A compatible almost complex structure is a smooth endomorphism $J:TX\to TX$ satisfying $J^2=-I$, preserving $\omega$, and making $g_J(u,v)=\omega (u,Jv)$ a positive-definite inner product. Preservation and $J^2=-I$ make this bilinear form symmetric.

To construct one, choose any Riemannian metric $h$ and define $A$ by $h(Au,v)=\omega (u,v)$. The endomorphism $A$ is skew-adjoint and invertible. Consequently $-A^2$ is positive definite and self-adjoint. Let $B=(-A^2{)}^{1/2}$ be its positive square root and setThe spectral theorem gives $B$ fibrewise, and the positive-square-root operation is smooth on positive-definite matrices, so this defines a smooth endomorphism. Since $A$ commutes with $B$, $J^2=-I$. Also $J^{\ast }=-J$, so $J^{\ast }J=I$. Commutation then givesThis is the metric construction of a compatible almost complex structure, proving nonemptiness.

If we start with $h=g_J$ for an already compatible $J$, then $h(Ju,v)=\omega (u,v)$, so $A=J$, $B=I$, and the construction recovers $J$. For two compatible structures, linearly interpolate their metrics,and apply the construction to $h_t$. This is a continuous smooth-in-$t$ path with endpoints $J_0,J_1$. Thus the space is nonempty and path connected. Interpolating instead to a single fixed auxiliary metric actually gives the contractibility of compatible almost complex structures.

For the cofinite topology ${\tau }_1$, the empty set and $\mathbb{N}$ are included. A nonempty union of cofinite sets is cofinite because its complement is contained in the finite complement of any one member. A finite intersection is cofinite because its complement is a finite union of finite sets; intersections involving the empty set are empty. This proves all topology axioms.

For the tail topology on the natural numbers ${\tau }_2$, $\mathbb{N}=I_1$. Any nonempty collection of tails has a smallest starting index, so its union is that tail; a finite intersection of tails is the tail with largest starting index. Empty sets and empty-family conventions supply the remaining cases. Thus ${\tau }_2$ is also a topology.

We classify continuous maps between cofinite and tail topologies. Continuity is equivalent toIf $f$ has unbounded image, each $E_k$ is nonempty, hence cofinite. For every threshold $k$, eventually $f(n)\ge k$, which is exactly $f(n)\to \infty$. If the image is bounded, it is a nonempty finite subset of $\mathbb{N}$ and has a maximum $N$. Then $E_N$ is nonempty and cofinite, but $f(n)\le N$ always, so $f(n)=N$ except at finitely many indices. These exhaust the two possibilities and prove necessity. Their sufficiency is checked separately below.

For integers $m,n$ not both zero, their greatest common divisor is the greatest positive integer dividing both. For the degenerate pair $(0,0)$ use the conventional value zero. A number $d$ divides both $m,n$ exactly when it divides both $m+kn,n$: one direction uses closure of divisibility under sums; the reverse subtracts $kn$. The positive common-divisor sets are therefore identical, provingThis also holds for $(0,0)$ under the stated convention and does not depend on the signs of the integers.

Bezout identity states that there exist integers $a,b$ such that $am+bn=\gcd (m,n)$. If a prime $p$ divides $mn$ but does not divide $m$, then $\gcd (p,m)=1$, so choose $a,b$ with $ap+bm=1$. Multiplying by $n$ gives $apn+bmn=n$. Both terms on the left are divisible by $p$, so $p\mid n$. Thuswhich proves the Euclid lemma from the requested identity. The separately numbered entries use these arithmetic facts for the Fibonacci numbers.

An equivalence relation on a set $X$ is a binary relation that is reflexive, symmetric and transitive: every $x\in X$ satisfies $x\sim x$; $x\sim y$ implies $y\sim x$; and $x\sim y$, $y\sim z$ imply $x\sim z$. Each of these quantifiers ranges over the entire specified set. The separate entries below test these three properties.

For the homogeneous solution, expand in the independent eigenvectors of $A$. Each coefficient obeys $c_j^{\prime }={\lambda }_j{c}_j$, so the complementary function isIf ${\mathbf{x}}_p$ is any particular integral, subtracting it from any other solution gives a homogeneous solution. Hence $\mathbf{x}={\mathbf{x}}_p+{\mathbf{x}}_c$ is the general solution. Complex-conjugate modes can be combined to produce real solutions when the forcing is real.

For a real two-by-two matrix, the characteristic polynomial is ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A$. Under the stated distinct-eigenvalue assumption, all nonzero homogeneous modes are purely oscillatory exactly when the eigenvalues are $\pm i\omega$ with $\omega >0$. Their sum and product imply $\mathrm{tr}A=0$, $\det A={\omega }^2>0$. Conversely these trace and determinant conditions force those eigenvalues, so there is no exponential growth, decay or Jordan-block secular growth.

