Proof. Our goal is to construct a maximal ${\mathcal {R}_d^t}$ -independent subgraph $G'$ of G with minimum degree $td$ . As we will see, the existence of such a subgraph quickly implies the lemma.

Fix an orientation $\vec {G}$ of G such that $\delta _{\vec {G}}(v)\geq \lfloor \deg _G(v)/2\rfloor $ for each $v\in V$ . It is a well-known result that such an orientation exists (see, for example, [Reference Frank4, Theorem 1.3.8]). Then $\delta _{\vec {G}}(v) \geq t\cdot 5d(d+1)$ for each $v\in V$ . For a subset F of E, let $\vec {F}$ denote the corresponding set of oriented edges. Recall that $N^+_{\vec {F}}(v)$ denotes the set of vertices $u\in V$ for which $vu\in \vec {F}$ .

Let U be a random subset of V such that each $v\in V$ is in U independently with probability $1/2$ . For each $j \in \{1,\dots , t\}$ , we recursively define a random subgraph $H_j=(U,F_j)$ of $G[U]$ as follows. Suppose that $H_1,\dots , H_{j-1}$ are already given. Let

$$\begin{align*}E_j= E(G[U]) - \left(F_1 \cup \ldots \cup F_{j-1}\right).\end{align*}$$

Consider a uniformly random ordering $\pi _j$ of the vertices in U. For each $v\in U$ , let $A_j(v)=\big \{u\in N^+_{\vec {E}_j}(v): u \text { precedes } v \text { in } \pi _j\big \}$ , and fix a subset $B_j(v) \subseteq A_j(v)$ of size $\min (d,|A_j(v)|)$ . Finally, let

$$\begin{align*}F_j=\bigcup_{v\in U}\{vu:{u\in B_j(v)}\}.\end{align*}$$

Let $F=\bigcup _{j=1}^t F_j$ and $H=(U,F)$ . For each $v\in V-U$ , add v to H and connect it with $\min (td,|N^+_{\vec { G}}(v)\cap U|)$ vertices of $N^+_{\vec { G}}(v)\cap U$ . Let the resulting graph be denoted by $G_0=(V,E_0)$ .

Proof. For convenience, we define $n=|V|$ and $k=5d(d+1)$ . Let $\eta \leq \frac {1}2$ be a parameter to be chosen later, and let $r=(1-\eta )\frac {k}{2}$ .

Let us fix a vertex $v \in V$ . Using the Chernoff bound (Theorem 2.3), we obtain that

(2) $$ \begin{align} {\mathbb{P}}\Big(|N^+_{\vec{ G}}(v)\cap U|\leq tr \Big) = {\mathbb{P}}\left(|N^+_{\vec{ G}}(v)\cap U|\leq (1-\eta) \frac{tk}{2} \right) \;\leq \; e^{-\eta ^2tk/4 }. \end{align} $$

Let Q denote the event that $v\in U$ and $|N^+_{\vec { G}}(v)\cap U|>tr$ . If Q holds, then $|N^+_{\vec { E}_j}(v)|\geq tr-td> d$ for every $j\in \{1,\dots ,t\}$ . Note that $\delta _{\vec {F}_j}(v)=|B_j(v)|=\min (d,|A_j(v)|)$ . Thus, it follows from Lemma 2.4 that

$$\begin{align*}{\mathbb{E}}\Big(\delta_{\vec{ F}_j}(v) \;\Big|\; Q \Big) \geq d-\frac{1}{2}\cdot \frac{d(d+1)}{tr-td+1}\geq d-\frac{d(d+1)}{2t(r-d)},\end{align*}$$

and hence,

$$\begin{align*}{\mathbb{E}}\Big( \delta_{\vec{ F}}(v) \;\Big|\; Q \Big) = {\mathbb{E}}\Big( \sum_{j=1}^t\delta_{\vec{ F_j}}(v) \;\Big|\; Q \Big) = \sum_{j=1}^t {\mathbb{E}}\Big( \delta_{\vec{ F_j}}(v) \;\Big|\; Q \Big) \geq td-\frac{d(d+1)}{2(r-d)}.\end{align*}$$

