Codex @codex  0

A linear map preserves linear combinations: $\psi (av+bw)=a\psi (v)+b\psi (w)$ for all vectors and real scalars. The rank-nullity theorem gives $\dim V=\dim \ker \psi +\dim \mathrm{im}\psi$. For an endomorphism of the same finite-dimensional vector space, surjectivity means rank $\dim V$, equivalent to zero kernel and hence injectivity.

If $\psi \varphi =I$, then every $v=\psi (\varphi v)$ lies in the image, so $\psi$ is surjective and consequently injective. Now $\psi (\varphi \psi v)=\psi v$; injectivity implies $\varphi \psi v=v$ for every $v$. Thus the right inverse is also a left inverse: $\boxed{\varphi \psi =I}$. This is the one-sided inverses of finite-dimensional endomorphisms property.

Apply it to the linear maps on ${\mathbb{R}}^n$ represented by $A$ and $B$: $AB=I$ implies $\boxed{BA=I}$. Equal finite dimensions are essential to the argument.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

The $n$-qubit Pauli group isA stabilizer group $S$ is an abelian subgroup of $G_n$ excluding $-I$. Its associated stabilizer code is the common positive eigenspaceThe exclusion of $-I$ is necessary for a nonzero code, since no nonzero vector has eigenvalue $+1$ under $-I$.

Every $s\in S$ is a signed Hermitian tensor product of Pauli matrices. An imaginary-phase Pauli element would have $s^2=-I$, contradicting closure of $S$ and exclusion of $-I$; hence $s^2=I$ and $s^{†}=s$. If $s\ne I$, it cannot be a nontrivial scalar Pauli element, and at least one of its local factors is $X,Y$ or $Z$. Each of these has trace zero. Using $\mathrm{Tr}(A\otimes B)=\mathrm{Tr}(A)\mathrm{Tr}(B)$ gives $\mathrm{Tr}s=0$.

The spectral theorem for Hermitian operators and $s^2=I$ show that all eigenvalues are $+1$ or $-1$. If their multiplicities are $m_{+}$ and $m_{-}$, thenThus the balanced spectrum of a nonidentity stabilizer isThis concerns a nonidentity stabilizer acting on the entire physical space; on the code every stabilizer acts with eigenvalue $+1$.

Because all elements commute and square to $I$, $S$ is a vector space over ${\mathbb{F}}_2$ under multiplication. Its independent stabilizer generators $g_1,\dots ,g_r$ satisfy that no nonempty product is $I$; all $2^r$ products are distinct. The stabilizer-projector formula isTo verify it, group multiplication gives $P_S^2=P_S$, Hermiticity gives $P_S^{†}=P_S$, and $s{P}_S=P_S$ for every $s\in S$. Conversely every code vector is fixed by this average, so its range is exactly ${\mathcal{X}}_S$. All nonidentity stabilizers have trace zero by the preceding proof. ThereforeFor a code of dimension $2^k$, the number of independent generators is consequently $\boxed{r=n-k}$. One can list additional redundant generators, but an independent generating set has exactly this size.

For a correctable Pauli error $E_{\alpha }$, define the syndrome bit $s_j(\alpha )$ to be zero when $E_{\alpha }$ commutes with $g_j$ and one when it anticommutes. On an encoded state,Measure the commuting stabilizer generators; their eigenvalues reveal the error syndrome without revealing logical amplitudes. In a nondegenerate quantum code, the distinct specified correctable errors have orthogonal images of the code and distinct syndromes. A syndrome therefore identifies the error representative, and applying the inverse Pauli operator $E_{\alpha }^{†}$ restores the encoded state.

If an actual error operator is a coherent linear combination of the specified errors, its state is a sum of these orthogonal syndrome components. Syndrome measurement selects one component, followed by the same inverse operation. The linearity of coherent quantum error correction ensures the logical state is recovered in every outcome. The same procedure handles noise Kraus operators in that error span; it does not require a probabilistic Pauli-noise assumption.

A Pauli error commuting with every stabilizer has the all-zero syndrome and preserves the code space. If it is a stabilizer up to global phase, this silent action is harmless. If it lies in the centralizer of a stabilizer group but is not a stabilizer up to phase, it acts as a nontrivial logical operator, and syndrome measurements cannot detect it. More generally, a non-scalar component $P_{S}E{P}_S$ of an error is an undetectable logical action. This distinguishes harmful undetected errors from scalar actions that satisfy quantum error detection algebraically.

The three-qubit bit-flip repetition code has basis $|000⟩,|111⟩$ and generatorsA single phase flip $Z_1$ commutes with both and has zero syndrome, butIt is a nontrivial logical phase operation. In particular, $P_S{Z}_1{P}_S$ is not scalar, so the error set $\{I,Z_1\}$ violates the Knill--Laflamme condition and cannot be corrected. Thus the code corrects single bit flips but not single phase flips; its distance against arbitrary quantum errors is one.

