The subspace filtration and quotient filtration in a short exact sequence are good when the associated graded algebra is Noetherian. Their cumulative Hilbert functions add in every degree. Positive leading coefficients make the degree of the sum the larger degree; when both degrees agree, factorial times the leading coefficients add. Thus equal-dimensional nonzero factors consume positive multiplicity of a filtered module. With a standard graded algebra normalization these multiplicities are integral.

Let $G_N(B)=E(20N[1+2B(1-B)],21N)-B$. The proportional scaling limit of the Erlang loss formula makes its limits at $B=1/100$ and $B=1/8$ respectively $-1/100$ and $7/520$. Since $G_N(0)>0$ and $G_N(1)<0$, sufficiently large $N$ give alternating signs at these four points. The intermediate value theorem supplies three distinct zeros in the intervening intervals. These are multiple solutions of the Erlang fixed point approximation, not multiple stationary laws of an exact finite irreducible Markov chain.

For an elliptic curve over a number field $K$, the Mordell-Weil theorem writes $E(K)$ as a finite torsion subgroup plus a free abelian group of finite rank of an abelian group $r$. This integer is the rank of the elliptic curve over $K$. It can increase under field extension. A two-isogeny descent supplies rank bounds from rational square-class images; the canonical height pairing can certify independence of rational points.

The saturation of a subgroup $H$ of the Mordell-Weil group consists of all points with a positive multiple in $H$. Saturation includes the full torsion subgroup. For a full-rank $H$ containing that torsion, it equals $E(K)$, but the index of $H$ may exceed one. Checking that $\ell P\in H$ forces $P\in H$ for every prime divisor of a certified index bound proves that $H$ is already the whole group. A canonical height pairing proves independence of found points; saturation answers the separate question of whether a full-rank subgroup is the full group.

For the monodisperse circular ash runout, integrate the local particle deposition flux ${\rho }_p{w}_{s}c(t)$ from first arrival to infinite time. Here $s=\max (s_0,r^2/R_{\infty }^2)$ for $0\le r<R_{\infty }$ and $s_0=R_0^2/R_{\infty }^2$. The displayed $\Sigma$ is deposited particle mass per horizontal area, and vanishes outside the limiting current. It is constant under the initial disk, then decreases to zero; integrating $2\pi r\Sigma$ gives the initial ash mass ${\rho }_p{c}_0\mathcal{V}$. A deposit thickness additionally requires its packing fraction or bulk deposit density.

For a compact string coordinate of radius $R$, choose $p_L=n/R+wR/{\alpha }^{\prime }$ and $p_R=n/R-wR/{\alpha }^{\prime }$, where the left-moving oscillators are denoted by tildes. The two classical zero-mode Virasoro constraints give $M^2=p_L^2+4\tilde{N}/{\alpha }^{\prime }=p_R^2+4N/{\alpha }^{\prime }$. Their difference yields the displayed shifted closed-string level matching condition. Their average is $M^2=(n/R{)}^2+(wR/{\alpha }^{\prime }{)}^2+2(N+\tilde{N})/{\alpha }^{\prime }$. Classically the center-of-mass momentum can be continuous; the integer momentum number belongs to compact-coordinate quantization. Quantum intercepts shift the mass formula but cancel out of level matching.

In the critical closed bosonic string, compactification of one coordinate gives $M^2=n^2/R^2+w^2{R}^2/{\alpha }^{\prime 2}+2(N+\tilde{N}-2)/{\alpha }^{\prime }$ and $\tilde{N}-N+nw=0$. At oscillator level zero, level matching forces $nw=0$. Thus a pure momentum ground state becomes a massless scalar field at $R=|n|\sqrt{{\alpha }^{\prime }}/2$, while a pure winding ground state does so at $R=2\sqrt{{\alpha }^{\prime }}/|w|$. These are threshold states descended from the bosonic tachyon; they are additional to the neutral graviton, two-form, vectors, dilaton and radius modulus. At the self-dual circle, the four charges $(\pm 2,0),(0,\pm 2)$ contribute to the enhanced massless scalar field spectrum.

Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower trace. The resulting representation has highest weight $p{\omega }_1+q{\omega }_2$. Its dimension isThe subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial inner product, is multiplication by the nonzero polynomial ${\sum }_i{x}_i{y}_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its Weyl dimension formula dimension, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed tensor.

