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The symmetric group has $n!$ elements, not the printed order $n$. For $n\ge 2$ the identity and the $n-1$ adjacent transpositions have probabilities $1/n$, so the specified transition matrix is normalized. Its stationary distribution is uniform and it satisfies detailed balance, since each transposition is its own inverse. Adjacent transpositions generate the symmetric group, making this an irreducible Markov chain; the positive holding probability makes it an aperiodic Markov chain.

For a fixed label $i$, right multiplication swaps positions, so $f(\sigma )={\sigma }^{-1}(i)$ is the label's position. Under the uniform distribution on a finite set it is uniform on $\{1,\dots ,n\}$. ThereforeFor each generator $s_j=(j,j+1)$, the squared change $[f(\sigma s_j)-f(\sigma ){]}^2$ is one if the label occupies either of the two positions, and zero otherwise. Its stationary expected value is $2/n$, givingThus the tagged-label Rayleigh quotient for adjacent transpositions gives the corrected bound

The printed factor six cannot be obtained: the hint's variance of a uniform distribution is wrong. Directly, $\mathrm{Var}(U)={\int }_0^1{u}^{2}du-({\int }_0^{1}udu{)}^2=1/12$. Bounded convergence in distribution of $U_n$ does imply convergence of its first two moments, hence convergence of its variance, but the limit is $1/12$.

More strongly, the claimed asymptotic bound itself is false for this kernel. The Aldous spectral gap theorem states that the interchange process on a finite connected weighted graph has the same spectral gap as the single-label random walk with identical edge rates; see Theorem 1.1 and its proof. Here $P-I$ is that interchange process generator on a path graph with edge rates $1/n$. For the tagged-label generator, direct substitution of $g_{\ell }(k)=\cos (\ell \pi (k-1/2)/n)$ gives eigenvalues $-2(1-\cos (\ell \pi /n))/n$, $0\le \ell <n$, including the reflecting endpoint equations. These $n$ distinct eigenvalues exhaust the tagged-label function space. Consequently the theorem yieldswhich exceeds $6/n^3$ asymptotically. This use of a substantial external theorem is only to certify the printed claim's failure; the corrected factor-twelve upper bound above follows directly from the requested test function and the variational formula.

The most plausible cause is strong dependence between the latent variables and covariance hyperparameters in their joint posterior distribution. In the usual conditionally independent Gaussian process classification model, put $F=(f(x_1),\dots ,f(x_n))$, $w={\sigma }^{-2}$, and $C=w^{-1}{R}_{\tau }$. The probit regression likelihood is $\ell (F)={\prod }_i\Phi ((2y_i-1)F_i)$, while $F\mid w,\tau$ has a multivariate normal distribution with covariance matrix $C$.

For distinct inputs and positive length scales, $R_{\tau }$ is positive definite. Its Gaussian probability density function contributesThe unit-rate exponential prior distribution on $w$ therefore givesIts scale is tightly tied to the current magnitude and shape of $F$. Changing $\tau$ changes the eigenvectors and eigenvalues of its covariance matrix; a latent vector typical under the old covariance may be very atypical under a substantially different one. Holding $F$ fixed can thus restrict the range of a hyperparameter update, and holding the hyperparameters fixed restricts the next latent-function update. Even exact sampling of each full conditional distribution can move slowly along a narrow ridge of the joint posterior distribution.

Binary observations provide limited information about the absolute size of large correctly signed latent values. This can leave substantial scale uncertainty and strengthen the dependence. Long correlation lengths can also make the covariance matrix nearly singular and the latent coordinates highly dependent. These are plausible explanations for a large mixing time, not proofs that every data set causes slow convergence. A blocked Gibbs sampler removes dependence within its chosen blocks, but not dependence between them. Repeated identical inputs should be represented by one shared latent value; otherwise the nominal Gaussian matrix is singular and the displayed inverse formula must be replaced by a representation on its support.

Insert ${\Omega }_{jk}=-{ϵ}_{jkl}{\varphi }^l$ and the result for $S^{jk}$ into the Spinor representation of the Lorentz group. The contraction of two Levi-Civita symbols ${\sum }_{j,k}{ϵ}_{jkl}{ϵ}^{jkm}=2{\delta }_l{}^m$ givesThe matrix exponential of this block diagonal matrix is the stated $O$, so ${\psi }^{\prime }(x)=O\psi ({\Lambda }^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.

