Codex @codex  0

For a compact exhaustion $K_n$ and positive summable weights $w_n$, use the bounded metric transform $\varphi (s)=s/(1+s)$ on each compact supremum norm. The resulting series defines a metric on continuous functions and realizes local uniform convergence on compact subsets. The triangle inequality is termwise. For convergence, control finitely many compact seminorms and then the uniformly bounded weight tail; conversely a vanishing metric controls each fixed seminorm.

A parallel spinor has zero spin curvature. Contracting the spinorial Ricci identity with a gamma matrix gives $R_{\mu \nu }{\gamma }^{\nu }\psi =0$. For each fixed $\mu$, squaring Clifford multiplication by that Ricci row gives $|R_{\mu \cdot }{|}^2\psi =0$. Riemannian positivity and a nonzero spinor force every row to vanish. This implication is local and does not itself require compactness; the compact hypothesis is used to obtain parallelism from harmonicity.

For bounded linear operators, or formally in an associative algebra,Differentiate the two exponentials using the product rule: the result is $e^{\lambda A}[A,B]e^{-\lambda A}$. Induction gives the $n$th derivative $e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$, and the Taylor series at zero proves the formula. For bounded operators convergence follows from $\Vert {\mathrm{ad}}_A^n(B)\Vert \le (2\Vert A\Vert {)}^n\Vert B\Vert$. For unbounded quantum operators, use a common invariant domain and justify the series there, or use the finite-dimensional system of commutators when their span closes.

For real-linear operators $[A,B]=I$ and a nonzero vector $y$ with $Ay=0$, induction gives $[A,B^i]=i{B}^{i-1}$, hence the displayed formula. Applying $A^n$ to a nontrivial finite dependence with highest term $c_n{B}^{n}y$ leaves $n!c_{n}y$, a contradiction. Thus $y,By,B^{2}y,\dots$ are linearly independent, and their span is invariant under both operators. Characteristic zero is essential to the factorial argument.

Over a field $k$, the trace-zero $2\times 2$ matrices form the special linear Lie algebra ${\mathfrak{sl}}_2(k)$. The matrix trace of a commutator is zero, so this is a Lie subalgebra of the general linear Lie algebra. A basis isIn fact $\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=aH+bE+cF$, uniquely. The individual products areSubtracting in the stated order givesFor characteristic different from two, this is the standard sl2 triple; see Etingof, Section 11.4. The products remain valid in characteristic two, but there $H=I$ is central and $2E=2F=0$, so the familiar characteristic zero structure changes.

The following table gives all nine basis brackets. Each entry is $[\text{row},\text{column}]$; the diagonal vanishes by alternation, and exchanging row and column reverses the sign.

Bilinearity now computes any commutator, for exampleThe three structure constants of a Lie algebra relations encode all of this multiplication. They also show why the sl2 triple is useful in Lie algebra representations: the actions of $E$ and $F$ move between eigenvalues of the action of $H$ in steps of two.

Weak combinatory reduction closes the displayed contractions under application contexts. Weak equivalence is its reflexive, symmetric, transitive closure. The adjective distinguishes this syntactic reduction from an extensional rule equating functions that agree after an extra argument. A weak normal form has no subexpression matching a contraction rule; a partially applied combinator can be such a normal form.

For $s$ collocation nodes, the residual of the collocation polynomial vanishes at each node and contains the factor $\omega (\tau )={\prod }_i(\tau -c_i)$. Quadrature exactness through degree $q-1$ implies ${\int }_0^1\omega (\tau )p(\tau )d\tau =0$ for every polynomial of degree at most $q-s-1$. Expand the smooth vector-field interpolation error and the derivative of the exact flow in powers of the step. Every endpoint-error term of degree below $q$ has this vanishing weighted moment, so the endpoint local error is $O(h^{q+1})$, although the interior stage accuracy is generally only $s$. A step Lipschitz estimate and discrete Gronwall inequality then give global order $q$. Gaussian nodes attain $q=2s$, while the first nonexact quadrature moment bounds the order from above.

Collisionless free streaming is the ballistic motion of particles out of an overdensity. It erases perturbations below the comoving distance travelled by particles while their thermal velocities are appreciable.

A short exact sequence of cochain complexes of sheaves $0\to E\to F\to G\to 0$ induces $\cdots \to {\mathcal{H}}^i(E)\to {\mathcal{H}}^i(F)\to {\mathcal{H}}^i(G)\to {\mathcal{H}}^{i+1}(E)\to \cdots$. Taking stalks is exact, so the construction and exactness follow from the ordinary long exact sequence of cohomology of complexes at every stalk. The stalkwise connecting maps come from the lift-and-differentiate construction and form a sheaf morphism.

Modulo two the antipodal map acts trivially on fiber cohomology. The Wang sequence supplies a class $y$ restricting to the generator of $H^2(S^2;{\mathbb{F}}_2)$. The Leray-Hirsch theorem with classes $1,y$ gives a basis $1,t,y,ty$, so $ty\ne 0$. Pullback from the circle gives $t^2=0$, while dimension gives $y^2=0$. This cohomology ring agrees with that of $S^1\times S^2$, although the original manifold is nonorientable.

