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A relativistic membrane sweeps out a three-dimensional worldvolume with induced worldvolume metric $h_{ij}={\partial }_i{X}^m{\partial }_j{X}^n{\eta }_{mn}$. Its leading geometric action is tension times its invariant volume. In static gauge in four-dimensional spacetime, one transverse field $Y$ gives $h_{ij}={\eta }_{ij}+{\partial }_{i}Y{\partial }_{j}Y$, and the matrix determinant lemma gives $S=-T\int \sqrt{1+{\partial }_{i}Y{\partial }^{i}Y}$. Expanding supplies the translation-zero-mode kinetic term. A domain wall of finite thickness also has higher-curvature corrections outside the long-wavelength limit.

Let $V_t=V_{t-1}+x_t{x}_t^T$ with positive-definite matrix $V_0$, and suppose $u_t=x_t^T{V}_{t-1}^{-1}{x}_t\le 1$. The matrix determinant lemma gives $\det V_t/\det V_{t-1}=1+u_t$. Since $2\log (1+u)-u$ has nonnegative derivative $(1-u)/(1+u)$ for $0\le u\le 1$, summation proves the displayed bound. In particular, $V_0=\lambda I$, $\lambda \ge 1$, and $\Vert x_t{\Vert }_2\le 1$ ensure $u_t\le 1$. Without that restriction a general form is ${\sum }_t\min \{1,u_t\}\le 2\log (\det V_n/\det V_0)$: for $u\ge 1$, $2\log (1+u)\ge 2\log 2\ge 1$. The bound connects widths of confidence ellipsoids to growth of a determinant.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.

For an invertible matrix $A\in {\mathbb{R}}^{n\times n}$ and real vectors $u,v\in {\mathbb{R}}^n$, the matrix determinant lemma statesThe expression $v^T{A}^{-1}u$ is a scalar, so a rank-one matrix update changes the determinant by a single multiplicative factor. One need not expand the updated determinant entry by entry.

For the preceding example with $A=I$, the dot product is $v^{T}u=3-2=1$. Hence the updated determinant is $1+1=2$. Direct evaluation confirmsMore generally, let $B(t)=I+tu{v}^T$ for a real scalar $t$. The same calculation gives $\det B(t)=1+t$. The determinant is one at $t=0$, two at $t=1$, and zero at $t=-1$.

A compact proof explains the scalar reduction. Put $w=A^{-1}u$ and factor $A+u{v}^T=A(I+w{v}^T)$. If $w\ne 0$, choose a basis starting with $w$. In that basis, $(I+w{v}^T)w=(1+v^{T}w)w$, while every other basis vector changes only by a multiple of $w$. Its matrix is triangular with diagonal $1+v^{T}w,1,\dots ,1$, giving the required determinant. If $w=0$, the correction vanishes. This uses multiplicativity of the determinant to restore the factor $\det A$.

The denominator in the Sherman–Morrison formula is exactly this factor. The rank-one update singularity criterion therefore identifies when the updated matrix inverse ceases to exist. Batson, Spielman and Srivastava use both identities for the symmetric update $A+v{v}^T$ in Twice-Ramanujan Sparsifiers, section 2.3, Lemmas 2.1–2.2.

Work for an arbitrary realization of the sequence, and set $u_t=X_t^T{V}_{t-1}^{-1}{X}_t$. In the Loewner order,By the spectral theorem for real symmetric matrices, every eigenvalue of $V_{t-1}$ is at least $\lambda$. The matrix inverse therefore has eigenvalues at most ${\lambda }^{-1}$, andFor $0\le u\le 1$, put $g(u)=2\log (1+u)-u$. Since $g(0)=0$ and $g^{\prime }(u)=(1-u)/(1+u)\ge 0$, we have $u\le 2\log (1+u)$. The matrix determinant lemma now givesTaking logarithms and using a telescoping sum,This elliptical potential bound holds for every realization and every $n$, without a probabilistic exceptional event. The condition $\lambda \ge 1$ is what ensures $u_t\le 1$.

