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Write $J_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ for the Dirac current and $J_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ for the axial current. Under parity symmetry in quantum field theory, the Dirac spinor phases cancel andThe extra sign for the axial current follows from ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. Combining these transformations with part (c), both currents transform under CP symmetry as $J_{V,A}^{\mu }(x)\mapsto -{\mathbb{P}}^{\mu }{}_{\nu }{J}_{V,A}^{\nu }(x_P)$. The vector relativistic quantum field transforms as $V^{\mu }(x)\mapsto {\eta }_C{\mathbb{P}}^{\mu }{}_{\nu }{V}^{\nu †}(x_P)$. The two parity matrices cancel in the tensor contraction, givingFor a real vector relativistic quantum field with ${\eta }_C=-1$, the interaction is invariant under CP symmetry. A complex vector relativistic quantum field instead maps to the conjugate-field interaction, which must be compared with its Hermitian conjugation partner.

The unit sphere and the kernel $e^{-c|y-x{|}^2}$ are invariant under every rotation in three dimensions fixing $y$. Therefore $T(y)$ has the same axisymmetry, so part (a) gives

Choose spherical polar coordinates with polar axis $y$ and put $u=\cos \theta =x\cdot y$. Since $|x|=|y|=1$,The tensor contraction giving the trace isSimilarly,At $c=0$ these expressions have the continuous limiting values $4\pi$ and $4\pi /3$. Writing them as $A=T_{kk}=3\alpha +\beta$ and $B=T_{ij}{y}_i{y}_j=\alpha +\beta$ gives

The Cartesian tensor law is $T_{ik}^{\prime }=Q_{ia}{Q}_{kb}{T}_{ab}$. In every complete tensor contraction, summation over a transformed index contracts a pair of $Q$ factors to a Kronecker delta. In detail,Thus all three are invariant under rotations. They are $\mathrm{tr}T$, $\mathrm{tr}(T^2)$ and $\mathrm{tr}(T^3)$. For a real symmetric tensor, orthogonal diagonalization givesExpanding ${\prod }_i(\lambda -{\lambda }_i)$ and using Newton's identities to recover the characteristic polynomial from traces of powers yieldsComparison identifies $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

The tensor contraction of a symmetric tensor with an antisymmetric one vanishes. Rename the two summed indices and use their opposite exchange symmetries:Hence $R_{ij}{A}_{ij}=0$. Using the decomposition from part (iii) givesThis is orthogonality of the real symmetric and antisymmetric matrix subspaces under the Frobenius inner product. It is a summed scalar identity, not an assertion about the unsummed products of individual entries.

Start from the three-index transformation law and contract the first two Cartesian indices:Orthogonality supplies ${\sum }_i{Q}_{ia}{Q}_{ib}={\delta }_{ab}$, where $\delta$ is the Kronecker delta. The remaining factor is exactly the vector transformation rule. Therefore $C_k$ is a vector.

This tensor contraction removes two indices by pairing them through the Euclidean metric. In general nonorthonormal coordinates, contraction must use the appropriate metric or pair a covariant and a contravariant index; the bare repeated-lower-index sum relies on Cartesian orthonormal coordinates.

Let $M_{ijkl}={\int }_V{x}_i{x}_j{x}_k{x}_l{d}^{3}x$. It is an isotropic tensor and is invariant under every permutation of its four indices. Thus the three coefficients in the general rank-four isotropic form must coincide:A double tensor contraction givesHere the three contracted Kronecker delta products contribute $9$, $3$ and $3$. Contracting instead with the four unit vectors yieldsThis agrees with subtracting two instances of the isotropic fourth moment of a ball.

A complete pairwise tensor contraction of copies of $A$ is represented by closed index cycles. Each cycle is a trace $\mathrm{tr}(A^r)$, possibly with a sign from reversing an index pair. If a contraction is internal to one copy, it contains the already vanishing $\mathrm{tr}A$. Otherwise an odd total number of copies forces at least one cycle of odd length, since a sum of even lengths is even. That trace vanishes by the transposition argument in part (ii). Therefore every such complete contraction is zero. This proves vanishing odd complete contractions of an antisymmetric matrix without assuming that all contractions form a single trace.