Equation (2) and the fact that the two sub-events in the definition of Q are independent together imply ${\mathbb {P}}(Q) \geq \frac {1}{2} (1-e^{-\eta ^2tk/4})$ . Hence,

$$\begin{align*}{\mathbb{E}}\big(\delta_{\vec{ F}}(v)\big)\geq {\mathbb{E}}\big(\delta_{\vec{F}}(v) \big| Q) \cdot {\mathbb{P}}(Q) \geq \frac{1}{2} \left(1-e^{-\eta ^2tk/4}\right)\left(td-\frac{d(d+1)}{2(r-d)}\right).\end{align*}$$

After summing over the vertices, we get

$$ \begin{align*}{\mathbb{E}}(|F|) = {\mathbb{E}}\bigg(\sum_{v \in V}\delta_{\vec{F}}(v) \bigg) = \sum_{v \in V}{\mathbb{E}}\big(\delta_{\vec{F}}(v) \big)\geq \frac{1}{2}\left(1-e^{-\eta ^2tk/4}\right)\left( td-\frac{d(d+1)}{2(r-d)}\right)n.\end{align*} $$

Let D denote $E_0-F$ . Then

$$ \begin{align*}{\mathbb{E}}\bigg(\delta_{\vec{D}}(v)\;\bigg|\; v\notin U \;\land\; |N_{\vec{ G}}(u)\cap U|>tr \bigg)= td.\end{align*} $$

Thus, by (2), we have

$$\begin{align*}{\mathbb{E}}\big(\delta_{\vec{D}}(v)\big)\geq \frac{1}{2}\left(1-e^{-\eta ^2tk/4}\right)td,\end{align*}$$

and by summing over the vertices, we obtain

$$\begin{align*}{\mathbb{E}}(|D|)\geq \frac{1}{2}\left(1-e^{-\eta ^2tk/4}\right) tdn.\end{align*}$$

It follows that

$$ \begin{align*} {\mathbb{E}}(|E_0|)={\mathbb{E}}(|F|)+{\mathbb{E}}(|D|) &\geq \left(1-e^{-\eta ^2tk/4}\right)\left( td-\frac{d(d+1)}{4(r-d)}\right)n\\ &\geq tdn-\left( e^{-\eta ^2tk/4}\cdot td + \frac{d(d+1)}{4(r-d)} \right)n\\ &\geq tdn-\left( e^{\frac{-\eta ^2tk}{4}+\frac{td}{3}} + \frac{d(d+1)}{4(r-d)} \right)n, \end{align*} $$

where the last inequality follows from the fact that $td\geq 6$ , and thus, $td<e^{td/3}$ .

Let us now fix the value of $\eta $ to be $0.45$ . With this choice, it is easy to verify that

$$\begin{align*}\frac{\eta ^2tk}{4} - \frac{td}{3} = \left(\frac{\eta^2}{4}-\frac{1}{5(d+1)\cdot 3}\right)t\cdot 5d(d+1)\geq \left(\frac{0.45^2}{4}-\frac{1}{60}\right) 120 = 4.075,\end{align*}$$

where for the inequality we used, again, the assumption that $d \geq 3$ and $t \geq 2$ . Note that $d \geq 3$ also implies $d \leq \frac {1}{4}d(d+1)$ , and hence

$$\begin{align*}r-d=0.55\cdot \frac{1}{2}\cdot 5\cdot d(d+1)-d \geq 1.375\, d(d+1) -0.25\, d(d+1)= 1.125\, d(d+1).\end{align*}$$

Thus,

$$ \begin{align*}{\mathbb{E}}(|E_0|)\geq tdn-\left(e^{-4.075}+\frac{1}{4\cdot 1.125}\right)n\geq \left(td-\frac{1}{4}\right)n,\end{align*} $$

which completes the proof of the claim.

