The observed death fraction isIn the independent two-exponential competing risks model, $\pi =\lambda /(\lambda +\gamma )$. Thus the observed fraction estimates the share of the combined rate attributable to death. Combining it with an estimate of $s=\lambda +\gamma$ yields $\hat{\lambda }=\hat{\pi }\hat{s}$ and $\hat{\gamma }=(1-\hat{\pi })\hat{s}$. This uses the event-type information that the death-only time analysis ignored.

Assume individual incubation delays are independent of the Inhomogeneous Poisson process of infections and of one another. This marking assumption is needed in addition to specifying a marginal incubation density.

In the discrete cohort model let $N_i\sim \mathrm{Poisson}(h_i)$ be the independent infection counts. Mark each infection by its eventual onset interval. For cohort $i$, the categories have probabilities $q_{ik}$ for observed intervals, with the remaining probability assigned to onsets outside the window. Poisson thinning makes the counts $N_{ik}$ in these categories mutually independent Poisson random variables with means $h_i{q}_{ik}$. One direct proof is their probability generating function:Different cohorts are independent. Summing their counts therefore gives independent Poisson onset counts in disjoint intervals:The same argument applies exactly in continuous time by the Independent marking theorem for Poisson point processes and mapping each marked infection to its onset time. The endpoint model approximates its means; its independent-Poisson conclusion is exact within that discrete model. A fixed cohort size would instead induce negatively correlated onset-bin counts, so the Poisson process infection assumption matters.

A nonzero finite-dimensional quiver representation is a brick module when ${\mathrm{End}}_Q(X)=k{1}_X$. If $X=U\oplus V$ with both summands nonzero, projection onto $U$ is an idempotent endomorphism different from zero and the identity. That is impossible in $k{1}_X$. Hence every brick is indecomposable.

For the requested converse counterexample, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\epsilon )$, and it is not simple because this is a nonzero proper ideal. The module is indecomposable: its endomorphism ring is $A^{\mathrm{op}}\cong A$, which has no nontrivial idempotents. But this ring has dimension two over $k$, so the regular module is not a brick. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.

Let $K=\ker f_{\rho }$ and $C=Y_j/\mathrm{im}{g}_{\rho }$. Consider the map from the arrow term of the extension complex of quiver representations to ${\mathrm{Hom}}_k(K,C)$ that restricts its $\rho$-component to $K$ and then takes the quotient in $Y_j$. This map is surjective: a map $K\to C$ can be lifted to $Y_j$ and extended from $K$ to $X_i$.

Every coboundary is killed by this map, since for $x\in K$,It therefore induces a surjectionBoth vector spaces in the final Hom functor are nonzero, so its dimension is positive. Hence ${\mathrm{Ext}}_Q^1(X,Y)\ne 0$. This is the kernel-cokernel obstruction to splitting a quiver extension and remains valid for loops and repeated arrows elsewhere in the quiver.

Finite rearrangement and the Von Mangoldt divisor identity giveFor $m=\lfloor x\rfloor$, integral comparison of the increasing logarithm gives $\log (m!)=m\log m-m+O(\log (m+1))$. Moving $m$ to real $x$ changes the main term by $O(\log (x+1))$. Thus the integrated Chebyshev sum hasFor $x\ge 2$, chooseThese are logarithmic Möbius function weights; they need not be upper-bound sieve weights. For $n\le x$, the Möbius inversion identities $\mu \ast 1=ϵ$ and $\mu \ast \log =\Lambda$ giveHence the expression suggested in the question isWe will use the elementary Möbius harmonic logarithmic moments, $A_j(x)={\sum }_{d\le x}\mu (d){\log }^j(x/d)/d$, for whichHere is a proof of the needed estimates. From ${\sum }_d\mu (d)\lfloor x/d\rfloor =1$, replacing floors by $x/d+O(1)$ gives $A_0=O(1)$. Also ${\sum }_d\mu (d)H_{\lfloor x/d\rfloor }/d=1$, because $\mu \ast 1=ϵ$. Substituting the permitted harmonic number estimate $H_{\lfloor t\rfloor }=\log t+\gamma +O(1/t)$ gives $A_1+\gamma A_0=1+O(1)$, so $A_1=O(1)$.

For $A_2$, the weighted Dirichlet hyperbola method gives the harmonic divisor sum expansionExplicitly, for $r=\lfloor \sqrt{t}\rfloor$, $B(t)=2{\sum }_{a\le r}{H}_{\lfloor t/a\rfloor }/a-H_r^2$. Substitute the harmonic estimate and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$, obtained by unit-interval integral comparison. This proves the expansion without a prime-distribution theorem. Since $\tau =1\ast 1$, we have $\mu \ast \tau =1$, and thereforeThe summed error is bounded by $x^{-1/2}{\sum }_{d\le x}{d}^{-1/2}\log (2x/d)=O(1)$, by integral comparison. Thus $A_2=2\log x+O(1)$.

