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First use the actual PDF boost bracket $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$; replacing the final $J_k$ by $K_k$, as the TeX transcription does, would be a different and incorrect algebra. Expanding the three brackets yieldsThusThe two sets have SU(2)-type brackets. Their compact real form exponentiates to $SU(2)\times SU(2)$, whereas their full complexification exponentiates to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$. This does not turn the physical real Lorentz group into a compact product. With Hermitian $J,K$, $A^{†}=B$, and the two factors are tied by the Lorentz reality condition; the connected physical spin cover is $SL(2,\mathbb{C})$.

For the right-handed spinor representation, the supplied sigma formula gives ${\bar{\sigma }}^{ij}={ϵ}_{ijk}{\sigma }_k/2$ and ${\bar{\sigma }}^{0i}=i{\sigma }_i/2$. ThereforeIt is the $(0,1/2)$ representation in this definition of $A,B$. These are finite-dimensional representation matrices; the boosts are not Hermitian for a positive-definite spinor inner product, as expected for a noncompact group. They are distinct from the operator-component commutator coefficients discussed in Question 1.

For the group map, encode a real Minkowski vector by the Hermitian matrixThe trace identity $\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }{\sigma }_{\nu })={\delta }^{\mu }{}_{\nu }$ recovers its contravariant components. For $N\in SL(2,\mathbb{C})$, $X^{\prime }=NX{N}^{†}$ is Hermitian and has the same determinant. Expanding $X^{\prime }$ in that basis givesThe coefficients are real because the trace of a product of Hermitian matrices is real. Equality of determinants for all $x$ proves ${\Lambda }^T\eta \Lambda =\eta$. The action for $N_1{N}_2$ is the composition of the two actions, proving the homomorphism law.

Since $SL(2,\mathbb{C})$ is connected and the identity maps to the identity, its image has determinant $+1$ and preserves time orientation. More directly, positive-definite matrices representing future timelike vectors remain positive definite under $X\mapsto NX{N}^{†}$. The kernel is $\{I,-I\}$: if every Hermitian $X$ is fixed, $X=I$ first makes $N$ unitary, and then commutation with all Hermitian matrices makes it a scalar with determinant one.

Conversely, $SU(2)$ gives every spatial rotation, and positive Hermitian determinant-one matrices $\exp (\mathbf{b}\cdot \mathbf{\sigma }/2)$ give every pure boost. Rotation-boost decomposition yields every proper orthochronous Lorentz transformation. Thus the formula gives the Lorentz spinor double coverIt is a map into $SO(1,3)$ as asked, with precisely its identity component as image; it does not produce disconnected time-reversing transformations.

Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.

Raising the dotted index only changes the oscillator basis by the unitary antisymmetric epsilon matrix, so $\{q_a,q_b^{†}\}=2m{\delta }_{ab}$. Since $q_a^{†}|\Omega ⟩=0$,The positive convention would formally label this state with $j_3=+1/2$ while $J_{+}$ acts nontrivially on it, contradicting the highest-weight ladder rule. This supplies an independent state-level check of the needed sign correction.

The two-charge state is a rotation singlet. Indeed, using $q_1^2=q_2^2=0$ and $q_2{q}_1=-q_1{q}_2$, the product rule gives $[J_i,q_1{q}_2]=0$ for every $i$: only the sum of the two diagonal spinor coefficients survives, and their trace is zero. Hence the requested commutators and quantum numbers areIts squared norm is $(2m{)}^2$ before multiplication by $N^4$, by the two oscillator anticommutators.

The spin-zero massive N=1 supermultiplet therefore consists of $|\Omega ⟩$, $N{q}_1|\Omega ⟩$, $N{q}_2|\Omega ⟩$ and $N^2{q}_1{q}_2|\Omega ⟩$. Their spins are $0,1/2,1/2,0$, all with the same mass $m$. Starting with an even vacuum, the two spin-zero states are bosonic and the two one-charge states fermionic. They are the two real scalar degrees of freedom and the two on-shell fermion polarizations of a massive chiral multiplet. Any product of three creation charges repeats an index and vanishes; annihilation charges reduce to these four states by the canonical anticommutation relations. There are exactly four independent states.

