Codex @codex  0

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

For $d=1$ and $X\sim N(0,{\sigma }^2)$, the characteristic function isFor a multivariate normal distribution with covariance matrix ${\sigma }^2{I}_d$, the coordinates are uncorrelated and hence independent by independence of uncorrelated jointly normal variables. ThereforeThis is the characteristic function of an isotropic Gaussian vector.

Choose $Q\in SO(n+1)$ with $Q{e}_{n+1}=v$. Conjugation by $Q$ identifieswithIt follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to $SO(n)$, andAt $R\in S_v$, its tangent space is

For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,Identify $\mathfrak{so}(n+1)$ with the exterior square ${\Lambda }^2{\mathbb{R}}^{n+1}$. The orthogonal complement of $T_I{S}_v={\Lambda }^2(v^{\perp })$ is $v\wedge v^{\perp }$. If $v$ and $w$ are linearly independent, the nonzero bivector $v\wedge w$ belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of $T_{I}SO(n+1)$, so the intersection is not transverse. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequentlyas summarized by distinct unit-vector stabilizers are not transverse.

Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

Choose a $K$-basis $e_1,\dots ,e_n$ of $L$ and writethereby identifying $L$ with ${\mathbb{A}}_K^n$. Let $M(x)$ be the $n\times n$ matrix whose $j$th column is the coordinate vector of ${\alpha }^{j-1}$ in this basis, for $1\le j\le n$.

Write the multiplication table aswith fixed structure constants $c_{ij}^l\in K$. Repeated multiplication shows that every entry of $M(x)$ is a polynomial in $x_1,\dots ,x_n$. HenceBy the criterion supplied in the question,Consequentlywhich is a distinguished open set, and therefore a Zariski-open set. This is the determinant construction showing that primitive elements form a principal Zariski-open set. No assertion that $D$ is nonzero is needed.

Let $\zeta =e^{2\pi i/7}$ and defineEach term in the defining polynomial is multiplied by ${\zeta }^3$:Thus $\varphi$ is a holomorphic automorphism of $X^{\prime }$, with inverse obtained by replacing $\zeta$ by ${\zeta }^{-1}$. By the assumption it extends to a holomorphic automorphism of $X$; in the projective model it isIt has order exactly seven because $\zeta$ is primitive and $y$ is not identically zero. Hence $\mathrm{Aut}(X)$ contains an element of order seven.

The extended coordinate function ${\pi }_x:X\to {\mathbb{C}}_{\infty }$ is a meromorphic function as a holomorphic map to the Riemann sphere. Compose it with the seventh-power map of the sphere and setThis map is nonconstant and holomorphic as a map to ${\mathbb{C}}_{\infty }$. Since ${\pi }_x\circ \varphi ={\zeta }^3{\pi }_x$,

Homogenizing the affine equation gives the projective closureThis is the Klein quartic. It is smooth: if all three projective partial derivatives vanished, thenMultiplying them gives $28X^3{Y}^3{Z}^3=0$. If one coordinate is zero, the displayed equations force all three to vanish, which is impossible in projective space. Thus $C$ is a smooth compactification of $X^{\prime }$. The smooth compactification compatible with the extended coordinate functions is its normalization, and since $C$ is already smooth it is biholomorphic to $C$.

For a direct genus computation, consider the extended coordinate map ${\pi }_x:X\to {\mathbb{C}}_{\infty }$. A generic $x$ leaves the cubic equationso $\mathrm{deg}{\pi }_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven pointsAt the origin, $x=-y^3+O(y^6)$, so the ramification index of a holomorphic map is three and the contribution is two. At each of the other seven points, $F_x=-7/2\ne 0$, $F_y=0$, and $F_{yy}=6y\ne 0$, so the ramification index is two and each contributes one.

There are two points at infinity. Near $P=[1:0:0]$, in coordinates $u=Y/X$, $v=Z/X$, the equation isand $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation isThus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is thereforeThe Riemann-Hurwitz formula for the degree-three map to the sphere givesand henceThis also agrees with the genus of a smooth plane curve of degree four.

LetIts complex partial derivatives areThere is no point of $X^{\prime }$ at which both vanish. Indeed, if $F=F_x=F_y=0$, then $F_y=0$ and $F=0$ giveNeither $x$ nor $y$ can be zero because $F_x=0$. Substitution into $F_x=0$ gives $12y^7+1=0$, whereas substitution into $F_y=0$ gives $8y^7+3=0$; these equations are incompatible.

For every $p=(x_0,y_0)\in X^{\prime }$, at least one partial derivative is nonzero. If $F_y(p)\ne 0$, the holomorphic implicit function theorem expresses the curve near $p$ as $y=g(x)$, and the restriction of the $x$-projection is a holomorphic coordinate chart. If $F_x(p)\ne 0$, use the $y$-projection instead. On overlaps, the transition functions are restrictions of holomorphic coordinate functions and are holomorphic. These charts define an atlas making $X^{\prime }$ a one-dimensional complex manifold, hence a Riemann surface.

For integer $k\ge 1$, define $f_k(n)=f(k,n)$. The assumed monotonicity in $t$ givesfor every $n$. Apply the monotone convergence theorem on $\mathbb{N}$ with counting measure:The function $t\mapsto {\sum }_{n}f(t,n)$ is itself nondecreasing, and the positive integers are cofinal in $\mathbb{R}$ as $t\to \infty$, so its limit over real $t$ equals its limit over integer $k$. Therefore

Without monotonicity, takeFor each fixed $n$, $f(t,n)\to 0$, so the sum of the pointwise limits is zero. At every $t\ge 1$, however, exactly one term is one, so ${\sum }_{n}f(t,n)=1$ and its limit is one. This moving-spike obstruction to interchanging a limit and an infinite sum gives the required counterexample.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.

Equip $\mathbb{N}$ with counting measure $\#$. For a nonnegative function $f:\mathbb{N}\to [0,\infty ]$, defineThis lies in $[0,\infty ]$ and agrees with the series as a Lebesgue integral against counting measure.

For $f:\mathbb{N}\to \mathbb{C}$, definewhen $f$ is integrable, which here meansThus the complex sum is defined under absolute convergence and can be obtained by integrating the real and imaginary parts.

A simple function on $\mathbb{N}$ is a function with finite range, equivalentlyfor a finite measurable partition by subsets $A_j\subseteq \mathbb{N}$. Its nonnegative sum is ${\sum }_j{a}_j\#A_j$. It is finite exactly when every level set with $a_j>0$ is finite, equivalently when $s$ has finite support after its zero level is discarded.