Finally substitute the asymptotic for $T$ into $F$. The main term isThe error sum is $O(x)$: partition the integers $d$ into $x/2^{j+1}<d\le x/2^j$; each interval contributes $O(x(j+1{)}^2/2^j)$, and the resulting series converges. Using the moment estimates proves the Selberg symmetry formula:In particular, the factor two comes from the quadratic logarithmic moment, rather than from an assumption of the Prime number theorem.

Write $f(n)=3^{\omega (n)}$, where $\omega$ is the prime omega function. If $p^a\Vert n$, its contribution to the Dirichlet convolution $(f\ast \Lambda )(n)$ isSince $3(a-1+1/3)\ge a$, summing over the prime factors proves the log-weighted convolution bound for three to the prime omega, $f(n)\log n\le 3(f\ast \Lambda )(n)$. On $\sqrt{D}\le n\le D$, $\log n\ge \frac{1}{2}\log D$, henceThis is the required inequality. The standard Chebyshev estimate $\psi (t)\ll t$ bounds its right side by $D(\log D{)}^{-1}{\sum }_{m\le D}f(m)/m$. To bound the latter sum, use multiplicativity and extend to all numbers with prime factors at most $D$:The Mertens second theorem states ${\sum }_{p\le D}1/p=\log \log D+O(1)$. Taking logarithms of the product, with a summable $O(p^{-2})$ remainder, gives $3\log \log D+O(1)$. Therefore the product is $O({\log }^{3}D)$, proving the summatory bound for three to the prime omega in the required range:

Let $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $\mathfrak{g}=\mathbb{C}(I+N)\subseteq {\mathfrak{gl}}_2(\mathbb{C})$. This one-dimensional abelian Lie algebra is not a semisimple Lie algebra. For $X=I+N$, its Additive Jordan decomposition is $X_s=I$ and $X_n=N$. Neither belongs to $\mathfrak{g}$: comparing diagonal entries and then the upper-right entry rules out both scalar multiples. Therefore nonsemisimple matrix Lie subalgebras need not be closed under Additive Jordan decomposition.

Use the complex-bilinear extension of the Nijenhuis tensor expression to complex vector fields. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, soFor $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V{]}^{1,0}$. For inputs of opposite types, the two leading Lie brackets cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the type decomposition of the complexified tangent bundle to both inputs consequently givesThus $N=0$ precisely when the Lie bracket of any two sections of either eigenbundle remains in that same eigenbundle. In other words, $N=0$ if and only if both tangent-type distributions are involutive distributions. This proves the equivalence directly and does not assume that an arbitrary almost complex structure already has holomorphic coordinates.

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.

Extend $J$ complex-linearly to the complexification of a real vector bundle $T_{\mathbb{C}}X=TX{\otimes }_{\mathbb{R}}\mathbb{C}$. Since $J^2=-\mathrm{id}$, its $+i$ and $-i$ eigenbundles have smooth projectionsEach has complex rank $n$: complex conjugation interchanges them, and together they have rank $2n$. These are the type decomposition of the complexified tangent bundle. For an arbitrary almost complex structure, they are smooth complex vector bundles; a holomorphic vector bundle structure requires integrability.

When $J$ comes from a holomorphic atlas, write a holomorphic coordinate as $z_j=x_j+i{y}_j$. Its Wirtinger derivatives areThe induced almost complex structure has $J{\partial }_{x_j}={\partial }_{y_j}$ and $J{\partial }_{y_j}=-{\partial }_{x_j}$. Hence the displayed vector fields are respectively $+i$ and $-i$ eigenvectors. They are linearly independent, and each collection has $n$ elements. Thus they give local frames for $T^{1,0}X$ and $T^{0,1}X$ respectively, with $T^{1,0}X$ the holomorphic tangent bundle. The repeated $T^{0,1}X$ in the first sentence of the printed item must be read as the two complementary eigenbundles.

Small data global regularity for wave maps also holds in three spatial dimensions. For localized compatible data small in high weighted Sobolev norms, the key is the classical null condition for wave equations. Each derivative contraction is a null form for wave equations, vanishing for parallel null derivatives. The vector field method for wave equations exploits derivatives tangent to the light cone and weighted energy estimates to close the bootstrap argument; ordinary $(1+t{)}^{-1}$ decay alone is insufficient.

Small data global regularity for wave maps holds in four spatial dimensions. Smallness is measured in sufficiently high weighted Sobolev norms relative to a constant map, with localized compatible Cauchy data. The vector field method for wave equations gives derivative decay $(1+t{)}^{-3/2}$. This is time-integrable, so commuted wave energy estimates close a small-data bootstrap argument for the derivative-quadratic semilinear wave equation. Higher regularity persists. No smallness of energy alone is asserted.

