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Triangulate the sphere by the boundary complex of the cross-polytope with simplicial vertices $\pm e_0,\dots ,\pm e_n$: a face contains at most one simplicial vertex from each antipodal pair. Take its barycentric subdivision. Its simplicial vertices are the nonempty faces $F$ of the original complex, and its simplices are strictly nested chains $F_0\subset \cdots \subset F_k$.

Form a quotient complex whose simplicial vertices are pairs $\{F,-F\}$, and whose simplices are the images of these chains. No chain contains both $F$ and $-F$. Moreover a chain of face-orbits lifts uniquely up to simultaneous negation: after choosing a largest face, every smaller face has at most one representative contained in it, because an original face contains no antipodal simplicial vertices. Thus the quotient is an honest simplicial complex, without two different simplices with the same simplicial vertex set. Its realization is homeomorphic to the sphere modulo its antipodal action. This provides a triangulation of Real projective space ${\mathbb{RP}}^n$. The subdivision avoids the duplicate-simplex problem of naively identifying simplicial vertices in the unsubdivided cross-polytope boundary.

Write $n=2m-1$. On ${\mathbb{R}}^{2m}$ define the orthogonal linear mapIt satisfies $J^T=-J$ and $J^{T}J=I$. Hence $⟨x,Jx⟩=0$ and $\Vert Jx\Vert =\Vert x\Vert$. Restricting to the unit sphere givesThis continuous unit tangent field is available exactly because the ambient dimension is even.

Here ${\mathcal{O}}_K=\mathbb{Z}[s]$ with $s=\sqrt{-14}$ and field discriminant $-56$. The imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so each ideal class has an integral representative of ideal norm at most four. Representatives therefore use prime ideals above two and three. PutThe prime two ramifies, with ${\mathfrak{p}}^2=(2)$, and three splits into $\mathfrak{q}\overline{\mathfrak{q}}=(3)$. No element has field norm two, because $N(a+bs)=a^2+14b^2$; thus $[\mathfrak{p}]$ is nontrivial of order two. Norm-four ideals represent $[\mathfrak{p}{]}^2=1$ and add no generator, so the class group is generated by $[\mathfrak{p}],[\mathfrak{q}]$.

The element $2-s$ has field norm 18. It belongs to $\mathfrak{p}$ and $\mathfrak{q}$, but not $\overline{\mathfrak{q}}$, since reduction there gives $s\equiv 1(\mathrm{mod}3)$. Its ideal therefore factors asThe exponents follow from the ideal norm 18, or from the root $s\equiv 2(\mathrm{mod}9)$ defining ${\mathfrak{q}}^2$. Hence $[\mathfrak{q}{]}^2=[\mathfrak{p}]$, so $[\mathfrak{q}]$ has order four and already generates all possible classes. ThereforeAs a separate check, the reduced forms of discriminant of a binary quadratic form $-56$ are $[1,0,14],[2,0,7],[3,2,5],[3,-2,5]$, giving exactly four classes.

For a square-free integer $\delta \equiv 2$ or $3(\mathrm{mod}4)$, the ring of integers of the quadratic field is ${\mathcal{O}}_{\mathbb{Q}(\sqrt{\delta })}=\mathbb{Z}[\sqrt{\delta }]$. Square-freeness is the usual convention underlying the first request: congruence alone would not suffice, for example $\delta =27$ gives the same field as $\delta =3$.

Now take the stipulated square-free $\delta \equiv 3(\mathrm{mod}4)$ and put $r=\sqrt{2}$, $s=\sqrt{\delta }$. For any integral $\alpha =a+br+cs+drs$, relative traces to the three quadratic subfields show $2a,2b,2c,2d\in \mathbb{Z}$. Those subfields all have integral bases $1$ and their square root, including $\mathbb{Q}(\sqrt{2\delta })$. Write $\alpha =(A+Br+Cs+Drs)/2$ with integer coefficients.

Its relative field norm to $\mathbb{Q}(s)$ isThus $AC$ is even, and the constant numerator is divisible by four. Since $\delta \equiv 3(\mathrm{mod}4)$, these conditions force $A,C$ both even: one odd makes the numerator odd, and both odd contradicts $AC$ even. They then force $B,D$ to have the same integer parity. Every algebraic integer consequently belongs to the lattice generated byConversely the last generator $w$ satisfies $w^2=(1+\delta )/2+\sqrt{\delta }$ and the monic integer equationAll four generators are integral and linearly independent over $\mathbb{Q}$, so this lattice is exactly ${\mathcal{O}}_L$, proving this integral basis of an odd biquadratic field containing square root of two without an unproved index assumption.

In $V_n\otimes V_n$, interchange of the two factors commutes with SU2. On the highest-weight vector $D^r{x}_1^{n-r}{x}_2^{n-r}$ it acts by $(-1{)}^r$. Each irreducible summand in the Clebsch-Gordan decomposition has multiplicity one, so this sign is its scalar action on the entire summand. The symmetric square is the $+1$ eigenspace and the exterior square the $-1$ eigenspace. ThereforeFor instance, $S^2{V}_1=V_2$ and ${\Lambda }^2{V}_1=V_0$. The dimensions agree respectively with $(n+1)(n+2)/2$ and $n(n+1)/2$.

The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.

These irreducible submodules have distinct highest weights and form a direct sum: a sum of earlier ones cannot contain an irreducible module inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an inner product over the compact group. Assuming $m\le n$, their total dimension iswhich equals the dimension of the tensor product. Hence they exhaust it, proving the theorem.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

The cyclotomic polynomial ${\Phi }_7=1+X+\cdots +X^6$ is irreducible: replacing $X$ by $X+1$ gives an Eisenstein polynomial at seven. Hence the extension has degree six, andThere is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb{Q}(\eta )$; the whole group fixes $\mathbb{Q}$. The order-two subgroup $\{1,6\}$ is complex conjugation and fixes $\mathbb{Q}(\eta +{\eta }^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta +{\eta }^{-1})X+1$ and is not real.

