2026-10-01 · chapter Yang–Mills and the mass gapPart 1 Mathematical preparationwhat is special about four dimensions

Detour: what happens only in four dimensions — rereading eight cases from the Yang–Mills side

The starting point of the article "What happens only in four dimensions" was the count that, while there are $n$ directions to move in, the number of planes one can rotate in is

In three dimensions planes correspond one-to-one with perpendicular axes, and a rotation can be described by its axis. In four dimensions what is perpendicular to each of the 6 planes is again a plane, and this 6 splitting into 3 + 3 is the starting point of many sections. The way it splits is written concretely in sections 02 and 03, as multiplication of quaternions on the left and on the right.

The table shows where the eight cases of the article appear in this book. The left column links to the headings of the article, the right column to sections of this chapter and to the chapters with the main treatment. Topics not in the article (scale and concentration, the winding number at infinity) are added in sections 04–06. The numbers on the left of the table are the numbers of the cases in the article, separate from the section numbers of this chapter. This chapter itself is a detour, so you may pick out and read only the rows that interest you.

Read a point of four dimensions as a single number. Quaternions are numbers made by adding the symbols $i,j,k$ to the real numbers, and the rules are the following one line. Multiplication is associative but not commutative ($ij=k$, $ji=-k$).

The quaternions of length 1 form $S^3$, and under multiplication they are the same group as SU(2) (that SU(2) is $S^3$ is in section 05 of 1-02) Known. Multiplying by $q,p$ of length 1 on the left and on the right gives a rotation of four dimensions, and every rotation can be written in this form. Only the pairs $(q,p)$ and $(-q,-p)$ give the same rotation Known. Unlike $SO(3)\cong SU(2)/\{\pm1\}$ in three dimensions (1-02), there are two copies of SU(2).

Example. If $q=\cos a+i\sin a$, $p=\cos b+i\sin b$, the $(x_1,x_2)$ plane turns by the angle $a-b$ and the $(x_3,x_4)$ plane by the angle $a+b$. The two sliders of case 01 (the speeds of the xy plane and the zw plane) are the difference and the sum of the left and right angles. We checked with 200 random pairs that (3) is an orthogonal matrix of determinant 1 Computed.

The field strength $F$ is at each point an antisymmetric array $F_{\mu\nu}=-F_{\nu\mu}$, with 6 independent components attached one to each plane. For electromagnetism, these 6 components are the 3 of the electric field $E$ and the 3 of the magnetic field $B$ ((13) of 1-06). In Yang–Mills each component is an element of a Lie algebra (1-03), but in this section we take them to be real numbers. The Hodge star $*$ is the operation taking each plane to the perpendicular plane; with the totally antisymmetric symbol of $\varepsilon_{1234}=1$,

The $*$ in $n$ dimensions sends 2-forms to $(n-2)$-forms, so only for $n=4$ does it return to 2-forms. Since $**=1$ (1-06), it can be split as $F^\pm=\frac12(F\pm *F)$ ($|F|^2=\sum_{\mu\lt\nu}F_{\mu\nu}^2$), and the density of the action and the density of topology (section 06) can be written with the same two. $F=\pm*F$ are called self-dual and anti-self-dual.

Using (4), $F^\pm$ has three independent components each, and arranged so that the squared length is $|F^\pm|^2$ they take the following form. The three axes of Figure 1 are in this order.

The key point is the relation with (3). The rotation $x\mapsto qx\bar p$ turns the 3 components of $F^+$ by $q$ only and the 3 components of $F^-$ by $p$ only, and does not mix the two Known.

Which one turns $F^+$ is fixed by convention. Changing the orientation or the assignment of coordinates to quaternions swaps left and right. This book follows (2) and $\varepsilon_{1234}=1$. In this convention, the matrix multiplying by $i$ on the left is proportional to $dx_1\wedge dx_2+dx_3\wedge dx_4$ and self-dual, and the matrix multiplying on the right is proportional to $dx_1\wedge dx_2-dx_3\wedge dx_4$ and anti-self-dual Computed.

