z): z\in C\}$ is linearly independent, and so we can use Theorem 2.9 to construct a $\Gamma $ -framework $(G,\ast ,q)$ that is infinitesimally rigid in $\mathbb R^d$ .

Proof. Choose $ y \in \mathbb R^d$ such that $y \not \in \mathrm {lin}\{p(u) - p(w): w \in C_1\}\cup \mathrm {lin}\{\tau ^{-1}(p(v) - p(x)): x \in C_2\}$ . For $t \in \mathbb R$ , let $q_t: V(G') \rightarrow \mathbb R^d$ be defined by $q_t(w) = p(w), w \in V(G)$ , $q_t(u') = p(u)+ty$ , $q_t(v') = p(v) +\tau (ty)$ . Let L be the matrix obtained from $R(G',q_0)$ by replacing the $uu'$ row with the row that has y in the u column-set and $-y$ in the $u'$ column-set, and replacing the the $vv'$ row with the row that has $\tau (y)$ in the v column-set and $-\tau (y)$ in the $v'$ column-set.

Now let $M_t$ be the matrix obtained from $R(G,q_t)$ by multiplying the $uu'$ and $vv'$ rows by $\frac 1t$ , and we complete the proof of the theorem in exactly the same way as the proof of Theorem 2.9.

Proof. Let $(G',\ast ,p)$ be a generic $\Gamma $ -framework. Then $(G',p)$ is infinitesimally rigid by assumption. Let $I = \{xv: v\in C\} \cup \{x^*w: w \in D\}$ . Observe that $X_{C \cup \{x\}} = X_C$ and $X_{D \cup \{x^*\}} = X_D$ . Since $(G',\ast ,p)$ is a generic $\Gamma $ -framework and $|X_C|, |X_D| \leq 2$ , it follows that both $\{p(w) - p(x): w \in C\}$ and $\{p(z) - p(x^*): z \in D\}$ are linearly independent. This also implies that the set of rows of $R(G',p)$ labelled by I is linearly independent. Choose a maximal independent row-set that contains the row-set labelled by I and let J be the corresponding set of edges of G. Since $(G',p)$ is infinitesimally rigid, it follows that J spans $V(G')$ and $|J| = d|V(G')| - \binom {d+1}2$ . Let $G'[J]=(V(G'),J)$ . We apply Theorem 2.12 (with $\tau $ being the non-identity element of $\Gamma $ ) to the framework $(G'[J],p)$ to obtain a $\Gamma $ -framework $(G",\ast ,q)$ , where $G"$ is obtained from $G'[J]$ by splitting x from y along C and then splitting $x^*$ from $y^*$ along D. By Theorem 2.12, $(G",q)$ is infinitesimally rigid. Since $G"$ is a spanning subgraph of G, $(G,\ast , q)$ is the required infinitesimally rigid $\Gamma $ -symmetric realisation of $(G,\ast )$ .

Proof. Suppose that $\mathcal {S}$ is not a simplicial k-circuit. Then some simplicial k-circuit $\mathcal {T}$ is properly contained in $\mathcal {S}$ . Since $\mathcal {S}$ is a $\mathbb Z_2$ -irreducible k-cycle, $\mathcal {T}^* \neq \mathcal {T}$ . It follows that $\mathcal {T} \triangle \mathcal {T}^*$ is a nonempty $*$ -invariant simplicial k-cycle contained in $\mathcal {S}$ , so $\mathcal {T} \triangle \mathcal {T}^* = \mathcal {S}$ , whence $\mathcal {S} = \mathcal {T} \sqcup \mathcal {T}^*$ . If $\mathcal {T}$ is a trivial simplicial k-circuit, then $K = V(\mathcal {T}) \cap V(\mathcal {T}^*)$ is a face of $\mathcal {S}$ and $K^* = K$ . Since $*$ is a free involution, it follows that $K = \emptyset $ and so $\mathcal {S}$ is a trivial $\mathbb Z_2$ -irreducible k-cycle, contradicting the hypothesis. Hence, $\mathcal {T}$ is nontrivial.

