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Write $q_{ij}=c{m}_{ij}$ for $i\ne j$, with a common positive constant $c$, and put $d_i={\sum }_{j\ne i}{m}_{ij}$. The infinitesimal generator matrix is symmetric because the graph is undirected. Thus the uniform distribution ${\pi }_i=1/|I|$ satisfies detailed balance, and the continuous-time random walk is reversible when started uniformly.

Its jump chain chooses a link uniformly from those incident to its current vertex:The flux ${\hat{\pi }}_i{\hat{P}}_{ij}=m_{ij}/{\sum }_j{d}_j$ is symmetric, so this random walk on a multigraph is also reversible. The continuous-time equilibrium is uniform because higher-degree vertices have shorter holding times; at jump times the equilibrium is degree-weighted. If the intended proportionality constant depends on the starting vertex, $q_{ij}=c_i{m}_{ij}$, reversibility still holds with ${\pi }_i\propto 1/c_i$, and the same jump chain results.

Use the equilibrium correspondence from the preceding part. For $i\ne j$,Consequently the detailed balance equations for the jump chain hold exactly when those for the continuous-time Markov chain hold. The reversibility characterizations already given prove the equivalence. Both chains here are started in their respective stationary distributions, which generally differ.

The jump chain records the successive states visited at actual jumps of the continuous-time Markov chain. Put $q_i=-q_{ii}$. For an irreducible chain with at least two states, all $q_i>0$, and its transition probabilities are ${\hat{P}}_{ij}=q_{ij}/q_i$ for $j\ne i$, with ${\hat{P}}_{ii}=0$. The holding time in state $i$ is exponential with mean $1/q_i$.

If $\pi Q=0$, set $C={\sum }_i{\pi }_i{q}_i$ and ${\hat{\pi }}_i={\pi }_i{q}_i/C$. ThenConversely, $\hat{\pi }\hat{P}=\hat{\pi }$ implies that ${\pi }_i=c{\hat{\pi }}_i/q_i$, normalized to sum one, satisfies $\pi Q=0$. ThereforeJump observations overrepresent states with short holding times. A one-state chain has no actual jumps; its trivial equilibrium can be handled separately, without using this division by $q_i=0$.

A Discrete-time Markov chain with initial distribution $\hat{\pi }$ is reversible if $(Y_0,\dots ,Y_n)$ and $(Y_n,\dots ,Y_0)$ have the same law for every $n$. Its detailed balance equations are ${\hat{\pi }}_i{\hat{P}}_{ij}={\hat{\pi }}_j{\hat{P}}_{ji}$. These equations are equivalent to reversibility. Indeed, reversing every factor in the probability ${\hat{\pi }}_{i_0}{\prod }_{r=1}^n{\hat{P}}_{i_{r-1}{i}_r}$ gives the reverse path probability; the converse follows by taking $n=1$.

Summing detailed balance over $i$ givesHence $\hat{\pi }$ is an stationary distribution.

Let $\pi$ be a probability vector satisfying detailed balance. Since each row of the infinitesimal generator matrix sums to zero,Thus $\pi Q=0$, and $\pi e^{tQ}=\pi$ for all $t\ge 0$. This proves that $\pi$ is an stationary distribution. The diagonal term causes no problem because its detailed balance equation is an identity.

A continuous-time Markov chain started with distribution $\pi$ is reversible if its finite-dimensional distributions are unchanged by reversing time: for every $T>0$, $(X_t:0\le t\le T)$ and $(X_{T-t}:0\le t\le T)$ have the same law. In particular, reversibility includes stationarity. The detailed balance equations areFor a finite continuous-time Markov chain, these equations are equivalent to reversibility with initial distribution $\pi$. Equivalently, they imply ${\pi }_i{P}_{ij}(t)={\pi }_j{P}_{ji}(t)$ for every transition matrix $P(t)=e^{tQ}$.

For $n=1$, the singular set is the point $\{0\}$, a zero-dimensional submanifold. For $n\ge 2$, every elementary matrix $E_{ij}$ is singular and the smooth curve $t\mapsto t{E}_{ij}$ lies in $X$ through zero. If $X$ were a submanifold, all $E_{ij}$ would belong to $T_{0}X$ and span $M(n)$. Its dimension at zero would therefore be $n^2$, making $X$ locally open there. But arbitrarily small matrices $tI$ are invertible, a contradiction. ThusThe failure is at the zero matrix; fixed-rank pieces of the singular set can separately be smooth submanifolds.

