Codex @codex  0

Constructing the subsheaf. On an integral scheme, every nonempty open subset is an irreducible topological space and hence a connected space. A locally constant function from such an open subset to $F$ is constant. Thus the constant sheaf $\underline{F}$ has sections $F$ on every nonempty open subset, and the structure sheaf embeds in it by restriction to the generic point. Empty members of the given open cover may be discarded.

For every open subset $W\subseteq X$, defineThis is a subsheaf of the sheaf of modules $\underline{F}$: its condition is local, and compatible sections glue in $\underline{F}$, with the regularity conditions gluing in ${\mathcal{O}}_X$. On $U_i\cap U_j$, the hypothesis says that $f_j/f_i$ is a section of the sheaf of units of the structure sheaf. Therefore multiplication by it preserves ${\mathcal{O}}_X$, andIt follows that $\mathcal{L}{|}_{U_i}=f_i^{-1}{\mathcal{O}}_{U_i}$: the condition with index $i$ implies every other condition on an overlap, and the converse is part of the definition. Multiplication by $1/f_i$ is an isomorphism ${\mathcal{O}}_{U_i}\to \mathcal{L}{|}_{U_i}$, with inverse multiplication by $f_i$. Hence $\mathcal{L}$ is an invertible sheaf, or equivalently a line bundle, andThese local descriptions also give uniqueness as a subsheaf of $\underline{F}$. This construction is the divisor line bundle for the local rational equations $f_i$.

The tensor product. Refine the two open covers to $W_{ij}=U_i\cap V_j$. The ratiosare sections of the sheaf of units of the structure sheaf on the corresponding overlaps, so the products give a line bundle $\mathcal{M}\subseteq \underline{F}$ by the construction just proved. Multiplication in $F$ defines a sheaf morphismOn each $W_{ij}$ it sends the local generator $(1/f_i)\otimes (1/g_j)$ to the local generator $1/(f_i{g}_j)$. It is thus an isomorphism on an open cover, and hence an isomorphism globally. In particular,

The sign for projective twists. Write homogeneous coordinates $[x_0:\cdots :x_n]$ on projective space ${\mathbb{P}}_{\mathbb{Z}}^n$, and let $U_i=D_{+}(x_i)$ be its standard affine open cover. For any integer $m$, take the nonzero elements of the function fieldTheir ratios are $f_i/f_j=(x_j/x_i{)}^m$, which are sections of the sheaf of units of the structure sheaf on $U_i\cap U_j$, including when $m<0$. The twisting sheaf on projective space $\mathcal{O}(m)$ has local generator $x_i^m$ on $U_i$. Divide a rational section of $\mathcal{O}(m)$ by $x_0^m$ to embed it in the constant sheaf of the function field; the local generator becomes $(x_i/x_0{)}^m=1/f_i$. This identifies the constructed line bundle with $\mathcal{O}(m)$ and fixes the sign explicitly.

For $n\ge 1$, $\mathrm{Pic}({\mathbb{P}}_{\mathbb{Z}}^n)\cong \mathbb{Z}$, with generator $[\mathcal{O}(1)]$; the displayed data gives all its powers, with $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$ by the product rule above. If $n=0$, projective space is $\mathrm{Spec}\mathbb{Z}$, its Picard group is zero, and the same formula gives only the trivial line bundle.

Put $b=\Delta {\omega }_b>0$ and $s=t-t_m$. At the peak time, $e^{-ip\Delta \omega t_m}=1$, so the exact comb sum becomesFor many occupied normal modes, $b\gg \Delta \omega$, replace the sum by an integral with density $1/\Delta \omega$. Setting $x=\omega -{\omega }_0$ and evaluating the Fourier transform of a Gaussian givesThus the isolated-optical pulse approximation hasThe optical pulse has a Gaussian function envelope and no chirp. Its carrier angular frequency is ${\omega }_0$, and its light intensity full width at half maximum is $2\sqrt{2\ln 2}/b$. Successive envelope maxima have different carrier-envelope phases, consistent with b.

Approximation in the printed assertion. A discrete optical frequency comb gives a train, not one exact isolated Gaussian pulse. Applying the Poisson summation formula to the ideal sum over all integers instead givesNear $t_m$, the $j=m$ term dominates when $bT\gg 1$. The other terms are exponentially small but not identically zero. A single Gaussian function used globally would decay between $0$ and $T$, whereas the exact light intensity has equal maxima there; this supplies a concrete reason the isolated form cannot be an exact global identity. The extension to negative frequencies is the usual narrow-positive-band analytic-field approximation when ${\omega }_0\gg b$.

