At $t=t_c$, the piston velocity reachesThe simple-wave relation then gives $c=0$ and $\rho =0$. For later times the formal continuation would require negative sound speed, so the gas can no longer follow the piston. By vacuum formation behind a receding piston, a vacuum region opens between $X_p(t)$ and the gas edgeThe earlier rarefaction solution remains to the right of $x_v(t)$, while $X_p(t)<x<x_v(t)$ is vacuum.

The undisturbed gas has $u=0$, $c=c_0$, and hence $R_{-}=0$. The disturbance from the receding piston is a right-moving simple wave, so this invariant remains zero throughout it:

Let $s$ be the time at which the relevant $C_{+}$ characteristic leaves the piston. The boundary condition is $u={\dot{X}}_p(s)$ there. Both $u$ and $c$ are constant along this characteristic, whose speed isHence $s=s(x,t)$ is determined implicitly by the receding-piston rarefaction wave equationWithin $X_p(t)<x<c_{0}t$, the solution is thereforeFor $x\ge c_{0}t$, the gas remains undisturbed: $u=0$ and $\rho ={\rho }_0$. The speed bound keeps $c$ and $\rho$ positive, and monotonic increase of $|{\dot{X}}_p|$ makes this an expanding rarefaction wave rather than a characteristic-crossing compression.

Assume (iii). Fix $n\ge 0$ and $A\in {\mathcal{F}}_n$, and defineThis is a stopping time: $\{T\le n\}=A\in {\mathcal{F}}_n$. Applying (iii) at time $n+1$ first to $T$ and then to the deterministic stopping time $n+1$ givesSubtracting yieldsBy the defining characterization of conditional expectation, this says $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus $M$ is a martingale, proving (i). Together the three implications establish the characterization of a martingale by stopped expectations.

Assume (ii). A martingale has constant expected value by the tower property of conditional expectation, so for every stopping time $\tau$ and every $n$,This is (iii).

Assume (i), and let $\tau$ be a stopping time. The increments of the stopped process satisfyBecause $\{\tau >n\}\in {\mathcal{F}}_n$, taking conditional expectation given ${\mathcal{F}}_n$ makes the right-hand side zero. Integrability follows froma finite sum of integrable random variables. Hence $M^{\tau }$ is a martingale, proving (ii).

Because $J_m\to \infty$ almost surely, no infinite portion of the jump chain is hidden inside a finite time interval. After either chain leaves a state $i$, the event that it ever returns is therefore the same event for both chains. Hence irreducibility and recurrence pass between a CTMC and its jump chain gives

The jump chain of $(X_t)$ is exactly $(Y_m)$: from $i$ it visits $j$ next with probability $p_{ij}$. Consequently the two chains have exactly the same directed paths of positive-probability transitions. Thus

Combining the preceding parts,This is the characteristic function of $N(0,2\alpha I_d)$. The uniqueness theorem for characteristic functions therefore givesWhen $\alpha =0$, this means the degenerate distribution concentrated at the origin. This proves the Maxwell characterization of the normal distribution.

First, $f(0)=1$. For every $r\ge 0$,It cannot vanish: if $f(r_0)=0$ for some $r_0>0$, thenwould imply $f(r_0/n)=0$ for every positive integer $n$, contradicting continuity at zero. Thus $f(r)>0$. Since every characteristic function has modulus at most one,

Now set $h(r)=-\log f(r)$. It is continuous, nonnegative, and additive:The classification of a continuous additive function on the nonnegative real numbers gives $h(r)=\alpha r$, where $\alpha =h(1)\ge 0$. Consequently

Independence of the coordinates gives factorization of their joint characteristic function of a random vector. Orthogonal invariance under coordinate permutations makes all coordinate marginals identical. Hence, for $r_1,r_2\ge 0$ and $d\ge 2$,Therefore

For an orthogonal matrix $A$, invariance in distribution givesThe orthogonal group is transitive on each Euclidean sphere, so ${\phi }_X(u)$ depends only on $\Vert u{\Vert }^2$. DefineThe continuity of a characteristic function makes $f$ continuous. Taking $A=-I$ also gives $X\stackrel{d}{=}-X$, and henceso $f$ is real-valued. Thusas for every orthogonally invariant random vector.