Now suppose that the second alternative in the lemma holds and $\mathcal {U}$ is a simplicial k-circuit properly contained in $\mathcal {S}$ . Then $\mathcal {S} = \mathcal {U} \sqcup \mathcal {U}^*$ by the same argument as the first paragraph, and $\mathcal {U} \triangle \mathcal {T} = (\mathcal {U}^* \cap \mathcal {T}) \cup (\mathcal {U} \cap \mathcal {T}^*)$ is a $*$ -invariant simplicial k-cycle contained in $\mathcal {S}$ . Therefore, either $\mathcal {U} \triangle \mathcal {T} = \emptyset $ and so $\mathcal {U} = \mathcal {T}$ or, $\mathcal {U}\triangle \mathcal {T} = \mathcal {S}$ and so $\mathcal {U} = \mathcal {S} \setminus \mathcal {T} = \mathcal {T}^*$ .

Proof. We have $|V(\mathcal {S})| \geq 2k+2$ since for any $U \in \mathcal {S}$ , we have $U \cap U^* = \emptyset $ , and hence, $|V(\mathcal {S})| \geq |U \sqcup U^*|=2k+2$ .

Suppose that $|V(\mathcal {S})| = 2k+2$ . Let $V(\mathcal {S})=\{x_i,x_i^*:1\leq i\leq k+1\}$ . If $\mathcal {S}$ contains two copies of some k-face W, then it follows easily that $\mathcal {S} = \{\{ W , W, W^*, W^*\}\}$ , and so $\mathcal {S}$ is a trivial $\mathbb Z_2$ -irreducible k-cycle. Thus, we may assume that each $W\in \mathcal {S}$ has multiplicity one. Then, for each $W \in \mathcal {S}$ and $w \in W$ , we have $W-w+w^* \in \mathcal {S}$ since $W-w$ is contained in at least two k-faces of $\mathcal {S}$ and $*$ is free. It follows that every transversal of $\{ \{x_i,x_i^*\}: 1\leq i\leq k+1\}$ is a k-face of $\mathcal {S}$ , and so, $\mathcal {S} \cong \mathcal {B}_k$ .

Proof. Choose $V_1, V_2 \subset V(\mathcal {S})$ such that $V_1 \cap V_2 = X$ , $G(\mathcal {S}) = G(\mathcal {S})[V_1] \cup G(\mathcal {S})[V_2]$ . Let $\mathcal {T}_i = \{W \in \mathcal {S}: W \subset V_i\}$ . Since X is a vertex separator of $G(\mathcal {S})$ , it follows that $\mathcal {S} = \mathcal {T}_1 \cup \mathcal {T}_2$ and that $\mathcal {T}_1, \mathcal {T}_2 \neq \mathcal {S}$ . The hypothesis that X is $*$ -invariant and the fact that $|X| \leq 2k$ imply that, for any $W\in \mathcal {S}$ , $W\not \subseteq X$ ; hence, $\mathcal {T}_1 \cap \mathcal {T}_2 = \emptyset $ . Thus, $\mathcal {S} = \mathcal {T}_1 \sqcup \mathcal {T}_2$ , $\partial \mathcal {T}_1 = \partial \mathcal {T}_2=:\mathcal {U}$ and $V(\mathcal {U}) \subset X$ .

Suppose $\mathcal {U} = \emptyset $ . Then $\mathcal {T}_i$ is a simplicial k-cycle for $i=1,2$ . Lemma 5.1 now implies that $\mathcal {T}_1$ is a simplicial k-circuit, $\mathcal {T}_2=\mathcal {T}_1^*$ and $\mathcal {S}=\mathcal {T}_1\sqcup \mathcal {T}_1^*$ . The minimality of X now gives $V(\mathcal {T}_1)\cap V(\mathcal {T}_1^*)=X$ . Hence, we may assume that $\mathcal {U} \neq \emptyset $ .

The definition of the boundary operator implies that $\mathcal {U}$ is a simplicial $(k-1)$ -cycle and each of its facets has multiplicity one. Hence, $\mathcal {U} \triangle \mathcal {U}^*$ is a $\mathbb {Z}_2$ -symmetric $(k-1)$ -cycle that does not contain a trivial simplicial $(k-1)$ -circuit. Since $V(\mathcal {U})\subseteq X$ and X is $*$ -invariant, $\mathcal {V}:=\mathcal {U} \triangle \mathcal {U}^*\subseteq 2^X$ . We can now apply Lemma 5.2 and the hypothesis that $|X| \leq 2k$ to $\mathcal {V}$ to deduce that either $\mathcal {V} = \emptyset $ or $\mathcal {V} \cong \mathcal {B}_{k-1}$ .

