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For fixed cosmological density parameters, the redshift drift has the form $\dot{z}=H_{0}F(z)$. Multiplying the Hubble constant changes the vertical scale of all three sketches but leaves their zeros unchanged. Thus the drift-zero redshift is independent of $H_0$ at fixed density fractions. Every model has the trivial $z=0$ zero; only the specified mixed model has a further positive zero, $z_{\ast }\simeq 2.089$. If instead physical matter and vacuum densities are held fixed while $H_0$ changes, their density fractions change too, so that is a different family of models.

We first derive the recurrence needed for the partition of unity. Denote an order-$r$ partition-normalized spline by $N_{i,r}$. For $r\ge 2$, set $h_t(u)=(u-t{)}_{+}^{r-2}$, so $(u-t)h_t(u)=(u-t{)}_{+}^{r-1}$. The elementary product identities for divided differences areThey follow from the Leibniz rule for divided differences because the divided differences of the linear factor vanish above first order. Use the first identity on $t_i,\dots ,t_{i+r-1}$ and the second on $t_{i+1},\dots ,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ then givesThis is the Cox-de Boor recurrence, derived here from the specified normalization rather than assumed as a partition theorem.

Extend the distinct knot sequence in both directions to a strictly increasing sequence $(t_i{)}_{i\in \mathbb{Z}}$ tending to $\pm \infty$. For order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order splines are supported on $[t_i,t_{i+r}]$, making the following sums locally finite. Reindexing the second term of the recurrence yieldsInduction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each coefficient is nonnegative wherever its associated lower-order spline is nonzero. Removing splines therefore gives the subpartition of unity for B-splines on the whole line.

For $t_k<t<t_{n+1}$, a spline with index $i\le 0$ ends no later than $t_k$, and one with $i\ge n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\dots ,n$. For $k\ge 2$ the splines are continuous and vanish at their support endpoints, so the equality extends to the closed basic knot interval:If order one is allowed, the standard half-open indicator convention gives equality on $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^{\infty }$ norms. For the continuous splines of order at least two no such convention is necessary.

With spatial flatness, the scalar as the only source, and $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$, the Friedmann equation isThe scalar equation gives $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2$, which is the energy-conservation equation $\dot{\rho }+3H(\rho +P)=0$. Differentiating the Friedmann equation on the expanding branch yields $6M_{\mathrm{Pl}}^{2}H\dot{H}=-3H{\dot{\varphi }}^2$, so $\dot{H}=-{\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2)$. Combining this with $\ddot{a}/a=H^2+\dot{H}$ gives the Friedmann acceleration equation:Thus cosmic inflation occurs exactly when $V(\varphi )>{\dot{\varphi }}^2$ in this canonical scalar model. Equivalently $P<-\rho /3$, or the first Hubble slow-roll parameter ${ϵ}_H=-\dot{H}/H^2$ is smaller than one.

Construct the tensor-product weight diagram by placing one octet diagram at each triplet representation weight $u,d,s$ and adding multiplicities where the shifted diagrams overlap. Its highest weight isso remove the $(2,1)$ diagram by highest-weight character subtraction. Its unequal-sided hexagon has nine boundary representation weights of multiplicity one and three interior representation weights of multiplicity two, giving dimension $15$.

To see the remainder explicitly, measure diagram coordinates by $(a,b)=(2h_1,2\sqrt{3}{h}_2)$. After subtracting this $15$, the residual representation weights areThe next highest weight is $(h_1,h_2)=(0,2/\sqrt{3})=2{\omega }_2$, with Dynkin labels $(0,2)$. The conjugate sextet $\overline{6}$ is a triangular weight diagram containing each of these six representation weights once. Subtracting it leaves $(1,1),(-1,1),(0,-2)$ once each in the integer coordinates: precisely the triplet. The SU(3) triplet-octet decomposition is thereforeThe dimensions add to $15+6+3=24$, and the figure below verifies the equality at every representation weight, including overlapping interior representation weights.

For the Metropolis–Hastings algorithm, start at a state $\theta$ with positive target density, propose ${\theta }^{\prime }$ from $q({\theta }^{\prime }\mid \theta )$, and generate an independent uniform draw. Accept the proposal with probabilityOn rejection retain $\theta$, including repeated states in the chain; removing repeats would generally change its stationary law. The accepted-flow density satisfieswhich is symmetric in its two states. This proves detailed balance and hence invariance of the target distribution, including the holding probabilities on the diagonal. Unknown common normalization constants cancel in the ratio. Invariance alone does not establish convergence from arbitrary initial states. Choosing proposals that yield a positive Harris recurrent Markov chain with aperiodicity supplies the usual convergence guarantee, and the resulting sample is dependent. Use burn-in and account for autocorrelation when assessing Monte Carlo precision.