Let $f(x)=(f_1(x),\dots ,f_p(x){)}^T$ and let an approximate experimental design $\xi$ specify the proportions of observations allocated to points of the design region. Its normalized information matrix of an experimental design isFor $N$ independent equal-variance observations realizing those proportions, $X^{T}X=NM(\xi )$, and the least-squares estimator has covariance matrix ${\sigma }^{2}M(\xi {)}^{-1}/N$. A D-optimal design maximizes $\det M(\xi )$, equivalently minimizing the determinant of this covariance at fixed $N$. A G-optimal design minimizes the largest variance of a fitted regression mean over the design region, equivalentlywhere $d$ is the design sensitivity function. This is the variance of the estimated mean multiplied by $N/{\sigma }^2$, not the variance of a new noisy observation.

For continuous regressors spanning a $p$-dimensional parameter space on a compact design region, the general equivalence theorem for optimal design states that, among approximate designs with nonsingular information matrices, the following are equivalent: being a D-optimal design, being a G-optimal design, andAt every support point of an optimal design, $d=p$. To see the criterion, the sensitivity average is $\int dd\xi =\mathrm{tr}(M^{-1}M)=p$, so the supremum is at least $p$. Adding an infinitesimal allocation at $x$ has directional derivative $d(x,\xi )-p$ for $\log \det M$. At a D-optimum each derivative is nonpositive. Conversely, if all sensitivities are at most $p$, the tangent bound for the concave function $\log \det M$ shows that no competing design increases it. A nonsingular D-optimum exists by the spanning assumption and compactness of the information-matrix set, so the smallest possible G-value is also $p$. Equality at support points follows from the average. The theorem refers to the full approximate-design class; constrained integer allocations or mandated augmentations require separate comparison.

For the linear mean ${\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2$, take $f(x)=(1,x_1,x_2{)}^T$. Equal replication at the four corners givesAll odd corner sums and the sum of $x_1{x}_2$ vanish, while each squared regressor sum is $4m$. If ${\bar{Y}}_{ab}$ is the mean of the $m$ observations at $(a,b)$, $a,b\in \{-1,1\}$, the least-squares normal equations giveIn particular the intercept is the average of the four cell means, and each slope is the average of its sign-weighted cell means. Independence and equal error variance suffice for this covariance formula; normal errors are not needed for it.

The normalized corner design has $M=I_3$, so its sensitivity iswith equality at all four corners. Since $p=3$, the general equivalence theorem for optimal design proves that the equally weighted corner design is D-optimal and G-optimal.

For the augmentation, all $m$ added observations share the regressor $f=(1,x_1,x_2{)}^T$. The enlarged design matrix satisfiesBy the matrix determinant lemma,For the permitted nine points, the determinant is $80m^3$ at the center, $96m^3$ at an edge midpoint and $112m^3$ at a corner. ThereforeAll four corners tie for the best mandated single-point augmentation. This is optimality among the specified augmented designs, not unrestricted D-optimality with $5m$ observations: the resulting normalized determinant is $112/125<1$, whereas equal corner proportions have determinant one. The restriction that the extra $m$ runs all occur at one point is what prevents recovering equal proportions.

A canonical Wess–Zumino model with a cubic superpotential already supplies a microscopic membrane-like object. Choose positive real $\mu ,g$ andThere are two isolated supersymmetric vacua, ${\varphi }_{\pm }=\pm \mu /g$. A planar domain wall has a static scalar profile depending on the transverse coordinate $z=X^3$, with one vacuum at either end. Its energy per unit area isChoosing $\alpha =\arg \Delta W$ gives $T\ge 2|\Delta W|$. The real increasing solution saturates the bound:The tension follows either from integrating $2({\varphi }^{\prime }{)}^2$ or from $W(\mu /g)-W(-\mu /g)=4{\mu }^3/(3g^2)$. This is a supersymmetric domain wall, and the first-order equation also implies the second-order scalar equation. Its thickness is of order $1/\mu$.

The wall breaks transverse translations: changing $z_0$ gives a normalizable scalar zero mode proportional to ${\varphi }^{\prime }(z)$. It also preserves only half the four real supercharges. In the chiral-fermion transformation, $F=-\overline{W^{\prime }}=-{\varphi }^{\prime }$ and the derivative term combine into a rank-half condition on the constant supersymmetry parameter, for example $ϵ=i{\sigma }^3\overline{ϵ}$ in a compatible Weyl convention. The orthogonal two real parameters are broken and generate localized fermion zero modes. One such profile satisfies $({\partial }_z+2g\varphi )f=0$, giving $f\propto {\mathrm{sech}}^2(\mu (z-z_0))$. The scalar and a three-dimensional Majorana fermion therefore form the low-energy wall multiplet. The domain-wall charge in N=1 supersymmetry is a boundary/tensor charge contributing to the extended algebra; the preserved combinations saturate $T=2|\Delta W|$. This avoids falsely applying the unextended particle-algebra argument of Question 3 to a wall with nonzero tension.

