2026-10-01 · chapter Yang–Mills and the mass gapPart 1 Mathematical preparationProbability in infinite dimensions

Probability in infinite dimensions — there is no uniform weight, so we start from Gaussian measures

The lattice probability of 1-08 could be written as "density × volume element" because the variables were finitely many. Refining the lattice and enlarging the box makes the variables infinitely many. Physics books write the probability of a free scalar field of mass $m$ (Euclideanised, 4-02) like this.

Neither of the two factors on the right-hand side has a meaning on its own. The measure corresponding to $\prod_xd\varphi(x)$ does not exist (§02), and the integral in the exponent is $+\infty$ for almost every sample (§05, §06). Even so, $\mu$ as a product can be constructed as a Gaussian measure (§04). The problem statement too describes the measures produced by constructive field theory as 'non-Gaussian, Euclidean-invariant measures on spaces of generalized functionals' Known. That means measures on spaces of distributions (§07) that are not Gaussian and are unchanged by rotations and translations. This chapter treats the Gaussian case that is their starting point. Example: dividing the interval into $N$ equal parts with $d=1$, (1) is the Gaussian distribution in $N$ variables of 1-08 §07, and where $N\to\infty$ leads is the subject of this chapter.

Length and area (Lebesgue measure) are unchanged by translation and give finite weight to bounded sets. In infinite dimensions these two cannot both hold. We state this with the space $\ell^2=\{x=(x_1,x_2,\dots):\sum_kx_k^2\lt\infty\}$ of sequences with finite sum of squares, the length $\|x\|=(\sum_kx_k^2)^{1/2}$, and the sequence $e_n$ whose $n$-th component alone is 1.

Theorem Known If a measure on $\ell^2$ is unchanged by translation and gives finite weight to every open ball, it is 0.

Proof. If $n\ne m$ then $\|e_n-e_m\|=\sqrt2\gt1$, so the balls $B(e_n,\frac12)$ are mutually disjoint, and all lie in $B(0,\frac32)$.

Each term on the left is the same value $c$ by translation, so $c=0$. $\ell^2$ can be covered by countably many balls of radius $\frac12$ centred at sequences with rational components that are 0 except for finitely many, so the total weight is 0 too. ∎ If $\mathbb R^n$, only up to $3^n$ balls fit (the ratio of volumes), and there is no contradiction.

The density of the standard Gaussian $N(0,I_n)$ on $\mathbb R^n$ (1-08 (8)) is largest at the origin, but most of the weight lies on a thin shell far away. $|x|^2=\sum_{k\le n}x_k^2$ is a sum of $n$ independent terms, so by the law of large numbers (1-08 (12)), $|x|^2/n\to1$.

$E|x|/\sqrt n=0.975,\ 0.9975,\ 0.99975$ at $n=10,\ 100,\ 1000$; the standard deviation of $|x|$ is $0.698,\ 0.706,\ 0.707$, and the thickness of the shell settles at $1/\sqrt2$ regardless of $n$ Computed. Letting $n\to\infty$ with variance 1 gives $|x|\to\infty$, and the limit is not in $\ell^2$. So we reduce the variance $\sigma_k^2$. For a sequence $\xi_1,\xi_2,\dots$ of independent $N(0,1)$,

the finiteness of the sum on the right is necessary and sufficient for $x$ to be a Gaussian random variable with values in $\ell^2$ (Hairer Exercise 4.7) Known. Example: if $\sigma_k=1/k$, then $E\|x\|^2=\pi^2/6$. A sequence with variance 1 can also be built, on the space $\mathbb R^\infty$ of all sequences. What is used is Kolmogorov's extension theorem (the theorem that if the distributions of finitely many at a time are given consistently, the distribution of the infinite sequence is fixed uniquely; Durrett Theorem 2.1.21) Known, and this is the seed of the white noise of §07.

Since densities cannot be used, Gaussian measures are defined by their "shadows". A probability measure $\mu$ on a space $E$ ($\ell^2$ or a space of continuous functions) is a Gaussian measure if, for every continuous linear map $\ell:E\to\mathbb R$, $\ell(x)$ follows a one-dimensional Gaussian distribution (Hairer Definition 4.4) Known. With mean 0, the covariance $C(\ell,\ell')=E[\ell(x)\ell'(x)]$ determines $\mu$ (the form of 1-08 (8)).