Suppose $\mathcal {V} \cong \mathcal {B}_{k-1}$ . Then $|X|=2k$ , and $\mathcal {V}$ is the $\mathbb Z_2$ -irreducible $(k-1)$ -cycle consisting of all transversals of the following partition of X: $\{ \{x_1,x_1^*\},\dots ,\{x_k,x_k^*\}\}$ . Hence, every k-subset of X which contains no pair $\{x_i,x_i^*\}$ is contained in $\mathcal {V}$ . Since $*$ is free and $V(\mathcal {U})\subset X$ , this implies that $\mathcal {U} \subset \mathcal {V}$ . Since $\mathcal {U}\subset \mathcal {V}$ and $\mathcal {V}=\mathcal {U}\triangle \mathcal {U}^*$ , $\mathcal {V} = \mathcal {U} \sqcup \mathcal {U}^*$ , and hence, $\mathcal {V}$ is a trivial $\mathbb Z_2$ -irreducible $(k-1)$ -cycle. This contradicts the assumption that $\mathcal {V} \cong \mathcal {B}_{k-1}$ .

Hence, $\mathcal {V}=\mathcal {U} \triangle \mathcal {U}^* = \emptyset $ so $\mathcal {U}=\mathcal {U}^*$ . Then $\mathcal {U}$ is a $\mathbb {Z}_2$ -symmetric $(k-1)$ -cycle with at most $2k$ vertices that does not contain a trivial $\mathbb Z_2$ -irreducible $(k-1)$ -cycle. Lemma 5.2 now implies that $\mathcal {U} \cong \mathcal {B}_{k-1}$ , and hence, $G(\mathcal {S})[X]$ is isomorphic to $G(\mathcal {B}_{k-1})$ .

Proof. We adopt the definitions of $\mathcal {S}^{\prime }_i$ , $\mathcal {S}_i$ and $\mathcal {S}_i^\dagger $ given in our description of a $\mathbb {Z}_2$ -symmetric Fogelsanger decomposition. Note that for all $1\leq i\leq m$ , $\mathcal {S}_i \neq \mathcal {S}$ since $xy \not \in E(\mathcal {S}_i)$ . Suppose $F \in \partial \mathcal {S}_i$ . Then $F \cap \{x,y,x^*,y^*\} \neq \emptyset $ otherwise $F \in \partial S$ , contradicting the fact that $\mathcal {S}$ is a simplicial k-cycle. For $v \in \{x,y,x^*,y^*\}$ , let $\mathcal {U}_v = \{U \in \partial \mathcal {S}_i: v \in U\}$ . Since no edge in $E(\mathcal {S}_i)$ joins two vertices of $\{x,y,x^*,y^*\}$ , we have $\partial \mathcal {S}_i = \mathcal {U}_x \sqcup \mathcal {U}_y \sqcup \mathcal {U}_{x^*} \sqcup \mathcal {U}_{y^*}$ . Furthermore, since $\partial (\mathcal {S}_i/xy)/x^*y^* = \partial \mathcal {S}^{\prime }_i = \emptyset $ , it follows that

(5.2) $$ \begin{align} \text{the mapping } F \mapsto F-y+x \text{ is a bijection } \mathcal{U}_y \rightarrow \mathcal{U}_x, \end{align} $$

and

(5.3) $$ \begin{align} \text{the mapping } F \mapsto F-y^*+x^* \text{ is a bijection } \mathcal{U}_{y^*} \rightarrow \mathcal{U}_{x^*}. \end{align} $$

In particular,

(5.4) $$ \begin{align} \mathcal{S}^\dagger_i = \{F +y: F \in \mathcal{U}_x\}. \end{align} $$

Also, $*$ induces a bijection $\mathcal {U}_x \rightarrow \mathcal {U}_{x^*}$ , and hence,

(5.5) $$ \begin{align} |\mathcal{S}^\dagger_i| = |\mathcal{U}_x|= |\mathcal{U}_y|= |\mathcal{U}_{x^*}|= |\mathcal{U}_{y^*}|. \end{align} $$

Now if $xy \not \in \mathcal {S}_i^+$ , then $\mathcal {S}^\dagger _i =\emptyset $ , and it follows from (5.5) that $\partial \mathcal {S}_i = \emptyset $ , contradicting the fact that $\mathcal {S}$ is a $\mathbb Z_2$ -irreducible k-cycle and $\emptyset \neq \mathcal {S}_i \subsetneq \mathcal {S}$ . So $xy \in E(\mathcal {S}_i^+)$ , and since $\mathcal {S}_i^+$ is $*$ -invariant, $x^*y^* \in E(\mathcal {S}_i^+)$ , proving (1).