Promote the translation collective coordinate to a slowly varying worldvolume field $Y({\xi }^0,{\xi }^1,{\xi }^2)$. At wavelengths long compared with the wall thickness, its leading relativistic geometric action is the Dirac membrane actionIt is invariant under worldvolume reparametrizations and the target Poincaré group. In static gauge $X^i={\xi }^i$, $X^3=Y$, the matrix determinant lemma givesDirectly substituting the leading collective-coordinate profile into the scalar kinetic term gives the same coefficient, since $\int dz({\varphi }^{\prime }{)}^2=T/2$. Target Lorentz symmetry fixes the leading first-derivative nonlinear completion to the invariant area; finite-thickness physics can add higher-curvature terms. The area action is therefore a long-wavelength effective field theory, not an exact replacement for the microscopic wall at arbitrary momenta. Its embedding equation is ${\partial }_i(\sqrt{-h}{h}^{ij}{\partial }_j{X}^m)=0$.

For the four-dimensional supermembrane, extend the embedding to $(X^m(\xi ),{\theta }^{\alpha }(\xi ))$ with a four-real-component Majorana spinor. Pull back the invariant coframe of Question 7:Replacing $\partial X$ by $\Pi$ makes the area term invariant under rigid target supersymmetry, but does not yet give the appropriate fermionic gauge symmetry. Add a Wess-Zumino brane coupling to a superspace three-form potential $C_3$. In one orientation and matching normalization,The supermembrane closed four-form uses the distinct four-dimensional Fierz identity $(C{\gamma }_{mn}{)}_{(\alpha \beta }(C{\gamma }^n{)}_{\gamma \delta )}=0$. Differentiating $G_4$ reduces its coefficient to this symmetric tensor, so $d{G}_4=0$. A local potential therefore exists. Since the curvature is supertranslation invariant, the potential can vary by an exact form; its pullback integral is invariant modulo boundary terms. This is why the three-form $H_3$ of Question 7 is not itself the membrane field strength: integrating a membrane potential requires three-form degree and hence four-form curvature.

The required additional local fermionic gauge invariance is kappa symmetry. Define the worldvolume gamma matrices ${\gamma }_i={\Pi }_i^m{\gamma }_m$ andWith mostly-plus target signature, a local orthonormal worldvolume frame gives ${\Gamma }_{\kappa }={\gamma }_0{\gamma }_1{\gamma }_2$, so ${\Gamma }_{\kappa }^2=1$ and $\mathrm{tr}{\Gamma }_{\kappa }=0$. Thus $(1\pm {\Gamma }_{\kappa })/2$ are rank-two kappa symmetry projectors. The embedding variation gives $\delta {\Pi }_i^m=2i\delta \bar{\theta }{\gamma }^m{\partial }_i\theta$. The area variation is consequently proportional to $-2iT\sqrt{-h}\delta \bar{\theta }{\gamma }^i{\partial }_i\theta$. Contracting the displayed $G_4$ with the fermionic variation and using ${ϵ}^{ijk}{\gamma }_{ij}=2\sqrt{-h}{\Gamma }_{\kappa }{\gamma }^k$ gives the opposite Wess-Zumino contribution with an extra ${\Gamma }_{\kappa }$. Their sum is proportional toFor the boxed kappa variation, $\delta \bar{\theta }=\bar{\kappa }(1+{\Gamma }_{\kappa })$, so it vanishes by $(1+{\Gamma }_{\kappa })(1-{\Gamma }_{\kappa })=0$. Reversing membrane orientation reverses the Wess-Zumino sign and the matching projector. The cancellation fixes the relative charge magnitude to the tension; rigid supersymmetry alone would not fix it.

Worldvolume reparametrizations remove three of the four embedding scalars, leaving one transverse mode. Kappa symmetry removes half the four real embedding spinor components, leaving a two-component three-dimensional Majorana field, whose first-order equation has one on-shell fermionic polarization. The physical worldvolume modes are one scalar and one Majorana fermion, matching the wall zero modes. Without kappa symmetry the fermionic degrees would be doubled. A planar configuration preserves the target supersymmetries that can be compensated by the kappa projector, reproducing the wall's half-supersymmetry condition. This is the four-dimensional effective domain-wall interpretation of a supermembrane, rather than the eleven-dimensional M2-brane particle-content problem.