It follows from Claim 3.3 that

$$ \begin{align*}{\mathbb{E}}\left(|E_0|+\frac{|V-U|}{2}\right) = {\mathbb{E}}(|E_0|)+\frac{1}{4}n\geq tdn.\end{align*} $$

Hence, there exist a set of vertices $U\subseteq V$ and a corresponding spanning subgraph $G_0=(V,E_0)$ such that

$$ \begin{align*}|E_0|+\frac{|V-U|}{2}\geq tdn.\end{align*} $$

Since $E_0$ is $\mathcal {R}_d^t$ -independent by Claim 3.2, we can extend it to a maximal $\mathcal {R}_d^t$ -independent subgraph $G' = (V,E_0 \cup E_1)$ of G by adding a suitable set of edges $E_1 \subseteq E-E_0$ . Then $|E_0 \cup E_1| = r_d^t(G)$ , and thus,

$$\begin{align*}|E_1|= r_d^t(G) - |E_0| \leq tdn-t\binom{d+1}{2}-|E_0|\leq \frac{|V-U|}{2}-t\binom{d+1}{2}<\frac{|V-U|}{2},\end{align*}$$

where in the first inequality, we used the fact that $r_d^t(K_n) = tdn - t\binom {d+1}{2}$ , which follows from Lemma 2.1.

Hence, there is some vertex $v_0\in V-U$ that is not incident to any edge in $E_1$ . It follows from the construction of $G_0$ that $d_{G'}(v_0) \leq td$ , and thus,

$$\begin{align*}r_d^t(G-v_0) \geq r_d^t(G'-v_0) \geq r_d^t(G) - td.\end{align*}$$

However, since $\deg _G(v_0) \geq td$ , and the addition of a vertex of degree $td$ preserves ${\mathcal {R}_d^t}$ -independence, we must have $r_d^t(G) \geq r_d^t(G-v_0) + td$ . It follows that $r_d^t(G-v_0) = r_d^t(G) - td$ .

Proof of Theorem 1.6.

As we noted before, if $d\geq 2$ or $t=1$ , then the statement follows from Theorems 1.4 and 1.7. Hence, it suffices to prove in the case when $d\geq 3, t\geq 2$ . We prove the statement by induction on the number of vertices. Let $c = t\cdot 10d(d+1)$ . If $|V|=c+1$ , then G is complete and thus $\mathcal {R}_d^t$ -rigid. It follows from Lemma 2.1 that G contains t edge-disjoint d-rigid spanning subgraphs.

Suppose now that $|V|>c+1$ . By Lemma 3.1, there is some vertex $v_0\in V$ such that $r_d^t(G-v_0) = r_d^t(G)-td$ . Let us consider the graph $G' = G-v_0+K(N_G(v_0))$ . On the one hand, by Lemma 2.2 and the choice of $v_0$ , every pair of vertices $u,v \in N_G(v_0)$ is ${\mathcal {R}_d^t}$ -linked in $G-v_0$ , and hence, $G-v_0$ is $\mathcal {R}_d^t$ -rigid if and only if $G'$ is. On the other hand, $G'$ is c-connected: indeed, it arises from the c-connected graph $G+K(N_G(v_0))$ by deleting a vertex whose neighbor set is a clique, and it is easy to see that deleting such a vertex preserves c-connectivity (except for the complete graph on $c+1$ vertices). Hence, by induction, $G'$ is $\mathcal {R}_d^t$ -rigid, and thus so is $G-v_0$ .

Since $|V| \geq c+1 \geq 2td+1$ , Lemma 2.1 implies that $G-v_0$ contains t edge-disjoint d-rigid spanning subgraphs. Adding a vertex of degree at least $td$ to $G-v_0$ corresponds to adding a vertex of degree at least d to each of these subgraphs, an operation that preserves d-rigidity. Since $\deg _G(v_0) \geq c \geq td$ , we conclude that G contains t edge-disjoint d-rigid spanning subgraphs, as claimed.

Proof. We have to verify that the conditions of Theorem 4.1 are satisfied. Let $g=g_{d,R}$ . As $|V|\geq \binom {d+1}{2}\geq d$ , we have $|E|=d|V|-\binom {d+1}{2}=g(V)$ . Now consider a set $X\subseteq V$ with $|X|\geq d$ . By (1), we have

$$ \begin{align*}i_G(X)\leq d|X|-\binom{d+1}{2} \leq d|X|- |R\cap X| = g(X),\end{align*} $$

as required. Next, consider a set $X\subseteq V$ with $|X|\leq d-1$ . Then, since G is simple, we have $i_G(X)\leq \binom {|X|}{2}= \frac {|X|(|X|-1)}{2} \leq |X|(d-1)\leq g(X)$ , which completes the proof.