The definition of $\mathcal {S}_i^+$ implies that $(\mathcal {S}_i^+/xy)/x^*y^* = \mathcal {S}^{\prime }_i$ . Since $\mathcal {S}_i'$ is a $\mathbb Z_2$ -irreducible k-cycle, this gives (3).

Since $\mathcal {S}_i^+ = \mathcal {S}_i \triangle ( \mathcal {S}_i^\dagger \cup (\mathcal {S}_i^\dagger )^*)$ , Claim 5.5 implies that $\partial \mathcal {S}_i^+ = \emptyset $ , and hence, $\mathcal {S}_i^+$ is a ${\mathbb Z}_2$ -symmetric k-cycle.

Suppose $\mathcal {T}\subseteq \mathcal {S}_i^+$ is a $\mathbb Z_2$ -irreducible k-cycle. If $xy, x^*y^* \not \in E(\mathcal {T})$ , then $\mathcal {T} \subsetneq \mathcal {S}$ , contradicting the fact that $\mathcal {S}$ is a $\mathbb Z_2$ -irreducible k-cycle. So $xy,x^*y^* \in E(\mathcal {T})$ . Now $(\mathcal {T}/xy)/x^*y^*$ is a nonempty ${\mathbb Z}_2$ -symmetric k-cycle contained in the $\mathbb Z_2$ -irreducible k-cycle $\mathcal {S}^{\prime }_i$ . Hence, $(\mathcal {T}/xy)/x^*y^*=\mathcal {S}_i'$ and $\mathcal {S}_i\subseteq \mathcal {T}$ . Since $\partial \mathcal {T} = \emptyset $ , it follows that $\mathcal {T} \supset \mathcal {S}_i^\dagger $ , and so $\mathcal {T} = \mathcal {S}_i^+$ . Hence, $\mathcal {S}_i^+$ is a $\mathbb Z_2$ -irreducible k-cycle, which is nontrivial by (3). Since $\mathcal {S}_i^+\setminus \mathcal {S} \subset \mathcal {S}_i^\dagger \cup (\mathcal {S}_i^\dagger )^*$ , each $K \in \mathcal {S}_i^+\setminus \mathcal {S}$ is a clique of G that either contains $\{x,y\}$ or contains $\{x^*,y^*\}$ . This completes the proof of (4).

If $U\in \mathcal {S}$ contains neither $\{x,y\}$ nor $\{x^*,y^*\}$ , then U is an element of $\mathcal {S}_i$ for a unique i, proving (5).

We next prove (6). Suppose, for a contradiction, that $\mathcal {R}= \mathcal {S} \triangle \left (\triangle _{i=1}^m \mathcal {S}_i^+\right )\neq \emptyset $ and choose $U \in \mathcal {R}$ . By (5), either $\{x,y\} \subset U$ or $\{x^*,y^*\} \subset U$ . Suppose $\{x,y\} \subset U$ (a similar argument works for the other case). Now $\mathcal {R}$ is a simplicial k-cycle, and since $\mathcal {S}$ and $\mathcal {S}_i^+, 1\leq i\leq m$ , are all nontrivial $\mathbb Z_2$ -irreducible k-cycles, $\mathcal {R}$ does not contain any trivial simplicial k-circuit. It follows that there is some $U' \in \mathcal {R}$ such that $U \cap U' = U -x$ . By (5), $U'$ must contain $\{x^*,y^*\}$ , and so $yy^* \in E(\mathcal {R}) \subset E(\mathcal {S})$ , contradicting the assumption that $*$ is a simplicial involution on $\mathcal {S}$ . Therefore, $\mathcal {R}$ must be empty – this proves (6).

