The free uniform spanning forest is a factor of IID

The explanation

Draw a square: four corners, four sides. A spanning tree keeps enough sides to connect all four corners, with no loop. Here that means dropping exactly one side, so there are four spanning trees. Pick one of them at random, each with chance 1/4. Each side then survives with probability 3/4.

Now give the sides weights, and make heavy sides more likely to stay. Nudge one weight and every survival chance shifts a little. The paper uses a clean rule for the total shift. For a side kept with probability p, the summed responses equal 2p(1 − p), never more than 1/2Source 1. For the square, p = 3/4 gives 2 × 3/4 × 1/4 = 0.375. With uneven weights 1, e, e^(−0.5) and e², the first side's chance becomes about 0.683. Its total response is then about 0.433, again equal to 2p(1 − p).

On an infinite graph, such as an endless grid, a random spanning tree can be taken as a limit of finite ones. Grow larger and larger finite pieces, pick a uniform spanning tree in each, and look at the limitSource 2. When nothing is glued at the boundary, the result is the free uniform spanning forest. It is a forest, not a tree, because it may split into several pieces.

A factor of IID asks for a recipe. Give every vertex its own independent random number, called a label. Then one fixed rule turns all the labels into a choice of edges. The rule may not pick a starting vertex, and it must treat all vertices alike. Lyons discussed this question for the free forest on graphs of groups in 2013Source 3. For a related object, the wired forest, a recipe was known on transient graphsSource 4. A graph is transient when a random walk on it eventually leaves every region for good. Timár built a recipe for recurrent and invariantly amenable random graphsSource 5. Amenable roughly means the graph can be cut into big pieces with small boundaries. His December 2025 paper still listed the general question as openSource 6.

The paper proves the free uniform spanning forest is a factor of IID on every infinite connected graph with finitely many neighbors per vertexSource 7. One rule works for all such graphs at once, and it uses no rootSource 8. No amenability, no transience and no degree bound is assumedSource 9. The rule is measurable, but it need not decide an edge from a bounded region of labelsSource 10.

The obstacle is to draw one sample from the infinite forest, treating every vertex alike, while controlling all the edges togetherSource 11. The paper adapts an observation method that Nam, Sly and Zhang used for Ising spins on treesSource 12. Imagine watching each edge through slowly fading noise. Remove the part of what you see that you could already predict. What is left is fresh noise, called innovations, and it is independent across edgesSource 13. Labels can make such noise directly. The response rule above shows that predictions change only gently when the noise changes. That lets the paper run the process backwards from any noise inputSource 14. Running it for a long time then reveals which edges belong to the forestSource 15.

The same method gives a second result. Strongly Rayleigh processes are random sets whose generating polynomial never vanishes in a certain region. They include determinantal processes, which are built from a matrix. On every countable group, a strongly Rayleigh process that looks the same everywhere is a factor of IIDSource 16. This needs no amenabilitySource 17.

How sure is this? Lean, a computer proof checker, has formal versions of both results. The scope note says one equivariant rule works for all such graphs and uses no root. It also lists the group results for strongly Rayleigh and determinantal laws. No specialist review of the paper has been published.

What it doesn't show. This does not say how far away the rule must look to decide one edge.