For the final specified initial-value problem, direct multiplication gives $A^2=-9I$. The forcing vector $\mathbf{b}=(2,3i-1{)}^T$ satisfies $A\mathbf{b}=3i\mathbf{b}$, so the forcing is resonant and $t\mathbf{b}{e}^{3it}$ is a particular integral. The matrix exponential isSince the resonant particular integral vanishes initially, the initial-value solution isThis complex solution is appropriate to the complex forcing as written. Taking real parts gives the corresponding solution for the real part of that forcing.

The direction field has slope $1-y^2$, independent of $x$. Its line segments slope upward for $-1<y<1$, are horizontal at $y=\pm 1$, and slope downward for $|y|>1$. Thus $y=1$ is an attracting equilibrium in forward $x$, while $y=-1$ is repelling. The two constant solutions $y\equiv \pm 1$ must be included separately when separating variables.

The following original sketch shows the slope field and one solution in each requested region. The lower exterior solution ends at a finite pole; it cannot be continued across that pole as a real finite-valued solution.

Two nonzero vectors are linearly independent when $ax+by=0$ implies $a=b=0$. Their span then has dimension two. If they are linearly dependent, one is a scalar multiple of the other, so their span is a nonzero line and has dimension one. Thus the requested dimensions are two and one, respectively.

The dot product and corresponding Euclidean norm areThe Cauchy-Schwarz inequality states $\boxed{|x\cdot y|\le \Vert x\Vert \Vert y\Vert }$. If $y=0$ this is immediate. Otherwise the nonnegativity of a squared norm, with $t=(x\cdot y)/\Vert y{\Vert }^2$, givesMultiplying by $\Vert y{\Vert }^2$ and taking square roots proves the inequality. Equality holds precisely when the vectors are linearly dependent. Expanding another squared norm and using the inequality givesand hence the triangle inequality $\boxed{\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert }$.

In ${\mathbb{R}}^3$, the two vectors lie in a common plane. If their directions make angle $\alpha$, the law of cosines gives $\Vert x-y{\Vert }^2=\Vert x{\Vert }^2+\Vert y{\Vert }^2-2\Vert x\Vert \Vert y\Vert \cos \alpha$. Comparing with the dot-product expansion yieldsThe sketch shows the same relation as a projection: multiply the length of $x$ by the signed component of $y$ in its direction. That component is negative for obtuse $\alpha$.

A unit vector is a vector of Euclidean norm one, equivalently $u\cdot u=1$. In particular the dot product of two unit vectors is the cosine of their mutual angle.

There are nontrivial proper normal subgroups. The determinant defines a surjective group homomorphism $G\to \{1,-1\}$. Its kernel is the rotational symmetry group of a cube, a normal subgroup of order twenty-four. Also $\{I,-I\}$ is a central subgroup of order two: central inversion preserves the cube and commutes with every matrix. Both provide affirmative examples.

For the body diagonal through $(1,1,1)$, write a general cube symmetry as $D_{\epsilon }P$, with $P$ a permutation matrix and $D_{\epsilon }=\mathrm{diag}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$, each ${\epsilon }_j=\pm 1$. Since $P(1,1,1{)}^T=(1,1,1{)}^T$, preserving the unoriented diagonal requires $({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$ to be either $(1,1,1)$ or $(-1,-1,-1)$. ThusThe map $(\pi ,\epsilon )\mapsto \epsilon P_{\pi }$ from $S_3\times C_2$ to $H$ is a group homomorphism, because central inversion commutes with every coordinate permutation. It is surjective by the displayed description and injective because a permutation matrix cannot equal the negative of another permutation matrix. Consequently the body-diagonal stabilizer in the cube symmetry group isThis also agrees with the orbit-stabilizer theorem for the transitive action on the four body diagonals. The line is preserved as a set, so exchanging its opposite endpoints is allowed.

The radius of convergence of a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ is the unique $R\in [0,\infty ]$ for which the series converges absolutely when $|z|<R$ and diverges when $|z|>R$. When $R=\infty$ it converges for every complex $z$; when $R=0$ it converges only at the centre. The definition makes no assertion about convergence on $|z|=R$, so boundary points must be checked separately.

The integral test for convergence applies when $a_n=h(n)$ for all sufficiently large $n$, where $h$ is nonnegative, continuous and nonincreasing on $[N,\infty )$. It states that ${\sum }_{n=N}^{\infty }{a}_n$ converges if and only if the improper integral ${\int }_N^{\infty }h(x)dx$ is finite. Its comparison bounds areNo conclusion about an arbitrary nonnegative sequence follows without such a monotone comparison function; finite initial terms do not affect convergence.

