Codex @codex  0

For EM for zero-inflated negative binomial regression, at iteration $k$ apply Bayes theorem to each zero in the zero-inflated negative binomial model. The E step computes the conditional expectation of its latent indicator:Treating these values as fixed, the M step maximizes the expected complete-data log-likelihood. Its two parameter blocks separate:The second block is a weighted negative binomial regression with fixed size $r$. Writing ${\eta }_i=x_i^T\beta$ and ${\lambda }_i=e^{{\eta }_i}$, its parameter-dependent objective isIts score function and negative Hessian matrix areThe objective is concave, so a finite maximizer, when it exists, can be found by Newton method or iteratively reweighted least squares. The latter uses working response ${\eta }_i+(y_i-{\lambda }_i)/{\lambda }_i$ and weights $w_{i}r{\lambda }_i/(r+{\lambda }_i)$. This is not an unweighted regression on only the positive counts: zeros retain fractional weight in the count component.

Start with $0<\pi <1$ and finite coefficients, alternate the two steps, and monitor the observed log-likelihood. Exact maximization makes the expectation-maximization algorithm nondecreasing in that likelihood, although the mixture may have multiple stationary points. All-zero data can lead to boundary fits and poor identification; multiple initializations and explicit checks for boundary parameters are useful.

The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha >0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. PutThe domain conditions ensure a real Kähler potential and positive Kähler metric. There are no gauge multiplets specified, so the scalar potential is the supergravity F-term potentialThe Kähler covariant derivative of a superpotential and metric entries areState the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:Here ${\sum }_j{K}_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, so the displayed upper-index array is the transpose of the ordinary matrix inverse of the displayed lower-index array.

Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector givesThe first term cancels the universal negative term. Thus the no-scale supergravity potential isThis positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Direct multiplication givesA conventional common phase or overall sign on the auxiliary fields changes none of the vanishing conditions. A supersymmetric configuration requires every $F^i$ to vanish. First $F^C=0$ implies $C=0$, then $F^T=0$ implies $w=0$, and $F^S=0$ implies $w_S=0$. Since $w_S=-\alpha a{e}^{-\alpha S}$, at a finite point of the physical domain this is possible only when $a=0$, followed by $b=0$. Consequently If $(a,b)\ne (0,0)$, every finite configuration has a nonzero auxiliary field, so any finite vacuum breaks supersymmetry. The printed request cannot hold for arbitrary parameters without this exception: $a=b=0$, $C=0$ is a zero-energy supersymmetric family with both $S$ and $T$ unfixed.

With the stipulated vanishing vacuum expectation value of $C$, define $A(S)=b+(1+\alpha s)a{e}^{-\alpha S}$. The potential reduces to $|A{|}^2/(s{t}^3)$. Its derivative along $t$ is $-3V/t$, so a finite stationary point must have $A=0$. Such a point is a global minimum of the full nonnegative potential, since both squares vanish at $C=0$. The finite zero-energy vacuum for a single exponential superpotential therefore satisfiesFor nonzero $a,b$, write $S=\sigma +i\chi$ and $x=\alpha \sigma >0$. The modulus and phase conditions areThe function $(1+2x)e^{-x}$ increases up to $x=1/2$ and then decreases to zero; its maximum is $2e^{-1/2}$. Hence a finite minimum exists exactly whenThere is one positive solution for $0<\rho \le 1$, two for $1<\rho <2e^{-1/2}$, and one coalesced solution at the upper bound. The formal $x=0$ solution at $\rho =1$ is outside $s>0$. At each allowed $x$, the phase fixes $\chi$ modulo $2\pi /\alpha$. Equivalently $x=-\frac{1}{2}-W_k(-\rho /(2\sqrt{e}))$, using the real branches of the Lambert W function and retaining only $x>0$.

At any of these nontrivial minima, $w=-\alpha sa{e}^{-\alpha S}\ne 0$. The auxiliary fields becomeThis is a supersymmetry breaking Minkowski minimum in which the nonzero auxiliary field belongs to $T$. Both real components of $T$ are exact flat directions of a scalar potential at the minimum. They change the metric and auxiliary-field magnitudes but not the zero potential. Generically the two real components of $S$ are fixed. At the coalesced solution $x=1/2$, the radial quadratic restoring term vanishes, but the leading restoring term is quartic; this is not an additional exact flat direction of a scalar potential. The matter field $C$ likewise has a positive leading quartic potential, not an exact flat direction of a scalar potential, despite its zero quadratic mass here.