Let $\mathbf{u}$ be the boost velocity, and decompose the fields into components parallel and perpendicular to it. The parallel components are unchanged. The perpendicular components obeyAll vectors inside these expressions are perpendicular to $\mathbf{u}$. Thus $|\mathbf{u}\times {\mathbf{B}}_{\perp }{|}^2=u^2{B}_{\perp }^2$, and similarly for ${\mathbf{E}}_{\perp }$. In the difference of squares, the mixed terms cancel becauseIt follows thatAdding the unchanged parallel components provesFor the mixed dot product, expand directly. The terms ${\mathbf{E}}_{\perp }\cdot (\mathbf{u}\times {\mathbf{E}}_{\perp })$ and $(\mathbf{u}\times {\mathbf{B}}_{\perp })\cdot {\mathbf{B}}_{\perp }$ vanish. The remaining cross-product identity isHenceParallel-perpendicular mixed dot products are zero in both frames, soRotations also preserve these two expressions, completing the proof for proper Lorentz transformations. The second expression is a pseudoscalar under spatial parity, but is invariant under the boosts at issue here.

These results also follow from tensor contractions. In the convention used below, $F_{\mu \nu }{F}^{\mu \nu }=2(B^2-E^2/c^2)$ and $F_{\mu \nu }{\tilde{F}}^{\mu \nu }=-4\mathbf{E}\cdot \mathbf{B}/c$. The numerical sign of the latter depends on the chosen orientation of the dual electromagnetic field tensor, while its invariance under proper Lorentz transformations does not.

The required invariant is obtained by tensor contraction, rather than by an ordinary sum of squares of the entries of the electromagnetic field tensor. With ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, raising both indices givesThe electromagnetic field tensor is an antisymmetric second-rank tensor, so every independent off-diagonal entry occurs twice in its contraction. The three electric contributions each acquire a minus sign, whereas the magnetic contributions retain their sign. ConsequentlyAll indices are contracted, so this quantity is a Lorentz scalar. Since the speed of light is invariant,is Lorentz invariant. Both terms have the units of squared electric field, ${\mathrm{V}}^2{\mathrm{m}}^{-2}$ in SI units.

Let ${\mathbf{v}}_1=(c/2,0,0)$ and ${\mathbf{v}}_2=(c/(2\sqrt{2}),c/(2\sqrt{2}),0)$ in A. Each has Lorentz factor $g=2/\sqrt{3}$. The relative Lorentz factor of $S_2$ in the rest frame of $S_1$ follows from their four-velocity tensor contraction:Let $b=\sqrt{1-{\Gamma }^{-2}}$ be the relative speed divided by $c$. In the rest frame of $S_1$, take $\theta$ between the photon's propagation direction and $S_2$'s velocity. The observed angular frequency is the invariant $\omega =-k\cdot U$. Applying it to the emitter, whose rest angular frequency is ${\omega }_0$, gives ${\omega }_0=\Gamma {\omega }_1(1-b\cos \theta )$. Since ${\omega }_1={\omega }_0$,This is a relativistic Doppler effect calculation using both emitters' and receivers' rest clocks; the time delay of reception introduces no extra factor for constant velocities.

For the direction in A, write the wave four-vector as $k^{\mu }=({\omega }_A/c)(1,\mathbf{n})$, where $\mathbf{n}$ is a unit vector. Equality of the emitted and received angular frequencies gives $k\cdot (U_1-U_2)=0$. Since the two laboratory Lorentz factors are equal, their time components cancel:Thus the complete answer in three spatial dimensions isIf propagation is additionally assumed to lie in the stars' motion plane, this reduces to either direction along the internal angle bisector, $22.5^{\circ }$ or $202.5^{\circ }$ from $+x$. Reflection exchanging the stars' laboratory velocities explains the angle bisector by symmetry. The source does not specify coplanar propagation or the stars' positions, so it does not fix the sign or exclude an out-of-plane ray. For example $\mathbf{n}=(0,0,1)$ gives ${\omega }_1={\omega }_2=g{\omega }_A$ and satisfies the same no-shift condition. Initial positions can be chosen to make such a ray connect emission and reception. All these possibilities give the same rest-frame $\theta$ above.