Proof. Let W denote the set of in-neighbors of X in ${\vec {G}}$ and let $R_X=X\cap R$ . First, assume that $|X|\geq d$ . We have $\rho _{\vec {G}}(v)=d$ for all $v\in X-R_X,$ and $\rho _{\vec {G}}(v)=d-1$ for all $v\in R_X$ . Thus,

$$\begin{align*}d|X|-|R_X| = \sum_{v\in X}\rho_{\vec{G}}(v) \leq i_G(X\cup W)\leq d|X\cup W|-\binom{d+1}{2},\end{align*}$$

where the last inequality follows from (1). Hence, $d|W|\geq \binom {d+1}{2}-|R_X|\geq \frac {\binom {d+1}{2}}{2}$ , and thus, $|W|\geq \frac {d+1}{4}> k-1$ .

Next, assume that $|X|\leq d-1$ . Note that we have

$$\begin{align*}\rho_{\vec{G}}(X) = \Big(\sum_{v\in X} \rho_{\vec{G}}(v)\Big) - i_G(X) \geq d|X|-|R_X|- \binom{|X|}{2} = |X| \left(d - \frac{|X|-1}{2}\right)-|R_X|.\end{align*}$$

Since each in-neighbor of X can send at most $|X|$ edges to X in ${\vec {G}}$ , we also have $|X| |W|\geq \rho _{\vec {G}}(X)$ . It follows that

$$\begin{align*}|W|\geq d - \frac{|X|-1}{2} -\frac{|R_X|}{|X|} \geq d - \frac{d-2}{2} -1 = \frac{d}{2} \geq 2k - 2\geq k,\end{align*}$$

as desired.

Proof of Theorem 1.2.

Since the $k=1$ case is settled by the theorem of Robbins, we may assume that $k\geq 2$ . Let $d=4k-4$ . We shall prove that if a graph $G = (V,E)$ has two edge-disjoint minimally d-rigid spanning subgraphs, then G has a k-connected orientation. Note that Theorem 1.6 guarantees the existence of such subgraphs if G is $(320 k^2-560k+240)$ -connected.

Suppose that G has two edge-disjoint minimally d-rigid spanning subgraphs $G_1$ and $G_2$ . Let us fix a set $R \subseteq V$ of vertices with $|R|=\binom {d+1}{2}$ . We construct the orientation ${\vec {G}}$ of G by defining the orientations of $G_1$ and $G_2$ , and then orienting the remaining edges arbitrarily. The orientation of $G_1$ is chosen to be a $(d,R)$ -orientation. The orientation of $G_2$ is chosen to be a reversed $(d,R)$ -orientation (in other words, its out-degrees respect the $(d,R)$ -specification). By Lemma 4.2, these orientations exist.

It remains to show that the union ${\vec {G}}$ of these oriented spanning subgraphs is k-connected. Suppose not; then there is a subset $S\subseteq V$ with $|S|\leq k-1$ for which ${\vec {G}}-S$ is not strongly connected. This means that there is a set $X\subseteq V$ with $V-X-S\not = \varnothing $ and whose in-neighbors are all in S (and hence, the out-neighbors of the set $V-X-S$ are all in S). If $|X\cap R|\leq \frac {\binom {d+1}{2}}{2}$ , then Lemma 4.3, applied to the $(d,R)$ -orientation of $G_1$ , gives a contradiction. Otherwise, $|(V-X-S)\cap R|\leq \frac {\binom {d+1}{2}}{2}$ , and we can apply the lemma to the orientation of $G_2$ to obtain a contradiction.

Proof. We essentially repeat the proof of [Reference Villányi21, Lemma 3.2], which is the special case when $\ell = D$ and $k_0 = D(D+1)$ . Fix $v \in V$ , and let k denote $\deg _G(v)$ . Lemma 2.4 implies that

(3) $$ \begin{align} {\mathbb{E}}(\min({\deg_{G,\overleftarrow{\pi}}}(v),D)) = D - \frac{1}{2}\cdot \frac{D(D+1)}{k+1}. \end{align} $$

Let $H_1,\ldots ,H_r$ denote the vertex sets of the maximal cliques of $G[N_G(v)]$ . For each $i \in \{D+1,\ldots ,k\}$ , let

$$ \begin{align*}\mathcal{S}_i &= \left\{ S \in \binom{N_G(v)}{i}: S \text{ induces a clique in } G \right\} \\ &= \left\{ S \in \binom{N_G(v)}{i}: \exists j \in \{1,\ldots,r\}, S \subseteq H_j \right\}.\end{align*} $$

Then $|\mathcal {S}_k| = 0$ and, by Lemma 5.1, $|\mathcal {S}_i| \leq \binom {k-1}{i}$ for each $i \in \{\ell + 1, \ldots , k-1\}$ .

