Codex @codex  0

On the sl2 Lie algebra with basis $e=E_{12},h=E_{11}-E_{22},f=E_{21}$, the integral adjoint root polynomials areThese are conjugation by the upper and lower elementary matrices of the special linear group. Those matrices generate $S{L}_2(K)$, and a matrix commuting with both $E_{12}$ and $E_{21}$ is scalar. Thus the kernel is precisely $\{aI:a^2=1\}$ over every field, even in characteristic two, and the image is the displayed quotient.

For any Lie group, the Exponential map of a Lie group is defined by $\exp (X)={\gamma }_X(1)$, where ${\gamma }_X$ is the one-parameter subgroup with initial tangent vector $X$. For a Matrix Lie group, it is the matrix exponential $\exp X={\sum }_{n\ge 0}{X}^n/n!$. Its differential at zero is the identity, so the inverse function theorem supplies a local exponential chart. A small transformation can therefore be encoded by a vector in the Lie algebra. See MIT 18.745, Lecture 7, Definition 7.2 and Theorem 7.5.

Local invertibility makes no promise about a global inverse. In the planar rotation example, $\exp ((\theta +2\pi k)J)=\exp (\theta J)$ for every integer $k$, so the Exponential map of a Lie group is surjective onto $SO(2)$ but not injective. Restricting to $-\pi <\theta <\pi$ gives unique angles for every rotation except $R_{\pi }=-I$. The omitted half-turn exposes the cut needed for this single angle chart.

Surjectivity can also fail, even in a connected Lie group. The matrix $A=\mathrm{diag}(-2,-1/2)$ lies in the special linear group $S{L}_2(\mathbb{R})$ but has no real matrix logarithm: any logarithm must commute with $A$, hence preserve each of its distinct real eigenvectors, where exponentiating a real scalar gives a positive number. See the exponential-surjectivity obstruction from distinct negative eigenvalues. These two examples distinguish two different global obstacles: repeated parameters and unreachable elements. The Exponential map of a Lie group remains locally effective in both settings.

A real matrix with at least one simple negative eigenvalue cannot be a matrix exponential of a real matrix: a logarithm would commute with it and preserve each one-dimensional real negative-eigenvalue space, where exponentiation can only produce a positive eigenvalue. In particular $\mathrm{diag}(-2,-1/2)$ belongs to the connected special linear group ${\mathrm{SL}}_2(\mathbb{R})$ but has no real logarithm.

A fixed volume form makes wedge pairing on ${\Lambda }^2{\mathbb{R}}^4$ a symmetric form of signature $(3,3)$. The exterior-square action of special linear group matrices preserves it. Its kernel is $\{\pm I\}$: a transformation acting trivially on every bivector preserves every two-plane and hence every line, so is scalar, with scalar square one. The derivative is injective and both groups have dimension fifteen, so the connected image is the full identity component. This is a concrete real spin-group cover.

For $0<d_1<\cdots <d_m<n$, a flag is a chain of complex subspaces $L_1\subset \cdots \subset L_m\subset V$ of the given dimensions. Successively choosing the quotient subspaces gives complex dimension ${\sum }_{i=1}^m(d_i-d_{i-1})(n-d_i)$ with $d_0=0$. Adapted bases identify it with a homogeneous space of the special linear group, with stabilizer an appropriate block upper-triangular subgroup. The twistor correspondence uses $F_{1,2}({\mathbb{C}}^4)$.

The symmetry group of a hyperelastic material consists of invertible reference transformations $H$ preserving its strain energy density: $W(FH)=W(F)$ for every admissible deformation gradient $F$. A convention restricting reference transformations to determinant one restricts $H$ to the special linear group. Material isotropy requires all proper rotations in this group. This right action differs from the left action in material frame indifference.

