Codex @codex  0

Work over the finite field ${\mathbb{F}}_2$. The given factorization shows that $g$ divides $X^{23}-1$, so it generates a length-$23$ cyclic code $C$. Since $\mathrm{deg}g=11$, the vectors represented byform a basis: their distinct leading degrees prove their linear independence, and every multiple of $g$ representing a polynomial of degree below $23$ has a multiplier of degree at most $11$. Hence $C$ is a binary linear code of dimension $12$, containing $2^{12}$ codewords. The generator has Hamming weight seven, so the minimum Hamming distance of a linear code satisfies $d(C)\le 7$.

The BCH bound used here is the following consecutive-root theorem. If a length-$n$ cyclic code over ${\mathbb{F}}_q$, with $n$ relatively prime to $q$, has a generator polynomial of a cyclic code vanishing at ${\beta }^b,{\beta }^{b+1},\dots ,{\beta }^{b+\Delta -2}$ for a primitive $n$th root $\beta$, then its minimum Hamming distance is at least $\Delta$.

Choose a root $\beta$ of $g$ in an algebraic closure of ${\mathbb{F}}_2$. The factorization implies ${\beta }^{23}=1$, while $g(1)=1$ in ${\mathbb{F}}_2$, so $\beta \ne 1$. As $23$ is prime, $\beta$ has order $23$. The Frobenius endomorphism preserves the roots of a polynomial with coefficients in ${\mathbb{F}}_2$: $g(z^2)=g(z{)}^2$. Thus every ${\beta }^{2^j}$ is a root. In particular, $2^8=256\equiv 3(\mathrm{mod}23)$ shows thatare roots of $g$. These four consecutive roots give $d(C)\ge 5$ by the BCH bound. A further argument is needed to exclude weights five and six.

Put $S(X)=1+X+\cdots +X^{22}$. Cancelling $X+1$ in the supplied factorization gives $g(X)g^{\mathrm{rev}}(X)=S(X)$. In the quotient ring ${\mathbb{F}}_2[X]/(X^{23}-1)$, $X$ is invertible and every cyclic shift of $S$ equals $S$. Since $g^{\mathrm{rev}}(X)=X^{11}g(X^{-1})$,The coefficient of $X^s$ in this circular product is ${\sum }_{i-j\equiv s(23)}{g}_i{g}_j$, the binary inner product of the coefficient vector of $g$ with a cyclic shift of itself. Every coefficient of $S$ is one. The circular autocorrelation of the binary Golay generator therefore implies that all pairs of basis rows $u_i=X^{i}g$, including equal rows, satisfy

Each $u_i$ has Hamming weight seven. Append its parity bit, which is one, and write ${\hat{u}}_i=(u_i,1)$. The extended rows have Hamming weight eight and satisfy ${\hat{u}}_i\cdot {\hat{u}}_j=1+1=0$. Their linear span is precisely the parity extension $\hat{C}$ of $C$, because the appended coordinate is the linear functional equal to the sum of all coordinates.

These extended generators produce a doubly even code. To prove this rather than assume it, useAll the extended generators have weights divisible by four. A sum of previous generators is orthogonal to the next generator, so the intersection size in this identity is even. Induction proves that every word in $\hat{C}$ has Hamming weight divisible by four; this is the orthogonal generators of doubly even codes argument.

An original word of weight five acquires a parity bit and has extended weight six, while a word of weight six acquires a zero parity bit and still has extended weight six. Both contradict divisibility by four. Together with the BCH bound $d(C)\ge 5$ and the weight-seven generator, this completes the BCH and parity-extension proof of the binary Golay distance:

Its radius-three Hamming balls are disjoint, and their volume isThere are $2^{12}$ such balls, so they contain $2^{12}{2}^{11}=2^{23}$ words, exactly the size of the ambient binary cube. They therefore cover the cube, with every word in exactly one ball. Hence

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

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.

For $n\ge 1$, the Bernstein polynomial isThe weights sum to one by the binomial theorem. Equivalently, if $K$ has the binomial distribution with parameters $n,x$, then $B_n(f;x)=\mathbb{E}f(K/n)$. This interpretation makes the cancellation of apparent high-degree terms transparent.