The irreducible affine hypersurface $y^2-z^3=0$ in the $(y,z)$-plane has dimension one, while a closed point has dimension zero. Therefore

The curve $C$ has dimension one. Its tangent has dimension one at every point except the origin and dimension two at the origin. By the smooth locus of a variety,and the origin is its unique singular point of an algebraic variety. Each reduced point component has dimension zero and zero-dimensional tangent space, so both $P$ and $Q$ are smooth on their respective components.

Let $(a,b,c)$ be a tangent vector in the $x,y,z$ directions. At a point $(0,y,z)\in C$, the two isolated components are absent in a Zariski neighbourhood, so the Zariski tangent space is that of $C$. Differentiating $x$ and $y^2-z^3$ givesFor $(y,z)\ne (0,0)$ this is a line. At the cusp,which is two-dimensional.

Each of $P$ and $Q$ is a reduced isolated component with local ring $k$, soThis last conclusion must use all functions in $I(V)$, as required by the definition of the Zariski tangent space. The Jacobian of only the three displayed defining polynomials has an extra spurious tangent direction at $Q$, because those polynomials do not generate the radical ideal there; for example, $x(x^2-1)\in I(V)$ removes that direction.

WriteThe equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the cuspidal cubicIt is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.

If $y=0$, the other equations becomeTheir solutions are $(0,0,0)$, which already lies on $C$, and the two pointsThus the irreducible components areEquivalently, the radical ideal of the cusp with two isolated-point components is

When $n$ is odd there is no alternating sign mode. The same cosine-sine construction gives invariant irreducible planes $W_k$ foragain pairwise nonisomorphic because their generator traces are distinct. The real Fourier decomposition isIts dimensions sum to $1+2(n-1)/2=n$. Thus the real regular representation of a finite cyclic group has one one-dimensional irreducible summand and $(n-1)/2$ nonisomorphic two-dimensional irreducible summands.

Let $e_0,\dots ,e_{n-1}$ be the standard basis and let the generator $g$ send $e_j$ to $e_{j+1}$ modulo $n$. The constant vectorspans a trivial module. Since $n$ is even, the alternating vectorspans a sign module on which $g$ acts by $-1$.

For each $1\le k<n/2$, define the real discrete Fourier modesThe plane $W_k=\mathrm{span}\{c_k,s_k\}$ is invariant, and $g$ acts on it by rotation through $2\pi k/n$. Since this angle is neither zero nor $\pi$, $W_k$ is irreducible over $\mathbb{R}$. The traces $2\cos (2\pi k/n)$ are distinct for $1\le k<n/2$, so these modules are pairwise nonisomorphic. Orthogonality of the discrete Fourier transform givesThe dimensions sum to $1+1+2(n-2)/2=n$, so this is the whole regular module: two nonisomorphic one-dimensional summands and $(n-2)/2$ nonisomorphic two-dimensional summands.

The two irreducible real modules $U$ and $W$ are nonisomorphic, so the kernel-image argument from part (a), which is valid over any field, givesEvery endomorphism of the one-dimensional trivial module is scalar:

Identify the oriented plane $W$ with $\mathbb{C}$, so that $g$ acts by multiplication by $e^{2\pi i/3}$. A real-linear map commutes with this rotation exactly when it is multiplication by a complex number. Equivalently,where $J$ is rotation through $\pi /2$. Taking $J$ gives a nonscalar intertwining endomorphism of an irreducible real representation. This is the Failure of the scalar conclusion of Schur lemma over the real numbers.

Let $G=C_3=⟨g⟩$ and let $e_0,e_1,e_2$ be the standard basis of its real regular representation, with $g{e}_j=e_{j+1}$ modulo three. SetBoth are invariant, they are orthogonal, andThe line $U$ is the trivial representation. On the plane $W$, the generator acts as a rotation through $2\pi /3$. Such a rotation has no real eigenline, so $W$ has no nonzero proper invariant real subspace and is irreducible.

The maximum flow is not unique. For any $x\in [1,2]$, setand split the flow from $b$ byFlow conservation then requiresAll capacities are respected precisely throughout the displayed interval, and every member of this one-parameter family has value $8$. For example, $x=1$ and $x=2$ give distinct maximum flows.