On $\ell^2$ it can be written with $C(\ell,\ell')=\langle\ell,C\ell'\rangle$ and the covariance operator $C$; for (4), $Ce_k=\sigma_k^2e_k$. If a symmetric positive $C$ has finite trace (sum of eigenvalues), there is a Gaussian measure with it as covariance (Hairer Proposition 4.17) Known. Example: $\ell(x)=\sum_ka_kx_k$ is Gaussian with variance $\sum_ka_k^2\sigma_k^2$. The problem statement gives the free field as the Gaussian measure with mean 0 and covariance $(-\Delta+m_0^2)^{-1}$ on the space of distributions $\mathcal S'(\mathbb R^d)$ Known. The two factors of (1) have been replaced by a single piece of information, the covariance.

Two probability measures are equivalent if they have the same null sets (sets of weight 0). Conversely, when a set of probability 1 for one has probability 0 for the other, the two are said to be mutually singular. A finite-dimensional Gaussian measure acquires a density factor when shifted (in one dimension $N(h,1)$ is $e^{hx-h^2/2}$ times $N(0,1)$). In general a function $D$ with $\mu_h(A)=\int_AD(x)\,d\mu(x)$ for every set $A$ is called the relative density and written $d\mu_h/d\mu$. For $\mu_h$, obtained by shifting $\mu$ by $h$ in the first $n$ coordinates of (4), we have the following.

$\|h\|_H$ is "the length of the shift measured by the fluctuation in that direction". The degree to which the two can be told apart (the total variation distance) reduces to a one-dimensional problem in the direction of $C^{-1}h$, and is exactly the following formula ($\Phi$ is 1-08 (13)).

(7) was checked by numerical integration and with samples in 40 dimensions Computed. As $n\to\infty$, if $\|h\|_H$ is bounded then $d_{\mathrm{TV}}\lt1$; if it diverges then $d_{\mathrm{TV}}\to1$, and in the latter case the two measures become mutually singular.

The Cameron–Martin theorem Known $\mu_h$ is equivalent to $\mu$ only when $\|h\|_H\lt\infty$, and otherwise they are mutually singular (Hairer Theorem 4.44 and its proof; the original paper is Cameron–Martin 1944). The set $H$ of these $h$ is the Cameron–Martin space. Yet for samples of (4), $\|x\|_H^2=\sum_k\xi_k^2=+\infty$ (the law of large numbers), and in infinite dimensions $\mu(H)=0$ (Hairer Proposition 4.45) Known. Samples fall, with probability 1, outside the set $H$ of directions in which one may shift.

ComputedThe numbers of Figure 1 and Table 1 were computed on this machine (the figure is drawn on the spot from (7) and partial sums, agreeing with the table to 4 digits). If JavaScript is disabled, they can be read from Table 1.

At the borderline $a=3/2$, $\|h\|_H^2\approx\ln n+0.577$ grows only logarithmically, and even at $n=10^{20}$, $d_{\mathrm{TV}}=0.9994$ Computed.

The Gaussian measure with values in continuous functions on $[0,1]$, mean 0 and covariance $E[B(s)B(t)]=\min(s,t)$ is the Wiener measure, and its sample $B$ is Brownian motion (said to begin with Wiener 1923) Known. The paths are Hölder continuous with any exponent below $1/2$ (if $\gamma\lt1/2$, then with a constant $C$ depending on the path, $|B(t)-B(s)|\le C|t-s|^\gamma$ for all $s,t$), and with probability 1 they are differentiable at no point (Durrett Theorems 7.1.5, 7.1.6) Known. Durrett gives the former as a result of Wiener 1923 and the latter as a discovery of Paley–Wiener–Zygmund 1933.

To put it in the form (4), we look for the eigenfunctions of the covariance operator $(Cf)(s)=\int_0^1\min(s,t)f(t)\,dt$. $g=Cf$ satisfies $g''=-f$, $g(0)=0$, $g'(1)=0$, so the eigenvalue problem becomes the ordinary differential equation $-g''=\lambda^{-1}g$, and from $\cos\omega=0$ the following comes out.