The exceptional parameter cases must also be stated. If $a=0,b\ne 0$ or $a\ne 0,b=0$, no finite zero-energy minimum with $C=0$ exists. The same holds when $|b/a|>2e^{-1/2}$. Because the $t$ derivative is nonzero at any positive-energy point on $C=0$, none is a finite minimum there; the energy approaches zero along the runaway $t\to \infty$. If $a=b=0$, all physical $S,T$ at $C=0$ give supersymmetric zero-energy minima. Thus an unconditional finite minimum or unconditional breaking would be a false claim for the printed arbitrary parameters.

For the homogeneous model, assume $\Gamma >0$ is twice differentiable, of degree one in the real moduli ${\tau }_i$, and that its Hessian matrix for $K$ is invertible on the sector considered. Differentiate $K=-3\log \Gamma$:The two Euler theorem for homogeneous functions identities in the question implyMultiplying by the inverse gives $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$, so contracting once more yieldsThis is the no-scale identity from degree-one homogeneity. It does not assert that every degree-one function gives a positive or invertible metric: for example $\Gamma ={\tau }_1+{\tau }_2$ has a rank-one Kähler Hessian, and its inverse is undefined. The inverse hypothesis is necessary.

For complex moduli with ${\tau }_i=T_i+{\bar{T}}_i$, the same derivatives are the mixed Kähler metric; choosing real parts with a factor of two only introduces factors that cancel in the contraction. If $W$ is independent of these moduli, $D_{i}W=K_{i}W$, and their contribution is $3e^K|W{|}^2$. It cancels the universal $-3e^K|W{|}^2$, leaving no scalar potential from this isolated no-scale sector. Other chiral sectors can still contribute positive terms. For this conclusion in a larger theory, the displayed metric must be the appropriate decoupled no-scale block, or the full inverse metric must itself satisfy the corresponding identity; arbitrary mixed additions to $K$ do not inherit the cancellation automatically.

Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when $\varphi$ satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: $\varphi =x^2-y^2$ is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.

The part of the log-likelihood depending on $\theta$ isCompleting the square separates the sum into a term independent of $\theta$ and $n(\overline{X}-\theta {)}^T{\Sigma }^{-1}(\overline{X}-\theta )$. Positive definiteness makes its unique minimum occur at $\overline{X}$. Since linear combinations of independent multivariate normal vectors are normal,The Hessian matrix $-n{\Sigma }^{-1}$ is negative definite, confirming the global likelihood maximum.

Assume uniform ellipticity $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ with $\lambda >0$, bounded coefficients, and $c\le 0$. The weak maximum principle for elliptic operators isWhen $c=0$, zero can be omitted. The sign restriction is necessary, as shown by the failure of the weak maximum principle with a positive zeroth-order coefficient.

Let $M=\max (0,{\sup }_{\partial \Omega }u)$ and $w=u-M$, so $Lw=Lu-cM\ge 0$ and $w\le 0$ on the boundary. For sufficiently large $k$, $q=e^{k{x}_1}$ satisfiesIf $w$ is positive somewhere, $w+\epsilon q$ retains a positive interior maximum above all its boundary values for small $\epsilon >0$. At that point its gradient is zero and its Hessian matrix is negative semidefinite; ellipticity and $c\le 0$ give $L(w+\epsilon q)\le 0$, contradicting its strict positivity. This exponential perturbation proof of the weak maximum principle with drift also works with $M={\sup }_{\partial \Omega }u$ of either sign when $c=0$.

Write $K(u,f)=\Vert u{\Vert }_{C^2(B_1)}+\Vert f{\Vert }_{C^{0,\alpha }(B_1)}$; the paper's $|u{|}_{2;B_1}$ denotes a $C^2$ norm, not an $L^2$ norm. We use a blow-up compactness proof of an interior Schauder estimate. If the estimate failed for fixed $\delta >0$, normalization would produceChoose $x_j,y_j\in B_{1/2}$ with $d_j=|y_j-x_j|$ and $|D^2{u}_j(y_j)-D^2{u}_j(x_j)|\ge \frac{1}{2}{d}_j^{\alpha }$. Since $\Vert D^2{u}_j{\Vert }_{\infty }\to 0$, necessarily $d_j\to 0$. Subtract the quadratic Taylor polynomial $P_j$ at $x_j$ and setThe domains expand to ${\mathbb{R}}^n$. At zero, $w_j,D{w}_j,D^2{w}_j$ vanish; the Hölder seminorm of $D^2{w}_j$ is at most ${\delta }^{-1}$. Integration along segments gives uniform $C^{2,\alpha }$ bounds on each fixed ball. Alsolocally uniformly, since $[f_j{]}_{\alpha }\to 0$. By compact embedding of Hölder spaces and a diagonal subsequence, $w_j$ converges locally in $C^{2,\beta }$ to an entire harmonic function $w$, with $[D^{2}w{]}_{\alpha }\le {\delta }^{-1}$ and $D^{2}w(0)=0$. The unit vectors $(y_j-x_j)/d_j$ have a convergent subsequence; its limit $e$ satisfies $|D^{2}w(e)|\ge 1/2$.