Example. $q=\cos a+i\sin a$ turns $F^+$ about the first axis by the angle $2a$, and returns to the start at $a=180^\circ$ ($q=-1$). The double cover of 1-02 is visible on $F^+$. We checked with 200 random elements that (5) holds, that the components mixing the two are at most $2.2\times10^{-16}$, and that under left multiplication $F^-$ is fixed while the mirror $x_4\mapsto-x_4$ swaps the two Computed.

ComputedThe initial field is $F=dx_1\wedge dx_2+0.6\,dx_1\wedge dx_4+0.3\,dx_2\wedge dx_3$. The figure computes (2)(4) on the spot, agreeing with the python3 values (including the rows with the mirror). The dotted lines are orbits.

The mirror and the Minkowski signature

A mirror changes the sign of the right-hand side of (5) (case 04). The mirror image of an instanton is an anti-instanton, and this is also why the θ term (the term added to the action by multiplying the integral of the right-hand side of (5) by the coefficient $\theta$) changes sign under a mirror (3-05) Known. With the Minkowski signature, $**=-1$, so there are no real self-dual fields (One sign different); this is one of the reasons for thinking in imaginary time (4-02).

Keeping the shape of the field and multiplying only its size by $1/\lambda$ gives $A_\lambda(x)=\lambda A(\lambda x)$, $F_\lambda(x)=\lambda^2F(\lambda x)$ (a gauge field has the dimension of inverse length). Replacing $y=\lambda x$,

so only in four dimensions does the action not depend on size. This means the classical theory has no preferred length, and it is another face of the $[g^2]=4-d$ of 2-04 Known. The problem statement's 'the mass gap is not classically visible' (Jaffe–Witten §6.6) is also related to this (4-07).

One consequence. For a solution (stationary point) of finite action, the action does not change at first order under deformation of size either.

Hence if $d\ne4$ then $F=0$, and non-trivial solutions of finite action exist only in four dimensions (assuming enough decay for the deformation to be allowed; an argument of the same kind is attributed to Derrick 1964) Known.

Example. Seen on a lattice it is the same. On a $n$-dimensional lattice of spacing $\epsilon$, the sum of quantities attached to the small squares (plaquettes) of the lattice plays the role of the action, and it is multiplied by the inverse coupling $\beta$ (the coefficient in front of the action, corresponding to $1/g^2$; the constant factor depends on normalisation, and in Wilson's convention $\beta=2N/g^2$) (5-01). The plaquette sum is approximated by $\frac{\epsilon^{4-n}}{4}S_{\mathrm{YM}}(A)$ ($S_{\mathrm{YM}}$ is the integral of the action of (7) with $1/2g^2$ removed). So classically $\beta$ need only keep $\beta\,\epsilon^{4-n}$ constant to compensate for the shrinking of the sum as $\epsilon^{4-n}$ ($\beta\propto\epsilon^{n-4}$), and at $n=4$ it does not move. In three dimensions $\beta\propto1/\epsilon$, agreeing with $[g^2]=1$. In physics the four-dimensional $\beta$ is believed to move like $\log(1/\epsilon)$, but there are doubts too, and in mathematics it is unsolved (Chatterjee, arXiv:1803.01950 §3) Physics. This running is asymptotic freedom (4-06).

ComputedThe left is (9), the right the $\rho^{d-4}$ of (7) ($\rho=1/\lambda$), computed on the spot (agreeing with the ratios of integrals in python3). $d\ne4$ is the ratio when the same shape is shrunk, not a solution.

From here on we deal with fields whose values are in a Lie algebra, so we fix $|F|^2$. Write $F_{\mu\nu}=\sum_aF^a_{\mu\nu}t_a$ with $t_a=-iT_a$ ($\operatorname{tr}(t_at_b)=-\tfrac12\delta_{ab}$), the $T_a$ of 1-04 multiplied by $-i$, and

Since $t_a$ is anti-Hermitian, $\operatorname{tr}F^2$ is negative, so a minus sign is attached. (5) is also read as summed over each $a$.

The simplest self-dual solution (the BPST instanton, 3-05) has its action density gathered roundly at the origin. The action density $s=\tfrac12|F|^2$ in units of $1/g^2$ and the fraction of the action inside radius $R$ are as follows for the solution of size $\rho$ ($d^4x=2\pi^2r^3dr$; $2\pi^2$ is the area of the unit $S^3$) Known. The form of $s$ was recomputed from the BPST connection with the normalisation above Computed.