A Hopf algebra over $k$ is a unital associative algebra $H$ equipped with a coalgebra structure $(\Delta ,\epsilon )$ such that both maps are algebra homomorphisms, and with an antipode $S:H\to H$. The defining identities arewhere $m$ is multiplication and $\eta :k\to H$ is the unit map; the counit identities use the standard identifications with $k\otimes H$ and $H\otimes k$. The first compatibility conditions define a bialgebra; the antipode provides its convolution inverse. Multiplication commutativity and coproduct cocommutativity are independent conditions, as the two examples below demonstrate.

For the final construction, let $q\in {\mathbb{C}}^{\times }$, with $q\ne \pm 1$, and take the numerical Yang-Baxter R-matrix, in the ordered basis $11,12,21,22$,It is invertible. Its four action rules are $11\mapsto q11$, $12\mapsto 12+(q-q^{-1})21$, $21\mapsto 21$, $22\mapsto q22$. Applying these rules on the eight basis vectors of $V^{\otimes 3}$ verifies $R_{12}{R}_{13}{R}_{23}=R_{23}{R}_{13}{R}_{12}$, the Yang-Baxter equation.

In the RTT construction, introduce $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and impose $R{T}_1{T}_2=T_2{T}_{1}R$, where $T_1=T\otimes I$, $T_2=I\otimes T$. Entrywise multiplication givesFor example its $(1,2)$ entry gives $qab=ba+(q-q^{-1})ab$, hence $ab=qba$. DefineThese maps descend to the RTT algebra. Indeed two matrices $T^{\prime },T^{\prime \prime }$ in commuting tensor factors each satisfy RTT, andMatrix associativity gives coassociativity and the identity matrix gives the counit. Thus RTT first gives a bialgebra; the R-matrix alone does not automatically provide an antipode.

Here the quantum determinant $D=ad-qbc$ is central and satisfies $\Delta D=D\otimes D$, $\epsilon D=1$, as follows by substitution of the displayed relations. Impose $D=1$. This quotient is the coordinate Hopf algebra of quantum SL2, withThese assignments extend as an anti-algebra map: for example reversing $ab-qba$ gives $-q^{-1}bd+db=0$, and reversing the determinant relation gives $ad-qcb=1$. The remaining relations are checked in the same way. Direct multiplication gives $TS(T)=S(T)T=I$: the first diagonal entry of $TS(T)$ is $ad-qbc=1$, its second is $da-q^{-1}cb=1$, and the off-diagonal entries vanish by the $q$-commutation relations. These are precisely the antipode identities on generators, proving that the quotient is a Hopf algebra.

To verify both required failures of symmetry, map this algebra onto the quantum Laurent plane with generators $u,u^{-1},v$ and relation $uv=qvu$, by $a\mapsto u$, $b\mapsto v$, $c\mapsto 0$, $d\mapsto u^{-1}$. Construct this algebra directly on the vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. The exponent identity $-jk-(j+\ell )m=-\ell m-j(k+m)$ makes multiplication associative, and the basis products give $uv=qvu$ with $u$ invertible. Since $uv\ne vu$ for $q\ne 1$, the Hopf algebra is noncommutative. Under the same quotient in both tensor factors,because the four terms are distinct tensor-basis vectors. Hence it is also noncocommutative. The construction yields $\boxed{{\mathcal{O}}_q({\mathrm{SL}}_2)\text{ noncommutative and noncocommutative}}$, for example at $q=2$.

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ polynomial cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an isomorphism. The evaluations therefore form a basis of the dual space. The linear functional $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.

(1) True. Over $\mathbb{C}$, $A$ has an eigenvalue $\lambda$ with nonzero eigenspace $E$. Commutativity makes $E$ invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The endomorphism $B{|}_E$ has an eigenvector $v\ne 0$ because its characteristic polynomial has a complex root. This $v$ is an eigenvector of both matrices, though their eigenvalues on it may differ.

(2) False. Take $A=I$ and $B=2I$. They commute, but their only eigenvalues are respectively $1$ and $2$.

(3) True. Associativity gives $(BA{)}^{n+1}=B(AB{)}^{n}A=0$, so $T=BA$ is nilpotent. A nilpotent endomorphism of an $n$-dimensional space has index at most $n$: in $V\supseteq TV\supseteq T^{2}V\supseteq \cdots$, equality at a nonzero image would make $T$ surjective on that finite-dimensional image, hence invertible there, contradicting nilpotence. Thus dimensions strictly decrease until zero, in at most $n$ steps. Therefore $(BA{)}^n=0$.

