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An index-$h$ handle is $D^h\times D^{n-h}$, attached along $S^{h-1}\times D^{n-h}$. Its attaching sphere is $S^{h-1}\times \{0\}$ with its attaching framing of an embedded sphere, and its belt sphere in the new boundary is $\{0\}\times S^{n-h-1}$. A handle decomposition is an ordered sequence of such handle attachments.

A handle slide of one $h$-handle over another replaces its framed attaching sphere by a band sum with a parallel framed copy of the second attaching sphere, along a band in the boundary of the previously attached handles. The second handle stays fixed. The band transports the framing of an embedded sphere. The union of the two handles and their attaching collars can be reidentified by a diffeomorphism, so this changes the decomposition, rather than the underlying manifold. A sequence of these moves, allowing the usual isotopies of attaching data, gives handle decompositions related by slides. Reading the decomposition backwards interchanges attaching spheres and belt spheres; a slide of dual handles is a slide of the corresponding original handles.

For the standard handle-calculus assertion, take $1\le k\le n-2$, so both the relevant attaching and belt spheres are connected. We need the local geometric move behind handle-slide isolation of a cancelling pair. Write $p=A_1\cap B_1$. If $q$ is an intersection of some $A_j$, $j>1$, with $B_1$, choose an arc on $B_1$ from $q$ to $p$, and take a narrow parallel band to the unique sheet of $A_1$ at $p$. Slide $A_j$ over an appropriate parallel copy of $A_1$. Choose the copy's orientation so that the new sheet at $p$ has the opposite local smooth intersection number to the sheet at $q$.

Within a thin neighbourhood of the arc, the old sheet at $q$ and the added sheet at $p$ are the two ends of one band. An isotopy pushes that band through the neighbourhood of $B_1$ and removes both ends. Outside this neighbourhood the original $A_j$ is unchanged, apart from the added copy of the parts of $A_1$ away from $B_1$. Since $A_1$ meets $B_1$ only at $p$, this copy contributes no further intersections with $B_1$. Thus one slide followed by an isotopy removes one original geometric intersection with $B_1$. This is a local band move, not an appeal to handle cancellation or to an algebraic cancellation of signs.

The arcs and bands may be chosen one at a time; in the one-dimensional sphere case, start with an intersection adjacent to $p$, so that the arc interior contains no other intersection. Perturb to make all intersections transverse. The relevant spheres are compact, so there are only finitely many intersections. Repeating the move for every $j>1$ leaves $A_1$ and $B_1$ fixed and givesThe bands can avoid the remaining attaching data; any consequent change to intersections with other belt spheres is allowed at this stage.

For the second stage, view the same interface from the dual decomposition. Now $B_i^{\prime }$ are attaching spheres and $A_j^{\prime }$ are belt spheres. Apply the same local move to each intersection of $B_i^{\prime }$, $i>1$, with $A_1^{\prime }$, sliding that dual handle over the one with attaching sphere $B_1^{\prime }$. This changes the original $k$-handles. It keeps $A_1^{\prime }$ and $B_1^{\prime }$ fixed, while removing the unwanted intersections with $A_1^{\prime }$. The parallel copies and bands may be taken disjoint from all $A_j^{\prime }$, $j>1$, because those spheres are already disjoint from $B_1^{\prime }$ and from $A_1^{\prime }$. Hence their intersections with $B_1^{\prime }$ remain empty. We obtainNo handle has been removed.

There is an implicit index-range qualification in this geometric assertion. If arbitrary zero-handles are included, the second conclusion need not hold: two zero-handles joined by one one-handle have an attaching zero-sphere with one endpoint on each zero-handle belt sphere. Slides preserving this handle inventory cannot make its second endpoint avoid both belts. The dual obstruction occurs for top-index handles. Thus the argument uses the customary intermediate-index setting of the handle-slide lemma; the unqualified assertion at the extreme indices would require additional hypotheses or additional moves.

For the numerical formula, fix coherent orientations for the integer smooth intersection numbers, and letKeep the orientation at the surviving point, so its number remains $\delta$. A slide of a $(k+1)$-handle adds or subtracts the first row of the intersection matrix. Removing all geometric intersections in the first column has net row operationEven if opposite-sign intersections require extra slides, their net signed coefficient is the displayed one. In particular,The dual slides then perform the column operationsSince $C_1^{\prime }=(\delta ,0,\dots ,0{)}^T$, these operations leave the lower-right block unchanged. Thus the final answer, including both stages of slides, isThe remaining block is the Schur complement of the unit pivot $\delta$. If the original orientation is chosen so that $\delta =1$, it is simply $m_{lk}-m_{l1}{m}_{1k}$. For unoriented handles use intersection numbers modulo $2$, with the same formula interpreted in ${\mathbb{F}}_2$.

