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Using the Levi-Civita symbol and the epsilon-delta identity,ThereforeOne particularly simple vector potential for $u$ isDirect differentiation gives $\nabla \cdot A=0$ and $\nabla \times A=(y^2,z^2,x^2)=u$.

Parametrize the curved cone byThe outward oriented surface element isSubstitution and integration over $0\le z\le h$, $0\le \phi <2\pi$ give

Because $\nabla \cdot u=0$, the divergence theorem predicts that the two outward fluxes sum to zero. On the top disk $S_2$, $d\mathbf{S}=(0,0,1)dxdy$ and $u_z=x^2$, so, with $R=h\tan \alpha$,as predicted.

The outward orientation on $S_2$ induces the stated anticlockwise orientation on $C$, whereas the outward orientation on $S_1$ induces the reverse orientation. Since $u=\nabla \times A$, Stokes theorem predictsOn $C$, put $(x,y,z)=(R\cos \phi ,R\sin \phi ,h)$. Thenverifying both identities.

Stokes theorem givesFor the upward orientation, the outer circle $r=2,z=1$ is counterclockwise from above and the inner circle $r=1,z=4$ is clockwise. On a counterclockwise circle of radius $R$ and height $z$,The outer circle contributes $-16\pi$. The inner circle would contribute $-7\pi$ counterclockwise and hence contributes $+7\pi$ with its induced clockwise orientation. Thusagreeing with the direct calculation.

For $n=(a,b,c)$ and $r=(x,y,z)$, the integrand is $\frac{1}{2}(n\times r)\cdot dr$. Sincethe Stokes theorem gives the signed planar area, and its absolute value gives the ordinary area.

The $xy$ projection of the given curve is the astroid $x=2{\cos }^{3}t$, $y={\sin }^{3}t$, whose area isThe plane $z=\sqrt{3}y$ stretches area from its $xy$ projection by $\sqrt{1+3}=2$. Hence

Since the magnetic field is $B=\nabla \times A$, Stokes theorem gives, for any oriented surface $S$ bounded by $C$,Under a gauge transformation $A\mapsto A+\nabla \chi$, the integral changes bybecause $C$ is closed. The flux expression is therefore gauge independent.

The integral of an exact differential form over the closed curve $S^1$ is zero by Stokes theorem, so the map is well defined on de Rham cohomology. It is surjective because the angular form $d\theta$ has integral $2\pi$. If a closed one-form $\alpha$ has zero integral, defineThe zero period makes this definition $2\pi$-periodic, and $dF=\alpha$. Thus the kernel is zero andis an isomorphism.

For the curvature two-formuse the normalizationfor the Second Chern form. Graded cyclicity of the trace gives $\mathrm{Tr}(A^{\wedge 4})=0$ andOn the other hand,andThus the coefficient in the Chern-Simons 3-form must beand

For the gauge transformation $A^{\prime }=gA{g}^{-1}-dg{g}^{-1}$, use $d(g^{-1})=-g^{-1}(dg)g^{-1}$ and the Maurer-Cartan equation. Expanding $d{A}^{\prime }+A^{\prime }\wedge A^{\prime }$ makes the terms linear and quadratic in $dg{g}^{-1}$ cancel, leavingInvariance of the matrix trace under conjugation then gives

A Yang-Mills instanton on ${\mathbb{R}}^4$ has finite Euclidean action, smooth curvature in the interior, and $F\to 0$ sufficiently rapidly at infinity. Its connection therefore approaches a pure gauge on the asymptotic three-sphere,up to a decaying correction, for a map $g:S_{\infty }^3\to SU(2)$. By Stokes theorem,Writing $\theta =dg{g}^{-1}$ gives $d\theta =\theta \wedge \theta$ and, for $A=-\theta$,Hence the instanton number as a winding number at infinity isUnder $SU(2)\cong S^3$, this integer is the degree of a map between oriented manifolds $g:S^3\to S^3$. Reversing the trace or orientation convention reverses the displayed sign.