To verify (7), we suppose, for a contradiction, that there exists a proper subset I of $\{1,\dots ,m\}$ such that $\mathcal {S}_j^+ \cap \triangle _{i \in I} \mathcal {S}_i^+ = \emptyset $ for all $j \in J = \{1,\dots ,m\} \setminus I$ . Then, by (6), $\mathcal {S} = \triangle _{i=1}^m \mathcal {S}_i^+ = \triangle _{i \in I} \mathcal {S}_i^+ \sqcup \triangle _{j \in J} \mathcal {S}_j^+$ . This yields a partition of $\mathcal {S}$ into two ${\mathbb Z}_2$ -symmetric k-cycles, contradicting the hypothesis that $\mathcal {S}$ is a $\mathbb Z_2$ -irreducible k-cycle. This proves (7).

We can now complete the proof of the lemma by verifying (2). Observe that if $uv \in E(\mathcal {S})$ and $uv\not \in \{xy,x^*y^*\}$ , then since $\mathcal {S}$ is a nontrivial $\mathbb Z_2$ -irreducible k-cycle and $xy^*,x^*y\not \in E(\mathcal {S})$ , there is some $U \in \mathcal {S}$ such that $u,v \in U$ and $\{x,y\}, \{x^*,y^*\} \not \subset U$ . By (5), $uv \in E(\mathcal {S}_i^+)$ for some i. Using this, together with (1), we see that $\bigcup _{i = 1}^m E(\mathcal {S}_i^+) = E$ . Since $G(\mathcal {S})$ has no isolated vertices, this implies that $\bigcup _{i=1}^m V(\mathcal {S}_i^+) = V$ . This proves (2).

Proof. Since $\mathcal {S}_j'=(\mathcal {S}_j/xy)/x^*y^*$ , the hypothesis that $|V(\mathcal {T}') \cap V(\mathcal {T}^{\prime *})| \leq 2k-2$ implies that $G(\mathcal {S}_j^+)$ has a $*$ -invariant vertex separator X with $V(\mathcal {T}') \cap V(\mathcal {T}^{\prime *})\subseteq X\subseteq V(\mathcal {T}') \cap V(\mathcal {T}^{\prime *})+y+y^*$ . In particular, $|X|\leq 2k$ . By Lemma 5.3, either $\mathcal {S}_j^+=\mathcal {T}_j\sqcup \mathcal {T}_j^*$ for some simplicial k-circuit $\mathcal {T}_j$ with $V(\mathcal {T}_j)\cap V(\mathcal {T}_j^*)=X$ or $G(\mathcal {S}_j^+)[X]$ is isomorphic to $G(\mathcal {B}_{k-1})$ . In the former case, we have the statement. So let us assume the latter case. Then we have $|X|=2k$ , and hence, $|X|=|V(\mathcal {T}') \cap V(\mathcal {T}^{\prime *})|+2$ , so $\{x,y,x^*,y^*\}\subseteq X$ . Together with $G[X] \cong G(\mathcal {B}_{k-1})$ , this implies that $xy^* \in E$ , contradicting an hypothesis of the lemma.

Proof. To prove sufficiency, we suppose $h \geq \max \{2(k-t),k+1,2t\}$ . Let $*_T$ be the non-adjacent vertex pairing induced on $V(\mathcal {T})\cap V(\mathcal {T}^*)$ by $*$ . Since $h\leq 2k$ , we can apply Theorem 4.5 to $(G(\mathcal {T}),\ast _T)$ to deduce that $(G(\mathcal {T}),\ast _T)$ is $\Gamma $ -rigid. Symmetrically, $(G(\mathcal {T}^*),\ast _T)$ is $\Gamma $ -rigid. Now Theorem 2.8 (the gluing theorem) implies that the union of $(G(\mathcal {T}),\ast _T)$ and $(G(\mathcal {T}^*),\ast _T)$ is $\Gamma $ -rigid if $h\geq (k+1)+h-1-\min \{h/2,d-t\}-\min \{h/2-1,t\}$ , or equivalently, $\min \{h/2,k+1-t\}+ \min \{h/2-1,t\}\geq k$ . By an elementary calculation, one can check that this is equivalent to $h\geq \max \{2(k-t),k+1,2t\}$ .