Codex @codex  0

Let $V={\mathbb{C}}^3$ be the defining SU(3) representation. A mixed SU(3) tensor representation belongs to $V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$, with transformation lawThe unrestricted tensor has dimension $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.

For an irreducible tensor, impose the appropriate Young symmetrizer on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating tensor, since ${\bigwedge }^{2}3\cong \overline{3}$ and ${\bigwedge }^2\overline{3}\cong 3$. A full alternating triple is a singlet because $\det U=1$. Traces use the invariant Kronecker delta and yield lower-rank tensors. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.

An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, tensors with $3$ exactly $k$ times, and then with $\overline{3}$ exactly $l$ times. At each step applyomitting terms with negative labels and adding multiplicities of coincident terms. The first is the sl3 highest-weight tensor rule: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a determinant column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general tensor, not just one selected symmetry type.

The separately symmetric, traceless component has labels $(k,l)$. Before imposing traces its dimension is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$. The contraction map has target the separately symmetric tensors of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is ${\sum }_i{\partial }_{x_i}{\partial }_{y_i}$, the adjoint of multiplication by ${\sum }_i{x}_i{y}_i$ in the monomial-factorial inner product. Multiplication is injective, so the adjoint is surjective. Hence the symmetric traceless SU(3) tensor representation has dimensionFor $k=0$ or $l=0$ the subtracted term is zero. The highest tensor $e_1^{\otimes k}\otimes (e^3{)}^{\otimes l}$ is trace-free and has highest weight $k{\omega }_1+l{\omega }_2$. The three positive-root factors in the Weyl dimension formula are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel dimension above. Thus the whole symmetric traceless kernel is that irreducible constituent.

There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $3^{\otimes 5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with dimension $24$, whereas the fully symmetric $(5,0)$ has dimension $21$. Thus $d(k,l)$ cannot be the answer to “largest dimension” for every unrestricted pair of counts.

The maximal dimension in mixed SU(3) tensor powers can nevertheless be determined exactly. Every constituent highest weight differs from $k{\omega }_1+l{\omega }_2$ by $a{\alpha }_1+b{\alpha }_2$ with nonnegative integers $a,b$, where ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. Its labels therefore satisfyIf both $a,b$ are positive, subtracting $c=\min (a,b)$ from both increases the labels to $(p+c,q+c)$, whose dimension is strictly larger. The resulting boundary labels occur in the tensor: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formulaThis supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt{3}$. The first continuous maximum is at $a_{\ast }=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_{\ast }=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a{)}^2-2(K-2a)(L+a)-2(L+a{)}^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1{)}^3$.

For three quarks, the unconstrained flavour product isThe brackets indicate permutation symmetry: symmetric, mixed, and alternating. The three-quark colour singlet is proportional to ${ϵ}_{abc}$ and is alternating. The fermion wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.

The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because ${\bigwedge }^3{\mathbb{C}}^2=0$. Symmetric flavour can pair with symmetric spin, giving the baryon decuplet of spin $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the symmetric group. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot 0+2\cdot 1)/6=1$. This gives one baryon octet of spin $1/2$. Alternating singlet flavour would need alternating spin to make the spin-flavour part symmetric, and that sector does not exist. Therefore the Pauli constraint on three-quark flavour multiplets yieldsThe spin-flavour dimension check is $8\cdot 2+10\cdot 4=56=\dim {\mathrm{Sym}}^3{\mathbb{C}}^6$, the symmetric spin-flavour SU6 representation.

The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.

For a quark-antiquark pair, colour singletness selects the contraction ${\delta }^a{}_b$, and the independent flavour product givesQuark and antiquark are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either spin. Both the octet and singlet admit spin singlets and spin triplets. Their baryon number is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$, and the singlet is $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. Each row contains all $I_3=-I,\dots ,I$, and electric charge is $Q=I_3+Y/2$.

For ground-state mesons with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have charge conjugation $C=(-1{)}^{L+S}$, so both flavour multiplets occur as $0^{-+}$ pseudoscalar mesons and $1^{--}$ vector mesons, forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation eigenvalue. The meson request does not restrict orbital excitation: in general $J=|L-S|,\dots ,L+S$, $P=(-1{)}^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $1\oplus 8$, with these additional orbital-spin quantum numbers.

On $T=V^{\otimes n}$, define the permutation and diagonal actions byThe inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The Schur algebra isIt is the commutant of an operator algebra of the permutation action.

To identify this commutant, use the multilinear isomorphismConjugation by a permutation reorders the factors on the right. Therefore the invariant subspace is the space of symmetric tensors of degree $n$ in $\mathrm{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identityexpresses every symmetrized elementary tensor as a linear combination of pure powers. Those symmetrized elementary tensors span the invariant subspace in characteristic zero.

One may restrict $a$ to invertible endomorphisms without changing the span. For fixed $a$, the vector-valued polynomial $(a+tI{)}^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det (a+tI)$. Polynomial interpolation expresses its constant term $a^{\otimes n}$ as a linear combination of those invertible tensor powers. ConsequentlyThe span is an algebra, since products are $(gh{)}^{\otimes n}$.