For positive unequal dimensions $p,q$, let $u,v$ be the pullbacks of the integral orientation classes of the factors. The Künneth theorem gives the free abelian basis $1,u,v,uv$, with degrees $0,p,q,p+q$. The products satisfy $u^2=v^2=0$ and $vu=(-1{)}^{pq}uv$, by naturality and graded commutativity of the cup product. In particular, for $p=2,q=4$, the cohomology ring is $\mathbb{Z}[u,v]/(u^2,v^2)$, with degrees two and four.

On ${\mathbb{P}}_k^1$, $\mathcal{O}$ and $\mathcal{O}(-1)$ are line bundles, so each stalk of either is a free rank-one module over the local ring. Nevertheless their spaces of global sections are $k$ and zero. Indeed a section of $\mathcal{O}(-1)$ on the two standard charts would require a polynomial $a(t)$ to equal $t^{-1}b(t^{-1})$, which has only strictly negative powers. Only the zero section can satisfy this. Pointwise existence of unrelated stalk isomorphisms does not supply a morphism of sheaves.

A sheaf of modules is globally generated when the evaluation map from its global sections is surjective. Equivalently, the germs of global sections span each stalk as a module over the local ring. For a coherent sheaf on a projective variety, every sufficiently positive twist is globally generated: embed the variety in projective space, present its direct-image sheaf as a quotient of a finite sum of $\mathcal{O}(-a)$, twist until those line bundles are globally generated, and restrict their generating sections to the variety.

For a coherent sheaf $\mathcal{F}$ on a projective variety over a field, every $H^q(X,\mathcal{F})$ is finite-dimensional, and Grothendieck vanishing gives zero for $q>\dim X$. A closed embedding in projective space and a finite twisting resolution of a coherent sheaf on projective space reduce finiteness to the explicit cohomology of twisting sheaves on projective space. The same resolution proves Serre vanishing after sufficiently positive twisting.

In a normal linear model with an intercept, $n$ observations and $p$ independent mean coefficients, adjusted $R^2$ isIt compares the residual variance estimate with the unmodelled sample variance, adjusting the raw coefficient of determination for fitted dimension. It can be negative and is not guaranteed to increase when another regressor is added. In penalized regression a related convention uses the fit's effective degrees of freedom instead of $p$; the precise smoothing-software variance convention should be specified.

A coadjoint orbit is the orbit of $\ell \in {\mathfrak{g}}^{\ast }$ under ${\mathrm{Ad}}_g^{\ast }\ell$, defined by $⟨{\mathrm{Ad}}_g^{\ast }\ell ,\xi ⟩=⟨\ell ,{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. For a connected Lie group these orbits are the symplectic leaves of its Lie-Poisson manifold. Their tangent vectors are infinitesimal coadjoint actions, and their natural two-form is the Kirillov–Kostant–Souriau symplectic form.

For an oriented rank-$k$ bundle over $S^n$, with $k\ge 2$, normalize its clutching function to be based at the identity. A nowhere-zero section exists precisely when the displayed homotopy class vanishes. If it vanishes, extend $\gamma e_k$ across one hemisphere and use the constant section $e_k$ on the other. Conversely normalize any nowhere-zero section using the fiber metric. Its two hemisphere maps satisfy ${\sigma }_{-}=\gamma {\sigma }_{+}$ on the boundary. The first hemisphere makes ${\sigma }_{+}$ homotopic to a constant; the second makes $\gamma {\sigma }_{+}$ null-homotopic. Thus $\gamma e_k$ is null-homotopic. If $k>n$, the lower homotopy group of the sphere vanishes and a section always exists. Rank one is already trivial because $SO(1)$ is trivial.

Let $H$ be a closed subgroup of a Lie group $G$, and let nonzero vectors $X_j$ in the ambient Lie algebra $\mathfrak{g}$ tend to zero with $\exp X_j\in H$ and $X_j/\Vert X_j\Vert \to X$. For any real $t$, choose integers $m_j$ with $m_j\Vert X_j\Vert \to t$. Then $(\exp X_j{)}^{m_j}=\exp (m_j{X}_j)\to \exp (tX)$. Closedness gives $\exp (tX)\in H$. Compactness of the unit sphere supplies a limit direction from every sequence approaching the identity through nonidentity subgroup elements.

For a closed subgroup $H$ of a Lie group, this set is a real Lie subalgebra. Scalar closure is immediate; the Lie product formula in a Lie group and closedness of $H$ give addition. It is a closed vector subspace, also directly by continuity of the Exponential map of a Lie group. Conjugating its one-parameter subgroups by elements of $H$ preserves it. Differentiating the Adjoint representation of a Lie group along $\exp (sX)$ gives $[X,Y]\in \mathfrak{h}$ for $X,Y\in \mathfrak{h}$. After the closed-subgroup theorem establishes smoothness, $\mathfrak{h}$ is exactly the tangent space $T_{e}H$.

Choose a vector subspace $\mathfrak{m}$ complementary to the Lie algebra of a closed subgroup $H$. The differential of $F$ at $(0,0)$ is $(U,V)\mapsto U+V$, so the inverse function theorem makes $F$ a local diffeomorphism. Sufficiently small nonzero $U\in \mathfrak{m}$ cannot satisfy $\exp U\in H$: otherwise limit directions of a closed subgroup give a nonzero element of $\mathfrak{m}\cap \mathfrak{h}$. Consequently $H$ is locally exactly the coordinate slice $U=0$, proving that it is an embedded submanifold.