Let Q denote the event that ${\deg _{G,\overleftarrow {\pi }}}(v) \geq D+1$ and ${N_{G,\overleftarrow {\pi }}}(v)$ does not induce a clique in G. If ${\deg _{G,\overleftarrow {\pi }}}(v) = i \geq D+1$ , then Q occurs if and only if ${N_{G,\overleftarrow {\pi }}}(v) \notin \mathcal {S}_i$ . Hence, for $i \geq \ell + 1$ , we have

$$\begin{align*}{\mathbb{P}}(Q|{\deg_{G,\overleftarrow{\pi}}}(v) = i) = 1 - \frac{|\mathcal{S}_i|}{\binom{k}{i}} \geq 1 - \frac{\binom{k-1}{i}}{\binom{k}{i}} = 1 - \frac{k-i}{k} = \frac{i}{k}.\end{align*}$$

It follows that

(4) $$ \begin{align} {\mathbb{P}}(Q) &\geq \sum_{i = \ell + 1}^k {\mathbb{P}}({\deg_{G,\overleftarrow{\pi}}}(v)=i){\mathbb{P}}(Q|{\deg_{G,\overleftarrow{\pi}}}(v)=i) \nonumber \\ &\geq \sum_{i = \ell + 1}^k \frac{1}{k+1}\frac{i}{k}\\ &= \frac{1}{k(k+1)} \cdot \left( \binom{k+1}{2} - \binom{\ell+1}{2}\right) \nonumber\\ &= \frac{1}{2} - \frac{1}{2} \cdot \frac{\ell(\ell+1)}{k(k+1)}.\nonumber \end{align} $$

If Q does not occur, then ${\deg _{G_\pi ^D,\overleftarrow {\pi }}}(v) = \min ({\deg _{G,\overleftarrow {\pi }}}(v),D).$ If Q occurs, then ${\deg _{G_\pi ^D,\overleftarrow {\pi }}}(v) = D+1 = \min ({\deg _{G,\overleftarrow {\pi }}}(v),D) + 1.$ Hence, by combining (3) and (4), we obtain

(5) $$ \begin{align} \begin{split} {\mathbb{E}}({\deg_{G_\pi^D,\overleftarrow{\pi}}}(v)) &= {\mathbb{E}}(\min({\deg_{G,\overleftarrow{\pi}}}(v),D)) + {\mathbb{P}}(Q) \\ & \geq D + \frac{1}{2} - \frac{1}{2} \cdot \left(\frac{D(D+1)}{k+1} + \frac{\ell(\ell+1)}{k(k+1)}\right) \geq D, \end{split} \end{align} $$

where the last inequality follows from the assumption that

$$\begin{align*}k_0^2 + k_0(1-D(D+1)) - \ell(\ell+1) \geq 0.\end{align*}$$

Thus,

$$\begin{align*}{\mathbb{E}}(|E^D_\pi|) = {\mathbb{E}}\left(\sum_{v \in V}{\deg_{G_\pi^D,\overleftarrow{\pi}}}(v)\right) = \sum_{v \in V}{\mathbb{E}}({\deg_{G_\pi^D,\overleftarrow{\pi}}}(v)) \geq D|V|.\end{align*}$$

