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Split the array in its first two slots:For every symmetric second-rank tensor $s$, antisymmetry gives $d_{ijk}^a{s}_{ij}=0$. Hence the stipulated vector equals $d_{ijk}^s{s}_{ij}$.

Let $R$ change the orthonormal frame, so $s_{ab}^{\prime }=R_{ai}{R}_{bj}{s}_{ij}$. The vector transformation law givesfor every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric matrix forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal basis matrices. Multiplying by $R_{ai}{R}_{bj}$ and using orthogonality yieldsThis is exactly the rank-three tensor law. It is the symmetric-test criterion for a third-rank Cartesian tensor, a form of the quotient theorem for Cartesian tensors. Equivalently one may test $s_{ij}=(u_i{v}_j+v_i{u}_j)/2$ for arbitrary vectors $u,v$ and apply the quotient theorem twice.

True. Since $A$ and $A^{\prime }$ are antisymmetric, the matrix $D=R^T{A}^{\prime }R-A$ is antisymmetric. Invariance of contractions with every antisymmetric second-rank tensor gives $D:B=0$ for every antisymmetric test matrix $B$. Such a matrix can be freely prescribed in one basis and then transformed to define a valid test tensor. Choose $B=D$ in that basis. Thenso $D=0$ and $A^{\prime }=RA{R}^T$. Together with the given antisymmetry, this proves that $A$ is an antisymmetric second-rank tensor. The antisymmetric contraction test for an antisymmetric tensor works because the Frobenius inner product is nondegenerate on the antisymmetric subspace.

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.

For each fixed coordinate function $x^{\alpha }$, the expression $\square x^{\alpha }$ is a scalar: $\alpha$ labels four chosen functions and is not a free tensor index. The equation is therefore coordinate invariant when those functions are held fixed. However, the demand that the currently chosen coordinates themselves be harmonic coordinates is a chart condition, not a four-vector equation invariant under arbitrary changes of coordinate functions.

Indeed, for new functions ${\bar{x}}^{\alpha }=f^{\alpha }(x)$ the d'Alembert operator givesThe second term prevents a vector transformation law. An affine redefinition preserves harmonic coordinates; an arbitrary nonlinear one does not. For example, a Cartesian spatial coordinate $x$ in Minkowski spacetime has $\square x=0$, whereas $\square (x^2)=2$, and $x^2$ is a valid new coordinate on $x>0$. The equations are scalar equations for specified functions, while harmonic gauge is chart dependent.

The Levi-Civita connection is the unique affine connection that has both vanishing torsion tensor and metric compatibility. The affine connection decomposition shows exactly how an arbitrary affine connection departs from it: the contorsion tensor contributes the metric-compatible torsion correction, and the disformation tensor contributes the nonmetricity correction. The torsion and nonmetricity tensor, together with the metric tensor, determine the difference uniquely.

In the paper's convention the actual contorsion correction is $-K$, with the corrected half factor from part (i), and the disformation correction is $W$. Both corrections are tensors, although the separate connection coefficients $\Gamma$ and $S$ are not tensors. If both torsion and nonmetricity vanish, then $\boxed{\Gamma =S}$. Vanishing nonmetricity alone permits torsion, and vanishing torsion alone permits nonmetricity.

Use the connection convention ${\nabla }_{\alpha }{Y}^{\rho }={\partial }_{\alpha }{Y}^{\rho }+{\Gamma }^{\rho }{}_{\mu \alpha }{Y}^{\mu }$, with the derivative index last. This is consistent with the printed curvature formula and the final formula in part (iv). The paper's $T_{\mu \nu }{}^{\rho }={\Gamma }^{\rho }{}_{\mu \nu }-{\Gamma }^{\rho }{}_{\nu \mu }$ is then the negative of the geometric torsion tensor defined by $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. Keep the paper's component convention throughout this question.

Set $A_{\mu \nu \rho }=g_{\rho \lambda }({\Gamma }^{\lambda }{}_{\mu \nu }-S^{\lambda }{}_{\mu \nu })$. The difference of affine connections is a tensor. Since the Levi-Civita connection $S$ is symmetric and metric compatible, expanding the nonmetricity tensor givesHere $T_{\mu \nu \rho }=g_{\rho \lambda }{T}_{\mu \nu }{}^{\lambda }$ is antisymmetric in its first two slots, while $N_{\mu \nu \rho }$ is symmetric in its first two slots. Cyclically permuting these identities and eliminating the other components givesRaise the last slot to obtain the connection decompositionwhereThe PDF is missing the factor $1/2$ in its definition of $K$. With the printed $K$, the decomposition would instead contain $-K/2$. In these formulas $T^{\rho }{}_{\mu \nu }$ raises the first slot of the fully lowered tensor; it must not be confused with $T_{\mu \nu }{}^{\rho }$, which raises the last slot.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

Use signature $(-,+,+,+)$, so the future-directed unit normal obeys $n^{\mu }{n}_{\mu }=-1$. The spatial projection tensor and the induced metric areIndeed $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\rho }{P}^{\rho }{}_{\alpha }=P^{\mu }{}_{\alpha }$: this is an idempotent linear map onto the tangent space of the spatial hypersurface. For a covariant tensor both slots must be projected:Contracting either slot with $n$ supplies the vanishing factor $P^{\mu }{}_{\alpha }{n}^{\alpha }$. Thus both normal contractions vanish, without requiring symmetry of $T$.