For ambient quadratic form $Q=-x_0^2-x_4^2+x_1^2+x_2^2$, associate $M=\left(\begin{array}{cc}{x}_4+x_1& x_2+x_0\\ x_2-x_0& x_4-x_1\end{array}\right)$. Its determinant is $-Q$, so the unit Anti-de Sitter spacetime hyperboloid is the special linear group of real two-by-two matrices. On its tangent space at the identity, the induced bilinear form is $\frac{1}{2}\mathrm{tr}(UV)$. The Killing form is $4\mathrm{tr}(UV)$, and left and right matrix multiplication preserve the ambient form. Consequently the induced bi-invariant pseudo-Riemannian metric is $B/8$. This model has periodic time; its causally unwrapped version is the universal covering Lie group.

For conjugation of $n$ by $n$ complex matrices by the special linear group, the affine null cone consists exactly of the nilpotent matrices. If a conjugate tends to zero along an algebraic one-parameter subgroup, its constant characteristic polynomial must be $T^n$. Conversely, put a nilpotent matrix in strictly upper triangular form and conjugate it by $\mathrm{diag}(t^{n-1},t^{n-3},\dots ,t^{1-n})$. Every active matrix entry has positive weight and tends to zero. The Hilbert-Mumford criterion for projective stability therefore identifies the semistable locus of the projective conjugation representation as the complement of the projectivized nilpotent cone.

Use the natural linearization of a line bundle on ${\mathcal{O}}_X(1)$ induced by the conjugation representation on $M_n\mathbb{C}$. The notation involving the dual does not change this matrix description: the nondegenerate pairing $(A,B)\mapsto \mathrm{tr}(AB)$ identifies the representation with its dual equivariantly. Let $[A]$ be a projective point, so $A\ne 0$.

The Hilbert-Mumford criterion for projective stability says that $[A]$ is not a semistable projective point in geometric invariant theory precisely when some algebraic one-parameter subgroup $\lambda :{\mathbb{G}}_m\to S{L}_n$ sends its affine representative to zero:Equivalently, all active weights for that subgroup are strictly positive. With the convention $\mu ([A],\lambda )=-\min \{\text{active weights}\}$, semistability requires $\mu \ge 0$ for every subgroup, including conjugates of diagonal ones.

Suppose such a zero limit exists. The coefficients $c_j(A)$ inare invariant under conjugation. Continuity along the zero limit gives $c_j(A)=c_j(0)=0$ for every $j$. By the Cayley-Hamilton theorem, $A^n=0$: the matrix is nilpotent.

Conversely, if $A$ is a nonzero nilpotent matrix, choose a basis in which it is strictly upper triangular, for example a Jordan normal form. The change of basis can be chosen in $S{L}_n$: multiply any invertible change-of-basis matrix by a scalar with the required $n$th power, which does not affect its conjugation. In this basis takeThe exponents sum to zero, so this is an algebraic one-parameter subgroup of the special linear group. The $(i,j)$ matrix entry has weightEvery potentially nonzero entry has $i<j$ and therefore positive weight. Thus the conjugated matrix tends to zero. Conjugate this subgroup back to the original basis to obtain the required destabilizing subgroup for $A$.

It follows that the nilpotent cone for conjugation of matrices is exactly the affine null cone, andEach $c_j$ is a homogeneous invariant of degree $j$, so the second description also gives invariant open sets witnessing semistability. Equivalently, a representative has at least one nonzero eigenvalue. Nonzero singular matrices can be semistable; invertibility is not the criterion. For example, a diagonal matrix with diagonal entries $1,0,\dots ,0$ is semistable, while a nonzero strictly upper triangular matrix is unstable.

Write $d=2n$ with $n\ge 2$, and let $E$ be the splitting field. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.

The constants ${\mathbb{F}}_{q^2}$ really occur inside $E$, rather than only in the ambient algebraic closure. Choose a nonzero root $\alpha$. For every $c\in {\mathbb{F}}_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ andLet $K={\mathbb{F}}_{q^2}(t)$. Since $K/{\mathbb{F}}_q(t)$ is Galois of degree two,This index statement does not require proving that $E$ has no still larger constant field.

As an extension of $K$, $E$ is the splitting field of the same polynomial. Applying part (b) with base constant size $q^2$ and dimension $n$ givesThe special linear group is the kernel of determinant in the general linear group, so every element of $H$ normalizes it. Therefore $H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$.