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

The Maxwell equation $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$, integrated over a fixed spanning surface $S$, gives by Stokes theoremThus the electromotive force and magnetic flux obey Faraday's law

Finite energy requires $|\varphi |\to 1$ and $D\varphi \to 0$ on the circle at spatial infinity. Writing $\varphi \sim e^{i\chi }$ there gives $A\sim d\chi$. The vortex number is the winding numberBy Stokes theorem, the magnetic flux is quantized:

For the rotationally symmetric Abelian Higgs vortex ansatz, $A=f(r)d\theta$ and $\varphi =h(r)e^{ik\theta }$ giveAfter the angular integration, the Abelian Higgs model energy becomesCompleting the square in the two pairs of terms givesRegularity at the origin and approach to the vacuum at infinity requireMore precisely, $h(r)=O(r^k)$ and $f(r)=O(r^2)$ near the origin. The boundary term is $k$, soThe bound is saturated exactly when both squares vanish, giving the radial Bogomolny vortex equationsThe asymptotic value $f(\infty )=k$ also makes the flux $2\pi k$, so the ansatz has vortex number $N=k$.

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the vector identity and contraction with $n$ leavesThe boundary integrand satisfies $(n\times v)\cdot dx=v\cdot (dx\times n)$, proving the formula. The vector $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin \theta e_{\theta }$. The projected divergence is $2\cos \theta$, so the left side isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

A field is irrotational when $\nabla \times F=0$. In ${\mathbb{R}}^3$ definewhere $\gamma$ is any path from $x_0$ to $x$. The difference between two choices is the circulation around a closed loop. By the Stokes theorem it is the flux of $\nabla \times F$ through a spanning surface and hence zero. Thus $V$ is well-defined, $V(x_0)=0$, and differentiation gives $F=-\nabla V$.

If $\nabla \times A_0=B$, every solution isfor an arbitrary differentiable scalar $\chi$, since the difference is curl-free. A necessary condition is $\nabla \cdot B=0$, because the divergence of a curl vanishes.

For forms of the same degree, define the Hodge inner product byOn a $p$-form $\beta$, define the codifferentialequivalently $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ in dimension $n$. If $\alpha$ has degree $p-1$, Stokes theorem on the compact boundaryless manifold givesRearranging and using the definition of $\delta$ givesThus $\delta$ is the formal adjoint of $d$.

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

Put $\alpha =\delta \eta$. The hypothesis says that the exact two-form $\beta =d\alpha$ is anti-self-dual. Since $M$ is compact without boundary, Stokes theorem givesHence $d\alpha =0$ by the exact anti-self-dual form on a compact four-manifold argument. Since $\delta$ is the formal $L^2$ adjoint of $d$,Therefore $\boxed{\delta \eta =0}$.

The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

Stokes theorem is ${\int }_S(\nabla \times F)\cdot ndS={\oint }_{\partial S}F\cdot dx$, with the boundary orientation given by the right-hand rule relative to $n$.

With $D_i\varphi =({\partial }_i-i{A}_i)\varphi$ and $B=F_{12}$, the critically coupled Abelian Higgs model energy in the conventions of the question isFinite energy requires $|\varphi |\to 1$, $D_i\varphi \to 0$, and $B\to 0$ at spatial infinity.

For the Derrick theorem test, preserve gauge covariance by definingThe magnetic, covariant-gradient, and potential energies scale asStationarity at $\lambda =1$ requires $E_B=E_V$, which is possible for a nonconstant finite-energy configuration. The oppositely scaling magnetic and potential terms therefore evade the usual Derrick obstruction and allow vortex solitons.

At infinity write $\varphi \to e^{i\chi }$. The condition $D\varphi \to 0$ gives $A\to d\chi$, and the phase winding defines the Abelian Higgs vortex numberBy Stokes theorem,

Finite Euclidean Yang-Mills action requires $F\to 0$ sufficiently rapidly at spatial infinity. Consequently the connection approaches a pure gauge,under the convention of the question. Compactifying the asymptotic boundary identifies it with $S_{\infty }^3$, while $SU(2)$ is itself topologically $S^3$. Thus $g{|}_{S_{\infty }^3}$ has an integer topological degree.

By $C_2=dY$ and Stokes theorem, the instanton number isup to the common simultaneous choice of trace and orientation signs. Smoothness and finite action are imposed in the interior, and $F=\pm {}^{\star }F$ selects an instanton or anti-instanton representative of the topological sector.

For an oriented surface $\Sigma$ with piecewise smooth closed boundary $\partial \Sigma$, and a vector field $\mathbf{F}$ that is continuously differentiable on an open neighborhood of $\Sigma$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector. The orientation of $\partial \Sigma$ follows the right-hand rule: with the right thumb pointing along $\mathbf{n}$, the fingers curl in the positive direction. Equivalently, walking along the boundary with one's head along $\mathbf{n}$ leaves the surface on the left. Each boundary component receives this induced orientation.

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.