Write $V=T_p\mathcal{M}$ and $V^{\ast }$ for its dual space. A type-$(1,1)$ tensor is equivalently a linear map $A:V\to V$: the corresponding scalar-valued multilinear map is $(X,\eta )\mapsto \eta (A(X))$. Thus the composition definesIt is linear in $X$ because $A$ and $B$ are linear maps, and linear in the covector $\eta$ by evaluation. This proves directly that the composition is a tensor of type $(1,1)$.

Choose a basis $e_{\alpha }$ of the tangent space and its dual basis $e^{\beta }$. With the paper's component order $A(e_{\alpha })=A_{\alpha }{}^{\gamma }{e}_{\gamma }$,Equivalently, in the usual output-first matrix notation, $C^{\beta }{}_{\alpha }=B^{\beta }{}_{\gamma }{A}^{\gamma }{}_{\alpha }$: the matrix product is $BA$, not $AB$. This composition of mixed tensors contracts one output index with one input index. Under the tensor component transformation law, the two Jacobian factors at that contracted index cancel, leaving exactly one factor for each free index.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

Work in positive dimension $n\ge 1$. A cap set has no distinct zero-sum triple. Over the finite field ${\mathbb{F}}_3$, if two elements of a zero-sum triple agree then all three agree, since $2=-1$. Thus a zero-sum triple in a cap set must be diagonal. We prove the Ellenberg–Gijswijt cap-set bound by the polynomial method in combinatorics, keeping the exponential constant explicit and removing the fixed prefactor.

First prove the needed slice rank of a diagonal tensor over any field. The slice rank of a three-variable tensor is the least number of terms in a decomposition of the forms $a(x)b(y,z)$, $c(y)d(x,z)$, and $e(z)h(x,y)$. A diagonal tensor on a set of cardinality $m$, with every diagonal entry nonzero, has slice rank at most $m$ by one slice for each entry. For the reverse inequality, suppose a decomposition has $r$ slices in the $x$ direction, $s$ in the $y$ direction, and $t$ in the $z$ direction, with $r+s+t<m$. LetThe rank-nullity theorem gives $\dim W\ge m-r$. The maximum-support vector in a finite-dimensional subspace lemma supplies $u\in W$ with at least $m-r$ nonzero coordinates: choose $u$ of maximum support of a vector; restriction $W\to F^{\mathrm{supp}u}$ must be injective, or adding a nonzero kernel of a linear map element would strictly enlarge that support of a vector. This argument is valid over finite fields, without any assumption that a generic vector avoids all coordinate hyperplanes.

Contract the proposed tensor decomposition in $x$ against $u$. The diagonal tensor becomes a diagonal matrix of matrix rank at least $m-r$. All $x$ slices vanish. Each remaining $y$ or $z$ slice becomes a matrix of matrix rank at most one, so subadditivity of matrix rank gives matrix rank at most $s+t<m-r$, a contradiction. Therefore the slice rank of a diagonal tensor is exactly $m$.

Now, for a cap set $A\subseteq {\mathbb{F}}_3^n$, consider the polynomialEach factor is one when its argument is zero and zero otherwise, because the two nonzero elements of ${\mathbb{F}}_3$ have square one. Thus $P$ is the indicator function of $x+y+z=0$. Restricted to $A^3$, it is a diagonal tensor with every diagonal entry one, and its slice rank is $|A|$.

Every monomial in the expansion has individual exponents at most two and total polynomial degree at most $2n$. Consequently at least one of its $x$, $y$, or $z$ blocks has polynomial degree at most $\lfloor 2n/3\rfloor$. Assign each monomial to one such block, breaking ties in a fixed way, and collect assigned terms by the monomial in that block. Each collection is a slice. Ifthere are at most $m_n$ slices in each direction. Thus $|A|\le 3m_n$.

The low-degree monomial count for the cap-set bound follows directly from a weighted sum. Each exponent vector counted by $m_n$ satisfies $1\le 2^{2n/3-{\sum }_i{a}_i}$, soThis supplies its own elementary exponential estimate; no unstated probabilistic estimate is required. Since ${\theta }^3=343/16<27$, we have $\theta <3$, but $|A|\le 3{\theta }^n$ alone would leave a fixed prefactor.

Use Cartesian powers of cap sets to remove that prefactor. The $k$-fold Cartesian product $A^k\subseteq {\mathbb{F}}_3^{nk}$ is still a cap set: in each block, a zero-sum triple of its elements must have all three entries equal. Hence $|A{|}^k\le 3{\theta }^{nk}$. Taking $k$th roots and letting $k$ tend to infinity proves the removal of an exponential prefactor by Cartesian powers:The strict middle inequality follows by cubing: $343/16<(14/5{)}^3$. We can therefore takeFor every $n\ge 1$, a set with $|A|\ge C^n$ cannot be a cap set, and hence contains the required distinct triple. If zero dimension were included, the singleton ${\mathbb{F}}_3^0$ would be another small-case exception; positive dimension is the convention used here.