For centered square-integrable variables with positive variances, the correlation coefficient isPut $v=\sqrt{\mathbb{E}[X^2]}$ and $v^{\prime }=\sqrt{\mathbb{E}[(X^{\prime }{)}^2]}$. ThenThus $\boxed{\rho =1⟺X=(v/v^{\prime })X^{\prime }\text{ almost surely}}$, with positive proportionality factor. Conversely positive proportionality immediately gives correlation one. For variables with nonzero means, this criterion applies to their centered versions and yields an affine, not necessarily proportional, relationship. That distinction is essential for the positive stock prices below.

At each event time within a stratum, let $r_A,r_B$ be the group risk sets and let $d$ be the total number of observed failures. Under the null of equal hazard functions, the expected A count is $d{r}_A/(r_A+r_B)$. The log-rank statistic adds observed minus expected A counts over event times; the stratified log-rank statistic first makes these comparisons within each patient stratum, then adds across patients. Let $V$ be the sum of the corresponding null conditional variances. The standardized statistic $U/\sqrt{V}$ has an approximate standard normal distribution under the null, and its square has an approximate one-degree-of-freedom chi-squared distribution. Compare it with the appropriate quantile for a two-sided test, under the usual independent censoring and exchangeability assumptions.

For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

For the constructed output, its defining quadratic says $h(X)=Y$ identically. Since the preceding branch calculation proves that its law is $IG(\mu ,\lambda )$, any variable with that law has the same pushforward distribution:One can also verify this chi-squared transform of an inverse Gaussian variable directly. The density contributions from the inverse branches areThe final equality follows from $x_{-}{x}_{+}={\mu }^2$. This confirms the transformed law independently of which reciprocal root was selected.

The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.

Take $\mathbb{N}=\{1,2,\dots \}$. We first prove the Ellis–Numakura lemma in the one-sided continuity convention needed here. Let $S$ be a nonempty compact Hausdorff space with an associative multiplication for which each map $x\mapsto xa$ is continuous. Among its nonempty closed subsemigroups, there is an inclusion-minimal one, $M$: a decreasing chain has nonempty intersection by compactness, and that intersection is a closed subsemigroup, so Zorn's lemma applies.

Fix $a\in M$. The set $Ma$ is nonempty and compact, hence closed in the Hausdorff space, and is a subsemigroup becauseMinimality gives $Ma=M$. Thus the set $T=\{x\in M:xa=a\}$ is nonempty. It is closed by the stated continuity, and it is a subsemigroup, since $(xy)a=x(ya)=xa=a$ for $x,y\in T$. Minimality again gives $T=M$, so $a\in T$ and $\boxed{a^2=a}$, giving a semigroup idempotent. If one instead uses the opposite one-sided continuity convention, the same proof uses $aM$ and $\{x:ax=a\}$; no joint continuity is required.

Apply this lemma to the Stone-Čech compactification of the natural numbers with its given addition. In the addition on the Stone-Čech compactification of the natural numbers convention,The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply there exists an idempotent ultrafilter $p$ with $p+p=p$.

On the positive integers such a $p$ is nonprincipal: a principal ultrafilter at $n$ adds to itself to give the one at $2n$, which is different. Hence every cofinite set belongs to $p$. If zero is included in one's convention for $\mathbb{N}$, use the same compact semigroup argument on the closed space of nonprincipal ultrafilters: it is nonempty by compactness of the infinite discrete set's compactification, and the displayed addition sends two free ultrafilters to a free ultrafilter. This avoids the trivial principal idempotent at zero.

To deduce Hindman's theorem, let $A$ be the unique colour class belonging to $p$. DefineIdempotence gives $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma also gives $A^{\ast }-n\in p$ whenever $n\in A^{\ast }$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ givesIntersecting this set with $A-n$ produces exactly $A^{\ast }-n$.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_r$ have been chosen with every nonempty finite sum in $A^{\ast }$. The finite intersectionbelongs to $p$, and is therefore nonempty. Choose $x_{r+1}$ from it. The new sums are $x_{r+1}$ and $s+x_{r+1}$, and they all lie in $A^{\ast }$. Induction yieldsThis proves every finite colouring of the positive integers admits a monochromatic finite-sums set generated by a strictly increasing infinite sequence, the required Hindman theorem.

