Gauge connections and Yang-Mills equations

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Problem 4.82 from Steeb W., Tanski I., Hardy Y. "Problems and Solutions for Groups, Lie Groups, Lie Algebras with Applications" [STH12]More problems in Chapter 4 "Applications"

Consider the Lie algebra u(n)\mathfrak {u}(n) and Aj∈u(n)A_j \in \mathfrak {u}(n). Thus Aj∗=−AjA_j^* = -A_j. The curvature

FA=∑j,k=14Fjkdxj∧dxk F_A = \sum _{j,k=1}^{4} F_{jk}dx_j \wedge dx_k

of a u(n)\mathfrak {u}(n)-valued connection differential one-form

A=∑j=14Ajdxj A = \sum _{j=1}^{4} A_j dx_j

on R4\mathbb {R}^{4} is

Fjk=[∇j,∇k]=−∂∂xjAk+∂∂xkAj+[Aj,Ak] F_{jk} = [\nabla _j, \nabla _k] = -\frac{\partial }{\partial x_j}A_k + \frac{\partial }{\partial x_k}A_j + [A_j,A_k]

where

∇j:=∂∂xj−Aj. \nabla _j := \frac{\partial }{\partial x_j} - A_j .

The connection (matrix-valued differential one-form) AA is anti-self-dual Yang-Mills on R4\mathbb {R}^{4} if

∗FA=−FA *F_A = -F_A

where ∗* is the Hodge star operator with respect to the Euclidean metric tensor field

g=dx1⊗dx1+dx2⊗dx2+dx3⊗dx3+dx4⊗dx4. g = dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + dx_3 \otimes dx_3 + dx_4 \otimes dx_4 .

The Hodge star operator is ff-linear. We have ∗(dx1∧dx2)=dx3∧dx4*(dx_1 \wedge dx_2) = dx_3 \wedge dx_4, ∗(dx1∧dx3)=−dx2∧dx4*(dx_1 \wedge dx_3) = -dx_2 \wedge dx_4.

?
(i)

Write down the anti-self-dual Yang-Mills in coordinates.

(ii)

Show that anti-self-dual Yang-Mills equation has a Lax pair representation. Set

z=x1+ix2,w=x3+ix4. z = x_1+ix_2, \quad w = x_3+ix_4 .

Thus

∇z=12(∇1−i∇2)=∂∂z−Az,∇z‾=12(∇1+i∇2)=∂∂z‾−Az‾ \nabla _z = \frac{1}{2}(\nabla _1-i\nabla _2) = \frac{\partial }{\partial z} - A_z, \quad \nabla _{\overline{z}} = \frac{1}{2}(\nabla _1+i\nabla _2) = \frac{\partial }{\partial \overline{z}} - A_{\overline{z}}

and ∇w,∇w‾\nabla_w, \nabla_{\overline{w}}, analogously.

References

[STH12] Steeb W., Tanski I., Hardy Y. Problems and Solutions for Groups, Lie Groups, Lie Algebras with Applications. World Scientific Publishing, 2012.