Every finitely generated module with a grading over the polynomial ring $S=k[X_1,\dots ,X_n]$ has a finite graded free resolution of length at most $n$. The Koszul resolution of $k$ has length $n$, so ${\mathrm{Tor}}_i^S(M,k)=0$ for $i>n$. In a minimal graded free resolution, this Tor functor is $F_i/(X_1,\dots ,X_n)F_i$; the graded Nakayama lemma forces $F_i=0$ for $i>n$.

A nonzero ring $R$ has invariant basis number if $R^m\cong R^n$ as free modules implies $m=n$ for all positive integers $m,n$. The property is not automatic for arbitrary rings; the commutative-ring case is proved below.

Suppose a finite group $G$ acts freely on $S^n$, $n>1$. The quotient has fundamental group $G$ and higher homotopy groups equal to those of $S^n$. Attach one $(n+1)$-cell killing a generator of ${\pi }_n\cong \mathbb{Z}$, then attach higher cells successively to kill all remaining higher homotopy. This constructs a Eilenberg–MacLane space $K(G,1)$ with only one $(n+1)$-cell. The one-cell bound on next-degree cohomology gives dimension at most one in degree $n+1$. But if $G=C_p\times C_p$, the mod-p cohomology ring of a cyclic group and Künneth theorem give dimension $n+2$ there. This contradiction excludes that group, and also excludes such a subgroup of any group acting freely on a sphere.

For every nonidentity element of a free group action on $S^{2r}$, its action map has no fixed point and hence degree minus one by fixed-point-free sphere maps are homotopic to the antipodal map. Degree of homeomorphisms defines a homomorphism into $\{1,-1\}$, whose kernel is trivial, since no nonidentity action map can have degree one. Thus the acting group has order at most two. For $S^0$, freeness on its two points gives the same conclusion directly.

If a nontrivial finite group acted freely on ${\mathbb{R}}^n$, a prime-order cyclic subgroup $C_p$ would too. The quotient is an $n$-manifold with contractible universal cover. The singular version of cellular free resolution from a contractible universal cover identifies its cohomology with group cohomology of $C_p$. The periodic resolution of a finite cyclic group gives $H^{2j}(C_p;\mathbb{Z})=\mathbb{Z}/p$ for every $j\ge 1$. But Poincaré duality for noncompact manifolds makes the quotient cohomology zero above dimension $n$. This contradiction excludes the action, for arbitrary homeomorphisms, not only Euclidean isometries.

A radial observer released from rest at infinity has $dr/d\tau =-\sqrt{2M/r}$. Integration from $r=2M$ to $r=0$ gives $\Delta \tau ={\int }_0^{2M}\sqrt{r/(2M)}dr=4M/3$ in geometric units. In ordinary units it is $4GM/(3c^3)$. The endpoint is the curvature singularity, not a regular spacetime centre; this value belongs to that specified energy, not every possible infalling worldline.

For a Fredholm operator between Hilbert spaces, the index is the difference of the finite dimensions of its kernel and cokernel. Since the range is closed, its cokernel identifies with $\ker T^{\ast }$ by orthogonal projection. An invertible operator has index zero. Conversely, an index-zero Fredholm operator becomes invertible after adding a finite-rank operator: choose an isomorphism $J:\ker T\to \ker T^{\ast }$, extend it by zero on $(\ker T{)}^{\perp }$, and observe that $T+J$ maps the two summands bijectively onto $\mathrm{ran}T\oplus \ker T^{\ast }$.

A bounded operator is Fredholm if and only if it has a bounded two-sided inverse modulo compact operators. For the forward implication invert its restriction from the orthogonal kernel complement to its closed range and extend by zero. Conversely a parametrix makes the identity compact on the kernel, forcing finite dimension. A failure of a lower bound on the kernel complement would yield unit vectors $x_n$ with $T{x}_n\to 0$; the compact error then forces a norm-convergent subsequence, whose nonzero limit lies both in the kernel and its complement. This proves closed range; applying the other parametrix identity to the adjoint gives finite cokernel.

