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The imaginary quaternions form Euclidean ${\mathbb{R}}^3$. Conjugation by a unit quaternion acts byIt preserves the imaginary subspace and its norm, and continuity from $q=1$ gives determinant $+1$. The kernel consists of quaternions commuting with all imaginary quaternions, hence is $\{\pm 1\}$.

For a unit imaginary quaternion $n$, take $q=\cos (\theta /2)+n\sin (\theta /2)$. If $v$ is perpendicular to $n$, quaternion multiplication giveswhile the component parallel to $n$ is unchanged. This produces the rotation through angle $\theta$ about axis $n$, and every element of $SO(3)$ has such an axis-angle description. The map is surjective, yieldingThe half-angle also explains why $q$ and $-q$ give the same rotation.

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.

Geometric quantization starts from a classical symplectic manifold $(P,\omega )$ and seeks a Hilbert space and quantum observables that reflect its geometry. It has two distinct stages: prequantization, which represents the full Poisson algebra on sections of a line bundle, and a choice of polarization in geometric quantization, which reduces the excessive phase-space dependence of those sections.

Use the conventions ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. A prequantum line bundle is a Hermitian line bundle $L\to P$ with unitary connection of curvatureExistence requiresfor every closed integral two-cycle $\Sigma$, equivalently integrality of the symplectic cohomology class. Necessity follows from the integral curvature periods of a unitary line bundle; conversely an integral class defines such a bundle, and one adjusts its connection to give the specified curvature. This is a restriction on the classical system, not a choice of local coordinates. For an exact cotangent symplectic form $\omega =-d\theta$, with $\theta =p_{j}d{q}_j$, the bundle can be trivial and a connection is $\nabla =d-i\theta /\hslash$. On a sphere with symplectic area $4\pi s$, the condition becomes $2s/\hslash \in \mathbb{Z}$. The sign of the curvature convention changes if one changes the Hamiltonian-vector-field convention or uses the dual bundle; the integrality condition is unaffected.

The Kostant-Souriau prequantum operator for a real observable isIts derivative part describes the classical Hamiltonian flow and the multiplication term corrects the bracket. Indeed,Since $X_f(g)=-\{f,g\}$, expanding the operator commutator givesWith the Hermitian bundle metric and Liouville measure, these operators are formally symmetric for real $f$ because Hamiltonian flows preserve symplectic volume. On noncompact spaces, domains and self-adjointness still require attention; the algebraic identity initially holds on an appropriate common smooth domain.

Prequantization alone is too large. Sections depend on $2n$ phase-space variables, and the canonical commutation relations are represented reducibly. A polarization is an involutive maximal isotropic distribution $\mathcal{F}\subset T_{\mathbb{C}}P$ of complex rank $n$. Polarized sections satisfyThe curvature vanishes on pairs of polarization vectors because $\omega {|}_{\mathcal{F}}=0$, and involutivity makes the equations locally compatible. A real polarization is tangent to a Lagrangian foliation. A complex polarization in a Kähler manifold chooses one complex tangent type. With the curvature sign used here, choose the holomorphic tangent distribution, giving antiholomorphic wavefunctions. The conjugate line-bundle and sign convention gives the more usual holomorphic description.

For $T^{\ast }Q$, the vertical real polarization is spanned locally by ${\partial }_{p_j}$. Since the canonical one-form has no vertical component, its polarized sections are functions of $q$ alone. In the above trivialization,One does not simply integrate these states over every momentum to retain the prequantum norm: they are constant along noncompact polarization leaves and that integral would diverge. The polarized inner product instead uses the configuration-space measure or an intrinsic half-density description. This recovers the Schrödinger realization.

Only observables whose Hamiltonian flows preserve $\mathcal{F}$ act directly on the polarized space. For example, the free-particle flow generated by $p^2/2$ sends vertical tangent vectors into directions with a configuration-space component, so its raw prequantum operator does not preserve the vertical polarization. To obtain the usual kinetic operator one needs an additional procedure, such as transporting the polarization and pairing back, rather than pretending every classical observable restricts automatically. The Blattner-Kostant-Sternberg pairing and related projection methods address this problem. The impossibility of quantizing all classical observables with every desired algebraic property is another reason to specify the admissible observable algebra.