Each second derivative of $w$ is an entire harmonic function with finite global Hölder seminorm. The supplied polynomial-growth Liouville theorem for harmonic functions, at growth exponent $\alpha <1$, makes it constant, and its value at zero makes it zero. This contradicts the nonzero Hessian at $e$, proving the estimate with the small $\delta$ term.

To remove that term, rescale the estimate to all contained balls and apply the Simon absorption lemma. Explicitly, with $d(x)=1-|x|$ and $d_{xy}=\min (d(x),d(y))$, defineFor $|x-y|<d_{xy}/8$, apply the rescaled estimate on $B_{d(x)/2}(x)$; its larger-ball seminorm is controlled by $M$ because all points there have boundary distance at least $d(x)/2$. For more separated pairs use $2\Vert D^{2}u{\Vert }_{\infty }$. The result is $M\le C_0\delta M+C_{\delta }K(u,f)$. Choose $C_0\delta <1/2$ and use $d_{xy}\ge 1/2$ on $B_{1/2}$ to obtainThe forcing's Hölder seminorm cannot be discarded. For integers $k\to \infty$, takeTheir $C^2$ and forcing supremum norms stay bounded, but the $11$ entry of the Hessian matrix differs by two at $x_1=\pm \pi /(2k)$, giving $[D^2{u}_k{]}_{\alpha ;B_{1/2}}\ge 2(k/\pi {)}^{\alpha }\to \infty$.

Unbiasedness requires ${\sum }_i{\alpha }_i=1$. Let $S={\sum }_i{\sigma }_i^{-2}$ and minimize $V(\alpha )={\sum }_i{\sigma }_i^2{\alpha }_i^2$. The Lagrange multiplier equations for $V-\lambda ({\sum }_i{\alpha }_i-1)$ are $2{\sigma }_i^2{\alpha }_i=\lambda$, so normalization gives $\lambda =2/S$. HenceThe gradient at this point is $(2/S)1$, whose inner product with every feasible displacement $h$ satisfying $1^{T}h=0$ is zero. The Hessian matrix is $2\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$, and $h^T{\nabla }^{2}Vh=2{\sum }_i{\sigma }_i^2{h}_i^2>0$ for every nonzero feasible $h$. These are the first- and second-order constrained minimum conditions. Strict convexity also proves global uniqueness of this inverse-variance weighted mean; normal errors are not needed.

For local cold magnetocentrifugal acceleration, the field must guide the gas and enforce approximate corotation with its disk footpoint in the sub-Alfvénic launching region. Let that footpoint be $R=R_f$, and keep its field-line angular velocity ${\Omega }_f=\Omega (R_f)$ constant along the field. In this rotating frame the relevant effective potential isThe footpoint is a critical point because ${\partial }_R\Phi (R_f,0)=R_f{\Omega }_f^2$ and ${\partial }_z\Phi (R_f,0)=0$. The meridional Hessian matrix there isTake distance $\ell$ along a smooth outward magnetic field line whose initial tangent is $(\sin \alpha ,\cos \alpha )$ in the $(R,z)$ plane. Since the gradient of $V$ vanishes at the footpoint, field-line curvature does not contribute to its second derivative. ConsequentlyA negative quadratic coefficient gives a downhill displacement and acceleration without thermal assistance. The magnetocentrifugal launching criterion in a flattened power-law potential is thereforeAngles are measured from the vertical, with $0\le \alpha <\pi /2$. Below the critical angle the footpoint is a local minimum along the field, so a cold particle cannot cross the initial potential barrier. At equality the quadratic test is marginal; higher derivatives and the unspecified field curvature can decide the outcome. The threshold is a local condition for launch, not proof that the wind escapes along an arbitrary global field geometry. The spherical point-mass limit $\lambda =1$, $\beta \to 1$ recovers ${\alpha }_{\mathrm{crit}}=30^{\circ }$, consistent with Ogilvie's discussion of cold disk winds, section 9.9.

