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Write the total velocity as $\mathbf{U}=\Omega y\hat{\mathbf{x}}+\mathbf{u}$ and the magnetic field as $B_0\hat{\mathbf{x}}+\mathbf{b}$. For solenoidal fields the resistive induction equation is $({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{U}+\eta {\nabla }^2\mathbf{B}$. The first-order smoothing approximation drops the fluctuating terms $\mathbf{b}\cdot \nabla \mathbf{u}-\mathbf{u}\cdot \nabla \mathbf{b}$, retaining shear exactly. The uniform field is not stretched by the background shear because ${\partial }_x(\Omega y)=0$. The other retained stretching terms are $\mathbf{b}\cdot \nabla (\Omega y\hat{\mathbf{x}})=\Omega b_y\hat{\mathbf{x}}$ and $B_0{\partial }_x\mathbf{u}$. HenceThe same phase cancellation as in part (i), together with ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{x}}=-K^2{e}^{i\mathbf{k}\cdot \mathbf{x}}$, proves that a response can be taken as $\mathrm{Re}(\mathbf{c}(t)e^{i\mathbf{k}(t)\cdot \mathbf{x}})$, whereIf $\mathbf{k}\cdot \mathbf{c}=0$ initially, its derivative is $-\eta K^2\mathbf{k}\cdot \mathbf{c}$, so magnetic solenoidality is preserved.

For $j=y,z$ the response equation is ${\dot{c}}_j+\eta K^2{c}_j=i{k}_x{B}_0{v}_j$. Its integrating-factor solution isThe velocity amplitudes and wavevector vary on the shear timescale. In the fast-diffusion limit $\eta (k_x^2+k_y^2)\gg |\Omega |$, and after the free transient, the integral is concentrated within time $(\eta K^2{)}^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the quasistatic magnetic response of a helical shearing wave:Complex conjugation introduces a minus sign in $c_j^{\ast }$. With precisely the EMF normalization specified in the question,A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal{E}$.

Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal{E}(t)\simeq -B_0{H}_{0}A/[\eta (A+{\Omega }^2{k}_x^2{t}^2{)}^2]$. For the time integral, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg |\Omega |$ is a sufficient uniform condition. For $\Omega k_x\ne 0$, substitute $z=|\Omega k_x|t/\sqrt{A}$. Since ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$, this givesThe integral identity follows immediately from $z=\tan \chi$, reducing it to ${\int }_{-\pi /2}^{\pi /2}{\cos }^2\chi d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega =0$ but $H_0\ne 0$, the leading EMF is constant rather than time-integrable; the finite integral relies on nonzero shear.

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.

Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

An almost complex structure is a smooth real bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular the real dimension is even, say $2m$. After complexifying, the type decomposition of the complexified tangent bundle isThe corresponding differential forms of type (p, q) are smooth sections of ${\Lambda }^p(T^{1,0}M{)}^{\ast }\otimes {\Lambda }^q(T^{0,1}M{)}^{\ast }$, regarded as alternating forms by the wedge product of differential forms. A form of type $(p,q)$ vanishes unless its arguments include $p$ vectors of type $(1,0)$ and $q$ of type $(0,1)$.

Besides vanishing of the Nijenhuis tensor, two equivalent descriptions of an integrable almost complex structure are: the smooth sections of $T^{0,1}M$ are closed under Lie bracket; and every point has complex coordinates in which $J$ is multiplication by $i$, with holomorphic changes of coordinates. The equivalence with existence of these coordinates is the Newlander-Nirenberg theorem for smooth $J$. A further equivalent condition is that the exterior derivative has only the two type components $d=\partial +\bar{\partial }$, with no components of bidegree $(2,-1)$ or $(-1,2)$. Indeed evaluation of $d$ of a $(1,0)$-form on two $(0,1)$ vector fields detects the $(1,0)$ component of their bracket; complex conjugation detects the other component.

For the almost complex structure induced by a complex atlas, write $z^j=x^j+i{y}^j$ and put $J{\partial }_{x^j}={\partial }_{y^j}$, $J{\partial }_{y^j}=-{\partial }_{x^j}$. The differentials of holomorphic transition maps are complex linear, so these local definitions agree. In these coordinates $J$ has constant coefficients and all coordinate vector fields commute. The expression defining the Nijenhuis tensor is tensorial, so evaluating it on this frame proves it is zero everywhere. Thus the induced structure is integrable.

