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Put $\rho ={\rho }_0+\delta \rho$ and retain quadratic fluctuations in the smooth zero-winding sector. The quadratic density-phase action isAt wavelengths long compared with the healing length, drop the density-gradient term. Completing the square givesThe real Gaussian density integral, or its equivalent contour shift, contributes only a field-independent determinant. Absorb that normalization and the uniform saddle action into $S_0$. To this quadratic long-wave accuracy,The Gaussian extension of $\delta \rho$ to the full real line is a fluctuation approximation around positive ${\rho }_0$; it is not an exact replacement of the global density constraint. The compact field's vortex/winding sectors also lie beyond this smooth-phonon integral.

The Bose-gas phase-only action is a continuum harmonic chain with Euclidean inverse propagator ${\omega }_n^2/g+{\rho }_0{k}^2/m$. Continuing to real frequency givesRestoring $\hslash$ gives $E_k\sim \hslash c_s|k|$ for a wavenumber $k$. This is the phonon branch of the Bogoliubov spectrum. Keeping the omitted density-gradient term replaces $g$ by $g+k^2/(4m{\rho }_0)$ and yields $E_k^2=c_s^2{k}^2+k^4/(4m^2)$ in the adopted units, so the full quadratic spectrum is $\sqrt{{ϵ}_k({ϵ}_k+2g{\rho }_0)}$ with ${ϵ}_k=k^2/(2m)$.

The phase-only action also tests the condensate assumption. At zero temperature its equal-time phase variance has an infrared contribution $\int d^{d}k/|k|$, logarithmically divergent in one dimension; at positive temperature the zero Matsubara mode gives $\int d^{d}k/k^2$, divergent in dimensions at or below two. Thus this same low-energy theory exposes the regimes where the mean-field condensate cannot describe true thermodynamic long-range order. It permits a finite infrared fluctuation at zero temperature for $d\ge 2$ and at positive temperature for $d>2$, within the remaining weak-coupling assumptions.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.

Since the first $k$ basis vectors span the invariant subspace $W$, their images have no components in the last $n-k$ coordinates. ThusHere $A$ is the matrix of $T{|}_W$, $D$ is the matrix of the induced map on the quotient vector space $V/W$, and $B$ is unrestricted. The chosen complementary span need not itself be invariant, so $D$ should not in general be called a restriction to that complement. The block-triangular determinant gives

Define the determinant by the signed permutation sumIt is multilinear and alternating in the rows, and likewise in the columns. Row multilinearity follows because every product contains exactly one entry from each row; a swap changes the permutation sign, and equal rows give cancellation in pairs. Applying the same argument to the transpose proves the column properties.

Consequently, for either rows or columns, a swap multiplies the determinant by $-1$; multiplication of one row or column by $c$ multiplies it by $c$; addition of a multiple of another row or column leaves it unchanged. For the last rule, multilinearity separates the original determinant and a determinant with two equal rows or columns, which is zero. These arguments also cover singular matrices and zero scaling.

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.

Write $t(\tau )=\Delta (\tau )/\Delta (2\tau )$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(2)$, the integral matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 2b\\ c/2& d\end{array}\right)$ has determinant one and satisfies $2\gamma \tau ={\gamma }^{\prime }(2\tau )$. Both discriminants acquire the same factor $(c\tau +d{)}^{12}$, so $t$ is invariant. It has neither zeros nor poles in the half-plane.

Reduction modulo two gives index three, with two modular cusp classes, infinity and zero, of cusp widths one and two. These can also be found from the orbits of upper triangular matrices on primitive columns modulo two. Compactness follows from the finite-index argument in part (a).

At infinity $t=q^{-1}(1+O(q))$, so it has a simple pole. At zero use $S\tau =-1/\tau$ and the modular discriminant inversion law:Thus zero is a simple zero, not a pole, in its width-two modular cusp coordinate. The single-pole criterion for a spherical coordinate now givesThis discriminant-ratio coordinate on X0 2 is different from a ratio of two $j$-invariants.

A real Lie group is a group $G$ equipped with a finite-dimensional real smooth manifold structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.

Its tangent space at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_{e}G$, left translation $L_g(h)=gh$ determines the left-invariant vector fieldThe commutator of these derivations on smooth functions is another left-invariant vector field, so define the Lie bracketThis construction supplies the Lie algebra structure. In a Matrix Lie group, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.

For the special linear group, differentiate the determinant along a curve $g(t)$ through $I$. Expansion of the determinant, or its differential $D{\det }_A(H)=\det (A)\mathrm{tr}(A^{-1}H)$, givesThe determinant differential is surjective at $I$, so the level set $\det =1$ has tangent space equal to its differential's kernel. Equivalently, every tangent matrix has zero trace, and every zero-trace matrix supplies a curve $e^{tX}$ of determinant $e^{t\mathrm{tr}X}=1$. Hencewith the matrix commutator bracket. The same calculation over $\mathbb{C}$ gives the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$; as a real group, the complex group has that space viewed as a real Lie algebra.

The matrix exponential is the everywhere-convergent serieswhose inverse matrix is $\exp (-X)$. The matrix logarithm is locally defined near $I$ byThese maps are inverse on suitable neighbourhoods of zero and $I$, giving a logarithmic chart of a matrix Lie group. A logarithm is not a globally single-valued inverse of the exponential.

Every invertible complex matrix nevertheless has at least one matrix logarithm. Put it in Jordan normal form. For a block $J=\lambda I+N$, $\lambda \ne 0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and setThe finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda (I+N/\lambda )=J$. Combine the blocks and conjugate back. Thus the exponential map is surjective on ${\mathrm{GL}}_n(\mathbb{C})$, by existence of a logarithm for every invertible complex matrix.

A connected counterexample is ${\mathrm{SL}}_2(\mathbb{R})$. It is connected: polar decomposition of an invertible real matrix writes each element as $KP$, with $K\in \mathrm{SO}(2)$ and $P$ positive definite symmetric of determinant one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.

But $g=\mathrm{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real eigenvalues, direct commutation makes $X$ a real diagonal matrix. Its exponential has positive diagonal entries, a contradiction. ThereforeThis is an exponential-surjectivity obstruction from distinct negative eigenvalues; connectedness does not eliminate the obstruction.

Choose the orientation of the Berezin integral so thatHere the product has increasing $i$; this explicitly fixes the otherwise convention-dependent overall sign in the compact measure notation.

Since $A$ is a diagonalizable matrix, write $A=S\Lambda S^{-1}$ with $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$. Make the independent changes $\eta =S\xi$ and $\bar{\eta }=\bar{\xi }{S}^{-1}$. The Grassmann change-of-variables formula gives the two factors $(\det S{)}^{-1}$ and $\det S$, so the complete measure is unchanged. The exponent becomes ${\sum }_i{\lambda }_i{\bar{\xi }}_i{\xi }_i$. Each summand is even and squares to zero, and the different even summands commute. Consequently,Only the term containing every generator survives the Berezin integral. Hence the Grassmann Gaussian integral isZero eigenvalues give zero on both sides, so invertibility of $A$ is unnecessary. In fact the identity extends to all ordinary matrices: the top-degree coefficient of the exponential is the alternating determinant expansion. The assumption that $A$ is a diagonalizable matrix makes the proof especially transparent.

Write the horizontal velocity amplitudes as $(u,v)$ and magnetic amplitudes as $(b_x,b_y)$. The perturbation is horizontally uniform, divergence-free, and has no vertical velocity, so the unperturbed density and pressure are consistent at linear order. In the Keplerian shearing sheet, the horizontal components of the linearized ideal magnetohydrodynamic equations areUse the undivided equations at a zero of $F$. If $F^{\prime \prime }=-k^{2}hF$, the magnetic-force coefficient becomes $ik{v}_A^2$; the stratification cancels. The ideal magnetohydrodynamic induction equation similarly reduces toThe azimuthal induction term is field stretching by the background differential rotation. The four amplitudes obey the homogeneous systemIts determinant must vanish for a nonzero normal mode. Put $A=v_A^2{k}^2$. For nonzero $s$, elimination first gives $(s^2+A)u-2\Omega sv=0$ and $s(s^2+A)v+\Omega (s^2-3A)u/2=0$, whose solvability condition is $(s^2+A{)}^2+{\Omega }^2(s^2-3A)=0$. The original determinant extends the same polynomial to marginal $s=0$. ThusThis is the ideal magnetorotational dispersion relation with the midplane Alfvén speed $v_A=B_0/\sqrt{4\pi {\rho }_0}$. The vertical structure enters through the admissible eigenvalues $k$, rather than through a different horizontal dispersion polynomial. No division by a vanishing growth rate is required in the determinant derivation.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

Write a vector in the domain as $(u,v,w,t{)}^T$. The first and third equations defining the kernel of a linear map imply $v=2\alpha u-2t$. Substitution into the second equation then givesFor $\alpha \ne 1$, this forces $t=(\alpha +1)u$, $v=-2u$ and $w=3u$. HenceThe kernel of a linear map has dimension one, so the rank-nullity theorem gives rank three and proves the assertion about the image of a linear map. Alternatively, the minor formed from columns $2,3,4$ has determinant $2(\alpha -1)$, including a nonzero value at $\alpha =-1$.

For $\alpha =1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. ThusThe third row is the first row minus the second. The image of a linear map is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1{)}^T$ and $(1,0,1{)}^T$, are linearly independent, so the matrix rank is two and this containment is equality:This is the sole kernel jump in a parameter-dependent linear map; there is no further exceptional case at $\alpha =-1$.

The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.

The first index of the printed Levi-Civita symbol is fixed at $1$. Thus the matrix in the first orthonormal basis isThe specified matrix in the other basis is $I$, with determinant one. If this array were a Cartesian second-rank tensor, an orthogonal basis change would give $A^{\prime }=RA{R}^T$, andThe determinants disagree, proving the array is not a tensor. The fixed index matters: the TeX aid drops it, but the original PDF clearly prints ${ϵ}_{1ij}$.