For $T^{abc}$, split its first pair into symmetric and antisymmetric parts. The fully symmetric part $S^{abc}$ is a symmetric cubic representation of SU(3). Write $M^{abc}=T^{(ab)c}-S^{abc}$ and $O^a{}_d={ϵ}_{dbc}{M}^{abc}$. Then $O$ is traceless and $M^{abc}=({ϵ}^{bcd}{O}^a{}_d+{ϵ}^{acd}{O}^b{}_d)/3$. The antisymmetric part is $T^{[ab]c}={ϵ}^{abd}{N}^c{}_d$, with $N^c{}_d={ϵ}_{dab}{T}^{abc}/2$. Splitting $N$ into its traceless part and trace gives the second adjoint representation of SU(3) and the singlet. These maps use invariant tensors, so they commute with the group action and explicitly realize the four irreducible summands.

Let $M$ be a closed oriented smooth four-manifold with $b_2^{+}(M)>1$, and letbe its blowup, with exceptional sphere $E$. A Spin-c structure extending one on $M$ with exceptional evaluation $⟨c_1(\tilde{\mathfrak{s}}),E⟩=\pm 1$ has the same Mixed Heegaard Floer invariant, up to the conventional overall sign. In particular, nonvanishing is preserved.

More generally an evaluation $\pm (2\ell +1)$, with $\ell \ge 0$, makes the exceptional-piece Heegaard Floer cobordism maps on the minus, infinity and plus versions act by multiplication by $U^{\ell (\ell +1)/2}$. Putting the blowup on one side of an admissible cut and using the cobordism composition law supplies the corresponding mixed-invariant formula. This states the factor at the map level, where the placement of $U$ is unambiguous. For $\ell =0$ that factor is one.^[ref]

If ${\Phi }_{M,\mathfrak{s}}\ne 0$ and a connected closed oriented embedded surface $S$ has positive genus and $S^2\ge 0$, thenBlow up $S^2$ points on $S$ and choose the $\pm 1$ exceptional determinant classes which increase the Chern pairing on its proper transform by $S^2$. The Blowup formula for the mixed Heegaard Floer invariant preserves nonvanishing, while the proper transform has square zero. A cut disjoint from it allows the relevant cobordism map to pass through $H{F}^{\pm }(S\times S^1,\mathfrak{s}{|}_{S\times S^1})$. These groups vanish when the Chern pairing exceeds $2g-2$, so the mixed map would then vanish, a contradiction. Reverse the surface orientation for the absolute value. The positive-genus condition is essential: a small nullhomologous sphere has square zero and zero pairing and would violate the displayed inequality.

The Kleinian limit set of an infinite subgroup is nonempty. If it is normal, its limit set is invariant under the ambient group by base-point independence and conjugation. Minimality of the Kleinian limit set gives one inclusion, and subgroup containment gives the other. Neither properness nor finite generation of the subgroup is needed.

Let a discounted gain $Y$ take values $-1,1,10$ with physical probabilities $1/10,4/5,1/10$. Then $m=17/10$ and $C=801/100$, so the minimal martingale measure in a one-period market has signed density $Z_{\ast }(10)=-610/801$. Nevertheless $(109/200,89/200,1/100)$ is a strictly positive pricing probability with zero mean gain. Thus absence of arbitrage does not ensure positivity of the minimal density. The minimum-norm identity holds for signed pricing densities; interpretation as a dominated martingale measure requires nonnegativity.

For any torsion-free connection, expand the coordinate Lie derivative of a tensor field and cancel the symmetric connection terms to obtain the displayed identity. If the connection is metric-compatible, lowering the vector index commutes with the covariant derivative and it becomes ${\mathcal{L}}_{\xi }{g}_{ab}=2{\nabla }_{(a}{\xi }_{b)}$. Symmetry of the connection alone does not suffice: with $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the two expressions ${\mathcal{L}}_{\xi }g$ and $2{\nabla }_x{\xi }_x$ are $2e^{2x}$ and $4e^{2x}$, respectively.

Changing from linear stellar metallicity to its logarithm multiplies the metallicity distribution function by its Jacobian determinant. A flat distribution in linear metallicity rises exponentially in logarithmic metallicity. The commonly used iron abundance coordinate is equivalent to total-metallicity scaling only if relative element abundances are fixed.

