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With zero degree assigned to the zero element of the homomorphism group of elliptic curves, the degree parallelogram law makes this symmetric pairing additive in both arguments. Its diagonal is the degree of an isogeny; it is positive definite on every finite-dimensional rational span after extension to the real numbers. The integral polarization $2B(\varphi ,\psi )$ is integer-valued, while $B$ can be half-integral. For instance, if an elliptic curve has an automorphism $\omega$ with ${\omega }^2+\omega +1=0$, then $1+\omega =-{\omega }^2$ has degree one, so $B(1,\omega )=-1/2$.

If nonnegative integer-valued random variables $X_n$ have all fixed factorial moments converging to ${\lambda }^k$, then their convergence in distribution is to the Poisson distribution with mean $\lambda$. The first-moment bound gives uniform tightness, and bounds on higher moments give uniform integrability of each fixed lower moment. Every subsequential limit therefore has the indicated factorial moments. For such a limit $X$ and $t\ge 0$, the nonnegative binomial expansion and the monotone convergence theorem give $\mathbb{E}(1+t{)}^X={\sum }_{k\ge 0}{t}^k{\lambda }^k/k!=e^{\lambda t}$. This identifies its moment-generating function as that of the Poisson distribution, uniquely determining the limit.

A degree-$g$ line bundle with vanishing first sheaf cohomology has exactly one section up to scale, by Riemann-Roch theorem. Its zero divisor is unique and varies in families by cohomology and base change for line bundles on a curve. For a fixed divisor $B$ of degree $N-g$, the open subset of $C^{(g)}$ where $\mathcal{O}(E)$ has vanishing first cohomology gives a chart in degree $N$, by $E\mapsto \mathcal{O}(B+E)$. Different charts are glued by the unique residual divisor of $\mathcal{O}(B+E-B^{\prime })$ wherever that line bundle is again nonspecial. Normalizing the universal bundles along a fixed rational point gives their canonical compatible gluing and a representing scheme for the Picard functor of a curve.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

Set $p=t-\lambda r$ using part (a). At every node,If $\lambda =0$, the error vanishes. Otherwise any degree-at-most-$n$ polynomial $q$ with strictly smaller error would make $q-p$ have alternating nonzero signs at the increasingly ordered nodes (reversed globally if $\lambda <0$). It would then have at least $n+1$ distinct roots by the intermediate value theorem, impossible. Hence

The coefficient of $x^{n+1}$ in $r_j$ isWith the increasing enumeration, its sign is $(-1{)}^{n+2-j}$. Therefore all terms $(-1{)}^j{c}_j$ have the same sign $(-1{)}^{n+2}$, and $R:={\sum }_j(-1{)}^j{c}_j\ne 0$. The leading coefficient of $t$ is $T:={\sum }_{j}f(y_j)c_j$. Canceling the coefficient of $x^{n+1}$ gives the unique valueBoth interpolants have degree at most $n+1$, so this single cancellation is equivalent to $\mathrm{deg}(t-\lambda r)\le n$.

A recurrent state is null recurrent if its expected elapsed return-cycle time is infinite. Recurrence itself agrees with that of the jump chain, but positive recurrence need not agree, because large states can have very short holding times.

A state is recurrent if, after departure, its return probability is one. For an irreducible nonexplosive continuous-time Markov chain, this is equivalent to recurrence of its jump chain and is shared by all states.

A state of an irreducible continuous-time Markov chain is transient if, after departure, its return probability is less than one. Equivalently its jump chain visits that state only finitely often almost surely. This is a class property.

Perfect correlation means an affine relation almost surely: there are constants $a\ne 0,b$ such that $Y=aX+b$ almost surely. The sign of $a$ is the sign of the correlation coefficient. More precisely, equality in the Cauchy-Schwarz inequality gives

A survey depth $R\sim 400\mathrm{Mpc}$ makes wavelengths substantially larger than its volume poorly measurable. A characteristic inverse depth is $1/R\simeq 2.5\times 10^{-3}\mathrm{Mp}{\mathrm{c}}^{-1}$; a Fourier wavelength that fits across a dimension $L$ instead requires $k\gtrsim 2\pi /L$. For $L$ between $R$ and $2R$, this is roughly $0.008$--$0.016\mathrm{Mp}{\mathrm{c}}^{-1}$. A limited sky patch can have smaller transverse dimensions, so its survey window function raises the practical minimum in those directions. An indicative useful band for a broad survey is around $10^{-2}\mathrm{Mp}{\mathrm{c}}^{-1}$ and above, including scales near equality if its volume is large enough.