For $h(x)=x^{-\alpha }$ with $\alpha >0$, all the hypotheses hold. If $\alpha \ne 1$,whereas for $\alpha =1$ the integral is $\log R$. The first expression has a finite limit exactly when $\alpha >1$. Thus the p-series satisfies

For $d\ge 2$, write ${\hat{x}}_j$ for the $(d-1)$ coordinates other than $x_j$, and setBy Tonelli theorem, $\Vert S_j{\Vert }_1=\Vert s_j{\Vert }_1$. The key estimate is the functional Loomis-Whitney inequality:Here is a proof in every dimension. In dimension two, the left-hand side separates into a product of two one-dimensional integrals, so Tonelli theorem gives equality. For $d\ge 3$, first integrate $x_d$. Apply Hölder's inequality with $d-1$ factors to $S_1,\dots ,S_{d-1}$, which may depend on $x_d$, and leave the factor $S_d$ outside that integral. WriteEach $H_j$ depends on the $d-2$ coordinates other than $j,d$. With $x^{\prime }=(x_1,\dots ,x_{d-1})$, the full integral is at mostPut $G(x^{\prime })={\prod }_{j<d}{H}_j^{1/(d-2)}$. A second application of Hölder's inequality, with exponents $d-1$ and $(d-1)/(d-2)$, bounds this expression byBy induction in dimension $d-1$, $\int G\le {\prod }_{j<d}(\int H_j{)}^{1/(d-2)}$. Since $\int H_j=\int S_j$, substitution proves the claimed functional inequality. Nonnegative integrands justify all integrations; truncation gives the extended-valued version if necessary.

The pointwise hypotheses imply $f^d\le {\prod }_j{S}_j$, soApply the proved functional inequality and take the $(d-1)/d$ power to obtainFor the useful case all the right-hand norms are finite. If one of them is zero, its line integral vanishes for almost every transverse coordinate and forces $f=0$ almost everywhere; there is no division by that factor. In dimension one the corresponding direct assertion is $\Vert f{\Vert }_{\infty }\le \Vert s_1{\Vert }_1$.

Let $x_1,\dots ,x_s$ contain all free variables of $M$. The lambda term $M$ is a solvable lambda term if there are closed lambda terms $P_1,\dots ,P_k$ such thatAdditional arguments beyond the first $s$ are allowed. A head normal form is a term $\lambda u_1\cdots u_n.y{M}_1\cdots M_r$ with a variable at the head; its arguments need not be beta-normal.

If $M$ reduces to this head normal form, close its free variables and supply every resulting initial binder with the same closed projection(take $P=I$ when $r=0$). The head variable becomes $P$ and discards all $r$ argument terms, even divergent ones, leaving $I$. Thus a head normal form implies solvability.

Conversely suppose a closure and applications of $M$ yield $I$. The standardization theorem for beta reduction supplies a finite successful head-reduction sequence. Until $M$ exposes a head variable, its head contractions lie within $M$: a closing substitution cannot create a new head redex there, and supplied arguments cannot be used until an initial abstraction is exposed. Each such contraction commutes with the closing capture-avoiding substitution. Project the successful head computation back onto $M$, ignoring contractions in supplied arguments and the consumption of its initial binders. If this projected head computation did not reach a variable-headed spine, the full computation could not reach $I$ either. It therefore gives a finite head reduction of $M$ to a head normal form. ConsequentlyThe same head-trace argument shows that an unsolvable lambda term stays unsolvable under substitution and application. This also justifies the unsolvability conclusion in the preceding solution.

Use sequential semicolon indices: $X^a{}_{;kl}={\nabla }_l({\nabla }_k{X}^a)$, where the second covariant derivative also differentiates the lower index of ${\nabla }_k{X}^a$. For a torsion-free connection, direct expansion givesSubtract the expression with $k,l$ exchanged. Commuting partial derivatives cancel the first term; the first-derivative terms cancel in pairs; symmetry of the lower Christoffel symbol indices cancels both terms containing ${\Gamma }^p{}_{kl}$. What remains isThis is the Ricci identity with the curvature convention printed on the PDF's first page. In terms of derivative operators it is $[{\nabla }_l,{\nabla }_k]X^a=R^a{}_{bkl}{X}^b$; writing the operator order explicitly avoids reversing the sign.

The lower-index rule follows by applying the commutator to the scalar contraction $Y_a{X}^a$. Its second derivatives commute, so the product rule implies $Y_{a;kl}-Y_{a;lk}=-R^p{}_{akl}{Y}_p$. The derivative commutator is a derivation on tensor products, since the cross terms cancel. It therefore acts once on every index, with a positive sign on each upper index and a negative sign on each lower index. For an arbitrary tensor of valence $(r,s)$,Local sums of decomposable tensors establish the formula for every tensor field, rather than only for a single tensor product.

The renormalization group compares descriptions at successively longer length scales. First eliminate short-wavelength degrees of freedom, then rescale coordinates and fields to restore the reference cutoff. This changes the couplings, including interactions and additive free-energy terms generated during elimination. A renormalization-group fixed point is a scale-invariant description. Linearized directions growing under coarse graining are relevant, shrinking directions are irrelevant, and zero-eigenvalue directions are marginal and require higher-order analysis.

A diverging correlation length means that many coarse-graining steps can approach a renormalization-group fixed point before a relevant perturbation becomes large. Systems with different microscopic details but the same renormalization-group fixed point and relevant directions share universal critical behavior. For the quadratic model below, Fourier modes decouple, so eliminating a high-momentum shell produces a Gaussian determinant but does not couple the remaining modes. The anisotropic gradient kernel requires different length rescalings in the parallel and transverse directions. This is an anisotropic renormalization group, not an isotropic rescaling of all $d$ coordinates.