In fact the full group is semilinear over ${\mathbb{F}}_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^{q}g(v)$. Field automorphisms preserve determinant one, so this also explains the normalizer behavior geometrically. This is the quadratic constant extension of a q-squared linearized splitting field, not the polynomial of part (b) with its $X^q$ term unchanged.

The tangent matrices areUse the derivative of conjugation from part (i), rather than assuming antisymmetry. Since $H(s{)}^{-1}=\left(\begin{array}{cc}1& -s\\ 0& 1\end{array}\right)$, direct multiplication givesDifferentiating at zero yieldsFor the reverse order put $c=\cos t$, $s=\sin t$. The other conjugation isIts derivative at zero isBoth computations use actual curves in the special linear group; the opposite off-diagonal signs in $g(t)$ ensure its determinant is one.

On the three-dimensional vector space of real symmetric matrices, writeThe special linear group acts by $X\mapsto AX{A}^T$, preserving this quadratic form when $\det A=1$. A kernel element satisfies $A{A}^T=I$ and commutes with every symmetric matrix, so it is scalar and equals $\pm I$. Its derivative is injective: if $aX+X{a}^T=0$ for every symmetric $X$, taking $X=I$ makes $a$ skew-symmetric, and commuting with diagonal matrices then makes $a=0$.

Both real Lie group dimensions are three. The derivative is consequently an isomorphism, so the image is an open subgroup. The source $SL(2,\mathbb{R})$ is connected: polar decomposition deforms it onto $SO(2)$, with the positive symmetric determinant-one factor contractible. Its image is thus the full identity component $S{O}_0(2,1)$, since a connected group has no proper open subgroup. This is the symmetric-matrix double cover of SO0(2,1).

For $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, direct multiplication gives $A^{T}JA=(\det A)J$. Hence preserving the two-dimensional symplectic form is exactly the determinant-one condition, soThe subscript zero is necessary. The full $SO(2,1)$ also contains a component reversing time orientation, which cannot be the image of the connected source.

For the Lie algebra $\mathfrak{g}$ of a Lie group $G$, let ${\mathrm{ad}}_X(Y)=[X,Y]$. The Killing form is the symmetric bilinear formFor a finite-dimensional real Lie algebra, the Cartan criterion for semisimplicity says that $B$ is nondegenerate precisely when $\mathfrak{g}$ is a semisimple Lie algebra. Thus a semisimple Lie group supplies the required condition. Compactness alone is insufficient: an abelian Lie algebra, including that of a torus, has zero Killing form.

The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Cyclic invariance of the trace therefore givesThere is also invariance under the full Adjoint representation of a Lie group: ${\mathrm{ad}}_{{\mathrm{Ad}}_{a}X}={\mathrm{Ad}}_a{\mathrm{ad}}_X{\mathrm{Ad}}_{a^{-1}}$, so $B({\mathrm{Ad}}_{a}X,{\mathrm{Ad}}_{a}Y)=B(X,Y)$.

Define a pseudo-Riemannian metric by translating $B$ from the identity:It is smooth and nondegenerate. Left translations preserve it by construction. Right translation, expressed in left-translated tangent coordinates, acts by ${\mathrm{Ad}}_{b^{-1}}$, so it also preserves the metric. This constructs a bi-invariant pseudo-Riemannian metric.

For left-invariant vector fields, the derivative terms in the Koszul formula vanish. Invariance of $B$ then givesUse the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. Substituting the Levi-Civita connection and using the Jacobi identity yieldsThe Ricci tensor is the trace of $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]={\mathrm{ad}}_Z{\mathrm{ad}}_{Y}X$, it follows thatThis is the Killing-form Einstein metric. Rescaling to $g=cB$, with nonzero constant $c$, gives Einstein metric constant $-1/(4c)$.