There is no upper wavenumber determined by depth alone. Mean separation of observed galaxies, galaxy power spectrum shot noise, angular/redshift resolution, and nonlinear clustering set it. Straightforward linear inference is typically restricted to roughly $k\lesssim 0.1\mathrm{Mp}{\mathrm{c}}^{-1}$ at low redshift, with smaller scales requiring modelling. The magnitude limit does not specify the galaxy comoving number density, so a more precise sampling cutoff cannot be deduced from these data.

The two measurements are complementary. The cosmic microwave background samples a large volume at an early epoch when relevant perturbations are mostly linear, and is free of galaxy bias; its matter-spectrum inference is indirect and depends on projection, cosmological parameters, foregrounds, and cosmic variance. A galaxy survey gives three-dimensional late-time information and can reach smaller scales, but observes biased, discrete tracers: galaxy bias, nonlinear gravitational growth, redshift-space distortions, selection effects, and galaxy power spectrum shot noise must be controlled. Combining both constrains the initial spectrum and its subsequent growth more strongly than either alone, while their overlapping modes test the inferred evolution.

The initial-value problem is a Riemann problem. Both the one-dimensional Euler equations and the two constant initial states are unchanged under the simultaneous scaling $(x,t)\mapsto (\lambda x,\lambda t)$ for $\lambda >0$. There is no imposed length or time scale. Rescaling a physically admissible entropy solution therefore gives another solution with exactly the same initial data. Uniqueness of the physical Riemann solution makes the two coincide, soThe smooth pieces are consequently those in part (a), while shocks and contacts lie on rays $x=st$. Across shocks one uses the conservative mass, momentum and total-energy equations and the entropy admissibility condition, rather than extending the smooth entropy-advection equation through the discontinuity. Scale invariance by itself, without selecting the unique admissible solution, would not justify the uniqueness step.

Secure in the intended ideal setting, with a sound entanglement-purification procedure. Bob confirms receipt before Alice reveals either the random basis string or the randomly chosen check positions. Eve must therefore release the travelling qubits before knowing which basis to use and which positions to protect from disturbance. Bob can then decode because $H^2=I$.

The random check sample tests the same ensemble of encoded transmissions as the retained data. Random computational basis and Hadamard basis encodings make the test sensitive to complementary disturbances; Eve cannot use the targeted zero-error attack available in Protocol 1. For example, measuring every intercepted qubit in one fixed one of these two bases uses the wrong basis half the time, and then disagrees with Alice half the time, producing a bit error rate of $1/4$. A sound entanglement purification protocol uses the tested error bounds, corrects both relevant types of error, and aborts if the noise is too high; secrecy is not claimed for every possible attack without an abort option.

When the protocol succeeds in producing nearly perfect Bell pairs, purity decouples a subsystem from its purification explains why they are nearly independent of Eve, and matching computational basis measurements produce an approximately uniform shared secret key. The disclosed check outcomes are discarded rather than included in that key. This is the standard entanglement-based quantum key distribution ordering: receipt first, basis and sample disclosure second, testing and purification next, and private measurements last.

Impossible to implement as written: Bob is never given the basis string. Step 7 requires him to know which incoming qubits were acted on by a Hadamard gate, but the announcement contains only the check positions. Since $H^{-1}=H$, knowing that a transformation is its own inverse does not identify which qubits need it.

Bob cannot recover the missing information from the qubits alone. For each encoded Bell pair, his reduced density matrix isfor either $b=0$ or $b=1$. His entire received register has the same density matrix $I/2^{2n}$ for every string $b$, so every local measurement in quantum mechanics has statistics independent of $b$. Nor can one fixed unitary undo both possibilities: preserving $|{\Phi }^{+}⟩$ requires that unitary to be proportional to $I$, whereas decoding $(I\otimes H)|{\Phi }^{+}⟩$ requires it to be proportional to $H$. Supplying the missing basis announcement after receipt would repair the protocol and make it the third one, but that communication is absent from the printed sequence.