A metaplectic correction tensors $L$ with a square root of the polarization's canonical determinant bundle when such a square root exists. These half-forms make the pairing more intrinsic and supply corrections to operator transport. They account for familiar Maslov and zero-point shifts; their existence is an extra topological condition. For a flat phase plane, take $z=q+ip$ and replace $\theta =pdq$ by the gauge-equivalent ${\theta }_s=(pdq-qdp)/2$. ThenThe antiholomorphic function therefore has Gaussian norm, the complex-conjugate Bargmann-Fock space. This explicit calculation fixes which polarization matches our curvature sign. Choosing the opposite tangent type with the same sign would give growing Gaussian sections, not the desired Hilbert space. On compact positive Kähler examples, the corresponding holomorphic description (or its conjugate in these conventions) gives finite-dimensional state spaces.

Finding a suitable polarization is therefore a substantial part of the construction. A globally smooth real Lagrangian foliation may fail to exist; for example a regular real polarization on $S^2$ would require a line field that its tangent topology forbids. Singular foliations and noncompact leaves complicate both states and inner products. Although the restricted prequantum connection is locally flat on a Lagrangian leaf, a global nonzero parallel section requires trivial holonomy. In an integrable system this selects the Bohr-Sommerfeld quantization leaves; including the half-form correction gives the familiar action conditionin the usual Maslov-index convention. Complex polarizations require compatible global complex geometry and positivity. Different choices can lead to quite different realizations, and a natural unitary comparison is not automatic. Thus integral curvature provides prequantization; polarization and its global compatibility determine the actual quantum state space.

Set $a_n=\Vert x^n\Vert$ and $D={\inf }_{n\ge 1}{a}_n^{1/n}$. If some $x^m=0$, all later powers vanish, their root norms tend to zero, and $D=0$. Otherwise fix $m\ge 1$ and write $n=qm+s$, $0\le s<m$. Submultiplicativity givesSince $q/n\to 1/m$, it follows that ${\mathrm{limsup}}_n{a}_n^{1/n}\le a_m^{1/m}$. Taking the infimum over $m$ and using $a_n^{1/n}\ge D$ provesThis establishes existence of the limit and its infimum characterization before identifying it with the spectral radius $r(x)={\max }_{\lambda \in \sigma (x)}|\lambda |$.

For $|\lambda |>D$, the root test makes ${\sum }_{n\ge 0}{x}^n/{\lambda }^{n+1}$ converge absolutely. Multiplication of partial sums shows that its limit is $(\lambda 1-x{)}^{-1}$. Thus $r(x)\le D$. For the reverse inequality, take any $\rho >r(x)$ and the circle $|\lambda |=\rho$. The resolvent of an element is holomorphic off the spectrum, andTo justify this identity, first use a radius larger than $\Vert x\Vert$ and integrate the uniformly convergent Neumann series term by term. Then deform to radius $\rho$: the intervening annulus has no spectral points. The Cauchy theorem for these Banach-valued integrals follows by applying bounded linear functionals and using their point separation. Consequently, with $M_{\rho }={\max }_{|\lambda |=\rho }\Vert (\lambda 1-x{)}^{-1}\Vert$,Taking root limits gives $D\le \rho$; let $\rho \downarrow r(x)$. Hence the full spectral radius formula is

For the product comparison, if $\lambda \ne 0$ and $R=(\lambda 1-xy{)}^{-1}$, direct multiplication on both sides verifiesIndeed $(\lambda 1-yx)y=y(\lambda 1-xy)$, and the remaining terms cancel against the inverse equation. Interchanging $x,y$ proves the converse. Thus the nonzero spectra of products in opposite orders agree, soThe possible difference at zero has no effect on the maximum modulus, including when that maximum is zero.