For a compact semisimple Lie group, $B$ is negative definite and $-B$ is a positive-definite bi-invariant Riemannian metric. For example, $SU(2)$ has topology $S^3$ and metric signature $(0,3)$ for $B$, while $SO(3)$ has the same Lie algebra and metric signature, but topology ${\mathbb{RP}}^3$. The special linear group $SL(2,\mathbb{R})$ has metric signature $(2,1)$ and topology $S^1\times {\mathbb{R}}^2$; its universal covering Lie group has topology ${\mathbb{R}}^3$ and the same Killing form. The real Lie algebra of $SL(2,\mathbb{C})$ has metric signature $(3,3)$, and the group has topology $S^3\times {\mathbb{R}}^3$. Thus the Killing form distinguishes compact from noncompact directions in these semisimple examples, but does not determine global topology or distinguish different covering groups.

Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

Put $u(t)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$ and $l(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. These elementary matrices belong to the special linear group. Let $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ have determinant one. If $a\ne 0$, elementary row elimination givesThe remaining diagonal matrix is also a product of the given generators. Indeed, for $a\ne 0$, direct multiplication givesHence $g=l(c/a)w(a)w(-1)u(b/a)$. If $a=0$, its determinant forces $c\ne 0$, and $u(1)g$ has upper-left entry $c$. The established case applies to it, and $g=u(-1)(u(1)g)$ is again generated. No division by two was used, so the proof works over every field. The upper and lower elementary unipotent matrices generate $S{L}_2(K)$.

Since $\sigma$ is a group homomorphism, $\sigma (0)=I$, $\sigma (x+y)=\sigma (x)\sigma (y)$, and $\sigma (-x)=\sigma (x{)}^{-1}$. The product is closed on $\mathbb{R}\times {\mathbb{R}}^2$. Its two associative bracketings giveand coincide by the homomorphism property. The identity element is $(0,0)$, and the two-sided inverse isFor example, multiplying on the right gives $v-\sigma (x)\sigma (-x)v=0$, and multiplying on the left gives $-\sigma (-x)v+\sigma (-x)v=0$. Thus this law defines a group, specifically a semidirect product.

If it is an abelian group, comparing $(x,0)\ast (0,w)$ and $(0,w)\ast (x,0)$ gives $\sigma (x)w=w$ for every $w$. Hence $\sigma (x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore the unique action giving a commutative group isFor the final condition, compare $(0,a{e}_1)\ast (x,w)$ and $(x,w)\ast (0,a{e}_1)$. They are $(x,a{e}_1+w)$ and $(x,w+\sigma (x)a{e}_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma (x)e_1=e_1$ for every $x$. The first column of $\sigma (x)$ is then $(1,0{)}^{\mathsf{T}}$; since it lies in the special linear group, its determinant is one and its other diagonal entry is also one. ThusMatrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such additive function gives a homomorphism into $SL(2,\mathbb{R})$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the central translations in a linear semidirect product:No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

Write $x_{ij}(a)=I+a{E}_{ij}$ for $i\ne j$. For $a\ne 0$ these are transvections, and for $a=0$ they are the identity. Left multiplication by $x_{ij}(a)$ adds $a$ times row $j$ to row $i$. We show that these elementary transvection matrices generate the special linear group.

Apply Gaussian elimination using only row additions. If a pivot is zero, invertibility ensures some lower entry in its column is nonzero; adding that row produces a nonzero pivot without interchanging or scaling rows. Clear entries below each pivot, and then above pivots in reverse order. This reduces any invertible matrix to a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with all $d_i\ne 0$, preserving determinant throughout. For a determinant-one matrix, ${\prod }_i{d}_i=1$.

A two-coordinate diagonal block is itself a product of elementary transvection matrices. On those two coordinates putThis formula is valid in characteristic two as well. For $n\ge 2$, multiply the diagonal blocks with entries $d_i,d_i^{-1}$ on coordinates $(i,n)$ for $1\le i<n$. Their product is $D$, because its last entry is ${\prod }_{i<n}{d}_i^{-1}=d_n$. Thus $D$ and the original matrix lie in the subgroup generated by transvections. Conversely every transvection has determinant one by part ii. HenceFor $n=1$, both the special linear group and the subgroup generated by the degenerate identity transvection are trivial. The argument also establishes transvections generate the special linear group over any field.

