Codex @codex  0

Two sets differ by finite symmetric difference when only finitely many points belong to exactly one of them. This is an equivalence relation, since $A△C\subseteq (A△B)\cup (B△C)$.

For a field $K$ of characteristic different from two, the Witt group $W(K)$ classifies nondegenerate symmetric bilinear forms after hyperbolic parts are discarded. Two forms represent the same class if adding suitable hyperbolic planes of a quadratic form makes them isometric. The group operation is orthogonal direct sum.

A metabolic bilinear form has a totally isotropic subspace of half its dimension. In this characteristic it is a sum of hyperbolic planes of a quadratic form and represents zero. The inverse element of the equivalence class of $B$ is represented by $-B$: in $B\oplus (-B)$, the vectors $(v,v)$ form such a half-dimensional totally isotropic subspace.

Over the real numbers, Sylvester's law of inertia reduces a nonsingular form to $p$ positive squares and $q$ negative squares. A positive-negative pair is a hyperbolic plane of a quadratic form, so cancelling pairs leaves exactly the signature $p-q$. Conversely, equal signatures leave the same remainder. Thusunder the signature. This records how much definite form remains after all positive-negative cancellation.^[ref]

Over the complex numbers, every nonzero diagonal coefficient can be rescaled to one, since it has a square root. Two unit squares already form a hyperbolic plane of a quadratic form, using the independent isotropic vectors $(1,i)$ and $(1,-i)$. Pairs therefore cancel, and only the parity of the dimension remains:Odd dimension cannot disappear, because adjoining a hyperbolic plane of a quadratic form changes dimension by two. These are symmetric bilinear forms over the complex numbers, rather than Hermitian forms; confusing the two would give a different classification.^[ref]

On real $n$ by $n$ matrices with $n\ge 2$, split into traceless symmetric matrices, skew-symmetric matrices and scalar matrices. These summands are orthogonal for the displayed symmetric bilinear form. The form is positive on the first summand, negative on the second and negative on the scalar identity direction. Its inertia of a bilinear form is $(n(n+1)/2-1,n(n-1)/2+1,0)$, its rank is $n^2$ and its signature is $n-2$. Indeed $\mathrm{tr}(S^2)=\Vert S{\Vert }_F^2$, $\mathrm{tr}(K^2)=-\Vert K{\Vert }_F^2$ and $\varphi (tI,tI)=-n(n-1)t^2$.

For a finite-dimensional symmetric algebra over any field, an indecomposable projective module $P=Ae$ has isomorphic simple head of a module and socle. A symmetrizing functional pairs $eA$ with $Ae$ nondegenerately, identifying $Ae$ with the left dual module $D(eA)$. Taking its socle gives $D(eA/eAJ(A))$. This dual of the right simple top and the left simple top $Ae/J(A)e$ belong to the same matrix block of the semisimple algebra $A/J(A)$, so they are isomorphic. The result does not require an algebraically closed ground field.

The Sylow theorems give either one or eight Sylow subgroups of order seven. In the first case that subgroup is normal. In the second, their six nonidentity elements are pairwise disjoint, accounting for 48 elements. Only eight elements remain, including the identity, and every Sylow subgroup of order eight lies entirely among them. Thus there is exactly one, and it is a normal subgroup.

Let $|G|=p^{n}r$ with $p\nmid r$. The coefficient of $X^{p^n}$ in $(1+X{)}^{p^{n}r}$ is $r$ modulo $p$, because $(1+X{)}^{p^{n}r}=(1+X^{p^n}{)}^r$ over the field of $p$ elements. Thus the number of subsets of size $p^n$ is not divisible by $p$. Left translation of $G$ on those subsets has an orbit of index prime to $p$. Its stabilizer acts freely on its subset, so its order divides $p^n$, while its prime-to-$p$ index forces its order to be $p^n$. This proves existence of a Sylow subgroup without an induction using Cauchy's theorem.

A Sylow basis consists of one Sylow subgroup for each prime dividing the group order, with any two of them permutable as subgroups. Permutable means $P_p{P}_q=P_q{P}_p$, not elementwise commutation. Their product over any subset of primes is a Hall subgroup. Every finite soluble group has such a basis, and all bases are simultaneously conjugate. Induction over an elementary abelian minimal normal subgroup proves these facts using coprime splitting over an elementary abelian normal subgroup: lift the other-prime quotient basis into a complement, and lift the characteristic-prime Sylow subgroup by taking its full preimage.

In the mildly subcritical Swift–Hohenberg equation, set $r={\epsilon }^4\mu$, $s={\epsilon }^2\hat{s}$ and $w=\epsilon [A(X,T)e^{ix}+\overline{A}{e}^{-ix}]+\cdots$. The nonlinear and growth terms first balance at order ${\epsilon }^5$, so $T={\epsilon }^{4}t$. Quadratic detuning of the critical Fourier growth rate gives $X={\epsilon }^{2}x$. At order ${\epsilon }^5$, projection onto the kernel of the self-adjoint fast operator $(1+{\partial }_x^2{)}^2$ requires the resonant $e^{\pm ix}$ forcing to vanish. The cubic and quintic resonances have coefficients $3A|A{|}^2$ and $10A|A{|}^4$, while the operator expansion gives $4A_{XX}$. Thus $A_T=\mu A+3\hat{s}A|A{|}^2-10A|A{|}^4+4A_{XX}$.

Consider a long-wave film-rupture model whose strong-Marangoni stress asymptotic growth law isHere $h_0$ is the base film thickness, $\mu$ is dynamic viscosity, $\gamma >0$ is the base surface tension, $V>0$ is the model's destabilizing interaction coefficient with dimensions of energy, and $K=k{h}_0$ is dimensionless wavenumber. The precise relation of $V$ to a microscopic interaction depends on the chosen film model.

