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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 $\tilde{M}=M\#{\overline{\mathbb{CP}}}^{,2}$, and let $E$ be the exceptional sphere. Extend a Spin-c structure by $c_1(\tilde{\mathfrak{s}})=c_1(\mathfrak{s})\pm \mathrm{PD}[E]$. Then ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}={\Phi }_{M,\mathfrak{s}}$. More generally, an evaluation $⟨c_1(\tilde{\mathfrak{s}}),E⟩=\pm (2\ell +1)$ introduces the factor $U^{\ell (\ell +1)/2}$. This follows by putting the blowup on one side of an admissible cut: the punctured exceptional piece is negative definite, its minus and plus tower maps are multiplication by this power of $U$, and the cobordism composition law gives the formula. In the $\pm 1$ case the exponent is zero, so nonvanishing is preserved.

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.

For the ideal magnetic response to a cylindrical cyclonic event, the plane-integrated $x$-component of $\mathbf{u}\times \mathbf{B}$ at duration $t$ isThis follows from $(\mathbf{u}\times \mathbf{B}{)}_x=(rf{B}_z+g{A}_r)\cos \varphi -(g/r)A_{\varphi }\sin \varphi$ and azimuthal integration. If boundary terms vanish, integration by parts gives the equivalent form $\pi B_0{\int }_0^{\infty }{r}^2{g}^{\prime }[\sin (ft)+tf\cos (ft)]dr$. For $f=g=a^2-r^2$ on $r<a$, zero outside, the response isThe value at zero is its continuous limit zero. It is initially $-\pi B_0{a}^{6}t/3$, but changes sign repeatedly at the nonzero roots of $\tan x=x$, with oscillatory envelope of order $t^{-1}$. The flow's kinetic helicity density is $2f^2\ge 0$, showing that the sign of a finite-duration ideal response is not fixed by the instantaneous helicity sign.

Turbulent magnetic pumping is an effective transport of the mean magnetic field produced by correlations in inhomogeneous or anisotropic turbulence. Its contribution to the mean-field electromotive force is ${\mathcal{E}}_{\mathrm{pump}}=\mathbf{\gamma }\times \overline{\mathbf{B}}$. In the mean induction equation this enters as $\nabla \times (\mathbf{\gamma }\times \overline{\mathbf{B}})$, the same algebraic form as advection by a velocity $\mathbf{\gamma }$. Thus it supplements the mean material velocity without requiring an equal mean mass flow. Downward pumping can help retain or couple fields in a solar dynamo.

Suppose a steady solenoidal velocity satisfies $\Delta \mathbf{u}=-\mathbf{u}$, and spatial averaging permits integration by parts without boundary terms. For a uniform test magnetic field $\mathbf{B}$, the small-magnetic Reynolds number expansion gives ${\mathbf{b}}_1=(\mathbf{B}\cdot \nabla )\mathbf{u}$ and $\Delta {\mathbf{b}}_2=-\nabla \times (\mathbf{u}\times {\mathbf{b}}_1)$. Using the self-adjointness of the Laplacian inside the average,The second equality concerns the averaged mean-field electromotive force; it does not assert that the quadratic forcing itself is monochromatic.

// Target: fluid-mechanics.bigb

In the residual unsatisfied formula, choose a Boolean literal contained in the largest number $a$ of clauses and make it true. Its opposite occurs in at most $a$ clauses, since it too was an eligible literal. The step permanently satisfies $a$ clauses; only clauses containing the opposite can become empty, so it loses at most $a$ clauses. Delete satisfied clauses and false literals and repeat. No satisfied clause or newly empty clause is counted twice. Summing gives total lost nonempty clauses at most total satisfied clauses $G$. All initially nonempty clauses are eventually in one of these classes, so $G\ge N_{+}/2\ge \mathrm{OPT}/2$. Initial empty clauses can never be satisfied and are excluded from $N_{+}$. This proves the approximation algorithm guarantee even with tautological clauses or repeated literals.

Write $\psi =\sqrt{\rho }{e}^{i\theta }$ away from zeros, and ${\mathbf{v}}_s=\nabla \theta$. Its Madelung equations give ${\rho }_t+\nabla \cdot (\rho {\mathbf{v}}_s)=0$ and ${\theta }_t+|{\mathbf{v}}_s{|}^2/2=1-\rho +{\nabla }^2\sqrt{\rho }/(2\sqrt{\rho })-\zeta D_n\rho$. For an incompressible normal fluid, $D_n\rho =-\nabla \cdot [\rho ({\mathbf{v}}_s-{\mathbf{v}}_n)]$. Taking a gradient proves the displayed equation when each space-time derivative of the slowly varying amplitudes is $O(ϵ)$: the quantum potential force is $O({ϵ}^3)$, while the bulk term is $O({ϵ}^2)$.

With the normal fluid at rest, linearize about the unit-amplitude plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-k^{2}t/2)}$. For amplitude and phase perturbations proportional to $e^{i\mathbf{l}\cdot \mathbf{x}+\sigma t}$, the continuity equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})R=l^2\Theta /2$, and the phase equation gives $(\sigma +i\mathbf{k}\cdot \mathbf{l})\Theta =-(2+l^2/2+2\zeta \sigma )R$. Their determinant is the displayed dispersion relation. The damping contains the laboratory exponent $\sigma$, not the Doppler-shifted exponent, because the reference flow is the normal component.

The density-product proof of the Girsanov theorem makes drift-corrected zero-starting increments a local martingale under the new measure. Adding the initial random variable requires its integrability under that measure. This is automatic if the density is one at time zero, because the measures agree on the initial sigma-algebra. Without this normalization it can fail: take $P(n)=2^{-n}$, $Q(n)=c/n^2$, constant $M_t=\log (Q(n)/P(n))$ and constant $X_t=n$. These are P-martingales and the prescribed exponential density is valid, but $X_0$ is not Q-integrable.