An objective time derivative transforms as a physical tensor under a time-dependent rigid change of observer. If $x^{\ast }=Q(t)x+c(t)$, with $Q$ orthogonal and $A^{\ast }=QA{Q}^T$, objectivity requiresThis means the constitutive equation makes the same physical prediction in translating and rotating frames, rather than requiring individual tensor components to be invariant.

The printed plus commutator uses the derivative-index-first velocity gradient, $G_{ij}={\partial }_i{u}_j$. This is the transpose of the component-index-first convention $L_{ij}={\partial }_j{u}_i$ often used for the velocity gradient. In this notation $\omega =G-G^T$, and the rate is the Jaumann derivativeTo check the covariance of the Jaumann derivative, let $R=\dot{Q}{Q}^T$. Direct transformation givesThe observer-rotation terms cancel in the displayed corotational rate. Equivalently, in the $L$ convention the physical spin tensor is $W=(L-L^T)/2=-\omega /2$, and the same rate is $D_{t}A-WA+AW$. One must transpose the gradient and reverse the commutator together; using $L$ in the printed plus-sign expression would not be objective.

Another example is the upper-convected derivative, written consistently in the $L$ convention asThe extra terms in $L^{\ast }=QL{Q}^T+R$ cancel those in $D_t^{\ast }{A}^{\ast }$, so this also transforms covariantly. In contrast, the uncorrected material derivative $D_{t}A$ of a second-rank tensor is not objective: its transformation contains $R{A}^{\ast }-A^{\ast }R$. For instance a constant anisotropic tensor in a stationary fluid has zero material derivative in its original frame, but its components have a nonzero commutator derivative in a rotating observer's frame. The ordinary partial time derivative is likewise not an objective tensor rate.

The material derivative follows the interfacial surfactant. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ describes dilution by tangential expansion and concentration by compression. The term $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing surface area due to normal motion of a curved interface. Finally, $D_s{\nabla }_s^{2}C$ describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.

The relevant surface Péclet number isWrite $S=\mathbf{n}\cdot \mathbf{A}\cdot \mathbf{n}$ and ${\mathbf{u}}_s=\mathbf{Ax}-S\mathbf{x}$. Since $\mathbf{A}$ has zero trace and ${\nabla }_s\mathbf{n}={\mathbf{I}}_s/a$, the surface divergence isThere is no normal motion. To first order in ${\mathrm{Pe}}_s$, the steady surfactant balance becomes $D_s{\nabla }_s^2{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. The traceless quadratic $S$ is a degree-two spherical harmonic, so ${\nabla }_s^{2}S=-6S/a^2$. The mean of $C^{\prime }$ is zero by total surfactant conservation; consequentlyThe discarded advective term ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$ is smaller by ${\mathrm{Pe}}_s$. This is the quadrupolar surfactant distribution on a spherical interface.

With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and $\gamma ={\gamma }_0-K{\gamma }_{1}S$, together with ${\nabla }_{s}S=2{\mathbf{I}}_s\mathbf{An}/a$, givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.

The ambient Stokes flow has the symmetry of the symmetric traceless tensor $\mathbf{E}$. In the Unscaled Papkovich–Neuber representation, ${\chi }_{\infty }=\mathbf{x}\cdot \mathbf{E}\cdot \mathbf{x}/2$ is harmonic and yields $\mathbf{Ex}$. The decaying vector potential must have the form $\mathbf{E}\nabla (1/r)$, and the scalar disturbance must have the degree-two form $\mathbf{E}:\nabla \nabla (1/r)$. They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives $1+P-3Q=0$. Its tangential velocity gives $\alpha =1+2Q$. The tangential stress boundary condition then yieldsHence $\alpha =5/(5+2M)$, $Q=-M/(5+2M)$ and $P=-(5+5M)/(5+2M)$. The inviscid interior supplies only the constant pressure balancing $2{\gamma }_0/a$. Matching the remaining normal stress givesEliminating $P,Q,\alpha$ givesAs $M\to \infty$, $\alpha \to 0$: the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; $M$ can grow through increasing ${\gamma }_1$ without invalidating the linear concentration approximation.

Withthe electromagnetic field tensor is the antisymmetric tensorUnder a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, the four-potential transforms as a four-vector and thereforeThus $F$ is an antisymmetric rank-two tensor under Lorentz transformations.

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

The covariant derivative adds one covariant index, so $\nabla S$ has type $(2,2)$. The outer product of two tensors adds their index counts. Hence

A tensor of rank $n$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. In components it obeys one change-of-basis factor for each of its $n$ indices.

A rank-$k$ Cartesian tensor has components $T_{i_1\dots i_k}$ transforming under an orthogonal change of basis $R$ as $T{\prime }_{i_1\dots i_k}=R_{i_1{j}_1}\cdots R_{i_k{j}_k}{T}_{j_1\dots j_k}$.