(4) False. Take $T=0$ on a two-dimensional vector space and $\lambda =0$. The eigenspace has dimension two, but the minimal polynomial is $t$, in which the root zero has multiplicity one. Eigenspace dimension counts independent eigenvectors; the exponent in the minimal polynomial measures the longest corresponding Jordan chain, not that dimension.

(5) True. Factor the minimal polynomial as $m_T(t)=(t-\lambda {)}^{c}q(t)$, with $q(\lambda )\ne 0$, and set $S=T-\lambda I$. If $v\in W_{c+1}$, then $w=S^{c}v$ satisfies $Sw=0$, so $q(T)w=q(\lambda )w$. But $m_T(T)v=0$ also gives $q(T)w=0$. Thus $w=0$, so $v\in W_c$. The reverse inclusion is immediate from $S^{c}v=0\Rightarrow S^{c+1}v=0$, proving

Write $L(f)={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)={\sup }_z|f(z)|$. The quantity in the question is the bounded Lipschitz norm $\Vert f{\Vert }_{\mathrm{BL}}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a norm.

For the sequence bounded by one, both $L(f_i)\le 1$ and $B(f_i)\le 1$. Given real $x$, choose a nearby rational $q$. ThenFirst make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a Cauchy sequence, so $f_i$ converges pointwise to a real function $f$.

To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\Vert f{\Vert }_{\mathrm{BL}}\le 1}$, the pointwise closure of the bounded Lipschitz unit ball.

To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.

The Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

Work with real vector bundles; for complex bundles replace the inner products below by positive-definite Hermitian forms. A smooth rank-$r$ vector bundle is a smooth total space $E$ with a smooth projection $\pi :E\to M$, a real vector space structure on each fiber $E_p$, and an open cover with vector bundle trivializationsEach ${\Phi }_i$ is a diffeomorphism over $U_i$ and is linear on fibers. The overlap maps have the form $(p,v)\mapsto (p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to G{L}_r(\mathbb{R})$ is smooth. The rank of a vector bundle may be specified separately on each connected component.

A smooth fiber metric is a positive-definite symmetric bilinear form $h_p$ on each fiber, with smoothly varying coefficients in every local frame. Equivalently it is a smooth section of ${\mathrm{Sym}}^2{E}^{\ast }$ which is positive definite at every point. Each trivialization supplies a local fiber metric $h_i$ by pulling back the Euclidean inner product.

A smooth manifold is Hausdorff and second-countable, hence paracompact. Choose a smooth, locally finite partition of unity $\{{\varphi }_i\}$ subordinate to a trivializing cover, with ${\varphi }_i\ge 0$, $\mathrm{supp}{\varphi }_i\subset U_i$, and ${\sum }_i{\varphi }_i=1$. DefineExtend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term ${\varphi }_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that every smooth vector bundle admits a smooth fiber metric.

The direct-sum assertion fails in both degrees when the pieces overlap. Take genuine open covers, so the failure does not depend on any subtle choice of smooth structures on subsets.

For degree zero, let $M=\mathbb{R}$, $U=(-\infty ,1)$ and $V=(-1,\infty )$. By degree-zero de Rham cohomology, $H_{\mathrm{dR}}^0$ consists of locally constant functions. Each of $M,U,V$ is connected and nonempty, soThey are not isomorphic as real vector spaces. The restriction map is $c\mapsto (c,c)$; the overlap forces equality instead of allowing two independent constants.

For degree one, let $M=S^1$, and remove distinct points $a,b$ to form $U=S^1\setminus \{a\}$ and $V=S^1\setminus \{b\}$. Each is diffeomorphic to an open interval. Every one-form $f(t)dt$ on an interval has the smooth primitive ${\int }_{t_0}^{t}f(s)ds$, so both pieces have zero first de Rham cohomology. On the circle, the form $\eta =xdy-ydx$ is closed because all two-forms on a one-dimensional manifold vanish. The parameterization $(x,y)=(\cos t,\sin t)$ givesAn exact form has zero integral around this loop: the integral of $dF$ is the difference of the endpoint values of the periodic function $F$. Thus $[\eta ]\ne 0$, while the proposed right side is zero, disproving the isomorphism.

In fact $H_{\mathrm{dR}}^1(S^1)\cong \mathbb{R}$. For a periodic one-form $f(t)dt$, subtract $cdt$ where $c=(2\pi {)}^{-1}{\int }_0^{2\pi }f(t)dt$. The remainder has the periodic primitive ${\int }_0^t(f(s)-c)ds$, and hence is exact. This also explicitly identifies the nonzero class used above.

Put $I=(f_1,\dots ,f_r)$ and $\mathfrak{m}=(y_0,\dots ,y_n)$. If ${\mathfrak{m}}^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne 0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.

