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After leaving zero, the jump chain avoids every future reset with probabilityThis probability is positive exactly when ${\prod }_{i\ge 1}(1+{\rho }_i)<\infty$. If $q>0$, the probability of ever returning to zero is less than one, so the irreducible chain is transient. If $q=0$, a reset to zero occurs almost surely after each departure; by the Strong Markov property this happens repeatedly, so the chain is recurrent. Therefore

A reversible Markov chain must satisfy detailed balance. For every $i\ge 1$, the rate from $i$ to $i+1$ is positive, while the reverse rate from $i+1$ to $i$ is zero. No positive measure can satisfy ${\pi }_i{g}_{i,i+1}={\pi }_{i+1}{g}_{i+1,i}$, so the irreducible chain is not reversible.

The total jump rate is ${\lambda }_0$ at zero and ${\lambda }_i(1+{\rho }_i)$ at $i>0$. Hence the jump chain has transition probabilitiesand, for $i>0$,All other entries vanish.

Take independent random variables $U_1,\dots ,U_d$, each uniform on $[0,1/2]$, and independent Rademacher random variables ${\epsilon }_1,\dots ,{\epsilon }_{d-1}$. Defineand let $\eta$ be the joint law of $(X_1,\dots ,X_d)$.

Each $X_i$ is uniform on $[-1/2,1/2]$. If the $d$th coordinate is omitted, the remaining signs are the independent ${\epsilon }_i$. If coordinate $j<d$ is omitted, the remaining signs arewhich are again uniformly distributed over all $2^{d-1}$ sign vectors because this map from $({\epsilon }_1,\dots ,{\epsilon }_{d-1})$ is a bijection. The independent magnitudes then show that every $(d-1)$-coordinate marginal is Lebesgue measure.

The full law is not $d$-dimensional Lebesgue measure, becausealmost surely, whereas this product is equally likely to have either sign under independent Lebesgue coordinates.

On the probability space $(X^2,\eta )$, put $A=I\times X$ and $B=X\times J$. The marginal conditions give $\eta (A)=x$ and $\eta (B)=y$, while $A\cap B=I\times J$. The elementary Fréchet inequalities yieldHence

Let $U$ be uniformly distributed on $X$ and let $\eta$ be the law of $(U,U)$. Both marginals are $\mu$, becauseHowever, $\eta$ is concentrated on the diagonal $\{(x,x):x\in X\}$, which has two-dimensional Lebesgue measure zero. Thus $\eta$ is not Lebesgue measure on $X^2$.

Saying that $U$ has law $\mu$ means that $\mu$ is the pushforward measureFor a real random variable, $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

The pair $U,V$ consists of independent random variables whenfor every pair of Borel sets $A,B$; equivalently, their joint law is the product of their marginal laws.

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.

The fundamental theorem of regular space curves states that smooth functions $k>0$ and $\tau$ on an interval determine a smooth arc-length-parametrized curve in ${\mathbb{R}}^3$, unique up to a proper Euclidean motion. Consequently two such curves $\alpha$ and $\tilde{\alpha }$ are related parameterwise by a proper Euclidean motion if and only iffor every $s$.

If $\tau =0$ everywhere, the third of the Frenet-Serret formulas gives $B^{\prime }=0$, so $B$ is a constant vector. MoreoverThus $\alpha (s)\cdot B$ is constant, and the image of $\alpha$ lies in the plane perpendicular to $B$.

Put $T={\alpha }^{\prime }(s)$. Since $s$ is arc length, $|T|=1$, so $T^{\prime }\perp T$. The curvature of a space curve isWhere $k(s)\ne 0$, define the principal normal and binormal byand define the torsion of a space curve byThus torsion requires nonzero curvature, so that $N$ and $B$ are defined.

Differentiating the orthonormality relations for the frame $(T,N,B)$ shows that its derivative matrix is skew-symmetric. Since $T^{\prime }=kN$, differentiating $B=T\times N$ and using the definition of $\tau$ gives the Frenet-Serret formulas

Let $P(x)={\prod }_{i=1}^r(x-{\lambda }_i)$. Since both $x$ and $y$ are invertible on $X$, the formuladefines a regular map $F:X\to {\mathbb{A}}^1$.

Fix $t\in k$. To solve $F(x,y)=t$, setThe curve equation becomesThis is a nonconstant polynomial, so it has a root because $k$ is algebraically closed. At a root, neither $x$ nor $t-x$ can vanish, and the equation also forces $P(x)\ne 0$. The displayed value of $y$ is therefore defined and nonzero, satisfies $y^2=P(x)$, and obeys $F(x,y)=t$. Thus $F$ is surjective.

A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted Leibniz ruleThe Zariski tangent space isSince $p(f)\ne 0$, $p$ extends uniquely to the localization $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f{)}^n$. Restriction gives a linear mapConversely, every $D\in T_{p}A$ has the unique extensionThe quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.

View infinity as the puncture of the Riemann sphere and consider $g(1/z)$ near $z=0$. If this map had an essential singularity, the Great Picard theorem in a coordinate chart would make almost every nearby value occur infinitely often, contradicting the injectivity of $g$. The singularity is therefore removable as a map into the compact Riemann surface $R$, and $g$ extends to a nonconstant holomorphic mapThe extension is injective. A collision between $\tilde{g}(\infty )$ and $g(z_0)$, or local degree greater than one at infinity, would give nearby values two distinct preimages by the local normal form of a nonconstant holomorphic map, again contradicting injectivity on $\mathbb{C}$. Thus $\tilde{g}$ is a holomorphic bijection between compact Riemann surfaces and hence a biholomorphism. Therefore $R\cong {\mathbb{P}}^1$, whose genus is zero.

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence

For $|z|<1$, sum the geometric series and its derivative:ThereforeThis rational continuation is holomorphic across every point of the unit circle except $z=-1$, where it has a double pole. The set of singular points is consequently

Here “covering map” uses the course convention of a local homeomorphism; “regular” adds the evenly-covered product condition. The derivative of $p(z)=z^4$ is $4z^3$, which never vanishes on the upper half-plane $\mathbb{H}$. The holomorphic inverse function theorem therefore makes $p$ a local biholomorphism and hence a covering map in this sense.

It is not regular. The point $1$ has only the preimage $i$ in $\mathbb{H}$, whereas $e^{i\epsilon }$ for small positive $\epsilon$ has the two preimages with arguments $\epsilon /4$ and $\pi /2+\epsilon /4$. Thus the fibre cardinality is not locally constant near $1$, so no neighbourhood of $1$ has a preimage of the form $U\times \Delta$ for a fixed discrete set $\Delta$.