Here is a direct proof for the smooth submanifolds with invariant complex tangent spaces are complex assertion. At a point of $Y$, its tangent space is a complex-linear subspace, of real dimension $2k$. Make a complex-linear change of ambient holomorphic coordinates so that this tangent space is ${\mathbb{C}}^k\times 0$. Projection onto the first $k$ coordinates has invertible real differential on $Y$. The real inverse function theorem states that a smooth map between equal-dimensional manifolds with invertible differential is a diffeomorphism on sufficiently small neighbourhoods. It therefore writes $Y$ locally as a smooth graph $w=F(z)$ over an open set in ${\mathbb{C}}^k$.

A tangent vector to this graph is $(v,dF(v))$. Its image under $J$ is $(iv,idF(v))$. Tangent-space invariance impliesso $F$ satisfies the Cauchy-Riemann equations. Smooth functions satisfying these equations are holomorphic: restrict to each coordinate complex line, use the one-variable criterion, and apply the iterated Cauchy integral formula on a small polydisc to get jointly holomorphic coefficients. Hence the graph parametrization is holomorphic. Restrictions of ambient holomorphic coordinates supply holomorphic transition maps between these graph charts. The restricted structure on $Y$ is an integrable complex structure. This also proves the induced structure agrees with the original restriction, rather than merely producing some complex structure on $Y$.

A nonsingular analytic subvariety is a closed analytic subset which near each of its points is the zero set of a holomorphic submersion of constant rank; equivalently, it is locally a coordinate subspace in holomorphic coordinates. The holomorphic implicit function theorem says that if a holomorphic map to ${\mathbb{C}}^r$ has differential of rank $r$ at a zero, its nearby zero set is a complex submanifold of codimension $r$.

The homogeneous equation defines a subset of Complex projective space because multiplying a vector by $\lambda \ne 0$ multiplies its quadratic value by ${\lambda }^2$. Its gradient on ${\mathbb{C}}^4$ is $2Qz$, which is nonzero for $z\ne 0$. More explicitly, in an affine chart set $z_i=1$. If all derivatives of the restricted equation vanished at one of its zeros, all components of $2Qz$ except possibly the $i$th would vanish. The Euler homogeneous function theorem gives ${\sum }_j{z}_j{\partial }_{j}p=2p=0$, forcing the remaining component to vanish as well. This contradicts invertibility of $Q$. The holomorphic implicit function theorem thus proves $S$ is a smooth complex surface.

To see its geometry, reduce the nondegenerate symmetric bilinear form to diagonal form over $\mathbb{C}$. There is a vector of nonzero self-pairing, since otherwise polarization would make the entire form zero. Normalize such a vector using a square root of its self-pairing. Its orthogonal complement remains nondegenerate, so induction gives an orthonormal basis. Apply the same argument to the nondegenerate form $w_0{w}_3-w_1{w}_2$: the two forms are equivalent under an invertible linear change of variables. The allowed projective linear transformation therefore reduces $S$ toView its coordinates as the nonzero rank-one matrix $A=\left(\begin{array}{cc}{w}_0& w_1\\ w_2& w_3\end{array}\right)=u{v}^T$. At $[A]$, holding $[u]$ fixed and varying $[v]$ gives a projective line, and holding $[v]$ fixed and varying $[u]$ gives another. They are distinct and meet at $[A]$. In fact these are the only lines through it: if a second rank-one matrix is $a{b}^T$, the entire pencil has determinant zero only if $\det (u,a)\det (v,b)=0$, so one projective factor is fixed. These are the two rulings of a smooth quadric surface.

The Segre embedding is the holomorphic mapEvery nonzero rank-one matrix has such a factorization, unique up to reciprocal rescaling, so $\Phi$ is bijective. Its inverse is holomorphic on each open set where a matrix entry $a_{ij}\ne 0$: the projective column $[a_{0j}:a_{1j}]$ recovers $[u]$, and the projective row $[a_{i0}:a_{i1}]$ recovers $[v]$. These formulas agree on overlaps. Thus the Segre description of a smooth quadric surface gives the promised stronger conclusion

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives

An inversion in a circle on the Riemann sphere is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it isIt reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a Möbius transformation fixing at least three distinct points, hence the identity. Therefore the two inversions agree.

Every inversion has the form $(a\overline{z}+b)/(c\overline{z}+d)$ with $ad-bc\ne 0$. Composing two conjugate-linear fractional expressions cancels the complex conjugations and gives $(Az+B)/(Cz+D)$ with nonzero determinant. Induction proves that an even number of inversions gives a Möbius transformation.

Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_R{J}_1(z)=R^{2}z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto 1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne 0$, factor a general Möbius transformation asThis is a composition of affine maps and $1/z$. If $c=0$, it is already affine. Every Möbius transformation is therefore a composition of an even number of inversions, with the identity represented by zero inversions or by any inversion repeated twice.

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.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

The complexification of a Lie algebra $L_0$ is $L=L_0{\otimes }_{\mathbb{R}}\mathbb{C}$, with the Lie bracket extended complex-bilinearly. Equivalently write $L=L_0\oplus i{L}_0$, whereComplex conjugation is an antilinear map and a Lie algebra automorphism whose fixed subalgebra is $L_0$.

If the solvable radical of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the solvable radical $R$ of $L$ is preserved by complex conjugation, because it is the unique largest solvable ideal. Consequentlyfor $z\in R$, both $(z+\overline{z})/2$ and $(z-\overline{z})/(2i)$ belong to $R_0$. If $R\ne 0$, $R_0\ne 0$ is a solvable ideal of $L_0$. This proves $L_0$ is semisimple if and only if $L_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is semisimple. Consistently, the complex Killing form is just the complex-bilinear extension of the real one; its determinant in a real basis is unchanged by extending scalars.

Take the sl2R Lie algebra ${\mathfrak{sl}}_2(\mathbb{R})$ and the special unitary Lie algebra $\mathfrak{su}(2)$. Both complexify to ${\mathfrak{sl}}_2(\mathbb{C})$. This is immediate for the former; for the latter, the real basis $ih,e-f,i(e+f)$ consists of traceless matrices that are skew-Hermitian matrices and is also a complex basis of ${\mathfrak{sl}}_2(\mathbb{C})$.

They are not isomorphic as real Lie algebras. Their Killing forms are $B(X,Y)=4\mathrm{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On ${\mathfrak{sl}}_2(\mathbb{R})$ the basis $h,e+f,e-f$ has a diagonal Gram matrix with entries $8,8,-8$, so the signature of a quadratic form is $(2,1)$. A Lie algebra isomorphism preserves the Killing form and therefore its signature.

A real split semisimple Lie algebra has a Cartan subalgebra whose Adjoint representation of a Lie algebra is simultaneously diagonalizable over $\mathbb{R}$, so its root-space decomposition is defined over $\mathbb{R}$. The example ${\mathfrak{sl}}_2(\mathbb{R})$ is split: the diagonal Cartan subalgebra $\mathbb{R}h$ has eigenvalues $0,2,-2$ and real root spaces $\mathbb{R}e,\mathbb{R}f$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\mathrm{ad}x$ skew-adjoint for the positive definite inner product $-B$, so its eigenvalues are purely imaginary. If it were diagonalizable over $\mathbb{R}$, all these eigenvalues would be zero and $\mathrm{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split Cartan subalgebra exists. Thus ${\mathfrak{sl}}_2(\mathbb{R})$ is split and $\mathfrak{su}(2)$ is not.

Use the unitary discrete Fourier transform, with $e(x)=\exp (2\pi ix)$:For a unit $a$, substitute $m=an$ in the sum. The multiplicativity of the Dirichlet character gives $\chi (a^{-1}m)=\overline{\chi (a)}\chi (m)$, henceThe complex conjugation is present in the original PDF and lost in the converted TeX. It matters for nonreal characters.

Now let $q=p^k$ and let $\chi$ be primitive. Since it does not descend to $p^{k-1}$, there is a unit $u\equiv 1(\mathrm{mod}{p}^{k-1})$ with $\chi (u)\ne 1$. For $k=1$, reduction is to the unit group modulo one. If $p\mid a$, then $a(u-1)\equiv 0(\mathrm{mod}q)$, so multiplication of the summation variable by $u$ leaves its exponential factor unchanged. It follows that $\hat{\chi }(a)=\chi (u)\hat{\chi }(a)$, and therefore $\hat{\chi }(a)=0$. Also $\chi (a)=0$. This proves the formula at every nonunit as well as every unit, including $a=0$.

Let $q_0$ be the conductor of a Dirichlet character and ${\chi }_0$ the inducing primitive even character. Removing the Euler factors absent from $L(s,\chi )$ gives the imprimitive Dirichlet L-function Euler correctionThe primitive functional equation therefore givesEquivalently, replace the final primitive function by $L(1-s,\overline{\chi })/P_{\overline{\chi }}(1-s)$, interpreted as a meromorphic identity with removable values handled by continuation. It is the conductor of a Dirichlet character $q_0$, rather than the possibly inflated modulus $q$, that enters the gamma factor and root number. The complex conjugation bar in the original PDF is lost in the converted TeX.