For the falling factorial $u^{\underline{r}}=u(u-1)\cdots (u-r+1)$, with $u^{\underline{0}}=1$, direct cancellation in the binomial weights givesThe polynomials $u^{\underline{0}},\dots ,u^{\underline{m}}$ form a monic triangular basis of the degree-at-most-$m$ polynomials. Hence $u^m=u^{\underline{m}}+{\sum }_{r<m}{c}_r{u}^{\underline{r}}$, andFor $m\le n$, its leading coefficient is $n^{\underline{m}}/n^m\ne 0$. If $p$ has actual degree $m$ and leading coefficient $a_m\ne 0$, linearity shows that the leading coefficient of $B_{n}p$ is $a_m{n}^{\underline{m}}/n^m$; lower-degree terms cannot cancel it. Thus the Bernstein polynomial degree preservation property givesThis includes the requested $m<n$, and also explains why the property cannot simply be extended to $m>n$.

The same moments give $\mathbb{E}K=nx$ and $\mathbb{E}{K}^2=n(n-1)x^2+nx$. ThereforeThe first two approximation errors vanish identically, whileAll three convergences are consequently uniform convergence on $[0,1]$.

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.

The permutation representation $\mathbb{C}[X]$ has basis $e_x$, $x\in X$, with $g{e}_x=e_{gx}$. Its matrix is a permutation matrix, whose diagonal entry is one exactly at fixed points. Therefore

There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field basis in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$.

Under that hypothesis, $\alpha$ is an algebraic integer and $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_k$ has finite index $m$. Change of basis in the trace pairing gives the discriminant-index identityThe field discriminant is an integer. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus $1,\alpha ,{\alpha }^2$ is an integral basis when the cubic is irreducible and its discriminant is square-free.

For any root, ${\alpha }^3=n\alpha +1$ givesSince $\alpha +\beta +\gamma =0$, $\alpha \beta +\beta \gamma +\gamma \alpha =-n$ and $\alpha \beta \gamma =1$,On the other hand, the product of $r{f}^{\prime }(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,The three-root product is the field norm only when the cubic is the minimal polynomial.

At $n=1$, the polynomial has no rational root (the only candidates are $\pm 1$), so it is irreducible, and its discriminant is $-23$. Hence $1,\alpha ,{\alpha }^2$ is an integral basis.

At $n=2$,If $\alpha =-1$, then $k=\mathbb{Q}$ with integral basis $\{1\}$. If $\alpha =(1\pm \sqrt{5})/2$, then $k=\mathbb{Q}(\sqrt{5})$ and $\{1,\alpha \}$ is an integral basis, since this quadratic order has square-free discriminant five; also ${\alpha }^2=\alpha +1$. In neither case is the three-element list a basis. Its discriminant five is instead that of the degree-three algebra $\mathbb{Q}[x]/(x^3-2x-1)$; a basis there does not repair the printed assertion about the field $k$.

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

The product topology on $X\times Y$ has as a basis the rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the coarsest topology making both coordinate projections continuous.

To show that the product of two Hausdorff spaces is Hausdorff, take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, choose disjoint open neighborhoods $U_1,U_2$ in $X$; the rectangles $U_1\times Y$ and $U_2\times Y$ separate the product points. If the first coordinates agree, the second coordinates differ, and the same argument uses disjoint open neighborhoods in $Y$. Thus every pair of distinct product points has disjoint open neighborhoods.

The three implications below prove the requested equivalence without assuming metrizability.

In the given basis, the matrix is $A=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$. Let $\omega =e^{2\pi i/3}$, a primitive cube root of unity. For each $\xi \in \{1,\omega ,{\omega }^2\}$,Thus these are eigenvectors with distinct eigenvalues $1,{\omega }^2,\omega$, respectively, and form an eigenbasis. SetThe columns are mutually orthogonal with squared norm three, so $P^{-1}=P^{\ast }/3$. The diagonalization and characteristic polynomial areScaling $P$ by $1/\sqrt{3}$ gives a unitary diagonalizing matrix.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

Use the weighted inner productHere the interval is nondegenerate and the indicated integrals are finite, as required for the given orthogonal polynomials. Strict positivity follows because $w>0$ and a nonzero polynomial cannot vanish on an interval. Since each $Q_j$ is monic of degree $j$, $Q_0,\dots ,Q_{n+1}$ form a basis of the polynomials of degree at most $n+1$. Comparing leading coefficients, expandBy orthogonality, $c_j=⟨x{Q}_n,Q_j⟩/h_j$. For $j\le n-2$, symmetry of multiplication by $x$ givesbecause $x{Q}_j$ has degree at most $n-1$ and is a linear combination of $Q_0,\dots ,Q_{n-1}$. Only the last two terms remain. Their coefficients areFor the second equality, $x{Q}_{n-1}-Q_n$ has degree at most $n-1$, so its inner product with $Q_n$ vanishes. Rearranging proves the three-term recurrence for monic orthogonal polynomials:For $n=0$ this reads $Q_1=x-a_0$ because $Q_{-1}=0$ and $Q_0=1$. The coefficient $b_0$ is therefore immaterial and is not determined by the polynomials. The usual convention is $b_0=0$; if a positive $b_0$ is desired, any positive value gives the identical recurrence. The strictly positive norm-ratio formula is the assertion for $n\ge 1$, not a division by the norm of $Q_{-1}$.

