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On the simply connected domain ${\mathbb{R}}^3$, a continuously differentiable differential one-form is exact if its mixed partial derivatives agree, equivalently if its associated vector field has zero curl. Here one can verify exactness directly by observing thatsatisfiesThus $udx+vdy+wdz=dF$. By the fundamental theorem for line integrals, its integral is path independent and equalsConsequently, for every path with the stated endpoints,

For an arbitrary constant vector $\mathbf{c}$, the product rule for divergence givesThe divergence theorem therefore impliesSince this holds for every $\mathbf{c}$,

For $\varphi =x+y$, $\nabla \varphi =(1,1,0)$. The ball of radius $a$ has volume $4\pi a^3/3$, soOn the sphere, use $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$ and the stated outward oriented surface element. The first component of the surface integral isand symmetry gives the same second component and a zero third component. The two sides agree.

The orthogonal group isThe special orthogonal group is its determinant-one subgroup

Every eigenvalue of an orthogonal matrix has modulus one. A real three-by-three matrix has at least one real eigenvalue, and nonreal eigenvalues occur in a complex conjugate pair. If $Q\in SO(3)$ has such a pair $e^{\pm i\theta }$, their product is one, so the remaining eigenvalue is $\det Q=1$. If all three eigenvalues are real, each is $\pm 1$ and their product is one; an odd number of three signs with product one must include $+1$. Hence every element of $SO(3)$ has an eigenvector of eigenvalue one and represents a rotation in three dimensions about its span.

It is false that every element of $O(3)$ is either a rotation or a plane reflection. For example,is an improper orthogonal transformation combining a rotation with a reflection. Its eigenvalues are $e^{\pm i\theta }$ and $-1$, whereas a plane reflection has eigenvalues $1,1,-1$ and a rotation has determinant one.

Conjugating a cycle merely relabels its entries:Thus conjugate permutations have the same cycle type. Conversely, if $\sigma$ and $\tau$ have the same cycle type, match the entries of each cycle of $\sigma$ bijectively and in cyclic order with those of a cycle of $\tau$ of the same length. Extending these matches to a permutation $g$ gives $g\sigma g^{-1}=\tau$.

Let $C_{S_n}(\sigma )$ be the centralizer of $\sigma$. Its $S_n$-conjugacy class remains one $A_n$-conjugacy class exactly when $C_{S_n}(\sigma )$ contains an odd permutation. Indeed, in that case $C_{A_n}(\sigma )$ has index two in $C_{S_n}(\sigma )$, so the orbit-stabilizer theorem givesIf the centralizer contains only even permutations, the $A_n$ class has half the size and the $S_n$ class splits into two $A_n$ classes.

The even cycle types in $S_5$ areThe classes of the identity, a three-cycle, and a double transposition do not split: their centralizers contain an odd permutation. The centralizer of a five-cycle is its cyclic subgroup of order five, which lies in $A_5$, so that class splits in two. Therefore

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.