The zeros are those of $L(s,{\chi }_0)$ together with the zeros of the finite Euler product, and multiplicities add. Since $|{\chi }_0(p)|=1$ at each extra prime, an extra factor vanishes at the imaginary points determined byEach extra prime creates infinitely many such points. Its nonzero-imaginary points are not zeros of the primitive function: the functional equation and nonvanishing of Dirichlet L-functions on the line one exclude them. Thus the zero sets are identical precisely when every prime dividing $q$ already divides $q_0$, making $P=1$. Increasing prime-power exponents alone can make a character imprimitive without changing its L-function. If $q_0=1$, the primitive function is zeta; the same Euler correction applies, with its pole at one retained.

Under the Riemann hypothesis, every $\beta =1/2$, so for $\sigma >1/2$ the absolutely convergent formula givesThe zero set is nonempty: otherwise Hadamard factorization would make $\xi$ an exponential of a linear polynomial, and its functional symmetry would force it to be constant, contrary to gamma growth on the positive real axis. Thus the inequality is strict in the open right half-plane. Continuity at $\sigma =1/2$, including at boundary zeros, proves the claimed increasing modulus on the closed half-line.

Conversely, suppose the modulus is nondecreasing for every fixed $t$. If a zero $\beta +i\gamma$ had $\beta >1/2$, then nonnegativity and monotonicity would force $|\xi (\sigma +i\gamma )|=0$ throughout $1/2\le \sigma \le \beta$. The identity theorem would make $\xi$ identically zero, a contradiction. A zero left of the line reflects to one right of the line by the functional equation and complex conjugation symmetry. Hence every zero lies on the critical line. This proves the xi modulus criterion for the Riemann hypothesis. The two following roman headers refer to supplied asymptotic assumptions, not further questions, and require no Solution sections.

Assume Riemann hypothesis. First fix $0<\delta <1/4$ and work on ${\sigma }_0=1/2+\delta$. We prove the subpower zeta bound to the right of the critical line, then move back to the line by the functional equation and Phragmén–Lindelöf principle.

Put $x=\log t$ for large positive $t$ and integrate the supplied smoothed logarithmic derivative identity horizontally from ${\sigma }_0$ to two. There are no zeros on this path under Riemann hypothesis, so the Euler-product logarithm at $2+it$ continues along it. Each prime-power term contributes at most $\Lambda (n)n^{-{\sigma }_0}/\log n\le n^{-{\sigma }_0}$, and the smoothing weights are at most one. Hence the integrated prime terms are bounded byAll zeros have $\rho =1/2+i\gamma$. The local zero-count estimate from Question 3, with its reflected version for negative ordinates, gives uniformly for ${\sigma }_0\le \sigma \le 2$To see the uniformity, sum the $O(\log (2+|\gamma |))$ zeros in successive unit ordinate intervals against $({\delta }^2+|t-\gamma {|}^2{)}^{-1}$; the distant dyadic tails are summable. The zero-term numerator has modulus at most $x^{-2\delta }+x^{-\delta }$. Its integrated contribution is therefore at most $C_{\delta }{x}^{-\delta }\log t/\log x=o(\log t)$. The integrated supplied remainder is $O(x^{-1-\delta }\log t/\log x)$, also $o(\log t)$. Since $\log \zeta (2+it)$ is bounded, we obtainThus for every fixed $\delta >0$ and $\eta >0$, $|\zeta (1/2+\delta +it)|{\ll }_{\delta ,\eta }{t}^{\eta }$. Negative $t$ follow by complex conjugation.

The zeta functional equation and the gamma ratio give $|\zeta (1/2-\delta +it)|{\ll }_{\delta ,\eta }{t}^{\delta +\eta }$. Zeta is holomorphic throughout this strip, since its pole at one is outside it, and Euler summation supplies polynomial vertical growth. The strip convexity conclusion of Phragmén–Lindelöf principle therefore gives at the midpointFor a prescribed $\epsilon >0$, choose $\delta$ and $\eta$ with $\delta /2+\eta <\epsilon$; the bounded $t$ range is harmless. This provesThe explicit-formula estimate is deliberately first made a fixed distance to the right of the critical line. No divergent zero bound at $\sigma =1/2$ is used.