A SU(2) matrix has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Comparing this with the Pauli expansion givesConsequently all four Pauli coefficients must be real, and their norm condition isConversely, the Pauli matrix multiplication law gives $A{A}^{†}=(u_0^2+|\mathbf{u}{|}^2)I$ and $\det A=u_0^2+|\mathbf{u}{|}^2$, so these conditions imply unitarity and determinant one. For complex coefficients the determinant equation alone would not imply unitarity.

On either open hemisphere,Both signs are necessary globally. The two hemispheres meet along $u_0=0$, and the full four-coefficient constraint is exactly the unit sphere in ${\mathbb{R}}^4$. The coefficient map and its inverse are smooth, establishing SU(2) as the three-sphere as a manifold, not merely a dimension count.

For two real coefficient quadruples, expanding the matrix product with the Pauli matrix multiplication law givesFor an infinitesimal second factor, $v_0=1+O(|\mathbf{v}{|}^2)$. Thereforewhile $d{u}_0=-u_i{v}_i$. These variations are tangent to the unit sphere because $u_{0}d{u}_0+u_{i}d{u}_i=0$.

On a hemisphere, differentiation of the norm constraint gives ${\partial }_{u_i}{u}_0=-u_i/u_0$, and henceContracting with the Pauli-coordinate left-invariant vector fields on SU(2) and using ${\mu }_{ji}{u}_i=u_0{u}_j$ givesThe Pauli matrix multiplication law gives exactly the same expression for $-A{\sigma }_j$, soIt follows thatThis uses the Pauli matrix commutator identity. The commutator of two first-order vector fields is again a vector field, and the matrix entries of $A$ span the four coordinate functions $u_0,u_1,u_2,u_3$. Equality of their actions on these entries determines the tangent field. ThereforeThe real tangent field underneath $T_j$ is generated by the flow $A\mapsto A{e}^{it{\sigma }_j}$; left multiplication transports that tangent consistently, explaining its left invariance. The rescaled generators $-T_j/2$ have the usual positive-sign angular-momentum commutator.

To determine the Haar measure, compute the metric induced from the unit three-sphere. Since $d{u}_0=-u_{i}d{u}_i/u_0$, its line element isThe rank-one matrix determinant lemma givesThus the Pauli-coordinate Haar measure on SU(2) isup to an overall normalization. To check its group invariance, fix one unit coefficient quadruple and multiply by it. The product formulas define a linear transformation of the other quadruple. The norm of a product is the product of norms, as follows either from $A{A}^{†}$ or directly from the dot and cross products. Fixed multiplication by a unit quadruple is therefore an orthogonal transformation of ${\mathbb{R}}^4$. It preserves the round sphere metric and volume, on both the left and the right, proving that this is the invariant measure.

The same density can be checked directly against the vector fields. On a fixed hemisphere, $u_0/|u_0|$ is constant, andThe second term vanishes because its contractions involve either ${\delta }_{ik}$ or $u_i{u}_k$, both symmetric in $i,k$. Thus the density has zero divergence along every invariant tangent direction.

The hemisphere chart becomes singular at the equator, but its volume is finite:Both hemispheres give total volume $2{\pi }^2$, so probability-normalized Haar measure is $d^{3}u/(2{\pi }^2|u_0|)$ on each hemisphere. Finally, the three-sphere is closed and bounded in ${\mathbb{R}}^4$, hence compact; its smooth identification with SU(2) proves that $SU(2)$ is a compact group. The finite integral reflects this compactness, and the equatorial divergence is only a coordinate artefact.

Let $q=(\alpha ,\beta {)}^T$, letand recover the sufficient cross-products from the model1 normal equations by setting $c=A{\hat{q}}_0$ and $s=154.30+{\hat{q}}_0^{T}A{\hat{q}}_0$. Thenis the total squared residual about the fixed line. The squared sum of the ten residuals within each person, summed across people, isFor one person's ten observations, the marginal covariance matrix is ${\tau }^2{I}_{10}+{\sigma }^{2}11^T$. The matrix determinant lemma and Sherman–Morrison formula therefore give, up to an additive constant, twice the negative marginal log-likelihoodBecause the model was fitted with REML = FALSE, it minimizes this ordinary marginal maximum-likelihood objective. Thus $(V1,V2,V3,V4)$ belongs to the stated argmin over $alpha,\beta \in \mathbb{R}$ and $sigma,\tau >0$.