Proof. We prove that $F_i = G^{d+1}_\pi [\{v_1,\ldots ,v_i\}]$ is ${\mathcal {M}_d}$ -independent by induction on i. $F_1$ is a single vertex, which is ${\mathcal {M}_d}$ -independent. Let us thus suppose that $2 \leq i \leq n$ . If $F_i$ is constructed from $F_{i-1}$ according to rule (a) or (b), then it is obtained from $F_{i-1}$ by the addition of a vertex of degree at most $d+1$ , and hence, $F_i$ is ${\mathcal {M}_d}$ -independent. Suppose that $F_i$ is constructed from $F_{i-1}$ according to rule (c), and let $x,y$ be vertices as described in the rule. As $x,y$ are nonadjacent in G, they are not ${\mathcal {M}_d}$ -linked in G, and hence neither in $F_{i-1}$ . This means that $F_{i-1}+xy$ is ${\mathcal {M}_d}$ -independent. Since $F_i$ is obtained from $F_{i-1}+xy$ by a $(d+1)$ -dimensional edge split, it follows that $F_i$ is also ${\mathcal {M}_d}$ -independent, as claimed.

Proof. We show that $K_n$ contains edge-disjoint spanning subgraphs $G_0$ and T such that $G_0$ is d-rigid and T is a tree. We first assume $n = d+{a}+2$ . Let us label the vertices of $K_n$ as $\{u_1,\ldots ,u_{{a}+1},v_1,\ldots ,v_{d+1}\}$ , and let us define the integers $t_0 = 0$ , $t_i = \binom {i+1}{2}$ for $i \in \{1,\ldots ,{a}-1\}$ , and $t_{{a}} = d$ .

We construct a spanning subgraph T of $K_n$ by adding an edge between $u_i$ and $v_j$ for each $i \in \{1,\ldots ,{a}\}$ and $j \in \{t_{i-1} + 1,\ldots ,t_i\}$ , and then adding the edges $v_{d+1}u_{{a}+1}$ and $u_iu_{{a}+1}$ for each $i \in \{1,\ldots ,{a}\}$ . See Figure 1. It is easy to verify that T is a spanning tree of $K_n$ . Note that for $i \in \{1,\ldots ,{a}-1\}$ , $u_i$ has exactly i neighbors in T among $v_1,\ldots ,v_d$ , while $u_{a}$ has at most ${a}$ such neighbors.

Figure 1

The construction of T in Lemma 5.4 in the case when ${a} = 3$ and $d=6$ . The complement of T can be obtained from a complete graph $K_{d+1}$ on $d+1$ vertices by successively adding vertices of degree d.

Let $G_0$ be the complement of T in $K_n$ . Then $G_0$ contains the complete graph on $\{v_{1},\ldots ,v_{d+1}\}$ . Moreover, $u_1$ is adjacent with $\{v_2,\ldots ,v_{d+1}\}$ , and similarly, for each $i \in \{2,\ldots ,{a}+1\}$ , the vertex $u_i$ has at least d neighbors among $\{v_1,\ldots ,v_{d+1},u_1,\ldots ,u_{i-1}\}$ . It follows that $G_0$ can be constructed from a complete graph by the addition of vertices of degree d, and hence, it is d-rigid, as required.

The $n> d+{a}+2$ case follows from the observation that for such n, $K_n$ can be constructed from $K_{d+{a}+2}$ by the addition of vertices of degree at least $d+1$ .

Proof. Let $G = G_1 \cup G_2$ . By Lemma 5.4, $G_1 \cap G_2$ contains edge-disjoint copies of a spanning tree and d-rigid spanning subgraph. We can use these subgraphs and the fact that each vertex of $V(G) - (V(G_1) \cap V(G_2))$ has at least $d+1$ neighbors in $V(G_1) \cap V(G_2)$ to construct edge-disjoint copies of a spanning tree and a d-rigid spanning subgraph in G.

Proof of Theorem 1.8.

Since in the $d \in \{1,2\}$ case we have stronger bounds from the theorem of Nash-Williams [Reference St and Nash-Williams13] and Tutte [Reference Tutte20], and Theorem 1.4, respectively, we may suppose that $d \geq 3$ . (The proof also works for $d \in \{1,2\}$ , but the bound we obtain is slightly weaker.) Let $c = d^2+3d+5$ .