Zero really can differ. In $A=\mathcal{B}({\ell }^2({\mathbb{N}}_0))$, let $S$ be the unilateral shift operator, $S{e}_n=e_{n+1}$, and take $x=S^{\ast }$, $y=S$. Then $xy=1$, whereas $yx=1-P_0$, with $P_0$ the projection onto $e_0$. The latter is zero on $e_0$ and identity on its orthogonal complement, and these blocks also explicitly invert it away from zero and one. Hence

Finally, for $t>r(x)$ the root norms of $(x/t{)}^n$ tend to $r(x)/t<1$, so their norms eventually decay geometrically and the powers converge to zero. Conversely, if the powers converge to zero, choose $N$ with $\Vert (x/t{)}^N\Vert <1$. Homogeneity and the infimum formula implyso $t>r(x)$. The admissible positive $t$ are exactly $(r(x),\infty )$, proving the power-decay characterization of the spectral radiusThis includes quasinilpotent elements with radius zero; no positive endpoint must be attained.

First the C-star identity and submultiplicativity give $\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, hence $\Vert a\Vert \le \Vert a^{\ast }\Vert$ when $a\ne 0$. Applying this to $a^{\ast }$ gives equality, so the involution is isometric and continuous. Also $\Vert 1\Vert =\Vert 1{\Vert }^2$ and $1\ne 0$, yielding $\Vert 1\Vert =1$.

For a Normal element of a C-star algebra $x$, put $b=x^{\ast }x$, which is a self-adjoint C-star element. Normality gives $(x^{\ast }{)}^2{x}^2=b^2$, and the C-star identity applied to $x^2$ and to $b$ givesThus $\Vert x^2\Vert =\Vert x{\Vert }^2$. Every power of $x$ is normal because $x$ commutes with $x^{\ast }$, so iteration yieldsThe spectral radius formula has a limit along all positive integers, and along this subsequence its root norms are exactly $\Vert x\Vert$. ThereforeThis proves spectral radius norm equality for normal elements directly from the defining norm identity, rather than assuming a spectral representation in advance.

The B-spline collocation matrix in the cubic case is the finite tridiagonal matrixFor $n=1$, this is simply the one-entry matrix $(2/3)$.

To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Then $B_n=D{A}_{n}D=(4I-C)/6$, where $C$ has ones immediately above and below the diagonal and zeros elsewhere. Since $\Vert C/4{\Vert }_{{\ell }^{\infty }}\le 1/2$, the convergent Neumann serieshas nonnegative entries and proves invertibility. Since $A_n^{-1}=D{B}_n^{-1}D$, its absolute entries equal those of $B_n^{-1}$. The absolute row sums are therefore the entries of $u=B_n^{-1}1$.

The equation $B_{n}u=1$ becomesIts constant particular solution is $3$; the two characteristic roots of the homogeneous recurrence are $\rho =2-\sqrt{3}$ and ${\rho }^{-1}=2+\sqrt{3}$. Imposing the two boundary values givesThis can also be checked by direct substitution, since $\rho +{\rho }^{-1}=4$. The numerator is smallest at the central index or central pair of indices: its successive difference is $(1-\rho )({\rho }^{n-i}-{\rho }^i)$. Thus, with $j=\lfloor (n+1)/2\rfloor$, the exact inverse norm of midpoint cubic spline collocation isEquivalently, if $n=2m-1$ the fraction is $2{\rho }^m/(1+{\rho }^{2m})$, and if $n=2m$ it is $({\rho }^m+{\rho }^{m+1})/(1+{\rho }^{2m+1})$. The finite norm is not exactly three: it equals $3/2$ for $n=1$, two for $n=2$, and tends to three as $n$ grows. Hence three is the sharp dimension-independent bound for these inverse matrix norms.

Apply the general B-spline interpolation operator norm estimate in part (I) to the inverse norm from (b):and hence in particular $\Vert P_{\mathbf{x}}{\Vert }_{L^{\infty }}\le 3$, independently of $n$. The strict finite-dimensional estimate uses the actual finite collocation matrix; it does not require periodic closure, natural boundary conditions or endpoint knot repetitions.

There is also a direct shorter verification of the uniform bound. For an index $i$ maximizing $|z_i|$, the triangle inequality giveswith an even larger margin when a neighbor is missing. This is the inverse infinity-norm bound from diagonal dominance, and again gives $\Vert A_n^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$.