With constant rest mass, $\mathbf{p}=m\gamma \mathbf{v}$ and the chain rule gives $\dot{\gamma }={\gamma }^3(\mathbf{v}\cdot \mathbf{a})/c^2$. Hence the laboratory relativistic force isUse the vector triple product identity $\mathbf{v}\times (\mathbf{v}\times \mathbf{a})=\mathbf{v}(\mathbf{v}\cdot \mathbf{a})-v^2\mathbf{a}$, and ${\gamma }^3(1-v^2/c^2)=\gamma$, to obtainThe relativistic energy is $E=m\gamma c^2$, whose time derivative is $m{\gamma }^3\mathbf{v}\cdot \mathbf{a}$. Taking the dot product of the first expression for $\mathbf{F}$ with $\mathbf{v}$ gives the same quantity, because $\gamma +{\gamma }^3{v}^2/c^2={\gamma }^3$. Thus $dE/dt=\mathbf{F}\cdot \mathbf{v}$. Since $d/d\tau =\gamma d/dt$, the four-force isIts tensor contraction with the four-velocity vanishes: $F^{\alpha }{U}_{\alpha }=-{\gamma }^2\mathbf{F}\cdot \mathbf{v}+{\gamma }^2\mathbf{F}\cdot \mathbf{v}=0$. This orthogonality also follows by differentiating $p^{\alpha }{p}_{\alpha }=-m^2{c}^2$. Resolving the laboratory acceleration relative to $\mathbf{v}$ gives ${\mathbf{F}}_{\parallel }=m{\gamma }^3{\mathbf{a}}_{\parallel }$ and ${\mathbf{F}}_{\perp }=m\gamma {\mathbf{a}}_{\perp }$.

For a diffeomorphism, pull a type $(r,s)$ tensor field back by the inverse differential on vector slots and the dual differential on covector slots. Invertibility distinguishes this construction from pullback of purely covariant tensors along an arbitrary smooth map. This operation preserves tensor products and tensor contractions, supplying the flow definition of the Lie derivative of a tensor field.

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

A tensor derivation is a real-linear operation preserving every tensor type, obeying the tensor-product Leibniz rule, and commuting with every tensor contraction. A derivation $D$ of smooth functions and a real-linear operator on vector fields satisfying $D(fY)=fDY+(Df)Y$ extend uniquely to a tensor derivation. On a differential one-form it must satisfyThis expression is linear over smooth functions in $Y$. In a local frame $D{e}_a=A^b{}_a{e}_b$, the dual rule is $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Apply the product rule to every coefficient and frame factor to define the extension; these dual signs cancel under contraction. Frame changes agree by differentiating the inverse matrix. Cutoffs prove locality and hence uniqueness from local expansions.

A tensor field is a smooth section of a vector bundle formed from tensor products of the tangent bundle and cotangent bundle. With $k$ covector factors and $l$ vector factors it belongs toIts coefficients in any local frame vary smoothly. Different conventions list the two counts in different orders, so the factor order should be specified. Tensor contractions use the canonical evaluation of a covector on a vector and are independent of the frame.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

For a real antisymmetric matrix, transpose gives $\mathrm{tr}(A^{2j+1})=\mathrm{tr}((-A{)}^{2j+1})=-\mathrm{tr}(A^{2j+1})$, so this trace vanishes. Every complete pairwise tensor contraction of copies of $A$, using the Euclidean Kronecker delta, splits into closed index cycles. Each cycle is a trace of a power of $A$, up to a sign from index order. An odd total number of copies forces an odd-length cycle, so the whole contraction vanishes. This assertion concerns pairwise index contraction with the metric; inserting additional tensors would be a different claim.