The order-three subgroup $\{1,2,4\}$ fixes $u=\eta +{\eta }^2+{\eta }^4$. Its other conjugate is $v={\eta }^3+{\eta }^5+{\eta }^6$. Expanding products and using ${\sum }_{j=1}^6{\eta }^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this fixed field is $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the Galois correspondence proves they are the full fixed fields.

A cyclotomic field over $\mathbb{Q}$ is a field $\mathbb{Q}({\zeta }_m)$ generated by an $m$th root of unity of exact order $m$. It contains all roots of $X^m-1$, and is its splitting field. In characteristic zero this polynomial is separable, so the extension is Galois.

Every automorphism sends ${\zeta }_m$ to ${\zeta }_m^a$ with $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$, and is determined by that image. Composition multiplies the exponents modulo $m$, yielding an injective homomorphism into an abelian group. Thus the Galois group is abelian. Using irreducibility of the cyclotomic polynomial, every unit exponent occurs andIf “cyclotomic” is used more broadly for a subfield of such a field, it is still Galois and abelian, because every subgroup of an abelian Galois group is normal.

The Fundamental theorem of Galois theory applies to a finite Galois extension $L/K$ with group $G$. It gives inclusion-reversing inverse bijectionsThe fixed field $L^H$ consists of elements fixed by every member of $H$. Moreover $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois iff $H=\mathrm{Gal}(L/M)$ is normal, in which case restriction gives $\mathrm{Gal}(M/K)\cong G/H$. All intermediate $L/M$ are themselves Galois. Finiteness and normality/separability of $L/K$ are essential for this stated form.

If a regular graph has $c>1$ connected components, the indicator vector of each component is a $\Delta$-eigenvector. They are linearly independent, and the Laplacian argument shows that every such eigenvector is constant on each component. Therefore its multiplicity is exactly $c$, in particular greater than one.

For a $\Delta$-regular undirected graph, its graph Laplacian is $L=\Delta I-A$, and for any complex vector $v$,If $Av=\Delta v$, the left side is zero, hence $v_i=v_j$ on every edge. In a connected graph all coordinates coincide. Thus the eigenspace is exactly the constant vectors, of dimension one. Since the adjacency matrix of a graph is symmetric and diagonalizable,

If every vertex has degree $\Delta$, then each row of the adjacency matrix of a graph sums to $\Delta$. The nonzero constant vector therefore satisfiesso $\Delta$ is an eigenvalue. This uses the usual finite, nonempty graph convention.

Let $A$ be the adjacency matrix of a graph and $Av=\lambda v$ with $v\ne 0$. Choose a vertex $i$ maximizing $|v_i|>0$. ThenDividing givesThis argument works even with a complex eigenvector; for a finite undirected graph the matrix is real symmetric and all eigenvalues are real.

Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.

Use fit.3 if backward selection is intended to combine adjacent periods. Each cumulative indicator is a separate model term; deleting its coefficient enforces equality of the two adjacent period means separated by that boundary. This offers an interpretable reduced stepwise profile. Fit.2 treats the entire factor as one term, so ordinary term-wise backward selection retains all five contrasts or removes period altogether. Removing selected baseline contrasts by hand would instead privilege period one, rather than grouping neighbouring periods. If the scientific objective were simply whether any period effect exists, retaining or deleting the whole factor in fit.2 would be the appropriate different procedure.

The two design matrices span the same six-dimensional space of arbitrary log means for the six periods. Fit.2 uses the first period as a baseline, whereas fit.3 uses cumulative indicator steps. If $a_j$ is the log mean for period $j$, fit.3 has intercept $a_1$ and jump coefficients $a_{j+1}-a_j$. Thus both fits give identical fitted means and identical maximized likelihood functions. ThereforeDifferent coefficient tests arise because they test different contrasts, not because one full fit explains the observations better.

The Poisson deviance tests overall fit against the saturated Poisson regression, not whether the single slope is zero. Under fit.1, its large-count approximation is ${\chi }_{60-2}^2={\chi }_{58}^2$. The observed value is 261.72, far above the one-percent upper critical value (about 85.95); its upper-tail probability is about $1.14\times 10^{-27}$. Thus reject fit.1 at the one-percent level. The slope's reported $p$-value 0.421 only compares this inadequate log-linear trend with an intercept-only model; it is not a goodness-of-fit test.

The appropriate homotopy invariance of winding number is: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for each $s$, then all loops $t\mapsto H(s,t)$ have the same winding number. The basepoint may move; every intermediate curve must close and avoid zero.

Compactness gives $m={\min }_{s,t}|H(s,t)|>0$. Uniform continuity gives $\delta >0$ such that $|s-s^{\prime }|<\delta$ implies $|H(s,t)-H(s^{\prime },t)|<m$ for every $t$. Part (d) then makes the winding numbers equal for nearby $s,s^{\prime }$. Subdivide $[0,1]$ into steps smaller than $\delta$ and chain these equalities. ThereforeThis proves the theorem directly for continuous loops, without requiring an integral formula or smooth homotopy.

The continuous closed curve $q(t)={\gamma }_2(t)/{\gamma }_1(t)$ satisfies $|q(t)-1|<1$. It lies in the right half-plane, which admits a single continuous complex argument in $(-\pi /2,\pi /2)$. Since $q$ closes, this argument has equal endpoint values, giving $w(q)=0$. Additivity under multiplication then yieldsThe strict inequality ensures that the quotient stays in a region avoiding zero and admitting a single argument branch.