This addresses the general interpolation bound in part (I); the subsequent (a)–(c) address the cubic case in part (II). Write the interpolant in the B-spline basis asThe stipulated uniquely defined spline interpolation operator presupposes invertibility of the B-spline collocation matrix. For the standard increasing distinct sites, the conditions $N_i(x_i)>0$ give it by the Schoenberg–Whitney theorem. Thus $c=A_{\mathbf{x}}^{-1}Rf$.

Use the standard partition-normalized B-splines of the question: they are nonnegative and satisfy the subpartition of unity for B-splines, ${\sum }_j{N}_j(t)\le 1$. This also holds outside the basic knot interval; one may extend the knots and regard the finite family as a subfamily of a full partition of unity. The recurrence proof of the full partition is given in Question 6(c). Repeated knots retain the same nonnegativity and subpartition properties, with zero-denominator recurrence terms interpreted as zero or by knot limits. ConsequentlySampling has norm at most one, $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, and soHere the induced matrix norm is the maximum absolute row sum. The label $L^{\infty }$-normalized means the usual partition normalization, not rescaling each spline to have maximum value one. Such a rescaling would invalidate the specific coefficient bound and the cardinal values used below.

We first derive the recurrence needed for the partition of unity. Denote an order-$r$ partition-normalized spline by $N_{i,r}$. For $r\ge 2$, set $h_t(u)=(u-t{)}_{+}^{r-2}$, so $(u-t)h_t(u)=(u-t{)}_{+}^{r-1}$. The elementary product identities for divided differences areThey follow from the Leibniz rule for divided differences because the divided differences of the linear factor vanish above first order. Use the first identity on $t_i,\dots ,t_{i+r-1}$ and the second on $t_{i+1},\dots ,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ then givesThis is the Cox-de Boor recurrence, derived here from the specified normalization rather than assumed as a partition theorem.

Extend the distinct knot sequence in both directions to a strictly increasing sequence $(t_i{)}_{i\in \mathbb{Z}}$ tending to $\pm \infty$. For order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order splines are supported on $[t_i,t_{i+r}]$, making the following sums locally finite. Reindexing the second term of the recurrence yieldsInduction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each coefficient is nonnegative wherever its associated lower-order spline is nonzero. Removing splines therefore gives the subpartition of unity for B-splines on the whole line.

For $t_k<t<t_{n+1}$, a spline with index $i\le 0$ ends no later than $t_k$, and one with $i\ge n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\dots ,n$. For $k\ge 2$ the splines are continuous and vanish at their support endpoints, so the equality extends to the closed basic knot interval:If order one is allowed, the standard half-open indicator convention gives equality on $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^{\infty }$ norms. For the continuous splines of order at least two no such convention is necessary.

Fourier transform on the infinite spatial lattice giveswith polynomial roots $G_{\pm }=iA\pm \sqrt{1-A^2}$. When $|A|$ is uniformly less than one, the polynomial roots have modulus one and a uniform positive separation. Diagonalizing the companion matrix then bounds all its powers independently of $\theta$ and $n$. Parseval's identity proves stability in the discrete $L^2$ norm for arbitrary square-summable starting perturbations at both levels.

At $\mu =1/2$ the coefficients giveConsequently $\mu =1/2$ is stable.

At $\mu =3/2$,The growing polynomial root at this phase has modulusTo turn this into a Cauchy problem linear instability proof, rather than rely on a plane wave of infinite norm, choose a small interval around $\pi /3$ on which the growing polynomial root has modulus at least $1+\delta$. Choose an initial Fourier transform amplitude that is a nonzero square-integrable function supported there, and set the level-one amplitude equal to that polynomial root times the level-zero amplitude. The recurrence then multiplies by the same growing polynomial root at every step on the support. Parseval identity gives $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, whereas both initial norms are bounded independently of $h$. Real data can be obtained by adding the conjugate interval around $-\pi /3$. Since $n=T/(\mu h)$ tends to infinity at fixed $T$, $\mu =3/2$ is unstable.