A reduced crystallographic root system in a real Euclidean space is a finite spanning subset $\Phi$ not containing zero, with $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$, closed under $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$, and satisfying $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for all roots. A base of a root system is a vector-space basis $\Delta \subseteq \Phi$ such that every root has integer coefficients of one sign in that basis.

Choose $h$ not orthogonal to any root and call roots with $(h,\alpha )>0$ positive. Let $\Delta$ be the positive roots that cannot be written as a sum of two positive roots. Repeated decomposition terminates because each summand has strictly smaller positive $h$-value in a finite set. Thus every positive root is a nonnegative integer combination of $\Delta$.

For distinct $\alpha ,\beta \in \Delta$, $(\alpha ,\beta )\le 0$: otherwise the allowed difference-root lemma would make either $\alpha -\beta$ or $\beta -\alpha$ positive, decomposing one of them. In a real linear relation among $\Delta$, move negative coefficients to the other side, obtaining $u=v$ as positive combinations with disjoint supports. Then $\Vert u{\Vert }^2=(u,v)\le 0$, so $u=0$. But a nonempty positive combination has $(h,u)>0$, a contradiction. Thus the elements are linearly independent. They span every root and hence $E$, proving existence of a base of a root system. The difference-root lemma is used only for distinct roots; that qualification excludes the zero difference.

For the displayed set, there are eight coordinate roots, twenty-four roots with two nonzero unit entries, and sixteen half-coordinate roots: forty-eight in total. It spans ${\mathbb{R}}^4$, is reduced, and has squared root lengths one and two. Signed coordinate permutations preserve it. Reflections in coordinate roots are sign changes, and reflections in roots $e_i\pm e_j$ are signed coordinate permutations. Every half-coordinate root is obtained by a signed permutation from $h=(1/2,1/2,1/2,1/2)$, so it remains to check its reflectionIt sends $e_i$ to $e_i-h$, a half-coordinate root; it fixes $e_i-e_j$ and sends $e_i+e_j$ to the negative sum of the other two coordinate vectors. For a half-coordinate root, the sum of its four signs is $0,\pm 2$ or $\pm 4$; the formula sends it, respectively, to itself, a coordinate root, or its negative when parallel to $h$. Thus reflection closure holds. For a long root, the required Cartan integer is an integer dot product; for a coordinate root it is twice a half-integer; for $h$ it is ${\sum }_i{\beta }_i\in \mathbb{Z}$. Symmetry gives the remaining cases. Therefore this is the F4 root system.

In the printed base order, the Gram matrix isUsing the convention $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, the Cartan matrix isThe row-coroot convention transposes this matrix. The Dynkin diagram is the chain $1-2\Rightarrow 3-4$, with the double bond pointing toward the shorter third root:

The associated symmetric positive definite form, normalized as a squared length, isIndeed it equalswhich is strictly positive for nonzero $x$. The alternative Gram normalization $Q/2$ is the same positive definite form up to a positive scalar.

For an integer partition $\lambda =({\lambda }_1,\dots ,{\lambda }_r)$, its Young diagram has ${\lambda }_i$ left-justified cells in row $i$. A Young tableau of shape $\lambda$ is a filling of its cells. Under the usual standard numbering convention, a standard Young tableau uses each integer $1,\dots ,n$ once and increases strictly along every row and down every column. For a different fixed ordered set of distinct labels, the same inequalities define the corresponding ordered-label version, which becomes standard after replacing its labels by their ranks.

Young's rule states that in characteristic zero,where the Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and type $\mu$: rows are weakly increasing, columns strictly increasing, and label $i$ occurs ${\mu }_i$ times. The summands are irreducible Specht modules, and the coefficients are nonnegative integer multiplicities.

The standard polytabloid basis is defined over the integers and remains a basis after reduction modulo three. Thus $\dim S^{(3,2)}=f^{(3,2)}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)-\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$. The given submodule has dimension four by part (d), so the quotient is one-dimensional and part (c) leaves only the trivial or sign action.

A transposition has ordinary trace one on $S^{(3,2)}$: the Murnaghan–Nakayama rule has only the horizontal bottom-row two-strip, leaving shape $(3)$ with dimension one. Reduction of its integral action matrix preserves this trace in $F$. Its trace on $D^{(4,1)}=S^{(4,1)}$ is $3-1=2$, since the natural five-point permutation representation has three fixed coordinates and contains the trivial line. Therefore its trace on the one-dimensional quotient is $1-2=-1$, distinct from one in characteristic three. This gives the characteristic-three sign quotient of the Specht module of shape three two. The quotient is the sign representation:

With the convention applicable also to disconnected compact groups, the Weyl group is $W=N_G(T)/T$, where the normalizer consists of elements taking $T$ to itself by conjugation. Put $\mathfrak{t}=L(T)$ and $\Lambda =\ker (\exp :\mathfrak{t}\to T)$. This is a full lattice with $T\cong \mathfrak{t}/\Lambda$.