Use base-two logarithms, so the Von Neumann entropy is $S(\rho )=-\mathrm{Tr}(\rho {\log }_2\rho )$, with $0\log 0=0$. The quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$; this is the quantity denoted by $S({\rho }_A:{\rho }_B)$ in the question, using the given joint state.

Strong subadditivity of Von Neumann entropy states that for every tripartite density operator,Equivalently the conditional quantum mutual information $I(A:C\mid B)$ is nonnegative. We apply this property to the original state or to a dilation of the relevant quantum channel.

For the replacement condition, interpret “every element has order $p$” in the usual exponent-$p$ sense: every nonidentity element has order $p$, while the identity has order one. ConsiderEvery nonzero element has order $p$. It is infinite, but no finite collection generates it: the supports of finitely many group generators cover only finitely many coordinates, so a unit vector outside those coordinates is missing from their generated subgroup. Also $E$ is abelian, and one coordinate subgroup $C_p$ is a proper nontrivial normal subgroup. ThereforeThe stronger restriction on all proper subgroups, rather than just individual element orders, is what forces the conclusions in parts (i) and (ii).

The two scalings give $\rho ={\rho }_{r0}{a}^{-4}+{\rho }_{s0}{a}^{-2}$. Flatness and the Friedmann equation give $H^2=8\pi G\rho /3$, with $H_0^2=(8\pi G/3)({\rho }_{r0}+{\rho }_{s0})$. ThusOn the expanding branch, putting the Big Bang at zero, separation givesIts positive square root is the required solution. If $\beta >0$, its early behaviour is $a\propto t^{1/2}$, as for radiation. The values $a=1$ and $a^2=\beta$ giveFor $\beta =0$ there is only the string component, with linear expansion and no radiation era.

For any rule $d$, the average risk is at most its supremum, and the specified Bayes-optimality property then givesTherefore $d^{\ast }$ is minimax.

In spherical coordinates, $z=r\cos \theta$ and $x^2+y^2=r^2{\sin }^2\theta$, so the elliptic paraboloid has equationAway from the origin this is $r=\cos \theta /{\sin }^2\theta$, with $0<\theta <\pi /2$. The undivided equation also includes the origin, where angular coordinates are not unique.

In cylindrical coordinates, $x^2+y^2={\rho }^2$, so the elliptic paraboloid is

For the contrapositive of $(b)\Rightarrow (c)$, suppose $\mathrm{char}k=p>0$. ConsiderLet $x_i$ act by multiplication and $D_i$ by formal differentiation. Differentiation preserves the defining ideal, since $D_j(x_i^{p}f)=x_i^p{D}_{j}f$ in characteristic $p$. The relations $[D_i,x_j]={\delta }_{ij}$ hold before quotienting and hence hold on $M$. Its basis consists of the monomials $x^{\alpha }$ with $0\le {\alpha }_i<p$, so ${\dim }_{k}M=p^n>0$. This is the explicit finite-dimensional module for the Weyl algebra in positive characteristic.

Therefore condition (b) fails in positive characteristic and implies characteristic zero. Combining this with the preceding two implications proves $(a)⟺(b)⟺(c)$. The same representation also has nonzero proper kernel, since the infinite-dimensional Weyl algebra cannot inject into a finite-dimensional endomorphism algebra, confirming its failure of simplicity in positive characteristic.

This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.

This part proves $(c)\Rightarrow (a)$. Write the Weyl algebra generators as $x_1,\dots ,x_n,D_1,\dots ,D_n$, with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ form a basis over any field: one may construct the algebra by successively adjoining the commuting derivations $D_i$ to $k[x_1,\dots ,x_n]$, with multiplication $D_{i}f=f{D}_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge 1$.

Assume $\mathrm{char}k=0$ and let $I$ be a nonzero two-sided ideal. Choose a nonzero element $a=\sum c_{\alpha \beta }{x}^{\alpha }{D}^{\beta }\in I$, and a nonzero term of maximal total degree, say $({\alpha }_0,{\beta }_0)$. The relations giveApply $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $({\alpha }_0,{\beta }_0)$, which forces equality. Thus the resulting element of $I$ isThe factorials are nonzero in characteristic zero. This is a nonzero scalar, so $1\in I$ and $I=A_n(k)$. Therefore $A_n(k)$ is simple in characteristic zero. This scalar extraction by iterated Weyl commutators supplies the full simplicity proof.

Maximal right ideals of $R/J$ correspond to maximal right ideals of $R$, all of which contain $J$. Their intersection in the quotient is zero. Hence $R/J$ is a right Artinian ring with zero Jacobson radical, and part (a) makes it semisimple.