Suppose, for a contradiction, that $G = (V,E)$ is a c-connected graph that is not ${\mathcal {M}_d}$ -rigid. We may assume that G has the least possible number of vertices among all such graphs. We may also assume that G has the largest number of edges among all such graphs on $|V|$ vertices. Then, for each $v \in V, N_G(v)$ does not induce a clique in G, for otherwise deleting v would result in a smaller counterexample. (As we noted before, deleting a vertex whose neighbor set is a clique preserves k-connectivity unless the graph is a complete graph on $k+1$ vertices.) Furthermore, every ${\mathcal {M}_d}$ -linked pair is adjacent in G, for otherwise connecting a nonadjacent ${\mathcal {M}_d}$ -linked pair by an edge would result in a counterexample with more edges. In particular, the ${\mathcal {M}_d}$ -rigid induced subgraphs of G are complete.

Let $\ell = d + {a} + 2$ , where

$$\begin{align*}{a} = \left\lceil \sqrt{2d + \frac{1}{4}} - \frac{1}{2}\right\rceil.\end{align*}$$

A short calculation shows that $\binom {{a}+1}{2} \geq d$ . Consider a vertex $v \in V$ , and let $H_1,H_2$ be the vertex sets of two different maximal cliques of $G[N_G(v)]$ . Then $G[H_1 \cup H_2 \cup \{v\}]$ is non-complete and hence not ${\mathcal {M}_d}$ -rigid. It follows from Lemma 5.5 that

$$\begin{align*}|(H_1 \cup \{v\}) \cap (H_2 \cup \{v\})| \leq \ell - 1,\end{align*}$$

and thus, $|H_1 \cap H_2| \leq \ell -2$ .

Hence, we may apply Lemma 5.2 with $D = d+1, \ell = d + {a} + 2$ , and $k_0 = d^2 + 3d + 5$ . (It is a straightforward, although tedious, calculation to check that these numbers satisfy the condition in the statement of Lemma 5.2, under the condition that $d \geq 3$ .) It follows that if $\pi $ is a uniformly random ordering of V, then ${\mathbb {E}}(|E^{d+1}_\pi |) \geq (d+1)|V|$ . This implies that there exists some ordering $\pi _0$ of V for which $|E_{d+1}^{\pi _0}| \geq (d+1)|V|$ . Moreover, by Lemma 5.3, $G_{\pi _0}^{d+1}$ is ${\mathcal {M}_d}$ -independent. But this is impossible, since an ${\mathcal {M}_d}$ -independent graph on at least d vertices can have at most $(d+1)|V|-\binom {d+1}{2} - 1$ edges.

Hence, every c-connected graph is ${\mathcal {M}_d}$ -rigid. Combining this with Lemma 5.4 (using the observation that $c + 1\geq d+{a}+2$ ), we deduce that every c-connected graph contains edge-disjoint copies of a spanning tree and a d-rigid spanning subgraph, as required.

Proof. The proof follows the construction of Lovász and Yemini [Reference Lovász and Yemini11] for $5$ -connected graphs that are not $2$ -rigid. Let $G = (V_0,E_0)$ be a K-regular K-connected graph on $2s$ vertices, where s is a large integer to be determined later. Let $G = (V,E)$ be the graph obtained from $G'$ by splitting every vertex into K vertices of degree one, and then adding a complete graph $G_v$ on the K vertices corresponding to v, for each $v \in V_0$ . It is not difficult to verify that G is K-connected.

Let $\mathcal {M} = (E,r)$ denote the union of $\mathcal {R}_{d_i}(G)$ for $i \in \{1,\ldots ,k\}$ . Since we have $E(G) = E_0 \cup \bigcup _{v \in V_0} E(G_v)$ , we obtain

$$ \begin{align*} r(E) \leq |E_0| + \sum_{v \in V}r(E(G_v)) &\leq {sK} + 2s \sum_{i = 1}^k \left( d_iK - \binom{d_i+1}{2} \right) \\ &= 2sK\left( \Big(\sum_{i=1}^k d_i \Big) + \frac{1}{2} - \frac{1}{2} \cdot \frac{K+1}{K} \right) \\ &= |V|\left( \Big(\sum_{i=1}^k d_i \Big) + \frac{1}{2} - \frac{1}{2} \cdot \frac{K+1}{K} \right). \end{align*} $$

If s is sufficiently large, then the right-hand side is less than $\sum _{i=1}^k \left (d_i|V| - \binom {d_i+1}{2}\right )$ , and hence, G cannot contain edge-disjoint $d_i$ -rigid spanning subgraphs for $i \in \{1,\ldots ,k\}$ .