# Is every $f(k)$-vertex-connected graph the edge-disjoint union of two $k$-vertex-connected graphs?

1. Does there exist a function $$f: \mathbf{N} \to \mathbf{N}$$ such that every $$f(k)$$-vertex-connected graph $$G$$ can have its edges partitioned into two spanning subgraphs $$G_1$$, $$G_2$$ such that both of them are $$k$$-vertex-connected?
2. If so, can $$f(k)$$ be chosen linear in $$k$$?

For the record, I know the answer to both questions is "yes" when we ask for edge-connectivity instead. We may take $$f(k) = 4k$$. Every $$4k$$-edge-connected graph has $$2k$$ disjoint spanning trees (see, e.g. here). We define edge-disjoint subgraphs $$G_1$$, $$G_2$$ of $$G$$ such that each $$G_i$$ contains at least $$k$$ of the spanning trees; and we assign every remaining edge arbitrarily either to $$G_1$$ or $$G_2$$. Since each $$G_i$$ has at least $$k$$ disjoint spanning trees, it must be $$k$$-edge-connected.

Update: An open conjecture due to Kriesell states that there exists a function $$f(k)$$ such that every $$f(k)$$-vertex-connected graph has a spanning tree whose removal gives a $$k$$-vertex-connected graph. This is only known to be true for $$k \leq 2$$. That is, in general it is not even known how to decompose highly connected graphs into a $$1$$-connected and a $$k$$-connected spanning graphs; let alone two $$k$$-connected spanning subgraphs. So my question, if true, is an open problem. But it is still possible that my question might be answered in the negative with a simple counterexample, which would also be interesting.