For a smooth Hilbert-space map $F$ with $F(0)=0$ and Fredholm derivative $L$, split the source into $\ker L$ and a complement and the target into $\mathrm{im}L$ and a finite-dimensional cokernel. The implicit function theorem solves the range component uniquely as a graph over $\ker L$. The remaining equation is a smooth map $\kappa :\ker L\to \mathrm{coker}L$ with $\kappa (0)=d\kappa (0)=0$. Thus the local zero set is ${\kappa }^{-1}(0)$. Compact symmetry groups permit equivariant splittings.

// Destination: geometry-and-topology.bigb

Let $q$ be the depth-integrated volume flux, $E$ the entrainment rate, $R$ the upward export of pure ice, and $e=c_p(T-T_{\infty })-L\varphi$. Then $q^{\prime }=E-R$, $[\overline{C}q{]}^{\prime }=C_{\infty }E$ and $[eq{]}^{\prime }=-f-R[c_p(T-T_{\infty })-L]$. Export carries negative relative enthalpy, so it is not a heat-loss term with the same sign as atmospheric cooling. Total production is the increase in advected ice plus its exported amount. When both retained ice fraction and total processed fraction are small, equilibrium salinity and conservation of energy give production $f/(L+c_{p}m{C}_{\infty })$ even when retained and exported fractions differ.

Over an integral domain $A$ with fraction field $K$, an invertible fractional ideal is a finitely generated nonzero $A$-submodule $I\subseteq K$ for which another fractional ideal $J$ satisfies $IJ=A$. Here $I^{-1}=\{x\in K:xI\subseteq A\}$ equals $J$. Choose $1=\sum u_i{v}_i$ with $u_i\in I$, $v_i\in J$. On the distinguished open where $u_i{v}_i$ is invertible, $I$ is generated by $u_i$, since $x/u_i=(x{v}_i)/(u_i{v}_i)$. The maps $x\mapsto v_{i}x$ form a finite dual basis and prove that $I$ is a rank-one projective module. Conversely, a finitely generated rank-one projective module embeds in $K$ after choosing a basis over $K$; its dual identifies with $I^{-1}$, and local evaluation shows $I{I}^{-1}=A$.

Let $\mu$ be a finite positive surface measure and $|f|\le 1$. Expand $|\widehat{f\mu }(x){|}^{2k}$ and integrate against $e^{-\pi |x{|}^2/R^2}$. The Fourier transform of a Gaussian turns the oscillatory kernel into $R^d{e}^{-\pi R^2|{\sum }_{j\le k}{\omega }_j-{\sum }_{j>k}{\omega }_j{|}^2}$, which is nonnegative. Taking absolute values of the bounded amplitudes gives an upper bound by the same moment with $f=1$. This weighted positivity is valid even when $f$ is merely measurable; it transfers a scalar surface-decay estimate to a local bounded-density moment estimate.

With $|{\mathbf{k}}_i|=k$, elastic scattering into a propagating wavevector ${\mathbf{k}}_s$ samples Fourier transfer $\mathbf{\kappa }={\mathbf{k}}_s-{\mathbf{k}}_i$. These transfers satisfy the displayed shifted sphere relation. This reciprocal-space construction appears in crystallographic diffraction and the Fourier diffraction theorem.

Integrate the formal pseudodifferential residue over a period, or over the real line when the integral converges and the relevant boundary terms vanish. More algebraically, apply a linear functional on the coefficient algebra that annihilates derivatives. For $A=a{\partial }^p$, $B=b{\partial }^q$ and $r=p+q+1\ge 0$, the two residues are $\left(\genfrac{}{}{0ex}{}{p}{r}\right)a{b}^{(r)}$ and $\left(\genfrac{}{}{0ex}{}{q}{r}\right)b{a}^{(r)}$. Since $\left(\genfrac{}{}{0ex}{}{q}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{p}{r})$, integration by parts proves $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$. For $r<0$ both residues vanish. Bounded orders ensure only finitely many monomial pairs contribute to the residue of a product.