Consider an investment-consumption problem with an independent one-way rate change of intensity $\lambda$ from $r_1$ to $r_0$. For a stock of constant drift $\mu$ and positive volatility $\sigma$, set $B_j=r_j+(\mu -r_j{)}^2/(2R{\sigma }^2)$, ${\gamma }_0=[\rho -(1-R)B_0]/R$, and $D_1=\rho +\lambda -(1-R)B_1$. In the finite-value regime ${\gamma }_0,D_1>0$, the prechange consumption coefficient is the unique positive root displayed above. The left side increases continuously from zero to infinity. Before the change, the value is ${\gamma }_1^{-R}{w}^{1-R}/(1-R)$, the stock fraction is $(\mu -r_1)/(R{\sigma }^2)$, and consumption is ${\gamma }_{1}w$. At the change these become the ordinary Merton consumption-investment problem controls for $r_0$. The jump term in the Hamilton-Jacobi-Bellman equation is $\lambda (V_0-V_1)$; knowing only the marginal law of the change time does not justify that term without a conditional intensity assumption.

For $0<R<1$, common constant interest and volatility, and observed excess drift bounded in absolute value by $b_{+}>0$, let $V_{+}$ be the finite Merton consumption-investment problem value for excess drift $b_{+}$. In its Hamilton-Jacobi-Bellman equation, optimizing the stock holding contributes $-b^2(V_{+}^{\prime }{)}^2/(2{\sigma }^2{V}_{+}^{\prime \prime })$. Since $V_{+}^{\prime \prime }<0$, this contribution increases with $b^2$. Thus every admissible policy makes accumulated discounted CRRA utility plus discounted $V_{+}$ a nonnegative local supermartingale, hence a supermartingale. Dropping its nonnegative terminal term and using the monotone convergence theorem proves the value bound. The condition $[\rho -(1-R)(r+b_{+}^2/(2R{\sigma }^2))]/R>0$ is essential for the finite supermartingale construction.

In the infinite-horizon Merton consumption-investment problem with constant relative risk aversion utility, this is the optimal ratio of consumption to wealth. A finite value requires ${\gamma }_M>0$. The state-price budget constraint then fixes consumption's multiplier because $\mathbb{E}[{\xi }_t(y{e}^{\rho t}{\xi }_t{)}^{-1/R}]=y^{-1/R}{e}^{-{\gamma }_{M}t}$, whose time integral is $y^{-1/R}/{\gamma }_M$. When this integral diverges, the value is $+\infty$ for $0<R<1$ and $-\infty$ for $R>1$.

Consider irreversible optimal stopping of employment paying income $\epsilon >0$ and imposing disutility $\lambda >0$. Assume $r,\rho >0$, nonzero market price of risk $\kappa$, a positive Merton consumption constant, and solvency allowing borrowing against perpetual income up to $\epsilon /r$ before retirement. Let $m<0$ solve ${\kappa }^{2}m(m-1)/2+(\rho -r)m-\rho =0$. The retirement threshold is the displayed wealth level.

To derive it, apply the wealth-variable Legendre dual. The working dual equals the retired dual plus $h(z)=\epsilon z/r-\lambda /\rho +D{z}^m$. Excluding the growing homogeneous solution enforces the solvency asymptotic. Value matching and smooth fit at the retirement marginal value $z_{\ast }$ give $h(z_{\ast })=h^{\prime }(z_{\ast })=0$, hence $z_{\ast }=(\lambda r/(\epsilon \rho ))m/(m-1)$ and $D=-\epsilon z_{\ast }^{1-m}/(rm)>0$. The retired inverse marginal value gives $\overline{w}={\gamma }_M^{-1}{z}_{\ast }^{-1/R}$.

The root equation implies $z_{\ast }/(\lambda /\epsilon )=1+{\kappa }^{2}m/(2\rho )<1$. Thus the boundary exceeds the instantaneous gain-zero level ${\gamma }_M^{-1}(\epsilon /\lambda {)}^{1/R}$. A tighter borrowing constraint changes this free-boundary problem. Smooth fit matches the first derivative; matching the second derivative would incorrectly remove the income option.

For an observed finite-state continuous-time Markov chain with rate matrix $Q$, constant bank rate $r$, and state-dependent stock drifts ${\mu }_i$ and positive volatilities ${\sigma }_i$, constant relative risk aversion utility gives $V_i(w)=A_i{w}^{1-R}/(1-R)$. Set ${\kappa }_i=({\mu }_i-r)/{\sigma }_i$ and $d_i=\rho -(1-R)(r+{\kappa }_i^2/(2R))$. The Hamilton-Jacobi-Bellman equation reduces to the displayed nonlinear algebraic system with $A_i>0$. Its feedback controls are $c_i=A_i^{-1/R}w$ and ${\theta }_i=({\mu }_i-r)w/(R{\sigma }_i^2)$.

A damped Newton method in the logarithmic variables $A_i=e^{x_i}$ preserves positivity. Solving these equations produces a candidate value; admissibility, stochastic-integral bounds and the investment value transversality condition still need verification.