$\sigma_k=1/\omega_k\approx1/(\pi k)$ decreases at the same rate as the example of §05, and the covariance with 4000 terms agrees with $\min(s,t)$ to within $5.1\times10^{-5}$ Computed. The Cameron–Martin space is the set of functions with $h(0)=0$, $\|h\|_H^2=\int_0^1\dot h^2\,dt\lt\infty$ (Hairer Exercise 4.29) Known, and this $\|h\|_H^2$ is the content of the exponent $m=0$ in the $d=1$ of (1). For the partial sum $S_K$ of $K$ terms, $\int_0^1\dot S_K^2\,dt=\sum_{k\le K}\xi_k^2\approx K$: the path of Figure 2 settles, yet this "energy" keeps growing.

ComputedThe numbers of Figure 2 were computed on this machine (partial sums with 128 fixed random numbers $\xi_k$). At $K=1,16,64,128$, $\sum\xi_k^2=2.13,\ 13.4,\ 56.5,\ 132.8$, $\max_t|S_K-S_{128}|=0.589,\ 0.202,\ 0.090,\ 0$. If JavaScript is disabled, it can be read from these numbers and Table 2.

Seen on a lattice — quadratic variation

The same thing is seen when cut into a lattice. The sum of squares of the increments over $N$ equal parts (the quadratic variation) approaches 1 (Durrett Exercise 7.1.6) Known, while for smooth functions it approaches 0, so the discretised energy $\sum(\Delta B)^2/\Delta t$ diverges in proportion to $N$ (Table 2). The $e^{-\frac12\int\dot x^2dt}\prod_tdx(t)$ of the path integral (2-07) is a formal way of writing this measure. To be exact, neither the exponent nor $\prod_tdx(t)$ has a meaning on its own (§01); as a pair they denote the Wiener measure.

The formal derivative $\xi=\dot B$ is white noise. It has no pointwise values (the lower panel of Figure 2), so it is defined paired with functions. White noise is a Gaussian random variable $\xi(f)$ for each square-integrable function $f\in L^2[0,1]$, linear in $f$ and satisfying the following (Hairer Example 4.63) Known.

Example: if $\mathbf 1_{[0,t]}$ is the function equal to 1 on $[0,t]$ and 0 outside, the covariance of $\xi(\mathbf 1_{[0,s]})$ and $\xi(\mathbf 1_{[0,t]})$ is $\int_0^1\mathbf 1_{[0,s]}(u)\,\mathbf 1_{[0,t]}(u)\,du=\min(s,t)$, and this is Brownian motion (Hairer Exercise 4.64). The variance of the mean $\xi(\mathbf 1_{[t,t+\varepsilon]})/\varepsilon$ over an interval of width $\varepsilon$ is $1/\varepsilon$; for a box of side $\varepsilon$ in $d$ dimensions it is $\varepsilon^{-d}$, and it diverges when shrunk to a point.

White noise is the Gaussian whose covariance is the identity map (all $\sigma_k=1$), and as in §03 it is not in $L^2$. To measure, dimension by dimension, how far it misses, we now move the stage from $[0,1]$ to the $d$-dimensional torus $\mathbb T^d=[0,2\pi)^d$ (a box with opposite ends joined) (the integral of (9) is replaced by the integral over $\mathbb T^d$). A function on $\mathbb T^d$ is written $f(x)=\sum_{k\in\mathbb Z^d}c_k\,e^{ik\cdot x}$, and $c_k$ are called the Fourier coefficients. $-\Delta=-\sum_i\partial_i^2$ multiplies $e^{ik\cdot x}$ by $|k|^2$, so on coefficients it is the operation turning $c_k$ into $|k|^2c_k$. The Fourier coefficients of white noise are independent except for the pairs related by $c_{-k}=\overline{c_k}$, and $E|c_k|^2$ is the same constant for all.

Let $H^{-s}$ be the space of formal series $\sum_kc_ke^{ik\cdot x}$ given the length $\big(\sum_k(1+|k|^2)^{-s}|c_k|^2\big)^{1/2}$ (the larger $s$, the more lightly high frequencies are counted). For white noise the mean of the square of this length is a constant times $\sum_k(1+|k|^2)^{-s}$, and the shell of radius $R$ holds about $R^{d-1}$ values of $k$, so it is finite when $s\gt d/2$. In fact, exactly then can white noise be realised in $H^{-s}$ (Hairer Exercise 4.65) Known. Things that only give values when paired with functions are distributions; the general theory is in 1-11, and stochastic quantisation, which shakes a field with white noise, is in 6-09.