Global regularity for one-dimensional wave maps holds for arbitrary smooth localized compatible data. In retarded and advanced null coordinates $u=t-x$, $v=t+x$, the wave map equation says $D_v{\varphi }_u=D_u{\varphi }_v=0$, with $D$ the target covariant derivative. Hence $|{\varphi }_u{|}^2$ depends only on $u$, and $|{\varphi }_v{|}^2$ only on $v$. Initial derivative bounds persist. Compactness of the sphere and differentiated energy estimates then prevent finite-time loss of smoothness.

Use the radial reduction of the three-dimensional wave equation $w=r\varphi$ and extend $w$ oddly in $r$. Write $w=w_h+w_F$, where $w_h$ is the homogeneous wave equation solution with the prescribed Cauchy data, and $w_F$ has zero Cauchy data. With $w_0(r)=r{\varphi }_0(r)$ and $w_1(r)=r{\varphi }_1(r)$, the D'Alembert formula givesBoth odd initial profiles have compact support, soThe Duhamel principle, applied to $w_{tt}-w_{rr}=-rF$, givesNow write $r=t+a$ with $1/2\le a\le 1$. The lower integration limit is $s+a>0$, so this domain of dependence never crosses the axis. Intersecting its integration interval with the support of the source leaves a subset of $[s+a,s+1]$, of length at most one. On this interval,The outgoing shell source estimate for a radial wave therefore yieldsFinally $r\ge t+1/2\ge (1+t)/2$, and thusThis constant depends only on the Cauchy data and the fixed unit bound for the source. The logarithm comes from integrating the shell's accumulated forcing, whereas the factor $(1+t{)}^{-1}$ comes from division by $r$.

For a radial function, the Laplacian is $\Delta \varphi ={\varphi }_{rr}+2{\varphi }_r/r$. Introduce the retarded and advanced null coordinates and the radial reduction of the three-dimensional wave equation:The chain rule gives ${\partial }_t={\partial }_u+{\partial }_v$ and ${\partial }_r=-{\partial }_u+{\partial }_v$. Consequently the inhomogeneous wave equation isMultiplying the radial wave equation by $r$ removes its first-order radial derivative:Thus the particularly useful retarded and advanced null coordinates form isThese formulas hold for $r>0$. At the axis, smooth spherical symmetry imposes ${\varphi }_r(t,0)=0$ and $w(t,0)=0$; equivalently $w$ has a smooth odd extension across $r=0$.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

Choose an injection $j:A\to \mathbb{N}$. For each $b\in B$, its fiber under the surjection $f$ is nonempty, so defineTwo distinct fibers cannot contain the same label $j(a)$, because $j$ is an injection and $f(a)$ has a single value. Thus $h:B\to \mathbb{N}$ is an injection, proving that $B$ is a countable set. Taking the least label avoids any need for a separate choice of one element from each fiber. If $A$ is empty, surjectivity makes $B$ empty as well.

For the unheaded enumeration request, list pairs of positive natural numbers by their sum. An explicit Cantor pairing function isFor $d=m+n-2\ge 0$, the positive integers $n=1,\dots ,d+1$ give the consecutive labels $d(d+1)/2+1,\dots ,(d+1)(d+2)/2$. These blocks are disjoint, consecutive and cover $\mathbb{N}$. Consequently $\pi$ is a bijection from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$, so $\mathbb{N}\times \mathbb{N}$ is countable.

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus the double cosets partition the group.

Unlike single cosets, double cosets can have different cardinalities. For finite $G$, decompose $HgK$ into left cosets of $K$, or apply the orbit-stabilizer theorem to the action of $H$ on $gK$. Its stabiliser subgroup is $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.

Introduce the slow time $T=\epsilon t$ and treat $t,T$ independently in the method of multiple scales. WriteAt the next order,For fixed $R$ define the period average byHere $f$ inside the averages is evaluated at the leading position and velocity $R\cos \varphi ,-R\sin \varphi$. Orthogonality to both fundamental harmonics is the solvability condition in the method of multiple scales: it removes the resonant forcing that would otherwise generate a secular term. Since the squared sine and cosine averages are $1/2$, the amplitude-phase equations areEquivalently $\dot{R}=-\epsilon ⟨f\sin (t+\theta )⟩$ and $\dot{\theta }=-\epsilon ⟨f\cos (t+\theta )⟩/R$ at first order. These equations describe a bounded, weakly perturbed oscillation over $t=O({\epsilon }^{-1})$ while the amplitude remains in a range where the expansion is ordered. At zero amplitude the oscillation phase coordinate is singular; Cartesian harmonic coefficients or an equilibrium analysis should replace it.

For polytropic index three, the Lane-Emden mass formula cancels the central mass density:Substituting the preceding constant $K$ givesFor fixed composition this is $M\propto \sqrt{1-\beta }/{\beta }^2$, as required. Squaring and rearranging yields the Eddington quartic relation, $(1-\beta )/{\beta }^4\propto {\mu }^4{M}^2$. The larger masses in this model have smaller gas fractions and relatively more radiation support.