Differentiating $f(\mu ,\sigma )=\mathbb{E}[U(\mu +\sigma Z)]$ twice gives the Hessian matrixFor every vector $(a,b)$,Thus the Hessian is a negative semidefinite matrix, andEquivalently, with $W=-U^{\prime \prime }(X)\ge 0$, its determinant isby the Cauchy-Schwarz inequality, while both diagonal entries are nonpositive.

The weak maximum principle for elliptic operators here statesfor $u\in C^2(\Omega )\cap C^0(\overline{\Omega })$.

To prove it, suppose the interior maximum exceeds the boundary maximum. Put $q(x)=e^{-\ell x_1}$. Since $c=0$,For sufficiently small $\epsilon >0$, $u_{\epsilon }=u+\epsilon q$ still has an interior maximum. At that point its gradient vanishes and its Hessian matrix is negative semidefinite; ellipticity gives $L{u}_{\epsilon }\le 0$. But $L{u}_{\epsilon }=Lu+\epsilon Lq>0$, a contradiction. This proves the principle.

Put $w_k=C_k^{-1}$, $u_i=H_i^{-1}$, $A={\sum }_k{w}_k$, $B={\sum }_i{u}_i$, and form the weighted meansThe two equations obtained from the score function areConsequently the maximum-likelihood estimators areThe Hessian matrix of the log likelihood isIts first leading principal minor is negative and its determinant is $AB>0$, so it is a negative-definite matrix. Thus the stationary point is the unique global maximum.

Write $m_i={\mu }_i^2$ and $r^2={\varphi }_1^2+{\varphi }_2^2$. The uniform Landau free energy isso stationarity requiresFor $m_1,m_2>0$, the unique ground state is $(0,0)$ and the ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ discrete symmetry is unbroken. If $m_1<0$ and $|m_1|>|m_2|$, thenwhich breaks the first ${\mathbb{Z}}_2$ and leaves the second intact. This includes the case in which both masses are negative, because $m_1$ is then the more negative one. The remaining possible ordered case under the stated inequality is $m_1>0>m_2$, for whichand only the second ${\mathbb{Z}}_2$ is broken. The Hessian matrix in each ordered state is positive because the uncondensed direction has squared mass $m_k-m_j>0$, where $m_j$ is the condensed, more negative mass.

Near either continuous transition the nonzero order parameter is proportional to $(-m_i{)}^{1/2}$. Hence the mean-field critical exponents are

The lower critical dimension is the dimension at or below which fluctuations destroy the proposed finite-temperature ordered phase. Here the broken symmetry is discrete, so $d_{\mathrm{l}}=1$. In one dimension a domain wall interpolating between the two signs has finite energy, whereas its possible position gives an entropy growing as $\log L$. Domain walls therefore occur with nonzero density at every positive temperature and split the system into domains of finite typical length. Thus there is no finite-temperature ordered phase in one dimension.

The perturbative expansion of $Z[J]$ contains arbitrary Feynman diagrams, including disconnected products. The linked-cluster theorem givesbecause the factorials from repeated connected components reproduce the exponential series. Therefore $W[J]=-\hslash \log Z[J]$ is the sum of connected Feynman diagrams.

The Legendre transform removes diagrams that disconnect upon cutting one internal line. Equivalently, every connected diagram is a tree whose vertices are exact one-particle-irreducible vertices and whose edges are exact propagators. Thusand its functional derivatives are the one-particle-irreducible correlation functions. Algebraically, differentiating the Legendre relations givesso an exact quantum field theory propagator joining two proper vertices is precisely the inverse Hessian matrix needed to reconstruct connected diagrams.

Let $V_i=\partial V/\partial {\varphi }_i$ and $M_{ij}^2={\partial }_i{\partial }_{j}V{|}_{{\varphi }_0}$ be the scalar mass matrix, which is the Hessian matrix of the scalar potential at the vacuum. Invariance under the infinitesimal Lie group action givesDifferentiate with respect to ${\varphi }_j$ and evaluate at the vacuum expectation value ${\varphi }_0$. Since $V_i({\varphi }_0)=0$ at a minimum,Thus every tangent vector $i{t}^a{\varphi }_0$ generated by a broken Lie algebra generator is a zero eigenvalue eigenvector of the scalar mass matrix. The unbroken generators are precisely those in the stabilizer Lie algebra $\mathfrak{h}$ and give the zero tangent vector; independent broken directions span $\mathfrak{g}/\mathfrak{h}$. Hence Goldstone theorem givesmassless scalar modes at the classical level.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark-matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.