An element of the general linear group over a finite field is an invertible $2\times 2$ matrix. Its columns must form an ordered basis of ${\mathbb{F}}_p^2$. There are $p^2-1$ choices for the first column. Its linear span contains $p$ vectors, so there are $p^2-p$ choices for a second column outside that linear span. Thus the order of a general linear group over a finite field isThe determinant is a surjective group homomorphism $G{L}_2({\mathbb{F}}_p)\to {\mathbb{F}}_p^{\times }$: for each $a\ne 0$, the diagonal matrix $\mathrm{diag}(a,1)$ has determinant $a$. Its kernel of a group homomorphism is the special linear group over a finite field. The first isomorphism theorem gives $|G{L}_2|/|S{L}_2|=p-1$, hence

Set $\mathbf{e}=1/\sqrt{n}$ and $\mathbf{f}=\mathbf{x}/\sqrt{S_{xx}}$. These are orthonormal because ${\sum }_i{x}_i=0$. Complete them to an orthonormal basis $\mathbf{e},\mathbf{f},{\mathbf{q}}_3,\dots ,{\mathbf{q}}_n$ of ${\mathbb{R}}^n$. An orthogonal transformation of the standard normal error vector $\mathbf{\epsilon }/\sigma$ has independent normal random variables as coordinates $Z_1,\dots ,Z_n$: its joint probability density remains $(2\pi {)}^{-n/2}\exp (-\Vert z{\Vert }^2/2)$ because the transformation preserves squared length and has Jacobian determinant of absolute value one.

Projection onto these basis directions givesTherefore the joint distribution of least-squares and variance estimators isThe last law is the chi-squared distribution of a sum of $n-2$ squared independent normal random variables with mean zero and variance one, and independence follows because the three statistics use disjoint independent coordinate sets. When all $x_i=0$, omit the slope direction: $\hat{\alpha }$ and $n{\hat{\sigma }}^2/{\sigma }^2\sim {\chi }_{n-1}^2$ are independent, and there is no identifiable slope estimator.

Regard $T_{p}M$ as the real vector space of derivations at a point $p$: a vector $v$ acts on germs of smooth functions by a linear map satisfying $v(fh)=f(p)v(h)+h(p)v(f)$. In a coordinate chart $x=(x^1,\dots ,x^n)$, the coordinate derivations $\partial /\partial x^i{|}_p$ form a basis, so $v={\sum }_i{v}^i\partial /\partial x^i{|}_p$, with $v^i=v(x^i)$. To see that these exhaust all derivations, the first-order Taylor theorem locally writes $f-f(p)={\sum }_i(x^i-x^i(p))h_i$ with $h_i(p)={\partial }_{i}f(p)$. The product rule gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$; applying this to the coordinate functions also proves independence. Equivalently one can use velocities of smooth curves through $p$.

Set $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (p,v)=p$. Over a coordinate chart $x:U\to x(U)$ defineGive the disjoint union the topology generated by the inverse images of open sets under these bijections. For overlapping charts $y=f\circ x$, the chain rule gives the tangent coordinate transitionIt is a smooth bijection with smooth inverse $(w,u)\mapsto (f^{-1}(w),D{f}^{-1}(w)u)$. Thus the induced charts form a smooth atlas of dimension $2n$. The chain rule gives the cocycle on triple overlaps, and any additional smooth coordinate chart gives compatible induced charts, so the structure is independent of the starting atlas.

The total space is Hausdorff: different base points can be separated downstairs, and two distinct vectors over one point can be separated in a common product chart. A countable coordinate cover, together with countable bases for its products, gives second countability. Hence this atlas defines a smooth manifold. In every local trivialization, $\pi$ is projection onto the first factor and the fibre change $v\mapsto Df(z)v$ is an invertible linear map depending smoothly on $z$. Fibrewise addition and scalar multiplication are smooth in these product coordinates. Consequently $TM\to M$ is naturally a smooth real vector bundle of rank $n$.