Clear a common denominator first, so that all coefficients are nonzero integers. This changes neither the solutions nor partition regularity. We prove the one-equation criterion directly; no form of Rado's theorem is used as an assumption.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Colour a positive integer by its first nonzero base-$p$ digit: writing $x=p^{v_p(x)}u$ with $p\nmid u$, use the colour $u(\mathrm{mod}p)\in \{1,\dots ,p-1\}$. Suppose the equation has a monochromatic solution. Let $v$ be the smallest P-adic valuation of its entries and put $I=\{i:v_p(x_i)=v\}$, a nonempty set. Divide the equation by $p^v$ and reduce modulo $p$. All entries indexed by $I$ contribute the same nonzero digit $u$, and all other terms vanish. HenceIt follows that $p$ divides ${\sum }_{i\in I}{a}_i$. The absolute value of this sum is less than $p$, so $\boxed{\sum _{i\in I}{a}_i=0}$.

For sufficiency, take a nonempty zero-sum index set $I$ and put $S={\sum }_{i\notin I}{a}_i$. If $S=0$, the full coefficient sum is zero and the constant vector $(1,\dots ,1)$ is already a monochromatic solution in every colouring. Otherwise choose $j\in I$, put $t=|a_j|$, and set $b_j=-\mathrm{sgn}(a_j)S$, $b_i=0$ for $i\in I\setminus \{j\}$. ThenAdd the same sufficiently large nonnegative integer $H$ to all the $b_i$, obtaining $c_i=b_i+H\ge 0$. The sum is unchanged because ${\sum }_{i\in I}{a}_i=0$. Let $K={\max }_{i\in I}{c}_i$.

Apply the scaled Brauer progression theorem, proved above from the permitted Van der Waerden theorem, to obtain one-colour numbers $td$ and $a,a+d,\dots ,a+Kd$. DefineAll entries are positive and monochromatic, andTherefore the row is partition regular exactly when some nonempty coefficient subset sums to zero. Repetition of entries is allowed by the definition of partition regularity.

The final property concerns a pointwise periodic self-map. Such a map is a bijection: each point has a predecessor in its finite cycle, giving surjectivity; if $f(x)=f(y)$, choose a common multiple $N$ of periods of $x,y$ and apply $f^{N-1}$ to get $x=y$. Its finite cycles are therefore disjoint.

If $X$ is finite, choose the least common multiple of the finitely many cycle lengths. This positive integer $N$ gives $f^N={\mathrm{id}}_X$. The empty set satisfies the property with $N=1$.

Conversely, in the usual set theory with choice, every infinite $X$ contains a countably infinite set. Divide such a subset into disjoint finite blocks $B_j$ of sizes $j+1$, $j\ge 1$, make $f$ a cyclic permutation on each block, and fix every other point. Every point returns to itself, but an identity iterate would have to be divisible by every block length $j+1$. No positive integer has that property: take a block longer than that integer. Thus the uniform period criterion for pointwise periodic maps givesThe axiom of choice assumption is stated here because extracting a countably infinite subset of an arbitrary infinite set is not an unrestricted theorem in set theory without choice.

An equivalence relation $R$ on $X$ is a reflexive relation ($xRx$), a symmetric relation ($xRy\Rightarrow yRx$), and a transitive relation ($xRy$, $yRz\Rightarrow xRz$). Its equivalence class at $x$ is $[x{]}_R=\{y\in X:xRy\}$. Because $R$ is a reflexive relation, every class is nonempty and every $x$ belongs to its own class, so the classes cover $X$. If $[x{]}_R$ and $[z{]}_R$ share an element $y$, symmetry and transitivity give $xRz$. For every $w\in [x{]}_R$, transitivity then gives $zRw$, so $[x{]}_R\subseteq [z{]}_R$; exchanging $x,z$ gives equality. Thus distinct classes are disjoint. The equivalence classes form a set partition of $X$. On the empty set, the empty family is the corresponding partition.

A local odd mass density slope $\Sigma (x)-\Sigma (-x)=2{\Sigma }_0\beta x/r_0$ changes the cancelling one-sided torques into this net torque. A local outer cutoff $L$ adds the factor $1-(H/L{)}^2$. Positive slope transfers net angular momentum from satellite to disk. An infinite linear mass density law is only a formal local extension, since it cannot remain nonnegative on both sides.