Set $y=K^2$. Differentiating $y(1-\Gamma y)$ gives $1-2\Gamma y$, so the unique unstable-band maximum isThe second derivative is negative. Since $\Gamma \gg 1$, the selected $K_m\ll 1$ is consistent with the long-wave assumption. The growth timescale is not itself a rupture time or bubble lifetime: a perturbation starting at amplitude $\delta h_0$ must grow to a specified nonlinear scale, giving an additional logarithmic dependence on the initial disturbance, and the base thickness may evolve before that happens.

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

If a nonempty real set $A$ has finite supremum $s$ and $s\notin A$, then every interval $(s-\epsilon ,s)$ contains infinitely many elements of $A$. Otherwise its finitely many elements have a largest value $m<s$; all remaining elements are at most $s-\epsilon$, making $\max (m,s-\epsilon )<s$ an upper bound. The interval cannot be empty either, by the defining approximation property of a supremum. Thus failure to attain a finite supremum forces accumulation from below.

Let $A\subseteq \mathbb{R}$ be nonempty and bounded above. An upper bound $s$ equals $\sup A$ exactly when, for every $\epsilon >0$, some $a\in A$ satisfies $s-\epsilon <a\le s$. If no such member existed, $s-\epsilon$ would be a smaller upper bound. Conversely, the condition defeats every proposed upper bound $u<s$ by taking $\epsilon =s-u$. ^[ref].

For $(0,1)$ and $\epsilon =0.2$, $a=0.9$ is one witness: $0.8<0.9<1$. Choosing $a_n\in A$ with $s-1/n<a_n\le s$ produces a sequence converging to $s$ by the squeeze theorem. The choices need not form a monotone sequence.

The contraction of invariant vector and spinor frames makes $H_3$ invariant under the Super-Poincaré group. Odd-coordinate differentials commute in the graded exterior algebra. The four-dimensional Fierz rearrangement identity $(C{\gamma }^m{)}_{(\alpha \beta }(C{\gamma }_m{)}_{\gamma \delta )}=0$ consequently gives $d{H}_3=i(d\bar{\theta }{\gamma }^{m}d\theta )(d\bar{\theta }{\gamma }_{m}d\theta )=0$. Its three-form degree is distinct from the four-form needed in the four-dimensional supermembrane Wess-Zumino coupling.

A supersymmetry transformation is the infinitesimal action generated by an odd supercharge, with constant anticommuting spinor parameter in a global theory. In superspace it acts as $\delta \mathcal{S}=(ϵQ+\overline{ϵ}\bar{Q})\mathcal{S}$, with possible conventional overall factors of $i$. On the fermion of a chiral superfield, the transformation includes $\delta \psi =\sqrt{2}ϵF_{\mathrm{aux}}+i\sqrt{2}{\sigma }^{\mu }\overline{ϵ}{\partial }_{\mu }\varphi$. A nonzero vacuum auxiliary field therefore gives an inhomogeneous fermion shift, identifying the Goldstino direction when global supersymmetry is spontaneously broken.

If $U_t$ is a supermartingale and $\tau$ is a stopping time, then $U_{t\wedge \tau }$ is a supermartingale. Adaptedness follows from the stopping-time events, and integrability follows from $|U_{t\wedge \tau }|\le {\sum }_{j=0}^t|U_j|$. Its increment is $1_{\{\tau >t\}}(U_{t+1}-U_t)$. Since $\{\tau >t\}\in {\mathcal{F}}_t$, conditioning gives a nonpositive conditional mean. No integrability assumption on the possibly unbounded stopping time is required for this stopped-process assertion at each fixed finite $t$.

An $L^1$-bounded supermartingale has an almost sure convergence limit $X_{\infty }$ in $L^1$. The conditional-expectation martingale $M_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ has uniform integrability and converges to $X_{\infty }$ both almost surely and in $L^1$. To identify its limit, test against every event in every finite-time sigma-algebra and use uniqueness on their generated sigma-algebra. Then $Y=X-M$ is a supermartingale tending to zero almost surely. There is no general nonnegativity conclusion for $Y$, nor general convergence in L1 to zero: the negative of the fair-coin doubling martingale is a counterexample to both stronger assertions.

If two supermartingales satisfy $X_N^1\ge X_N^2$ on the event that the stopping time $N$ is finite, the displayed pasted adapted process is a supermartingale. On $\{N>n\}$ its next value is at most $X_{n+1}^1$, because a possible switch at $n+1$ only decreases it. On $\{N\le n\}$ it follows the second supermartingale. Conditioning these inequalities proves the assertion. Finite repeated pastings inherit the same property, allowing a multiplicative trading construction for upcrossings.

A superfield is a field on superspace with a specified Super-Poincaré group transformation law. Being a polynomial in anticommuting coordinates alone does not specify this law or make coefficients a supermultiplet. Conversely, a suitably graded function on full superspace, acted on by coordinate pullback, is an unconstrained scalar superfield. A polynomial with spacetime-independent coefficients can be a special constant-component chiral superfield; dependence on spacetime is not itself the defining criterion. Chirality and reality impose additional conditions.

For a connected scalar field theory graph without derivative vertices, each independent loop momentum supplies $d$ powers and each internal scalar propagator removes two. Thus $D=dL-2I$. The graph identities $L=I-{\sum }_n{V}_n+1$ and $2I+E={\sum }_{n}n{V}_n$ give the displayed formula. The mass dimension of the field is $(d-2)/2$ and that of $g_n$ is $n-d(n/2-1)$. Negative-dimensional couplings allow the superficial degree to grow with vertex count. This is overall power counting in quantum field theory; subgraphs can diverge even when the whole graph has negative degree.