In SI units, Gauss's law relates the outward electric flux through a closed surface to its enclosed charge:The outward unit normal specifies the sign of the electric flux.

The modified transition matrix $\tilde{P}$ describes the original Markov chain until it hits $B$, and then keeps it at the first state hit. Each state of $B$ is therefore an absorbing state, andFor a nonnegative superharmonic function for a Markov chain $g$ satisfying the boundary condition, part (c) and positivity of the transition matrix give by induction $g\ge {\tilde{P}}^{n}g$. Also $g\ge 1_B$. ConsequentlyLet $n\to \infty$ to obtain $g_i\ge h_i$. Since $h$ satisfies the conditions by parts (a),(b), $h$ is their pointwise minimal solution. This extends the hitting probability is the minimal nonnegative harmonic extension principle to the superharmonic inequality. Nonnegative sums justify the argument on a countable state space as well as a finite one.

Now suppose the Markov chain is irreducible and recurrent, and let $g$ satisfy condition (a). In such a chain every state is hit almost surely from every other state. To see this directly, recurrence gives infinitely many visits to a starting state $i$. Irreducibility supplies a finite path from $i$ to any target $j$ with positive probability $\alpha$, chosen with no return to $i$ before reaching $j$. At each successive return to $i$, the Strong Markov property gives another trial with conditional success probability at least $\alpha$. The chance of avoiding $j$ through $N$ such trials is at most $(1-\alpha {)}^N$, which tends to zero.

Apply the absorbing modification at the singleton $\{j\}$. Part (c) holds for arbitrary $g$ satisfying (a), soSince $g_j\ge 0$ and the probability tends to one, $g_i\ge g_j$. Interchanging $i,j$ yields equality. Thus every nonnegative superharmonic function is constant on an irreducible recurrent Markov chain.

Conversely, if an irreducible Markov chain is not recurrent, choose a state $j$ with return probability $r_j={\mathbb{P}}_j({\tau }_j^{+}<\infty )<1$. Its hitting probability $h$ for $\{j\}$ satisfies condition (a), and $h_j=1$. The first-step formula for a return after time zero gives $(Ph{)}_j=r_j<1$. Therefore $h$ cannot be constant: a constant function with value one at $j$ would have $(Ph{)}_j=1$. This supplies a nonconstant nonnegative superharmonic solution. HenceThe admissible constant solutions are precisely the nonnegative constants. This is recurrence characterized by constant nonnegative superharmonic functions.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

For a complex matrix, its Hermitian conjugate is obtained by transposing and taking entrywise complex conjugates:The defining conditions areFor a square matrix, either one of the two inverse identities in the unitary matrix condition implies the other. A Hermitian matrix may have complex entries, but its diagonal entries are real.

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A Hausdorff axiom is not built into this definition. For neighbourhood bases ${\mathcal{B}}_x$ at each point, the precise neighbourhood criterion for a topological group isNecessity follows from continuity: the inverse image of a neighbourhood of $xy$ contains a product neighbourhood of $(x,y)$, and the inverse image of a neighbourhood of $x^{-1}$ contains a neighbourhood of $x$. Conversely, these inclusions give exactly those neighbourhood conditions for continuity of multiplication and inversion, since the rectangles $V\times W$ form a basis for the product topology.

An equivalent identity-level formulation is useful. Bases must be obtainable by translation, ${\mathcal{B}}_x=x{\mathcal{B}}_e$, and for every $U\in {\mathcal{B}}_e$ one must haveThese conditions are necessary because translations are homeomorphisms, multiplication is continuous at $(e,e)$, inversion is continuous at $e$, and each conjugation is a homeomorphism. For sufficiency at $(x,y)$, first choose $D$ with $D^2\subseteq U$, then $V$ with $y^{-1}Vy\subseteq D$ and take $W=D$. This gives $(xV)(yW)\subseteq xyU$. For inversion near $x$, first choose $H$ with $xH{x}^{-1}\subseteq U$, then $V$ with $V^{-1}\subseteq H$; thus $(xV{)}^{-1}=x^{-1}(x{V}^{-1}{x}^{-1})\subseteq x^{-1}U$. The pointwise criterion is therefore satisfied everywhere. If Hausdorffness is also required, the additional identity-neighbourhood condition is ${\bigcap }_{U\in {\mathcal{B}}_e}U=\{e\}$: it gives separation of distinct points by translating sufficiently small neighbourhoods.