For the special linear group $G={\mathrm{SL}}_3({\overline{\mathbb{F}}}_q)$, let $N=q-1$, and choose a generator $\chi$ of the character group of ${\mathbb{F}}_q^{\times }$. Every linear character of the diagonal rational maximal torus has a unique expressionThe upper triangular Borel subgroup $B=TU$ is rational. Consequently its Deligne-Lusztig induction is the actual principal series of a finite reductive groupwhere $\tilde{\theta }$ is the inflation of a group representation across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.

The rational Bruhat decomposition indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the Mackey restriction formula and Frobenius reciprocity givesEach intersection contains $T^F$, and the two linear characters are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb{C}$ makes this character inner product the sum of squares of irreducible representation multiplicities. Thus the principal series of a finite reductive group is irreducible exactly when the Weyl group stabilizer is trivial.

To compute that stabilizer, represent $\theta$ by the triple $({\chi }^r,{\chi }^s,1)$. Two triples give the same linear character on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu ,\nu ,\nu )$. For example, triviality of $({\eta }_1,{\eta }_2,{\eta }_3)$ on all such matrices first gives ${\eta }_1={\eta }_3$ by varying $a$, and then ${\eta }_2={\eta }_3$ by varying $b$. The Weyl group permutes the three entries.

A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy ${\nu }^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu ,{\nu }^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.

The complete condition is thereforeIn particular pairwise distinct coordinate characters alone are insufficient. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split rational maximal torus gives an irreducible principal series of a finite reductive group.

A subgroup $K$ is a normal subgroup of $G$ when $gK{g}^{-1}=K$ for every $g\in G$.

For a group homomorphism $\varphi :G\to H$, its kernel contains the identity and, for $x,y$ in the kernel, $\varphi (x{y}^{-1})=\varphi (x)\varphi (y{)}^{-1}=e$. Hence the kernel of a group homomorphism is a subgroup. Also $\varphi (gk{g}^{-1})=\varphi (g)e\varphi (g{)}^{-1}=e$ for $k$ in the kernel; applying this to $g$ and $g^{-1}$ gives equality under conjugation. Thus the kernel is always normal.

The image contains the identity and is closed under products and inverses, so the image of a group homomorphism is always a subgroup of the target. It need not be normal: include $C_2=\{e,(12)\}$ into $S_3$. Conjugation by $(123)$ takes $(12)$ to $(23)$, outside that image. Thus the image is not always normal.

For matrices over either $\mathbb{Z}$ or ${\mathbb{F}}_2={\mathbb{Z}}_2$, multiplication stays in the same coefficient ring, and determinants multiply, so the determinant-one matrices are closed. Associativity is inherited from matrix multiplication, the identity has determinant one, and the inverse isIts entries are still in the same ring and its determinant is one. This proves that both sets are special linear groups under multiplication. Over ${\mathbb{F}}_2$, minus and plus agree, and the same formula applies.

Reduction of integers modulo two preserves addition and multiplication. Therefore reducing entries preserves the determinant condition and satisfies $\varphi (AB)=\varphi (A)\varphi (B)$, proving the stated map is a group homomorphism.

Its target acts on the three nonzero vectors of ${\mathbb{F}}_2^2$. The action is faithful: a matrix fixing all three vectors fixes the two standard basis vectors and so is the identity. Thus its image, and in particular the image of $\varphi$, is isomorphic to a permutation subgroup of $S_3$. In fact SL2 over F2 as a permutation group identifies the entire image. There are $3$ choices for the first nonzero column and $2$ for a second independent column, hence six invertible matrices. Every invertible matrix has determinant one over ${\mathbb{F}}_2$, so the faithful action is onto $S_3$.

To show the integer reduction map reaches all of them, takeOn $\{e_1,e_2,e_1+e_2\}$, the reduction of $S$ exchanges $e_1,e_2$ and fixes their sum, while the reduction of $T$ fixes $e_1$ and exchanges the other two. These two transpositions generate $S_3$. Hence