In this excellent lecture ("2d Percolation Revisited") Stanislav Smirnov mentioned the connection of the theory of percolation with the notion of the so called black noise—see at 29:42 (the notion introduced by Boris Tsirelson). I would like to understand various claims which were made in this lecture, in particular:

Black noise is a noise where there is no spectrum at all […] it is a random field, so it is Fock space and all of that, so it an object from quantum mechanics but which you cannot detect by any linear functional—so harmonic oscilator does not see this.

And also the following:

Now suppose that you look at percolation as a model where you have some information at every side and you look at connection probabilities—so that's your sigma algebra of events is connection probabilities. So the question is: can you pass to the limit? On a lattice you can do it but is there an object on a plane where at every point you have a bit of information and in the end you end up storing only connection probabilities?

As I understood correctly the answer is YES and the relevant object is the scaling limit of planar percolation which is a black noise.

So to summarize: my question is rather vague so what kind of answer I am expecting? Any comments and remarks clarifying the above two quotes, maybe an explanation "how to think about black noise correctly" and justification of the use of the term "noise" in this context (for example: white noise (for example on the line) is some sort of random function which is a derivative of a random continuous function (Brownian motion): in fact this derivative should be understood as a distribution due to the lack of differentiability of Brownian motion—nevertheless it is a (generalized) stochastic process. However for the black noise, as I understood correctly, it is defined only as a family of sigma fields not a family of random variables which is somehow weaker—is this lack of random variables related to the fact that "there is no spectrum at all"?).


1 Answer 1


Percolation 'noise' is generated by a perfectly good family of random variables, the 'quad-crossing' events. Basically, for every diffeomorphic image of the unit square, this random variable is $1$ if one can cross from the left to the right face traversing only open bonds and $0$ otherwise. Let's call this random variable $X_\phi$ where $\phi$ is the diffeomorphism in question. Now, for any open set $U$, we have a $\sigma$-algebra $F_U$ which is generated by all the $X_\phi$ such that $\phi([0,1]^2) \subset U$. This is a 'noise' in the sense that $F_U$ and $F_V$ are independent if $U \cap V = \emptyset$.

In this context, a 'linear' random variable is a random variable $Y$ such that, for every smooth enough closed curve $\Gamma$, writing $\Gamma_i$ and $\Gamma_e$ for its interior and exterior respectively, one has $$Y = \mathbf{E}(Y\,|\, F_{\Gamma_i}) + \mathbf{E}(Y\,|\, F_{\Gamma_e})\;.$$ The claim is that $0$ is the only linear random variable, which is the definition of 'black noise'.

This is in stark contrast to 'white noise', for which the space of linear random variables is sufficiently rich to generate the full $\sigma$-algebra.

  • $\begingroup$ that's interesting, but to follow up on the question of the OP, how is it related to quantum mechanics? $\endgroup$
    – Sascha
    Commented Dec 13, 2019 at 11:43
  • $\begingroup$ Not sure what Stas meant with that remark. The noise being black precisely means that the probability space doesn’t admit a random field that’s compatible with the noise structure... $\endgroup$ Commented Dec 13, 2019 at 12:37
  • $\begingroup$ It is good to have a concise explanation like this one. (I have no experience in this field and am just curious.) However, I cannot imagine how the event of horizontal (?) traversing can depend on a diffeomorphism, and how you define "left" and "right", especially for an arbitrary diffeomorphic image of the square. Can you detail this? Also, does your comment "the probability space doesn’t admit a random field that’s compatible with the noise structure" mean that there is no nonzero "linear" random variable? $\endgroup$ Commented Dec 13, 2019 at 13:26
  • $\begingroup$ The left and right faces would the images of the left and right faces of $[0,1]^2$ under $\phi$. Regarding the other comment, I meant that there is no non-zero distribution-valued random variable $\zeta$ such that $\zeta(\psi)$ is $F_{\supp \psi}$-measurable for every test function $\psi$. Modulo some technical analytical regularity requirements, this is essentially the same as not admitting any non-trivial linear random variables as in the answer. $\endgroup$ Commented Dec 13, 2019 at 15:26
  • $\begingroup$ @MartinHairer : Thank you for this response. I think I understand now what you mean by the left and right faces of a diffeomorphic image of the square. But I still cannot imagine how the event of crossing can be defined so that it may depend on the choice of a diffeomorphic image; shouldn't this event be invariant under diffeomorphisms? $\endgroup$ Commented Dec 13, 2019 at 16:13

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