When Stokes factors act on solution coefficient vectors, a full positive turn gives their ordered product equal to the inverse of formal monodromy for a polynomial system with no finite singularities. The true fundamental matrix is entire, so its actual monodromy is trivial, but its normalized formal branch changes by $M_f$. Indeed, write $Y_{j+1}=Y_j{S}_j^{-1}$; then $Y_{N+1}=Y_1(S_N\cdots S_1{)}^{-1}$, whereas normalized continuation after the positive turn gives $Y_{N+1}=Y_1{M}_f$. Cancelling the invertible fundamental matrix proves the identity. The uncorrected product is identity only when $M_f=I$. For a cubic-exponential two-by-two system with $U=1,V=-1,P=1,R=2,d=3/2$, the formal exponent is $\theta =1/2$, so $M_f=-I$ and the Stokes product is $-I$.

The order-$\epsilon$ part of associativity is $af(b,c)-f(ab,c)+f(a,bc)-f(a,b)c=0$, so $f$ is a Hochschild cocycle. A change of coordinates $T=1+\epsilon g$ changes $f$ to $f+\delta g$. Therefore equivalence classes of first-order associative deformations of a fixed underlying algebra are its second Hochschild cohomology. Unital deformations use normalized cochains.

The algebra over $k[t]$ defined by $yx-xy=t$ has ordered basis $x^i{y}^j$, by its differential Ore-extension construction. Its $t$-adic completion is a formal associative formal associative deformation of $k[x,y]$. In characteristic zero its first-order multiplication term is $f(a,b)={\partial }_{y}a{\partial }_{x}b$. The cocycle is not a coboundary: a degree-two coboundary on a commutative algebra is symmetric, whereas $f(y,x)-f(x,y)=1$.

With no forcing, the linear dispersion relation has positive phase velocity $\sigma k^2/(\beta {\alpha }_1)$ and group velocity three times this. A decaying permanent wave with $\xi =x+ct$ instead obeys $\sigma (A^{\prime }{)}^2=A^2(\beta {\alpha }_{1}c+{\alpha }_{2}A/3)$. For positive coefficients a nonzero homoclinic wave requires $c>0$ and is negative, travelling upstream. Its inverse width is $k_s=\frac{1}{2}\sqrt{\beta {\alpha }_{1}c/\sigma }$. The opposite directions reflect the nonlinear amplitude-dependent propagation speed.

For $H_X+\lambda (H^{-2}-H^{-3})=X$, small $\lambda$ gives $H_0=(b^2+X^2)/2$. In the large-domain limit its inverse-power integrals are $I_1=2\pi /b$, $I_2=2\pi /b^3$ and $I_3=3\pi /b^5$. The force-free condition is $I_2-I_3=C(4I_1/3-I_2)$, or $2b^2-3=C(8b^4/3-2b^2)$. Expanding the root near $b^{-2}=2/3$ proves the displayed minus sign and leading pressure drop. The absolute downstream pressure relative to the elastic reference pressure fixes $s$ and therefore closes the speed-pressure relation.

The amplitude equations $u^{\prime }=-2uv$, $v^{\prime }={\lambda }_2+v^2+u^2$ satisfy $u^{\prime }=F_v$, $v^{\prime }=-F_u$, hence preserve the displayed first integral. For ${\lambda }_2=-a^2<0$ and $u\ge 0$, the centre equilibrium is $(a,0)$ with $F=2a^3/3$, while the two boundary saddle equilibria are $(0,\pm a)$ with $F=0$. The level $F=0$ contains the boundary connection and the arc $u^2=3(a^2-v^2)$ joining the two saddle equilibria. Levels $0<F<2a^3/3$ form closed periodic orbits. Adding an independent rotating phase turns these closed amplitude trajectories into invariant tori.