For the standard fixed routing loss network with independent Poisson processes, the stationary occupancy law depends on independent holding-time distributions only through their means. The occupancy process alone need not be a Markov process when those holding times lack the memoryless property.

For a product affine connection, its curvature satisfiesExpand the defining derivative commutator on lifted vector fields. Opposite-factor fields commute and have zero mixed covariant derivatives, leaving exactly the two factor curvatures. Tensoriality gives the formula for arbitrary tangent vectors. Thus the product connection is flat if the two factors are flat; for a product Riemannian metric this applies to its Levi-Civita connection.

If a positive-definite matrix $Q=L{L}^T$ is the precision matrix of a multivariate normal distribution, draw independent $\xi \sim N(0,I)$ and solve $L^T\eta =\xi$. Then $m+\eta$ has covariance $L^{-T}{L}^{-1}=Q^{-1}$. This avoids forming the inverse and exploits a banded Cholesky decomposition when $Q$ is banded.

For the positive-domain power diffusion with $0<\alpha \le 1$, assuming the lifetime is the limit of the hitting times of $1/n$, it is finite almost surely exactly when $\alpha <1$. For $\alpha <1$, the Itô formula givesThe negative drift bounds this positive process above by $X_0^{1-\alpha }+(1-\alpha )B$, whose first hit of zero is finite by recurrence of one-dimensional Brownian motion. The lifetime must precede that hit. At $\alpha =1$, $X_t=X_0{e}^{B_t-t/2}$ is positive and finite on every compact time interval, so the lifetime is infinite. Approaching zero only as $t\to \infty$ is not a finite boundary lifetime.

The support of a positive operator $A$ is the closed subspaceThe range identity follows from image-kernel orthogonality for an adjoint and $A=A^{\ast }$. Equivalently, it is the range of the spectral projection $E_A((0,\infty ))$. In finite dimension, it is the span of eigenvectors with positive eigenvalues, and the range is already closed.

Closure matters in infinite dimension. On the l2 sequence space, $A{e}_n=e_n/n$ has zero kernel and hence full support, but is not onto: $(1/n)$ is square-summable and its formal inverse image $(1,1,\dots )$ is not. In quantum relative entropy, containment of the first state's support in the reference state's support is necessary for finiteness; finite dimensions also make it sufficient, while infinite dimensions can have divergence despite that containment.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

A qubit depolarizing channel shrinks every Bloch vector by the same factor: ${\Lambda }_p(\rho )=p\rho +(1-p)I/2$. In this retention-parameter convention it is completely positive for $-1/3\le p\le 1$; the usual mixture of unchanged and fully randomized states uses $0\le p\le 1$.

Use the Dupuit approximation: the saturated region is shallow enough that hydrostatic pressure is $p=\rho g(h-z)$ and flow is predominantly horizontal. Darcy's law then gives the horizontal Darcy velocity $u_x=-u_b(1+\beta z)h_x$, where $u_b=k_0\rho g/\mu$. Integrating through the saturated depth gives the volume flux per unit widthThe stored water volume per unit area is $\varphi h$, so mass conservation yields the unconfined aquifer with depth-dependent permeability equationDefine the positive river discharge by $Q(t)=-q(0,t)$. A useful check on all subsequent results is the integrated mass conservation law

For the stated axisymmetric flow, the diagonal components of the rate-of-strain tensor areTheir sum is $u_r+u/r+w_z=0$, the incompressibility condition. In the leading thin-sheet approximation, vanishing tangential traction makes $u$ independent of $z$. The normal stress boundary condition is ${\sigma }_{zz}=-p_{\mathrm{ext}}$, so the Newtonian fluid stress tensor givesThereforeFor the small annular sector, the inner and outer radial faces contribute $2\delta \theta {\partial }_r(rh{\sigma }_{rr})\delta r$ in the radial direction. The two azimuthal faces contribute $-2\delta \theta h{\sigma }_{\theta \theta }\delta r$: their hoop tractions have inward radial components. The combined radial force of the external pressure on the sloping upper and lower surfaces is $2\delta \theta r{p}_{\mathrm{ext}}{h}_r\delta r$.

Neglecting inertia, force balance is thusSubstituting the two stresses cancels the terms involving $p_{\mathrm{ext}}{h}_r$ and gives the axisymmetric viscous-sheet stretching equations:Finally, conservation of mass in the sector gives

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when its supports eventually lie in a single compact set and every derivative converges uniformly. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is its continuous linear dual; weak convergence means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. We use $⟨u^{\prime },\phi ⟩=-⟨u,{\phi }^{\prime }⟩$ and $⟨au,\phi ⟩=⟨u,a\phi ⟩$ for multiplication of a distribution by a smooth function.

The kinetic term fixes the mass dimension of a canonically normalized real scalar field in $d$ dimensions to $(d-2)/2$. In six dimensions $[\varphi ]=2$, and every term in the Lagrangian density has dimension six. Therefore $[g_n]=6-2n$, andFor positive integral $n$, these mean $n=1,2$, $n=3$ and $n\ge 4$, respectively. If $n=0$ is included, its vacuum-energy constant is also relevant. These are classical power counting in quantum field theory classifications of relevant couplings, marginal couplings and irrelevant couplings, not assertions that loop corrections preserve exact marginality. For the remaining parts write $g=g_3$, assume $m^2>0$ and $0<\Lambda <{\Lambda }_0$, and define a local quartic coupling by $\int g_4{\varphi }^4/4!$.

Use the valence formula for the modular group: for a nonzero modular form of weight $k$ on $S{L}_2(\mathbb{Z})$,The sum contains one representative of each other modular group orbit of zeros. The orders are nonnegative because $f$ is holomorphic on the complex upper half-plane and at infinity. The half and third weights account for the elliptic stabilizers of the modular group.

In weight two, the transformation under $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ at its fixed point gives $f(i)=i^{2}f(i)=-f(i)$. Thus any nonzero $f$ would have $v_i(f)\ge 1$. The valence formula for the modular group would give a left side at least $1/2$ and a right side $1/6$, which is impossible. Therefore the unheaded request is answered by vanishing of weight-two level-one modular forms: