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Reading the coefficients of the linear map givesUnder the inverse identification, the four standard real basis vectors map respectively to $(1,0)$, $(i,0)$, $(0,1)$ and $(0,i)$ in ${\mathbb{C}}^2$. Applying $A$ and taking real and imaginary parts therefore gives the columnsThus the real matrix isFor verification, write its blocks as $\left(\begin{array}{cc}{R}_{\theta }& I\\ I& R_{-\varphi }\end{array}\right)$, where $R_{\psi }$ is the planar rotation matrix. Block elimination gives $\det B=\det (R_{\theta -\varphi }-I)=2-2\cos (\theta -\varphi )$. Therefore the determinant identity isThis is an instance of the realification determinant identity. The identification is real-linear; it is not an identification of complex vector-space dimensions.

Expanding the commutator immediately gives $[A,A]=0$, $[A,B]=-[B,A]$, and $[A,\lambda B]=\lambda [A,B]$. For its matrix trace, writeConsequently every matrix commutator is traceless: $\boxed{\mathrm{tr}[A,B]=0}$. This is the trace of a matrix commutator identity, valid for arbitrary square matrices over the complex numbers.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

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.

Write a matrix in the upper unitriangular group as $M=I+A$. Here $A^3=0$, and its only potentially nonzero entry in $A^2$ is $(A^2{)}_{13}=ab$. The binomial theorem therefore gives the unitriangular matrix power formulaFor an odd prime number $p$, both $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)=p(p-1)/2$ vanish in the finite field ${\mathbb{F}}_p$. Thus $M^p=I$. A nonidentity $M$ has order dividing the prime $p$ (by the same exponent-division argument as in part (i)), and so has order exactly $p$.

For $p=2$, take $a=b=1$ and $x=0$. Then $M^2=I+A^2\ne I$, while $M^4=(I+A^2{)}^2=I$. This element has order $4$, not $2$. Hence every nonidentity element has order $p$ exactly for odd $p$. The counterexample at $p=2$ accounts for the quadratic term in the matrix power formula.

Identify the projective line with one-dimensional subspaces of ${\mathbb{F}}_p^2$: use $v_t=(t,1{)}^T$ for finite $t$, and $v_{\infty }=(1,0{)}^T$. This convention turns the matrix action into the stipulated Möbius transformation.

Since $x$ and $z$ are distinct projective points, $v_z,v_x$ are a basis. Write $v_y=A{v}_z+B{v}_x$. Both $A$ and $B$ are nonzero, because $y$ is different from $x,z$. The matrix with columns $A{v}_z$ and $B{v}_x$ is invertible and sends the lines of $(1,0{)}^T,(0,1{)}^T,(1,1{)}^T$ to those of $z,x,y$, respectively. This constructs the requested map without separate exceptional formulas at infinity.

A matrix fixing $0$ and $\infty$ is diagonal, say $\mathrm{diag}(a,d)$ with $a,d\ne 0$. To fix $1$ it must also satisfy $a=d$, so the stabilizer subgroup of $(0,1,\infty )$ is precisely $\{\lambda I:\lambda \in {\mathbb{F}}_p^{\times }\}$. Any two solutions differ on the right by one of these matrices. ThusEquivalently, after dividing out scalar matrices the projective general linear group action on the projective line is sharply three-transitive on a projective line.

The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all real linear functionals. For a finite basis $e_1,\dots ,e_n$, define the dual basis by $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$, so $e_i^{\ast }(e_j)={\delta }_{ij}$. Every linear functional satisfiesThis equality follows by evaluating both sides on each vector of the basis. It proves spanning directly. If ${\sum }_i{b}_i{e}_i^{\ast }=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.

For a vector subspace $U$, its annihilator of a vector subspace isExtend a basis $e_1,\dots ,e_r$ of $U$ to a basis of $V$. Then $U^{\circ }$ has basis $e_{r+1}^{\ast },\dots ,e_n^{\ast }$, so $\boxed{\dim U^{\circ }=n-\dim U}$.

The dual map of $\alpha :V\to W$ is ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, $\lambda \mapsto \lambda \circ \alpha$. In the corresponding dual bases its matrix is the transpose of the matrix for $\alpha$. A matrix and its transpose have the same rank, giving $\mathrm{rank}{\alpha }^{\ast }=\mathrm{rank}\alpha$. Directly,Thus $\boxed{\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\circ }}$. Also every $\lambda \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}{\alpha }^{\ast }\subseteq (\ker \alpha {)}^{\circ }$. Both spaces have dimension $\mathrm{rank}\alpha$ by the preceding annihilator of a vector subspace formula and rank-nullity theorem, provingAlternatively, a linear functional vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a basis.

For the infinite-dimensional polynomial ring, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic dual space. The linear functional $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus the $L_i$ do not form a basis of $V^{\ast }$. An infinite formal sum of coefficient functionals is not a finite linear combination.

The central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.

The matrix $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ fixes $\omega$. Weight-four transformation gives $E_4(\omega )=(\omega +1{)}^4{E}_4(\omega )$, and $(\omega +1{)}^4\ne 1$. Hence $E_4(\omega )=0$ and $j(\omega )=0$. Since $j$ is a degree-one coordinate, this is a simple zero on the quotient surface. Pulled back to the half-plane it has order three, by the analytic ramification index in part (a).

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.

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.

Reduction modulo $p$ maps ${\mathrm{SL}}_2(\mathbb{Z})$ onto ${\mathrm{SL}}_2({\mathbb{F}}_p)$, of order $p(p^2-1)$. The image of ${\Gamma }_1(p)$ is the upper unipotent subgroup of order $p$. Thus its index in ${\mathrm{SL}}_2$ is $p^2-1$, and, since $-I\notin {\Gamma }_1(p)$ for $p\ge 5$, its effective index isEvery element of ${\Gamma }_1(p)$ has trace congruent to two. An effective elliptic element of order two or three has trace zero or $\pm 1$ in a lift, so neither is possible for $p\ge 5$. Hence $r_2=r_3=0$.

Represent a modular cusp by a primitive column $(a,c)$, modulo sign. Its reduction is a nonzero vector in ${\mathbb{F}}_p^2$ modulo sign, and the unipotent subgroup acts by $(a,c)\mapsto (a+bc,c)$. For $c=0$, the nonzero values of $a$ give $(p-1)/2$ orbits. For $c\ne 0$, $a$ varies freely and $c$ modulo sign gives another $(p-1)/2$ orbits. Thus there are $p-1$ modular cusps. The reduction classification is sufficient as well as necessary: completing two primitive columns to determinant-one matrices and adjusting the second columns by a translation makes congruent columns related by $\Gamma (p)$.

More explicitly, if a determinant-one matrix has first column $(a,c)$, conjugating $T^w$ givesIts least allowable cusp width in ${\Gamma }_1(p)$ is one for $c\equiv 0$ and $p$ otherwise. Both types therefore number $(p-1)/2$, with cusp widths one and $p$; their cusp width sum is $d$. There are no sign-twisted modular cusp periods here, because trace two cannot be congruent to minus two for these primes. Substitution givesThese are the prime Gamma 1 cusp counts and widths.

For a pair $(L,t)$ choose an oriented basis $({\omega }_1,{\omega }_2)$ with $\mathrm{Im}({\omega }_1/{\omega }_2)>0$ and $t={\omega }_2/N$ modulo $L$. Such a basis exists because an exact-order point gives a primitive vector modulo $N$, which can be completed to a determinant-one basis. Define the marked-lattice model of a modular form byChanging to a basis with the same marked point uses a matrix with $c\equiv 0$, $d\equiv 1$ and therefore $a\equiv 1(\mathrm{mod}N)$, exactly ${\Gamma }_1(N)$. The weight-$k$ transformation of $f$ cancels the factor from ${\omega }_2^{-k}$, proving independence of the basis.

The resulting function has homogeneity $F_f(uL,ut)=u^{-k}{F}_f(L,t)$ for $u\in {\mathbb{C}}^{\times }$. Conversely evaluating at $(\mathbb{Z}\tau +\mathbb{Z},1/N)$ recovers $f$. Holomorphy in $\tau$ and holomorphy in the local parameters of degenerating lattices at every modular cusp characterize the functions arising from modular forms, rather than arbitrary homogeneous lattice functions.

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.