At the fitted hyperparameters, define $K=k_0(t,t)$, $K_{\ast \ast }=k_0(t^{\ast },t^{\ast })$, and $K_{\ast t}=k_0(t^{\ast },t)$, with the matrices formed by pairwise kernel evaluations. Conditional independence of observation noise and future latent values givesThe residual $f_{\ast }-K_{\ast t}{V}^{-1}y$ has zero covariance with $y$, so the joint normal law makes it independent of $y$. This derives the conditional multivariate normal distribution and the Gaussian process regression posterior:The covariance is a Schur complement and is positive semidefinite. When nonsingular, its joint density is $(2\pi {)}^{-M/2}|{\Sigma }_{\ast }{|}^{-1/2}\exp [-\frac{1}{2}(f_{\ast }-m_{\ast }{)}^T{\Sigma }_{\ast }^{-1}(f_{\ast }-m_{\ast })]$. Repeated phases can make it singular, in which case the displayed normal law is interpreted on its support.

This predicts the true latent light curve, so no future measurement variance is added. For noisy future observations, add their independent noise covariance to ${\Sigma }_{\ast }$. Fixing the hyperparameters omits their estimation uncertainty, as requested; integrating over their posterior would add it. The periodic model extrapolates by phase, even though all prediction times follow the observed times.

The Grassmann field $c$ is odd and the adjoint scalar field $\varphi$ is even. The two printed signs are consistent with a right-acting BRST differential, whose graded Leibniz rule isThe bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary Lie bracket $[c,\varphi ]=c\varphi -\varphi c$ givesThe second bracket here is graded because both its entries are odd. The graded Jacobi identity gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$. ThereforeThe side of the odd derivation is essential. With the usual left graded Leibniz rule and the same two printed signs, the result would be $-[[c,c],\varphi ]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent BRST symmetry formulas here use the right-acting convention.

Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}{T}^a$, an invariant positive invariant bilinear form on a Lie algebra, and the adjoint covariant derivative $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Couplings are absorbed into this convention; restoring $D_{\mu }={\partial }_{\mu }+g[A_{\mu },\cdot ]$ multiplies each ghost interaction vertex below by $g$. The associated BRST charge acts as $Q{A}_{\mu }=-D_{\mu }c$, $Q\bar{c}=ih$, and $Qh=0$.

Write $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$. For the gauge-fixing fermion $\Psi =\int {\bar{c}}^a(F^a-i{h}^a/2)$, the right graded Leibniz rule givesThe Gaussian functional integral over the Nakanishi-Lautrup field produces the positive gauge fixing term $F^a{F}^a/2$. The ghost operator is $-\partial \cdot D-(n\cdot \partial ){\mathrm{ad}}_{\varphi }/\sqrt{2}$.

Now use the canonical free kinetic terms $F_{\mu \nu }^a{F}_{\mu \nu }^a/4+(\partial {\varphi }^a{)}^2/2$. At nonzero momentum in Euclidean space $p$, put $\eta =n\cdot p$. The quadratic kernel for $(A_{\mu }^a,{\varphi }^a)$, per color, isThe transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2{\delta }_{\mu \nu }$. The scalar quantum field theory propagator is the inverse Schur complement, not merely the inverse of the scalar diagonal entry:Hence the free adjoint-scalar propagator in scalar-dependent gauge fixing isFor completeness the mixed quantum field theory propagator is $⟨A_{\mu }^a{\varphi }^b{⟩}_0=-{\delta }^{ab}\eta p_{\mu }/(\sqrt{2}{p}^4)$; ignoring this mixing would give an incorrect scalar answer.

The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$. For a unit $n$, $p^2=p_{\perp }^2+p_{\parallel }^2$, and the free scalar quantum field theory propagator still depends on $p_{\parallel }$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless zero mode in field theory at $p=0$ needs a separate infrared prescription.

Finally, integration by parts gives the ghost action in an unambiguous convention:Use $e^{ip\cdot x}$ for the Fourier transform of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the ghost vertices in scalar-dependent gauge fixingThe free Faddeev-Popov ghost field propagator is ${\delta }^{ab}/p^2$ and every closed ghost loop contributes a minus sign. There are no further ghost interaction vertices in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the Fourier transform and ghost-ordering conventions; the displayed ghost action fixes both here.