Part (b) gives the finite radical filtrationEach factor $J^i/J^{i+1}$ is annihilated by $J$, so it is a right module over $R/J$. Every module over a semisimple ring is semisimple: a free module is a direct sum of simple summands of copies of the regular module, and a quotient of a semisimple module is again semisimple, because the images of its simple summands are either zero or simple and span the quotient. A maximal independent family of these simple images spans: any simple image not contained in the sum intersects it trivially and could be added. Each factor is also Artinian, as a subquotient of $R_R$. A semisimple Artinian module has only finitely many simple summands, since an infinite direct sum would give a descending chain by successively deleting summands. Consequently every factor has finite module length.

Concatenating composition series of these finitely many factors makes $R_R$ a module of finite module length. Such a module satisfies both chain conditions, since every strict change consumes a positive amount of length. Therefore $R$ is right Noetherian. This is the Hopkins-Levitzki theorem, obtained from the two preceding results rather than invoked in their proofs.

The descending powers $J\supseteq J^2\supseteq \cdots$ stabilize because they are right ideals. Choose $m\ge 1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.

Suppose $I\ne 0$. Among the right ideals $K$ such that $KI\ne 0$, choose a minimal one; the set is nonempty since $RI=I\ne 0$. The right ideal $KI$ lies in $K$, and $(KI)I=K{I}^2=KI\ne 0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne 0$. Then $(kR)I=kI\ne 0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.

Since $K=KI$ and $I\subseteq J$, its generator satisfies $k={\sum }_{\nu }k{r}_{\nu }{i}_{\nu }=kj$ for an element $j={\sum }_{\nu }{r}_{\nu }{i}_{\nu }\in J$. Thus $k(1-j)=0$. The unit criterion for the Jacobson radical makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne 0$. Therefore $I=0$, andThe minimal-right-ideal proof of radical nilpotence is important here: one cannot apply the finite-generator form of Nakayama lemma directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary right ideal cyclic before using the unit argument.

Use the descending chain condition on the finite intersections of maximal right ideals. It gives a minimal such intersection $K=M_1\cap \cdots \cap M_r$. Intersecting $K$ with any other maximal right ideal cannot decrease it, by minimality. Thus $K$ lies in every maximal right ideal, and $K=J(R)$. When $J(R)=0$, the diagonal mapis injective, and each summand on the right is a simple module.

For completeness, a submodule $N$ of a finite direct sum of simple modules is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus (N\cap T)$, and induction again applies. This also shows finite module length. Apply it to the image of the diagonal map to obtainThis is the semisimplicity of a right Artinian ring with zero radical. It does not assume right Noetherianity in order to prove it.

Assume the coefficients are sufficiently differentiable in the region considered, with no material discontinuity or caustic at the front. Let $t=T(\mathbf{x})$ describe arrival time, $\mathbf{p}=\nabla T$ the slowness vector, and $\mathbf{n}$ the unit elastic wave propagation normal. For any field, its jump is the value immediately behind the front minus that immediately ahead. If the first nonzero jump occurs in the kth time derivative, $k\ge 2$, writeA smooth background may be added without changing these leading jumps. Since $F_j^{\prime }=F_{j-1}$, differentiation expresses spatial jumps in terms of $p_j$ and time-derivative jumps. The restriction $k\ge 2$ lets displacement field and its first time derivative remain continuous and avoids an imposed impulsive source at the front.

Insert this local expansion in $\rho {\partial }_t^2{u}_i={\partial }_j(c_{ijpq}{\partial }_p{u}_q)$. The coefficient of $F_{k-2}$ isFor isotropic elasticity the slowness acoustic matrix is $\mu |\mathbf{p}{|}^{2}I+(\lambda +\mu )\mathbf{p}{\mathbf{p}}^T$. Its longitudinal eigenvalue is $(\lambda +2\mu )|\mathbf{p}{|}^2$. The P branch therefore gives the eikonal equation and polarizationThe next coefficient, $F_{k-1}$, isThe matrix on the left is symmetric and annihilates $\mathbf{n}$. Project onto $n_i$ to obtain its solvability condition. Put $a=a^{[k]}$. The two projected terms can be evaluated without dropping gradients of mass density or moduli:These identities use $|\mathbf{n}{|}^2=1$, hence $n_q{\partial }_p{n}_q=0$, and $(\lambda +2\mu )/\alpha =\rho \alpha$. Their curvature terms cancel, leavingMultiplying by $a$ proves the requested P-wavefront discontinuity transport law,The wave amplitude here is real for a real jump; a complex wave amplitude uses the corresponding modulus-squared law. An elastic ray follows $d\mathbf{x}/ds=\mathbf{n}$, with arclength $s$ and $dt/ds=1/\alpha$. If $J$ is the infinitesimal cross-sectional area of a neighboring ray tube, $d\log J/ds=\nabla \cdot \mathbf{n}$. Thus $\rho \alpha a^{2}J$ is constant along that tube, the ray-tube conservation for a P-wave jump. This defines the geometrical spreading and shows why the wave amplitude changes with material properties even before attenuation is considered. At a caustic the simple ray-tube wave amplitude is singular and a uniform local description is needed.