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Use Euclidean coordinates $(x^1,x^2,x^3,x^4)$ and orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. For $D_{\mu }={\partial }_{\mu }+A_{\mu }$, the gauge curvature is $F_{\mu \nu }=[D_{\mu },D_{\nu }]$. The ASDYM equations are $F=-\ast F$, equivalentlyPut $w=x^1+i{x}^2$, $z=x^3+i{x}^4$, with ${\partial }_w=({\partial }_1-i{\partial }_2)/2$ and similarly for $z$. The bars label the complex conjugate coordinates on the Euclidean real slice; after complexification they can be treated as independent coordinates. A Lax pair for the anti-self-dual Yang-Mills equations isAt infinity rescale the operators and use the other affine coordinate. Their commutator isCompatibility for every $\lambda$ is therefore precisely $F_{wz}=F_{\bar{w}\bar{z}}=0$ and $F_{w\bar{w}}+F_{z\bar{z}}=0$. To verify the real component signs,The conjugate equation supplies the remaining real components. This is compatibility of an overdetermined system, not an additional field equation on $\Psi$ for only one value of the spectral parameter.

Impose invariance under translations in $x^1,x^2,x^3$, writing $t=x^4$. In a symmetry-adapted gauge the potentials depend only on $t$. A further $t$-dependent Yang-Mills gauge transformation sets $A_4=0$ locally, by solving the usual parallel transport ordinary differential equation; it does not reintroduce spatial dependence. ThenThe three real equations above become ${\dot{A}}_1=[A_2,A_3]$, ${\dot{A}}_2=[A_3,A_1]$ and ${\dot{A}}_3=[A_1,A_2]$, namelyThese are the Nahm equations. The orientation chosen above gives the requested plus sign; reversing orientation exchanges self-duality of gauge curvature and anti-self-duality of gauge curvature. Translation invariance here is a local dimensional reduction and is not a claim of finite four-dimensional action over all three translation directions.

To prove conservation, abbreviate $P=A_1+i{A}_2$, $Q=A_1-i{A}_2$ and $R=A_3$, so $A(\lambda )=P+2R\lambda -Q{\lambda }^2$ and $B(\lambda )=-iR+iQ\lambda$. Expanding the commutator, including the cancellation of the cubic term, givesThe Nahm equations giveThus the polynomial Lax representation of the Nahm equations isTake a finite-dimensional matrix representation of the Lie algebra and its complexification. For every positive integer $p$, differentiating the product and using cyclicity of the matrix trace givesThis proves the trace invariants of a Lax equation directly. Since $\mathrm{Tr}(A(\lambda {)}^p)$ is a polynomial of degree at most $2p$ in the affine coordinate $\lambda$, its derivative vanishes identically only if the derivative of every coefficient vanishes. Therefore every trace-polynomial coefficient is independent of $t$. At projective infinity these polynomials are sections of $\mathcal{O}(2p)$; they need not be constant functions of $\lambda$ on the whole projective sphere.

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

When $u$ is independent of $y$ and $z=x+iy$,PutThe zero-curvature equation becomesSincewe obtain the Isospectral Lax equationTherefore, for every positive integer $n$, the cyclic property of the matrix trace givesThese are the trace invariants of a Lax equation.

The PDE from part (b) reduces toWith $\varphi =2u$, this isTo extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication givesThe terms involving $\lambda$ are constant, and $p={\varphi }^{\prime }/2$. Henceis a first integral. Direct differentiation verifies it:

Let $L$ be the symmetric zero-diagonal Jacobi matrix with $L_{j,j+1}=L_{j+1,j}=a_j$, and let the skew-symmetric matrix $B$ haveThen $\dot{L}=[B,L]$ is equivalent to the Finite Volterra lattice. The corresponding similarity flow makes the eigenvalues of $L$, the coefficients of its characteristic polynomial, and its trace invariants first integrals.