Let $J=11^T$ be the all-ones matrix. Every feasible block matrix in the first formulation has $t\ge 1$, because its principal $2\times 2$ submatrix on indices $0,i$ is $\left(\begin{array}{cc}t& 1\\ 1& 1\end{array}\right)$ and is a positive semidefinite matrix. In particular, $t>0$.

By the Schur complement,Set $U=tZ-J$. Then $U⪰0$, $U_{ii}=t-1$, and $U_{ij}=-1$ on every edge. Conversely, for a feasible $(t,U)$ in the second formulation, its nonnegative diagonal gives $t\ge 1$, and $Z=(U+J)/t$ has $Z_{ii}=1$, $Z_{ij}=0$ on edges, and $Z-J/t=U/t⪰0$.

Both transformations preserve the objective $t$. Hence the two semidefinite programs defining the complement theta number have the same feasible objective values and the same minimum.

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.

Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

The assertion uses the positive parameters required for $M$ to be a preconditioner: assume $\alpha >0$ and $\beta >0$. For $x\in {\mathbb{R}}^n$ and $z\in {\mathbb{R}}^m$, complete the square:The matrix 2-norm bound $\Vert Ax{\Vert }_2\le \Vert A{\Vert }_2\Vert x{\Vert }_2$ shows thatfor $x\ne 0$ when $\alpha \beta >\Vert A{\Vert }_2^2$. If $x=0$ and $z\ne 0$, the square contributes $\beta \Vert z{\Vert }_2^2>0$. Thus $\boxed{M\text{ is positive definite}}$. Equivalently, the Schur complement of the lower-right block is $\alpha I-A^{T}A/\beta \succ 0$.

Taken literally without the positivity inherited from part b, the product condition alone is insufficient: $A=0$ and $\alpha =\beta =-1$ is a counterexample. Thus $\alpha ,\beta >0$ is a necessary implicit hypothesis.

Root the graph at $x_0$ and let $P^{(0)}$ be the transition matrix of simple random walk killed on hitting $x_0$, indexed by the remaining vertices. Its Green matrix is $G=(I-P^{(0)}{)}^{-1}$. Successively eliminating vertices in an order $x_1,\dots ,x_n$ takes Schur complements; the corresponding diagonal pivot at step $j$ is exactly $g_{D\setminus \{x_0,\dots ,x_{j-1}\}}(x_j)$. The product of the pivots is thereforewhich is invariant under the elimination order.

PutAfter ordering $I_1$ before $I_2$, the only nonzero off-diagonal entries between the two blocks occur at $(i,j)$ and $(j,i)$, where they equal $-a$. Thuswhere the coordinate vectors in the two blocks are understood. Expanding the determinant according to whether neither or both cross-block entries are selected givesThe minus sign is the sign of the transposition pairing the two cross-block entries. This formula remains valid when either diagonal block is singular, so no inverse or Schur complement is needed.

For additive Gaussian white noise and a time-even order parameter, a midpoint discretization of the Onsager–Machlup functional evaluates drift derivatives at the midpoint configurations. Reversing a trajectory visits the same midpoints in reverse order, so the noise-to-path Jacobian determinant is invariant under reversal. It may depend on the trajectory, and must be distinguished from the constant normalization of the noise measure.

For nonconserved order-parameter dynamics with $\dot{p}=-\Gamma \mu [p]+f$, the factor at one time step, apart from a path-independent power of the step size, isHere the fields have first been restricted to a finite spatial grid, and ${\mathcal{H}}_n$ is the Hessian matrix of the free energy.

For mixed conserved and nonconserved order-parameter dynamics, the required noises for a joint configuration and flux trajectory arewhere $\mathsf{D},\mathsf{G}$ represent the divergence and gradient, and $\Delta =\mathsf{DG}$ represents the Laplacian. The Jacobian matrix isTaking its Schur complement gives the joint factorFor periodic boundary conditions, discretizing the two spatial operators compatibly gives $\mathsf{D}=-{\mathsf{G}}^{\mathsf{T}}$, so $\Gamma I-M\Delta$ is the positive relaxation operator when $\Gamma ,M>0$. Reversal leaves ${\mathcal{H}}_n$ unchanged and reverses the sign of the flux. The product of the joint factors is consequently the same for both histories, justifying its cancellation from the joint path probability of an order parameter and its flux. The functional chain rule used in the action ratio holds in the continuum limit of this midpoint convention.