If $\psi$ is a torus automorphism, exponential compatibility gives $\psi (\exp X)=\exp (D{\psi }_{I}X)$. Therefore $D{\psi }_I$ sends $\Lambda$ into itself; applying the inverse proves it sends the lattice onto itself. In a lattice basis this derivative is an element of $G{L}_r(\mathbb{Z})$.

Conjugation by a connected subgroup $Q\subset N_G(T)$ induces a continuous family of these integer matrices. Since $G{L}_r(\mathbb{Z})$ is discrete, this family is constant and equals its value at the identity. Its action on $\mathfrak{t}$ is therefore trivial, and surjectivity of the torus exponential shows that $Q$ centralizes $T$.

Apply this to the identity component $N_G(T{)}^0$. It lies in $G^0$ and centralizes $T$. The allowed maximal-abelian property of $T$ in $G^0$ gives $N_G(T{)}^0\subseteq T$, while the connected group $T$ lies in that identity component. Thus $N_G(T{)}^0=T$. The normalizer is a closed compact Lie subgroup, so its component quotient is discrete and compact, hence finite. Consequently $W$ is finite and discrete. For disconnected $G$, its action on the lattice need not be faithful; the component argument is essential.

The distinction is recorded in Weyl group of a disconnected compact Lie group.

Schur orthogonality for irreducible complex characters of a compact group, with normalized Haar measure, isEquivalent representations clearly have equal characters, since trace is invariant under conjugation. Conversely complete reducibility gives ${\chi }_V={\sum }_{\alpha }{m}_{\alpha }{\chi }_{\alpha }$, where the nonnegative integer multiplicity is recovered by $m_{\alpha }=⟨{\chi }_V,{\chi }_{\alpha }⟩$. Equal characters thus give equal multiplicities of every irreducible constituent, hence equivalent representations.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

Take $S_1$ to consist of the three usual first-order sentencesThese assert reflexivity, symmetry and transitivity and therefore define exactly the equivalence relations.

For $S_2$, adjoin to $S_1$ the single first-order sentenceIts witness $y$ gives a second member of each class, while its final universal clause excludes all others. Hence every class has exactly two members, and conversely such a relation satisfies it. This remains a finite set of first-order sentences.

For each positive integer $n$, writeThenEvery class in a first-order model has at least $n$ elements for every $n$, so cannot be finite. Conversely every infinite equivalence class supplies each finite collection of distinct witnesses. The displayed infinite scheme therefore has exactly the requested first-order models.

A nonempty weakly linear function has a domain $D$ closed under rational scalars and addition, hence a vector subspace of $\mathbb{R}$ over $\mathbb{Q}$. Additivity first gives $f(0)=0$ and $f(-x)=-f(x)$, then $f(nx)=nf(x)$ for integers $n$. If $q=m/n$ with $n>0$, rational closure puts $qx$ in $D$ andThus $f(qx)=qf(x)$: it is a $\mathbb{Q}$-linear map. The empty function is not maximal because it extends to the function on $\{0\}$ with value zero.

Suppose a maximal $f$ had proper domain $D\ne \mathbb{R}$. Choose $t\notin D$ and any prescribed real value $a$, for instance $a=0$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$: if two expressions coincide and the coefficients differ, then $t$ would belong to $D$. DefineUniqueness makes this well-defined. The displayed logical formula and the $\mathbb{Q}$-linearity of $f$ show directly that $h$ is additive and has a domain closed under rational scalars. Hence it is a weakly linear function and extends $f$. It is a proper extension because $t$ is now in its domain, contradicting maximality. ThereforePart (a) and Zorn's lemma guarantee the existence of maximal extensions as well as this description of their domains.

Use the following group axioms in the given language:Commutativity supplies the other identity and inverse equations. These first-order sentences therefore specify precisely an abelian group with the named identity $0$; an inverse function symbol is unnecessary because inverses are asserted to exist by a quantified formula.

For each positive integer $n$, let $t_n(x)$ be the first-order term recursively defined by $t_1(x)=x$ and $t_{n+1}(x)=t_n(x)+x$. Add the torsion-free abelian group axiom schemeEvery first-order model of these first-order sentences is an abelian group with no nonidentity element of finite order. Conversely every torsion-free abelian group satisfies each of the displayed first-order sentences. This is a countably infinite first-order axiomatization of exactly the required class.

No finite axiomatization exists. Write $\Gamma$ for the axioms just constructed and suppose a finite set $\Delta$ of first-order sentences had exactly the same first-order models. Its logical conjunction $\delta$ is a single first-order sentence, and $\Gamma \models \delta$. By the compactness theorem there is a finite subset ${\Gamma }_0\subseteq \Gamma$ with ${\Gamma }_0\models \delta$: otherwise $\Gamma \cup \{\neg \delta \}$ would be finitely satisfiable and hence satisfiable, a contradiction.