For an invertible $A$, factor $\det (A+tB)=\det A\det (I+t{A}^{-1}B)$. The determinant's permutation expansion gives $\det (I+tC)=1+t\mathrm{tr}C+O(t^2)$. Differentiating at zero yieldsAt a determinant-one matrix choose $B=A/n$; the derivative is one. Thus one is a regular value of the smooth determinant map on $GL(n)$. The regular level set theorem makes $SL(n)$ a submanifold of codimension of a submanifold one, withThis includes $n=1$, when $SL(1)=\{1\}$ has dimension zero.

The determinant is continuous and $GL(n)={\det }^{-1}(\mathbb{R}\setminus \{0\})$ is open in $M(n)\simeq {\mathbb{R}}^{n^2}$. Every open subset inherits the ambient smooth charts, so it is a submanifold and

A geometric circle is an unknot. Compactify ${\mathbb{R}}^3$ to $S^3$, and carry the circle to $\{(z_1,z_2)\in {\mathbb{C}}^2:|z_1{|}^2+|z_2{|}^2=1,z_2=0\}$. Its complement has coordinates $z_1\in \Delta$ and $z_2=\sqrt{1-|z_1{|}^2}{e}^{i\theta }$, making it homeomorphic to $\Delta \times S^1$. It deformation retracts to $S^1$. Removing the additional point at infinity does not change the fundamental group: van Kampen for a small ball about that point uses a simply connected punctured ball, homotopy equivalent to $S^2$. ThusA meridian linking the circle once generates it. This uses “circle” in its geometric sense; the complement of an arbitrary knotted embedded copy of $S^1$ need not have this group.

Choose open neighborhoods of the sphere and circle in the wedge, each enlarged by a small contractible neighborhood of the attaching point in the other piece. They deformation retraction to $S^2$ and $S^1$ and have contractible intersection. Van Kampen therefore givesThe generator traverses the circle once; the sphere contributes no fundamental-group generator.

The van Kampen theorem says: if $X=U\cup V$ with $U,V$ open and path connected, $U\cap V$ path connected, and a base point in the intersection, then ${\pi }_1(X)$ is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of every element of ${\pi }_1(U\cap V)$. Equivalently it is the pushout of the two inclusion-induced fundamental-group maps.

The element $\omega$ satisfies ${\omega }^2-\omega +6=0$, so is an algebraic integer. Conversely, write an algebraic integer as $(a+b\sqrt{-23})/2$ with $a,b$ rational. Its trace gives $a\in \mathbb{Z}$, and its field norm implies $23b^2\in \mathbb{Z}$. If $b=p/q$ in lowest terms, $q^2$ divides 23, forcing $q=1$. The integral field norm condition modulo four then makes $a,b$ have the same parity. Therefore it equals $(a-b)/2+b\omega$, provingBoth reductions of $x^2-x+6$ modulo two and three split into the distinct roots zero and one. The index is one, so ideal factorization givesTheir norms are two and three respectively. The product ${\mathfrak{p}}_2{\mathfrak{p}}_3$ is generated by $6,2\omega ,3\omega ,{\omega }^2$. All are multiples of $\omega$, using $6=\omega \overline{\omega }$; and $\omega =3\omega -2\omega$ belongs to the product. HenceThe Minkowski bound is $(2/\pi )\sqrt{23}<4$, in fact about $3.05$. Every class therefore has a representative of ideal norm one, two or three. The displayed primes and their conjugates exhaust those possibilities. Their classes are all $1,c,c^{-1}$, with $c=[{\mathfrak{p}}_2]$, because $[{\mathfrak{p}}_3]=c^{-1}$ and conjugation inverts a quadratic ideal class. There are at most three classes.