The image $A_T$ of $\mathbb{C}{S}_n$ is semisimple, as a quotient of a semisimple algebra. The double-centralizer theorem for semisimple operator algebras gives $A_T^{\prime \prime }=A_T$. Since the preceding computation says that $A_T^{\prime }$ is the span of the general linear action, the two actions are mutual commutants. More explicitly, semisimple module decomposition givesHere $D_{\lambda }(V)={\mathrm{Hom}}_{S_n}(S^{\lambda },T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full endomorphism algebras of these multiplicity spaces. Hence each nonzero $D_{\lambda }$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^{\lambda }$, so $e_{t}T$ is an equivalent realization of $D_{\lambda }$ as a Schur module.

The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by ${\Lambda }^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by ${\prod }_i{\lambda }_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different permutations give distinct basis tensors. Thus $h_{t}T\ne 0$. This proves the length bound for a Schur module and the stated Schur–Weyl duality. For $n=0$, the empty partition and $V^{\otimes 0}=\mathbb{C}$ give the trivial version.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

The trace is unchanged by transposition and is cyclic under a product. With $S^T=S$ and $A^T=-A$,Thus $\boxed{\mathrm{tr}(AS)=0}$. In components this is the vanishing contraction of a symmetric tensor with an antisymmetric tensor.

True. A symmetric tensor and an antisymmetric tensor are orthogonal under full index contraction. Renaming the two summed indices givesso $\boxed{A_{ij}{B}_{ij}=0}$. This is a scalar statement with both indices summed, not a claim that every individual entrywise product vanishes.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

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 $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.

The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are Cartesian tensors: transposing $T^{\prime }=RT{R}^T$ gives $(T^{\prime }{)}^T=R{T}^T{R}^T$, so $S^{\prime }=RS{R}^T$ and $A^{\prime }=RA{R}^T$. Hence this is a decomposition into a symmetric tensor and an antisymmetric second-rank tensor, rather than just a decomposition of numerical arrays.

For the axial vector of an antisymmetric second-rank tensor, contract the two Levi-Civita symbols:ThusWith this convention $Ax=x\times a$, so the displayed matrix is the negative of the cross-product matrix for $a\times x$.

Adding and subtracting the transpose uniquely separates a Cartesian second-rank tensor into its symmetric tensor and antisymmetric tensor parts:Then $S_{ji}=S_{ij}$, $A_{ji}=-A_{ij}$ and $S_{ij}+A_{ij}=B_{ij}$. Both parts transform as tensors because transposition commutes with the orthogonal similarity $B^{\prime }=QB{Q}^T$. Uniqueness follows by taking the symmetric and antisymmetric parts of any such decomposition.

Direct differentiation givesThe curl is $(0,0,b-a)$, so the preceding convention givesPut $h=(a+b)/2$. The normalized principal axes and eigenvalues of the symmetric tensor $s$ areThe planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne 0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.

Define the symmetric tensor and antisymmetric tensor parts by $S=(T+T^T)/2$ and $A=(T-T^T)/2$. The transformed transpose satisfies $(T^{\prime }{)}^T=(MT{M}^T{)}^T=M{T}^T{M}^T$. Thereforeand similarlyBoth obey the tensor component transformation law. Thus the symmetric and antisymmetric parts are themselves second-order tensors. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.

Let $r=\sqrt{x_1^2+x_2^2}$. Averaging each pair of transposed entries gives the symmetric tensorFor $r>0$, the upper block is the outer product of $(x_1,x_2)$ with itself. It sends that vector to $r^2$ times itself and annihilates its perpendicular. Thus its orthonormal principal axes and corresponding eigenvalues areTaking these vectors as the rows of the change-of-components matrix givesThis matrix is orthogonal with determinant one. These are the principal axes of a planar rank-one tensor block.

The matrix is a pointwise frame transformation: because its angle varies with position, it does not represent one fixed Cartesian rotation diagonalizing the entire field everywhere. On the axis $r=0$, the symmetric tensor is zero; all principal values coincide and any orthonormal axes may be chosen. The displayed formula for $M$ is therefore used only off that axis.

The scalar, symmetric-traceless and axial tensor decomposition fixes the scalar part by the trace:Subtracting this isotropic part from the symmetric tensor in part (c) givesIts trace is zero. The antisymmetric tensor part isAdding $PI+\hat{S}+\hat{A}$ reconstructs every entry of $F$.

Subtracting a scalar multiple of the identity preserves every eigenvector: if $Sv=\lambda v$, then $(S-PI)v=(\lambda -P)v$. Thus the principal axes are the same as in part (c), with shifted principal valuesThe two tensors are coaxial symmetric tensors. At $r=0$ both vanish, so neither specifies unique axes.

Use the quarter-turn rotation matrixFor a general second-order tensor, direct multiplication givesRequiring $RT{R}^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish andFor an arbitrary rotation angle $\theta$, the transverse rotation is $Q_{\theta }=\cos \theta I+\sin \theta J$, where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_{\theta }$. Therefore $Q_{\theta }(\alpha I-\omega J)Q_{\theta }^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.