At $B=0$,with eigenvalues ${\lambda }_{+}=a+b=2\cosh (\beta J)$ and ${\lambda }_{-}=a-b=2\sinh (\beta J)$. Diagonalizing givesSubstitution in part (i), together with $Z={\lambda }_{+}^N+{\lambda }_{-}^N$, givesSince ${\lambda }_{-}/{\lambda }_{+}=\tanh (\beta J)$, the required One-dimensional Ising correlation function is

Insert $s_1{s}_i$ into the spin-basis expansion of the trace. Summing the spins along the first path from site $1$ to site $i$ produces $(M^{i-1}{)}_{st}$, while summing along the complementary path around the periodic chain produces $(M^{N-i+1}{)}_{ts}$. Therefore the two-point correlation function is

Assuming $\mu$ is $T$-invariant, $T$ is an ergodic transformation when every measurable set invariant modulo null sets is trivial:Equivalently, the invariant sigma-algebra is trivial modulo null sets.

A probability measure $\mu$ is $T$-invariant whenEquivalently, the pushforward measure satisfies $T_{\ast }\mu =\mu$, so $T$ is a measure-preserving transformation.

For an example satisfying $(\ast )$, consider the incomplete surface z equals r to three halvesA meridian of a surface of revolution reaches the missing origin in the finite lengthso $S_{\ast }$ is geodesically incomplete. The curvatures of a parametrized surface of revolution giveBy constant speed of an affinely parametrized geodesic, a geodesic with a finite endpoint has finite length and converges in ${\mathbb{R}}^3$. The finite-dimensional continuation criterion rules out an endpoint in a compact subset of $S_{\ast }$, and infinity is infinitely far away. It must approach the origin, so $(\ast )$ holds.

The answer to the second question is no. The punctured circular coneis incomplete because a generator reaches the missing vertex in finite time. It is inextendible: a proper connected smooth extension would have to add the origin, but the tangent planes approaching the origin depend on $\theta$ and therefore cannot be the continuous tangent planes of a smooth surface. Yetby the Gaussian curvature of a cone away from its vertex. Thus this incomplete inextendible surface does not satisfy $(\ast )$.

Suppose that a connected smooth surface $\tilde{S}\subset {\mathbb{R}}^3$ properly contains $S$. Since $S$ and $\tilde{S}$ have the same dimension, $S$ is open in $\tilde{S}$. Choose a point of its relative boundary. Inside a sufficiently small compact normal neighbourhood in $\tilde{S}$, choose $q\in S$ and a closest point $p\in \tilde{S}\setminus S$. A minimizing $\tilde{S}$-geodesic from $q$ to $p$ stays in $S$ before its endpoint, since encountering the complement earlier would contradict the choice of $p$.

The restriction of this curve is consequently a geodesic of $S$ with a finite maximal endpoint at $p$. On the other hand, the Gaussian curvature of $S$ is the restriction of the smooth Gaussian curvature of $\tilde{S}$, so it remains bounded near $p$. This contradicts condition $(\ast )$. Henceso $S$ is an inextendible embedded surface. This is the curvature-blowup criterion for inextendibility of a surface.

The weights of $V_2$ are $z^2,1,z^{-2}$. Degree-three monomials in these weights show that the character of the symmetric power $S^3{V}_2$ has weightsCharacters determine finite-dimensional $SU(2)$ representations, hence

Repeatedly using $V_r\otimes V_1\cong V_{r+1}\oplus V_{r-1}$ shows that the multiplicity of $V_{n-2j}$ in the nth tensor power satisfies Pascal's recursion. Consequentlywhere $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$. The multiplicity is the difference between the numbers of weight vectors of weights $n-2j$ and $n-2j+2$.

Applying the Clebsch-Gordan decomposition for SU2 givesandOn the summand $V_{6-2j}$, interchange of the two tensor factors acts by $(-1{)}^j$; this is the flip parity in the SU2 tensor square. The even-$j$ summands make the symmetric square, so

Above $T_B$, essentially all $N$ particles occupy the excited levels, soBelow $T_B$, the ground-state condensate carries no energy and $\mu \simeq 0$, henceDifferentiating gives the heat capacityThe displayed formula in the examination paper omits the factor $e^{\beta ϵ}$. Its stated expression follows after the additional high-temperature approximation $\beta ϵ\ll 1$, for which $e^{\beta ϵ}\simeq 1$.