For a rotation through $2\pi$ about $z$, the Pauli matrix has eigenvalues $\pm 1$, hence $e^{i\pi {\sigma }^3}=-I_2$. The spacetime rotation is the identity, whileThis is the spinor sign under a full spatial rotation. A vector returns to itself already at $2\pi$. The comparison concerns the exponentiated Lorentz group representations: a Lie algebra by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the Spin group.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

The fixed-point equation for $u$ factors asThe cubic derivative is $3u^2+2u+3>0$ for every real $u$; its values at zero and one have opposite signs. Thus the possible first coordinates in the stated domain are exactly $u=0,u_{\ast },1$.

For finite $v$, the second fixed-point equation is $v[v^3-(1+u^2{)}^2]=0$. The renormalization-group fixed points are consequentlyThere are six finite points and three boundary points at infinity. Infinity is interpreted in a compactified domain, not as an ordinary real number. The reciprocal coordinate at infinite coupling $w=1/v$ transforms as $w^{\prime }=(1+u^2{)}^2{w}^4$, making $w=0$ a well-defined boundary fixed point.

At any finite fixed point $(\tilde{u},\tilde{v})$, let $\delta u=u-\tilde{u}$, $\delta v=v-\tilde{v}$. The Jacobian matrix linearization isThe only point with both coordinates strictly interior is $\tilde{u}=u_{\ast }$, $\tilde{v}=(1+u_{\ast }^2{)}^{2/3}$. Using its fixed-point relations, the matrix becomesIts eigenvalues are $\rho$ and $4$. In particular, $u_{\ast }<1/\sqrt{3}$ because the increasing cubic is already positive there, so $\rho >1$ (numerically about $1.68$). The corresponding eigenvectors can be chosen as $(1,c/(\rho -4){)}^T$ and $(0,1{)}^T$. Both discrete multipliers exceed one, henceBoth perturbations are relevant directions of a fixed point under repeated coarse-graining, rather than one stable and one unstable direction. A length-rescaling factor was not specified, so these multipliers should not be assigned numerical critical scaling exponents without additional information. At $a=0,1$, the $u$ multiplier is zero; at $v=0$ or $w=0$ the other multiplier is also zero. Thus the four corner points attract locally within the domain, the two positive finite boundary points at $a=0,1$ are saddles, and the points $(u_{\ast },0),(u_{\ast },\infty )$ have one repulsive and one attractive direction.

Compute the coupling flow at zero magnetic field, or assume the source has support only in retained Fourier modes, so that the shell Gaussian measure is centered. First justify the shell propagator. For a positive quadratic kernel on the shell, the Gaussian functional integral gives covariance equal to the inverse kernel,with both momenta restricted to $\Lambda /b<|\mathbf{p}|\le \Lambda$. Inserting the Fourier transforms and using the Dirac delta function yields the Gaussian shell covarianceChanging $\mathbf{p}$ to $-\mathbf{p}$ gives the negative-exponent convention as well. The dependence is on the separation, as required by translation invariance. The printed numerator uses $\mathbf{x}$ alone: it is correct only if that symbol means the separation or if $\mathbf{y}=0$. For example at $\mathbf{x}=\mathbf{y}\ne 0$, the covariance must equal the constant coincident value $G_{>}(0)$, while the literal printed integral generally depends on $\mathbf{x}$. Thus the missing separation is a real source qualification, not a change of Fourier-sign convention.

Integrating out the shell gives $H_{\mathrm{eff}}[{\varphi }_{<}]=H_0[{\varphi }_{<}]-\log ⟨e^{-V[{\varphi }_{<}+{\varphi }_{>}]}{⟩}_0^{\mathrm{shell}}$, up to a field-independent constant. At first order in $u_0$, the cumulant expansion of a coarse-grained free energy retains $⟨V⟩$. Odd moments of the centered Gaussian measure vanish and Wick theorem gives $⟨{\varphi }_{>}^4⟩=3G_{>}(0{)}^2$. Writing $I_1=G_{>}(0)$,The last term affects only the constant; the second is a momentum-independent tadpole diagram correction. Before rescaling it changes $r_0$ to $r_0+u_0{I}_1/2$, leaves the quartic coefficient at $u_0$, and produces no gradient correction. Combining with part (a) gives the leading coupling recursionThis is one-loop shell mass renormalization in scalar quartic theory. Here $I_1={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}\alpha /(p^2+\alpha r_0)$. The shell Gaussian is well defined if $p^2+\alpha r_0>0$ there; near $r_0=0$ this follows from $\alpha >0$. A negative slow-mode mass may be stabilized by the retained quartic term and does not require pretending that the full unconstrained quadratic measure at negative mass is normalizable.