Insecure as a proposed test-and-purify protocol: the basis string and check positions are announced before receipt. Eve can intercept and store the travelling qubits until the announcement. For each data position with basis bit $b_j$, she measures the stored qubit in the basis $\{H^{b_j}|0⟩,H^{b_j}|1⟩\}$. If the outcome is $z_j$, the Alice–Bob pair becomesand Eve knows $z_j$. She forwards that qubit, and forwards every publicly identified check qubit untouched. After Bob's Hadamard gate decoding, every data pair is the known product state $|z_j{z}_j⟩$, while the unmodified check pairs remain perfect. Thus the check bit error rate is zero even though Eve knows all raw data bits. This is premature basis announcement in quantum key distribution.

The retained data are a separable quantum state with a classical label held by Eve; LOCC cannot turn them into private Bell pairs. Consequently the listed check does not justify the claimed success of the subsequent entanglement purification. A sound purification procedure that independently detects the missing entanglement must fail or abort on this attack. If Step 9 were instead read as an independently guaranteed production of private, nearly perfect Bell pairs, their subsequent key would of course be private; the flaw is that this protocol's test cannot establish that guarantee. Naming an EPP does not repair the information leak or the invalid sampling test.

A secure key can be produced, at one ideal secret bit per copy. Denote Alice's additional flag qubit by $F$. Alice can measure $F$ in the computational basis and, when its outcome is $1$, apply the Pauli Z gate to her half of the shared pair. The two flag outcomes each have probability $1/2$, and $Z_A|{\Phi }^{-}⟩=|{\Phi }^{+}⟩$; either branch therefore supplies the same pure Bell pair.

Equivalently, Alice applies the local controlled unitaryThis local flag correction of a Bell-state phase mixture changes the full density matrix intoThe corrected pair is pure, so every purification factors between $AB$ and $FE$. Eve may know the flag without knowing the key bit obtained from $AB$. Alice need not disclose the flag, but even disclosing it does not compromise the corrected pure Bell pair. Tracing out the flag before correction would give the nonprivate mixture in the preceding case; retaining Alice's local side information is essential.

No secure key can be produced. The off-diagonal terms in the two Bell state projectors cancel, giving the separable quantum stateA possible purification of a density operator is $(|00{⟩}_{AB}|0{⟩}_E+|11{⟩}_{AB}|1{⟩}_E)/\sqrt{2}$. Eve measures her register in the computational basis and learns whether Alice and Bob have $00$ or $11$. Their apparently shared random bit is therefore completely known to Eve. Conditioned on her record the Alice–Bob state is a known product state; public discussion and local private randomness cannot create a shared secret unknown to that record. The state has no distillable secret key in the stated purification model.

No secure key can be produced from this state alone. Factoring the printed amplitudes gives the normalized product stateEve is uncorrelated with the joint pure state, but Alice and Bob also have no correlations with each other: local outcome probabilities factor. They may generate independent private random bits locally, but local operations and classical communication cannot make those bits shared and secret using only public messages. A product state supplies no shared secret resource.

The normalized ray produces a secure key; the literal printed vector is not normalized. The squared norm of the coefficients in the PDF is $8/9+1/81=73/81$. Thus the physical pure state on that ray has normalized coefficients $6\sqrt{2}/\sqrt{73}$ and $1/\sqrt{73}$. This is a genuine entangled state with two nonzero Schmidt coefficients, and Eve is independent of it by purity decouples a subsystem from its purification.

For an explicit way to obtain uniform key bits, Alice applies a two-outcome local measurement in quantum mechanics with Kraus operatorsTheir squared products sum to the identity. The successful unnormalized state is $(|00⟩+|11⟩)/\sqrt{73}$, soAlice announces only success or failure. Success leaves an exact Bell pair, yielding one uniform secret bit; failure leaves $|00⟩$ and is discarded. With many copies the successful fraction is positive. Alternatively, direct computational basis measurements give a shared biased bit with probabilities $72/73$ and $1/73$, from which privacy amplification can extract uniform secret bits. The coefficient $1/9$ has not been silently replaced by $1/3$.

A secure key can be produced, at one ideal secret bit per copy. The shared Bell state is pure, so the preceding factorization shows that Eve's purifying system is independent of it. Measuring both qubits in the computational basis gives matching outcomes $00$ or $11$, each with probability $1/2$. Neither party announces these key outcomes. Eve's state is the same for both possibilities, so each copy supplies a shared uniform private bit.