For $T>T_B$, the ground-state occupation is negligible to leading order, so $N\simeq N_{ϵ}$. Comparing the number equation with its value at $T_B$ gives

For $T<T_B$, $\mu$ is close to zero andUsing $M=N(e^{ϵ/(k{T}_B)}-1)$ in $N_0=N-N_{ϵ}$ and solving for $\mu$ gives

At the critical temperature $T_B$, the excited states are just saturated and the condensate is about to form, soTherefore

At fixed temperature, the excited population is largest as $\mu \uparrow 0$, henceIf $N>N_{ϵ}^{\max }$, the excited levels cannot accommodate all particles. The excess $N-N_{ϵ}^{\max }$ occupies the zero-energy level macroscopically: this is Bose-Einstein condensation.

The Bose-Einstein distribution gives one occupation for the nondegenerate ground level and $M$ identical contributions from the excited levels:

Assume (iii). Fix $t\in \{1,\dots ,T\}$, a coordinate $i$, and an event $A\in {\mathcal{F}}_{t-1}$. Choose the bounded predictable risky strategyand choose the numéraire holding to make the strategy self-financing with $V_0=0$. Its terminal discounted value isCondition (iii) yieldsfor every $A\in {\mathcal{F}}_{t-1}$. The defining property of conditional expectation impliesEvery coordinate is therefore a $\mathbb{Q}$-martingale, so (iii) $\Rightarrow$ (i). This proves the martingale characterization by bounded self-financing strategies.

If $V$ is a $\mathbb{Q}$-martingale, thenbecause ${\mathcal{F}}_0$ is $\mathbb{P}$-trivial and equivalence makes it $\mathbb{Q}$-trivial as well. Thus (ii) $\Rightarrow$ (iii).

Condition (i) says that each discounted risky price $X^i$ is an integrable $\mathbb{Q}$-martingale. For a bounded predictable risky holding process $\theta$, self-financing gives the discounted wealth equation in discrete timeEach increment is integrable andHence $V$ is a $\mathbb{Q}$-martingale, proving (i) $\Rightarrow$ (ii).

Regard the set-theoretic inclusionas a linear map between Banach spaces. Its graph is closed. Indeed, if $f_n\to f$ in $L^q$ and $f_n\to g$ in $L^p$, part (i) gives an almost-everywhere convergent subsequence for the first convergence and then a further subsequence converging almost everywhere for the second; hence $f=g$ almost everywhere. The closed graph theorem therefore gives

Let $E_R=X\cap B(0,R)$. It has finite measure, and $\mu (E_R)\uparrow \mu (X)$. Applying the bound to $1_{E_R}$ gives, for finite $q$,so $\mu (E_R{)}^{1/p-1/q}\le C$. For $q=\infty$, it gives $\mu (E_R{)}^{1/p}\le C$. In either case the measures $\mu (E_R)$ are uniformly bounded. Continuity from below now yieldsThis proves the converse summarized by Lq inclusion implies finite measure on a Euclidean Borel subset.

First suppose $1\le p<\infty$. Choose a subsequence so thatBy Tonelli theorem,Hence the sum inside the integral is finite almost everywhere, which forces $|f_{n_k}(x)-f(x)|\to 0$ almost everywhere.

If $p=\infty$, choose representatives and remove the countable union of the null sets on whichOutside that null set, norm convergence bounds the pointwise difference and the entire sequence converges to $f$. Thus in every case an almost-everywhere convergent subsequence exists, as stated in Almost-everywhere convergent subsequence from Lp convergence.

For either ${\alpha }_j$, the same permutation-of-factors argument shows that every coefficient of $f(t,{\alpha }_j)$ lies in $L^G$.

They need not generate the whole fixed field. Assume $\mathrm{char}K\ne 2$, choose $a_1,a_2\in K$ with independent nonsquare classes, and putLet $G=\{1,\sigma \}$, whereThenso all their coefficients generate only $K$. However $\sqrt{a_1{a}_2}$ is fixed by $\sigma$, and the degree count in part (a) showsThus separate orbit polynomials can miss mixed invariants.