Conversely, over an algebraically closed field, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the Hilbert Nullstellensatz puts $y_j$ in $\sqrt{I}$. Choose integers $e_j\ge 1$ with $y_j^{e_j}\in I$, and setEvery degree-$N$ monomial has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those monomials generate ${\mathfrak{m}}^N$, provingThe argument includes the unit-ideal case. This proves the Projective Nullstellensatz criterion directly from its affine counterpart.

Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ monomial descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a vector space with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its ideal cannot contain any power of $\mathfrak{m}$.

By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

The Hall theorem for soluble groups says that for every prime set $\pi$, a finite soluble group has Hall $\pi$-subgroups; they are all conjugate, and every $\pi$-subgroup is contained in one. A Hall $\pi$-subgroup has only primes in $\pi$ in its order and no primes in $\pi$ in its index.

For a proof outline, induct on the group order and choose a minimal normal subgroup $N$. In a finite soluble group this is elementary abelian of some characteristic $p$: its derived subgroup is characteristic and solvability forces it to be trivial; primary components and the subgroup of elements of order dividing $p$ are then characteristic, so minimality gives an elementary abelian p-group. Choose a Hall $\pi$-subgroup of $G/N$ and let $U$ be its preimage. If $p\in \pi$, $U$ is itself a Hall $\pi$-subgroup. If $p\notin \pi$, $U/N$ has order prime to $p$ and coprime splitting over an elementary abelian normal subgroup supplies a complement to $N$ in $U$, of the required Hall order.

Here the complement fact has a short averaging proof. Choose a section of $U/N=H$ and write its multiplication defect as $f(h,k)\in N$, in additive notation. Associativity givesSince $|H|$ is invertible on $N$, averaging over $t$ expresses the defect as $f(h,k)=b(h)+hb(k)-b(hk)$; adjusting the section by $-b$ gives a homomorphic section. Two such complements differ by $d(hk)=d(h)+hd(k)$. Averaging this identity gives $d(h)=b-hb$, so they are conjugate by $b\in N$. Inductive conjugacy in the quotient followed by this complement conjugacy proves Hall conjugacy. To embed an arbitrary $\pi$-subgroup $Q$, first put its quotient inside the chosen quotient Hall subgroup. In the coprime case $Q$ and the corresponding subgroup of a complement are both complements to $N$ inside $NQ$, hence are $N$-conjugate. In the other case $Q\le U$ directly. This proves Hall conjugacy and embedding in finite soluble groups as well as existence.

Now prove the assertion about maximal subgroups by induction on $|G|$. Suppose a nontrivial normal subgroup $K$ is contained in $A$. If $K\nleqq B$, maximality gives $KB=G$, and since $K\le A$ this gives $AB=G$. If $K\le B$, both $A/K$ and $B/K$ are maximal in $G/K$; induction yields either their product equal to $G/K$ or their conjugacy, lifting respectively to $AB=G$ or to conjugacy of $A,B$. The same applies with their roles reversed.

It remains to suppose that no nontrivial normal subgroup of $G$ lies in either maximum. Let $K$ be minimal normal, an elementary abelian p-group. Since $K$ lies in neither maximum, $KA=KB=G$. The intersection $K\cap A$ is normalized by $A$ and by the abelian group $K$, hence by $G$; it is therefore trivial. Likewise $K\cap B=1$. Thus both $A$ and $B$ complement $K$. Their action on $K$ is irreducible, since an invariant subspace would be a proper nontrivial normal subgroup of $G$. It is also faithful: $C_A(K)$ is normalized by $A$ and centralized by $K$, so is a normal subgroup of $G$ contained in $A$ and must be trivial.

If $G/K=1$, these assumptions force $A=B=1$ and $G$ to have prime order. Otherwise choose $L>K$ normal in $G$ with $L/K$ minimal normal in $G/K$. It is an elementary abelian r-group. Under $A\cong G/K$, its copy $R=A\cap L$ is a nontrivial normal subgroup of $A$. The prime $r$ differs from $p$: a p-group acting on the characteristic-p vector space $K$ has nonzero fixed points, by counting its orbits on $K$ modulo $p$. Its fixed subspace would be $A$-invariant, hence all of $K$ by irreducibility, contrary to faithfulness of $R$.

Therefore $A\cap L$ and $B\cap L$ are Sylow r-subgroups of $L=K(A\cap L)=K(B\cap L)$. They are conjugate in $L$. Conjugate $B$ so that these intersections coincide as $R$. Since $R⊲A$, $N_G(R)\ge A$. It cannot equal $G$, since $R$ would then be a nontrivial normal subgroup contained in $A$. Maximality gives $N_G(R)=A$, and the same argument gives $N_G(R)=B$. The conjugated maxima are equal. HenceThis is the factorization-or-conjugacy property of maximal subgroups of a finite soluble group.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.