Only finitely many of the integers $n$ in the torsion-free scheme occur in ${\Gamma }_0$. Choose a prime number $p$ larger than all of them. The cyclic group $\mathbb{Z}/p\mathbb{Z}$ satisfies all the group axioms and each of these finitely many conditions from the torsion-free abelian group axiom scheme: multiplication by an integer $1\le n<p$ is invertible modulo $p$, so $nx=0$ implies $x=0$. It consequently satisfies ${\Gamma }_0$, then $\delta$, and therefore $\Delta$.

But its nonzero elements have finite order $p$, so it is not a torsion-free group, contradicting the assumed description of the first-order models of $\Delta$. This establishes the nonfinite axiomatizability of torsion-free abelian groups, ruling out every finite set of defining first-order sentences, not just finite subsets of the particular scheme in part (c).

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.

The expression is the Binet formula for the Fibonacci numbers. Compute the two roots and round the result to the nearest integer:

The returned object is an integer. This floating-point arithmetic implementation correctly rounds the requested range through $n=50$; it is not an arbitrary-precision method for indefinitely large $n$. An integer recurrence is a useful independent numerical check.

It performs a trial division test for a prime number. An integer below $2$ is rejected; otherwise the function searches for a divisor from $2$ through $\lfloor \sqrt{n}\rfloor$. A composite number has a prime factor no larger than its square root, so finding no such divisor proves primality. For example, mystery2(97) returns True, while mystery2(91) returns False, since $91=7\cdot 13$.

This describes the intended mathematical test. For extremely large integers, the floating-point computation n**0.5 can round or overflow; an implementation intended for arbitrary-size integers should use an exact integer square root.

With its default argument and a nonnegative integer input, it prints the decimal digits of $n$ in order, one per line. For example, mystery4(407) prints 4, then 0, then 7. Integer division removes the final digit, recursion handles the preceding digits, and the remainder gives the final digit. The function returns None.

There is a subtlety for a nondefault base: the recursive call omits m, so it resets to $10$. For $m\ge 2$ and $n\ge m$, the actual output is the decimal digits of $\lfloor n/m\rfloor$, followed by the remainder $n\mathrm{mod}m$. For example, mystery4(13, 2) prints 6, then 1, rather than the binary digits of $13$. Passing m on to the recursive call would give a consistent base-$m$ positional notation. The interpretation as digits assumes $n\ge 0$ and $m\ge 2$; arbitrary integer arguments need not satisfy those conditions.

Conditional predictions for observed birds and years add their fitted random intercepts to the fixed part of the Gaussian linear mixed model:These are predicted expected values, so they need not be integers. The first is likely more reliable: age three is within the displayed age range, whereas age seven requires extrapolation beyond ages one to six. A quantitative comparison would need the prediction intervals, including uncertainty in both fixed coefficients and random effects; the displayed coefficients alone do not supply them.

The regression diagnostics raise concerns about the mean and variance assumptions. The residual-versus-fitted smoother is curved rather than horizontal, and the scale-location plot suggests changing residual spread, contrary to homoscedasticity. Diagonal bands are expected when an integer-valued response is fitted by a continuous model, since each fixed response value gives $r=y-\hat{y}$. They are not, by themselves, proof of a failed mean model. The residual Q-Q plot is broadly linear but has tail departures and unusual observations, so exact normality is questionable.

The Cook's distance plot identifies observations worth checking for recording errors and influence on the age estimate. A rough $4/n\simeq 0.023$ screening threshold flags several points, but it is not an automatic deletion rule. Repeat the fit with and without influential observations as a sensitivity analysis and report material changes.

The bird random intercept Q-Q plot has a pronounced upper-tail departure; the year plot is closer to a line, although eleven groups give little power to assess normality. These plotted effects are estimated, shrunken conditional mode of Gaussian random effects, not independent observed draws from the latent normal distribution, so their appearance requires care.

The analysis is therefore provisional. Check a nonlinear age effect, the changing residual variance, dependence beyond the fitted random intercepts, and a response model appropriate for counts. If clutch size is known, a binomial distribution model for hatched eggs out of available eggs may be more meaningful. A bootstrap from the existing Gaussian model cannot validate the assumptions questioned by these plots. Distinguishing within-bird ageing from between-bird differences would also strengthen interpretation.

For $\delta \notin 2\pi \mathbb{Z}$, the finite geometric series givesThe exponential prefactor has modulus one. ThusAt an excluded phase every summand is one, which supplies the removable limiting value directly. Here $N$ is a positive integer.