Apply the matrix determinant lemma to $HP-\lambda I=(H-\lambda I)-Hu,u^T$: its determinant is ${\chi }_H(\lambda )[1-u^T(H-\lambda I{)}^{-1}Hu]$.

On each half-space the deformation is affine. Continuity on the separating plane makes the two affine constants equal and gives $(A-B)t=0$ for every tangent vector $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth determinants are positive, hence $\boxed{A^{-1}a\cdot N<1}$.

Write $a_1={\eta }_1$, $a_2={\eta }_2$, $S=a_1^2+a_2^2$ and $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.

It remains to test the material plane. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the plane is spanned by $t$ and $e_3$. Both deformation gradients send $e_3$ to $a_1{e}_3$. If the plane is not the twin plane, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-plane vectors $Bt$ and $At$ are parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts prefactor is nonzero, so coplanarity requires $p_1=\kappa p_2$, precisely $p\parallel (1,\kappa ,0)\parallel N$. Conversely, on that twin plane $(A-B)x=0$, so the two affine maps agree and its image is a plane. Thus $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

For an invertible matrix $A$, the rank-one matrix update $B=A+u{v}^T$ is a singular matrix exactly when $1+v^T{A}^{-1}u=0$. This follows directly from the matrix determinant lemma. At that value, $u\ne 0$ and the nonzero vector $w=A^{-1}u$ satisfiesEvery vector in the kernel is a multiple of $w$, since $Bx=0$ implies $x=-A^{-1}u(v^{T}x)$. The matrix rank therefore drops by exactly one.

For $B(t)=I+tu{v}^T$ with $u=(1,2{)}^T$ and $v=(3,-1{)}^T$, the update becomes a singular matrix at $t=-1$, and $B(t)u=(1+t)u$. At $t=-0.99$, this direction is compressed by a factor $0.01$ and the inverse amplifies it by a factor 100. The actual spectral norm of the matrix inverse, which maximizes amplification over every direction, is about $700.17853409$.

The exact formula can expose a large condition number in the updated linear system even when the original one had a small condition number. It does not guarantee numerical stability merely because its evaluation uses fewer arithmetic operations. The denominator condition certifies existence; the condition number and floating-point behavior address different questions about using the resulting answer.

The Sherman–Morrison formula computes the matrix inverse after a rank-one matrix update. If $A$ is an invertible matrix and $\delta =1+v^T{A}^{-1}u\ne 0$, thenThe inverse correction is itself an outer product, divided by the scalar $\delta$. The matrix determinant lemma shows that the nonzero-denominator condition is exactly the condition that the updated matrix remains an invertible matrix.

For $A=I$, $u=(1,2{)}^T$, and $v=(3,-1{)}^T$, we have $\delta =2$. Thus the computed inverse isMultiplication in either order gives the identity matrix. The result reuses the original inverse $I$ and the same correction vectors, rather than solving independently for all four inverse entries.

Often only the solution of an updated linear system is needed. If $A{x}_0=b$ and $Az=u$, thenprovided $1+v^{T}z\ne 0$. The two solves use the same LU decomposition of $A$. This expresses the computational benefit without requiring an explicitly stored matrix inverse. Strang's Lecture 14, transcript pp. 9–10 explains this reuse of the original factorization.

To see why the inverse correction works, first consider $A=I$ and put $s=v^{T}u$. The rank-one matrix $R=u{v}^T$ satisfies $R^2=sR$. Expandingalso works in the reverse order. The general case follows by factoring out $A$ and replacing $u$ by $A^{-1}u$. A matrix identity has reduced to cancellation of scalar coefficients.

Historically, Jack Sherman and Winifred J. Morrison studied updating an inverse when a single matrix entry changes in their ^[ref]. Such an entry change is the special rank-one matrix update $\alpha e_i{e}_j^T$, with standard basis vectors $e_i,e_j$. The reference and the earlier 1949 abstract on row or column changes are recorded in Henderson and Searle, On Deriving the Inverse of a Sum of Matrices (1981), references 40–41. This distinguishes the specific original setting from the more general $u{v}^T$ formula used here.