ComputedTable 1 Numerical integration of (9). The total does not depend on $\rho$, and half always lies inside radius $\rho$.

Example. As $\rho\to0$, the density collapses to a point while keeping the action at $8\pi^2/g^2$. In three dimensions or fewer, shrinking increases the action (Figure 2), so concentration stops, but in four dimensions it does not stop. That handling this concentration is the crux of the analysis of classical Yang–Mills in four dimensions is said to have been known since Uhlenbeck's work of 1982 Known. §2 of the problem statement writes 'Asymptotic freedom ensures a decisive regularity in the case when classical Sobolev inequalities are borderline.', stating flatly that asymptotic freedom "ensures" a decisive regularity. This chapter reads it as meaning that asymptotic freedom makes up for the classical borderline seen in (7) and Table 1 (estimates falling just short in four dimensions) (an interpretation).

The far region of $\mathbb R^d$ is a large sphere $S^{d-1}$. If the action is finite and the decay sufficient, far away $A$ approaches a form that can be written by a gauge transformation alone (pure gauge; 1-07), and the map of the gauge transformation remains (the same mechanism as the monopole of 1-07). To distinguish it from the coupling constant $g$, in this section we write that map as $h$. The classes of maps, with maps that can be continuously deformed into each other regarded as the same, form the homotopy group $\pi_{d-1}$.

In four dimensions both the far region and SU(2) are $S^3$, and maps between spheres of the same dimension have an integer, "how many times it covers" (the degree). To be exact, choose one general point of the target $S^3$, attach $+1$ to each point of its preimage where $h$ preserves orientation and $-1$ where it reverses it, and add them up (the same value whichever point is chosen). This is the instanton number $k$.

The inequality follows from (5). With the normalisation of section 05 one can write $k=\frac{1}{16\pi^2}\int(|F^+|^2-|F^-|^2)\,d^4x$, so $S=\frac{1}{2g^2}\int(|F^+|^2+|F^-|^2)\,d^4x\ge\frac{8\pi^2}{g^2}|k|$, with equality when self-dual or anti-self-dual. The BPST solution has $k=1$ (3-05) Known.

Example. According to the tables of homotopy groups, the classes $\pi_{d-1}(S^3)$ into SU(2) = $S^3$ are $0,0,\mathbb Z,\mathbb Z/2$ for $d=2,3,4,5$, and the classes into U(1) = $S^1$ are $\mathbb Z$ only for $d=2$ (the number of vortices; the same mechanism as the monopole of 1-07). For a compact simple Lie group $\pi_3$ is always $\mathbb Z$, and SU(3) too has $k$ Known.

As in case 05, two planes in four dimensions usually meet in one point. Two closed surfaces in a closed four-dimensional manifold also meet in finitely many points, and the sum of $\pm1$ with orientations included does not change when the surfaces are moved. This is the intersection form (a symmetric matrix of integers).

Example. The intersection form of $\mathbb{CP}^2$ is $(1)$, and that of $S^2\times S^2$ is $\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$ Known. Using the space of solutions for anti-self-dual connections ($F^+=0$ in (6); the same as self-dual connections with the orientation reversed, and for the convention of the primary sources see 3-06) as a tool, Donaldson arrived at the theorem that "the definite intersection form of an oriented closed smooth four-dimensional manifold can be diagonalised". He is said to have shown the simply connected case in 1983 and later extended it to a form without the assumption of simple connectivity (details in 3-06). Contrasted with Freedman's topological theorem, spaces homeomorphic but not diffeomorphic to $\mathbb R^4$ (exotic $\mathbb R^4$) appear, and Taubes is credited with first showing that there are continuum-many of them Known. The main treatment is in 3-06.

Detour: knots and Wilson loops

Multiplying the vertices of the 24-cell (two coordinates $\pm1$, two 0) on the left by $(1+i)/\sqrt2$ and scaling by $1/\sqrt2$ gives the following 24 unit quaternions Computed. They form a finite subgroup of SU(2) closed under multiplication (the binary tetrahedral group; its image in SO(3) is the 12 rotations of the regular tetrahedron) Known.