There is no element of field norm two: $N(a+b\omega )=a^2+ab+6b^2=(a+b/2{)}^2+23b^2/4$, which exceeds two for $b\ne 0$ and is a square for $b=0$. Thus ${\mathfrak{p}}_2$ is not principal. Moreover $N(\omega -2)=8$, and $\omega -2$ belongs to ${\mathfrak{p}}_2$ but not its conjugate, so $(\omega -2)={\mathfrak{p}}_2^3$. Consequently $c$ has order three, and

The ideal class group of a number field is the group of nonzero fractional ideals of its ring of integers under multiplication, modulo the subgroup of principal fractional ideals. For degree $n$ and $r_2$ pairs of complex embeddings, the Minkowski bound for ideal classes states that each class contains an integral ideal $I$ withThere are only finitely many integral ideals of bounded ideal norm: as sublattices of the free abelian group of integers, they have bounded index, and finitely many subgroups have any fixed index. Thus the ideal class group is finite.

The splitting field over ${\mathbb{F}}_2$ is ${\mathbb{F}}_{16}$, and it contains the unique subfield ${\mathbb{F}}_4$. Over ${\mathbb{F}}_4$, Frobenius is the square of the order-four ${\mathbb{F}}_2$ Frobenius, so it has two orbits of length two on the four roots. Thus $f$ factors into two irreducible quadratics, andgenerated by $x\mapsto x^4$ on ${\mathbb{F}}_{16}$.

Over a finite field ${\mathbb{F}}_q$, the splitting field is a finite extension ${\mathbb{F}}_{q^m}$, where $m$ is the least common multiple of the degrees of the distinct irreducible factors. Its Galois group is cyclic of order $m$, generated by $x\mapsto x^q$.

Over ${\mathbb{F}}_2$, the polynomial has no linear factor: its values at zero and one are both one. The only irreducible quadratic is $x^2+x+1$; reduction modulo it gives $x$, not zero. A reducible quartic without linear factors would split into quadratics, so $f$ is irreducible. Thus its splitting field is ${\mathbb{F}}_{16}$ and

True for finite connected simple graphs. average degree of a graph at most two implies $e\le v$. A spanning tree has $v-1$ edges, leaving at most one extra edge. Hence the graph is a tree or unicyclic. trees are planar, and a unicyclic graph is a cycle with trees attached to its vertices, which can also be embedded in the plane. Connectedness is essential to this argument: a disconnected nonplanar component can be diluted by isolated vertices.

False. Attach 594 pendant vertices to one vertex of $K_{3,3}$. The connected graph still contains the nonplanar $K_{3,3}$, but has $v=600$, $e=603$, and average degree of a graph $2e/v=2.01$. Attaching more leaves lowers the average further toward two without making the graph planar.

False. The complete bipartite graph $K_{3,3}$ has chromatic number two, but is nonplanar. For example a simple bipartite plane graph has $e\le 2v-4$, because all facial cycles have length at least four; $K_{3,3}$ has $v=6$, $e=9>8$.

Separate fluid conservation gives ${\dot{\rho }}_M=-3H{\rho }_M$, hence ${\rho }_M={\rho }_{M0}{a}^{-3}$ after normalization at today. For $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$, the conservation equation gives ${\dot{\rho }}_{\Lambda }=0$. ThusAssume both densities are positive and set $A=8\pi G{\rho }_{M0}/3$, $B=8\pi G{\rho }_{\Lambda 0}/3$. On the expanding flat branch, ${\dot{a}}^2=A/a+B{a}^2$. With $b=a^{3/2}$ this becomes $\dot{b}=(3/2)\sqrt{A+B{b}^2}$. Choosing the Big Bang at $t=0$ and integrating givesThe normalization $a(t_0)=1$ requires $\sinh (\beta t_0)=\sqrt{{\rho }_{\Lambda 0}/{\rho }_{M0}}$; equivalently $\alpha ={\sinh }^{-2/3}(\beta t_0)$. Thus the age $t_0$ and the two positive densities are not independent once this normalization and bang time are fixed.

As $t\downarrow 0$, $a\sim \alpha (\beta t{)}^{2/3}$, the matter-dominated law. As $t\to \infty$, $a\sim \alpha 2^{-2/3}{e}^{\sqrt{B}t}$ and $\ddot{a}/a\to B>0$, giving accelerated dark-energy expansion. Direct differentiation gives $\ddot{a}/a=B-A/(2a^3)$, so acceleration begins once $a^3>A/(2B)$. If dark energy vanishes, the limiting matter-only solution must be taken instead; negative dark-energy density would not have the stated late-time accelerating form.