It is an old result that every $6$-connected graph is rigid in $\mathbb{R}^2$:



> 
> Lovász, László, and Yechiam Yemini. "On generic rigidity in the plane." *SIAM Journal on Algebraic Discrete Methods* **3**, no. 1 (1982): 91-98.
> [DOI](https://doi.org/10.1137/0603009).
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It is natural to hope that sufficiently high connectivity implies rigidity in $\mathbb{R}^3$.



> 
> ***Q.*** Is it known that there is some $k$ such that any $k$-connected graph is
> generically rigid in $\mathbb{R}^3$?
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> 


Informally, $G$ is rigid if the distances between vertices connected in $G$
determine all the distances between vertices not connected in $G$.
More formally, $G$ is *generically rigid* in $\mathbb{R}^d$ if every generic representation in $\mathbb{R}^d$
is [infinitesimally rigid](https://en.wikipedia.org/wiki/Structural_rigidity)—no
infinitesimal length-preserving velocities (if they exist) can be extended.
An embedded graph representation (a *framework*) is *generic* if the coordinates of its configuration do not satisfy
any non-trivial algebraic equation with rational coefficients.


There have been recent advances in 3D rigidity and I am unclear on the current status
of this question ***Q***.



Is every 1-million-connected graph rigid in 3D?
I think this is still an open problem, but recent work of [Clinch, Jackson, and Tanigawa](https://arxiv.org/abs/1911.00207) (almost) shows every $12$-connected graph is generically rigid in $\mathbb{R}^3$.


In that paper, they prove that $12$-connectivity is sufficient to force rigidity in the $C\_2^1$-cofactor matroid (see the paper for precise definitions). In an [earlier paper](https://arxiv.org/abs/1911.00205), the same authors showed that the $C\_2^1$-cofactor matroid is the unique maximal abstract $3$-rigidity matroid. A long-standing conjecture in rigidity theory is that the unique maximal abstract $3$-rigidity matroid is in fact the generic $3$-dimensional rigidity matroid. If you believe this conjecture, then the answer to your quesiton is yes, with one million replaced with $12$.


**Acknowledgement.** This answer is entirely due to Katie Clinch.