The method of multiple scales applies when a small perturbation acts cumulatively over a long time, as in weakly perturbed harmonic oscillators or slowly modulated waves. A regular asymptotic expansion can contain secular terms, such as $ϵt\sin (\omega t)$, which cease to be small on the time scale of interest. Introduce independent fast and slow times, for example $t$ and $T=ϵt$, and allow the leading amplitude and phase to depend on $T$. The solvability condition in the method of multiple scales removes resonant forcing at each order and determines that slow evolution. For the Mathieu equation, these amplitude equations reveal parametric resonance and its instability intervals.

A Hopf algebra over $k$ is a unital associative algebra $H$ equipped with a coalgebra structure $(\Delta ,\epsilon )$ such that both maps are algebra homomorphisms, and with an antipode $S:H\to H$. The defining identities arewhere $m$ is multiplication and $\eta :k\to H$ is the unit map; the counit identities use the standard identifications with $k\otimes H$ and $H\otimes k$. The first compatibility conditions define a bialgebra; the antipode provides its convolution inverse. Multiplication commutativity and coproduct cocommutativity are independent conditions, as the two examples below demonstrate.

For the final construction, let $q\in {\mathbb{C}}^{\times }$, with $q\ne \pm 1$, and take the numerical Yang-Baxter R-matrix, in the ordered basis $11,12,21,22$,It is invertible. Its four action rules are $11\mapsto q11$, $12\mapsto 12+(q-q^{-1})21$, $21\mapsto 21$, $22\mapsto q22$. Applying these rules on the eight basis vectors of $V^{\otimes 3}$ verifies $R_{12}{R}_{13}{R}_{23}=R_{23}{R}_{13}{R}_{12}$, the Yang-Baxter equation.

In the RTT construction, introduce $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and impose $R{T}_1{T}_2=T_2{T}_{1}R$, where $T_1=T\otimes I$, $T_2=I\otimes T$. Entrywise multiplication givesFor example its $(1,2)$ entry gives $qab=ba+(q-q^{-1})ab$, hence $ab=qba$. DefineThese maps descend to the RTT algebra. Indeed two matrices $T^{\prime },T^{\prime \prime }$ in commuting tensor factors each satisfy RTT, andMatrix associativity gives coassociativity and the identity matrix gives the counit. Thus RTT first gives a bialgebra; the R-matrix alone does not automatically provide an antipode.

Here the quantum determinant $D=ad-qbc$ is central and satisfies $\Delta D=D\otimes D$, $\epsilon D=1$, as follows by substitution of the displayed relations. Impose $D=1$. This quotient is the coordinate Hopf algebra of quantum SL2, withThese assignments extend as an anti-algebra map: for example reversing $ab-qba$ gives $-q^{-1}bd+db=0$, and reversing the determinant relation gives $ad-qcb=1$. The remaining relations are checked in the same way. Direct multiplication gives $TS(T)=S(T)T=I$: the first diagonal entry of $TS(T)$ is $ad-qbc=1$, its second is $da-q^{-1}cb=1$, and the off-diagonal entries vanish by the $q$-commutation relations. These are precisely the antipode identities on generators, proving that the quotient is a Hopf algebra.

To verify both required failures of symmetry, map this algebra onto the quantum Laurent plane with generators $u,u^{-1},v$ and relation $uv=qvu$, by $a\mapsto u$, $b\mapsto v$, $c\mapsto 0$, $d\mapsto u^{-1}$. Construct this algebra directly on the vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. The exponent identity $-jk-(j+\ell )m=-\ell m-j(k+m)$ makes multiplication associative, and the basis products give $uv=qvu$ with $u$ invertible. Since $uv\ne vu$ for $q\ne 1$, the Hopf algebra is noncommutative. Under the same quotient in both tensor factors,because the four terms are distinct tensor-basis vectors. Hence it is also noncocommutative. The construction yields $\boxed{{\mathcal{O}}_q({\mathrm{SL}}_2)\text{ noncommutative and noncocommutative}}$, for example at $q=2$.

For a real continuous function on $[-1,1]$, the Chebyshev alternation theorem gives a unique best uniform approximation $p_n^{\ast }$ from the space of polynomials of degree at most $n$. If $f$ is not already in that space, a candidate $p$ is best if and only if there exist ordered points $x_0<\cdots <x_{n+1}$ and a sign $\sigma \in \{1,-1\}$ withThus a nonzero best error has at least $n+2$ alternating extrema of equal magnitude. If $f$ is already a polynomial of degree at most $n$, the unique best approximant is $f$ itself with zero error.

Existence does not require assuming that the infimum is attained: a minimizing sequence has uniformly bounded polynomial norms, and evaluation at $n+1$ fixed distinct nodes bounds its coefficients through an invertible interpolation matrix. A convergent coefficient subsequence then attains the infimum. The alternation criterion characterizes that minimum and gives uniqueness. In the following parts, $n\ge 1$ whenever $E_{n-1}$ occurs.