The free field of (1), considered on the torus, is white noise smoothed by multiplying its Fourier coefficients $c_k$ by $(|k|^2+m^2)^{-1/2}$. Writing this operation $(-\Delta+m^2)^{-1/2}$, the covariance is $(-\Delta+m^2)^{-1}$ (the form of the problem statement in §04), and the variance of the coefficients is proportional to $1/(|k|^2+m^2)$. By the count of §07, the condition to lie in $H^{-s}$ becomes $s\gt d/2-1$: one order smoother than white noise, but for $d\ge2$ it is not a function. The free field on a bounded region can be made as samples in $L^2$ if $d=1$, and cannot for $d=2$ (the remark after Sheffield Proposition 2.7) Known.

On the lattice $a\mathbb Z^d$ with lattice spacing $a$, we replace $-\Delta$ by differences: the $(\Delta x)_j=x_{j+1}-2x_j+x_{j-1}$ of 1-08 (10) added over coordinate directions and divided by $a^2$. $e^{ik\cdot x}$ are eigenfunctions of this difference Laplacian, and from $e^{ik_ia}-2+e^{-ik_ia}=-4\sin^2\frac{k_ia}2$ the eigenvalues are $-\sum_i\frac4{a^2}\sin^2\frac{k_ia}2$. Hence the variance at a point is the following integral ($k$ runs over the box that the lattice can distinguish).

For $a\to0$ the contribution of large $|k|$ is about $\int^{1/a}k^{d-3}\,dk$, and it diverges as $d\ge2$. Instead of the value at a point, look at the value $\varphi(\rho_\ell)$ paired with a Gaussian kernel $\rho_\ell(x)=(2\pi\ell^2)^{-d/2}e^{-|x|^2/(2\ell^2)}$ of standard deviation $\ell$ (formally $\int\varphi(x)\rho_\ell(x)\,dx$, the smoothed value); its variance is

(on the lattice, $k$ is restricted to the same box as in (10) and the denominator replaced by that of (10)). The factor $e^{-\ell^2|k|^2}$ suppresses large $|k|$, so it is finite for every $d$.

(10) has been checked against exact values for $d=1$ and against direct sums on lattices for $d=2,3$. $d=1$ converges to $1/(2m)$; for $d=2$ the difference from $\frac1{2\pi}\ln\frac1a$ settles at 0.276; for $d=3$, $a\,G_a(0)$ approaches about 0.253 (0.2515 at $a=1/64$); and for $d=4$, $a^2G_a(0)$ approaches 0.155 Computed. These two destinations are the values at the origin of the lattice Green's functions for $\mathbb Z^3$, $\mathbb Z^4$, and come out as 0.2527 and 0.1549 from a different one-dimensional integral ($d=3$ is half of Watson's integral) Computed. The variance of the smoothed value converges to a finite value for every $d$, and agrees with the limit to the third decimal place at $a=1/32$. The value of the field is "infinite at a point, finite once smoothed". A figure setting this quantity, whose borderline is 2, beside quantities that switch at four is the dimension slider of 1-12 §10.

ComputedThe numbers of Figure 3 and Table 3 were computed on this machine (the values for $a=2^{-p}$, $p=1,\dots,6$ are embedded in the figure as a table; Table 3 adds to them the row for $a=1$ and the row of growth). If JavaScript is disabled, they can be read from Table 3.

As in §08, for $d\ge2$ the mean of $\varphi(x)^2$ is infinite. So terms like $\varphi^4$ need treatment by subtraction (renormalisation, 4-04, 6-03), and products of distributions cannot in general be defined (1-11). The four-dimensional Yang–Mills field $A$ is expected to be, at short distances, as rough as the four-dimensional free field (asymptotic freedom, 4-06) Physics. To be exact, the expectation is that correlation functions at short distances approach those of the free field up to logarithmic corrections. The problem statement explains asymptotic freedom as 'at short distances the field displays quantum behavior very similar to its classical behavior' (§1), and requires that the correlation functions of the theory to be constructed 'should agree at short distances with the predictions of asymptotic freedom' (§4) Known. The $A\wedge A$ of the field strength $F=dA+A\wedge A$ (1-07) is a product of such distributions (6-10).

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