Linearizing about $(r_0,u_0)=(0,0)$ givesIts mass and quartic eigenvalues are $b^2$ and $b^{4-D}$. The off-diagonal term is an additive critical-mass shift: setting the bare $r_0=0$ is not generally the critical tuning when $u_0>0$. The mass remains relevant for every $D$; at $D=2$ the two linear eigenvalues coincide and the matrix can have a Jordan block, but both perturbations are still relevant.

For $D>4$, the quartic interaction is an irrelevant operator, and the Gaussian fixed point attracts weak quartic perturbations after the mass and magnetic field have been tuned. The quartic can nevertheless be a dangerously irrelevant coupling: its positive value stabilizes the ordered phase. For $D<4$, it is a relevant operator, so the Gaussian description is unstable toward interactions. At $D=4$ it is a marginal operator; first-order perturbation theory alone does not decide its fate.

To settle that borderline case, retain the leading quartic contribution from the second cumulant, at zero external momentum in the local expansion. The two-fast-field part of $V$ is $(u_0/4)\int {\varphi }_{<}^2{\varphi }_{>}^2$. Since $⟨{\varphi }_{>}^2(\mathbf{x}){\varphi }_{>}^2(\mathbf{y}){⟩}_c=2G_{>}(\mathbf{x}-\mathbf{y}{)}^2$, its connected second cumulant contributesKeeping the local quartic term and using Parseval identity gives $\int d^{D}y{G}_{>}(\mathbf{x}-\mathbf{y}{)}^2=I_2$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}[\alpha /(p^2+\alpha r_0){]}^2$. Thus the one-loop shell quartic renormalization isThe coefficient follows from multiplying the quartic density correction $-u_0^2{I}_2/16$ by $4!=24$. This local, zero-external-momentum coupling extraction is not a claim that the exact finite-shell effective action retains only its original polynomial; further operators are generated.

More explicitly, for a thin shell $b=e^{d\ell }$, let $c_D=\mathrm{area}(S^{D-1})/(2\pi {)}^D$, $R=\alpha r_0/{\Lambda }^2$ and $g=c_D{\alpha }^2{\Lambda }^{D-4}{u}_0$. With increasing length scale $\ell$, the leading flow isThis convention has the opposite scale direction to a renormalization-group beta function defined using increasing momentum. On the tuned critical surface at $D=4$, $R=O(g)$ and $dg/d\ell =-3g^2/2+O(g^3)$, so weak positive $g$ is marginally irrelevant, tending to zero with logarithmic corrections rather than remaining an exactly marginal parameter. Just below four dimensions, $D=4-\epsilon$ with $0<\epsilon \ll 1$, the same calculation yields the Wilson-Fisher fixed point $g_{\ast }=2\epsilon /3+O({\epsilon }^2)$, $R_{\ast }=-\epsilon /6+O({\epsilon }^2)$. It does not justify extrapolating a small-$\epsilon$ expansion to every lower dimension. The ordinary scalar quartic upper critical dimension is therefore

Use a real orthonormal basis of Fourier modes on the time circle:The eigenvalue of $L_{\omega }$ on the constant mode is ${\omega }^2$, and each sine/cosine pair has eigenvalue ${\omega }^2+{\nu }_r^2$. Retaining these modes giveswhere $C_N$ can depend on $N,\beta$ but is independent of $\omega$. Each real Gaussian integral contributes $\sqrt{2\pi /{\lambda }_r}$. Consequently the Fourier-cutoff oscillator functional determinant isA spectral cutoff between successive eigenvalue pairs selects precisely this finite subspace. In comparing two frequencies, the displayed regularization keeps the same number of modes $N$; holding an absolute spectral threshold fixed need not give the same $N$ at finite cutoff.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.