The 120 vertices of the 600-cell likewise form the binary icosahedral group $\bar I$.

Example. In lattice gauge theory (5-01) elements of a finite group can be placed on the links. Petcher–Weingarten 1980 computed with $\bar T$, $\bar O$, $\bar I$, and according to the abstract, for $\bar I$ the mean plaquette, Wilson loops and estimates of the string tension agreed with SU(2) over a wide range of couplings. At sufficiently weak coupling it is said to behave differently from SU(2) because of the gaps between elements Physics (sources and details in 5-10).

ComputedTable 2 All three checked to be closed under products. "Angle to the farthest point" is the maximum (an estimate) of the angle from 200,000 random points on $S^3$ to the nearest element.

The volume of the ball in case 08 comes from the formula for the area of the unit sphere. This formula makes sense even when $d$ is not an integer, and the integrals of the form first met in perturbation theory (4-03) also become functions of $d$.

$\Gamma$ is the gamma function, the extension of the factorial to real numbers ($\Gamma(n)=(n-1)!$, $\Gamma(\tfrac12)=\sqrt\pi$, $\Gamma(x+1)=x\Gamma(x)$). The integrand is $k^{-4}$ far away and the volume $k^{d-1}dk$, so it converges only for $d\lt4$, and four dimensions is the borderline where divergence begins, logarithmically. In the dimension where the coupling is dimensionless, the divergence is exactly logarithmic (4-04). Computing at $d=4-\epsilon$ (dimensional regularisation, 't Hooft–Veltman 1972, Bollini–Giambiagi 1972) and cutting off the momentum at $K$ give

and $1/\epsilon$ and $\ln K$ correspond with the same coefficient $2\pi^2=|S^3|$.

ComputedTable 3 The formula (13) and numerical integration.

Example. The last column approaches $2\pi^2=19.739209$, and for the cut-off, $I_K-2\pi^2\ln K=-9.869604$ at $K=10^4$ ($-\pi^2$). The volumes of balls in the article ($V_5=5.2638$ and so on) also agree with the same formula Computed.

Using $\Gamma(x+1)=x\Gamma(x)$, $(4-d)\,I(d)=2\pi^{d/2}\Gamma(3-\tfrac d2)$, and the right-hand side is smooth across $d=4$. The divergence is only in the factor $1/(4-d)$, and the remaining coefficient is $2\pi^2\Gamma(1)=2\pi^2$ at $d=4$. For the cut-off, $I_K=\pi^2\bigl(\ln(1+K^2)+\frac{1}{1+K^2}-1\bigr)$, and the growth $2\pi^2K^4/(K^2+1)^2$ for each increase of $\ln K$ by 1 approaches the same $2\pi^2$. Figure 3 sets these two side by side.

ComputedThe left computes $I(d)=\pi^{d/2}\Gamma(2-\tfrac d2)$ and $(4-d)I(d)=2\pi^{d/2}\Gamma(3-\tfrac d2)$, the right $2\pi^2K^4/(K^2+1)^2$ and $I_K$, on the spot (the gamma function by Lanczos's approximation; agreeing with the python3 values). The left slider moves the $u$ of $4-d=10^{-u}$.

We line up the $d$-dependences scattered through the book on a single slider. Each row divides by the sign of a power $p(d)$, and at the borderline $p=0$ the power vanishes, leaving a logarithm or invariance. The six rows of the four of spacetime reach their borderline at the same $d=4$. The remaining four rows switch somewhere other than four, or for a different reason. Known To be exact, the six rows line up because of the same counting of dimensions (the 2 of the propagator $1/k^2$, twice), and this does not mean that construction in four dimensions is impossible.

ComputedThe borderlines were computed on this machine; the sentences are copies from each chapter. If JavaScript is disabled, see Table 4.

Table 4 The rows of Figure 4.

What this chapter can say

Sources and reproduction

Next chapter: 2-01 Lagrangians and the action. It continues in 3-05, 3-06, 4-07 and 6-10.

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