A positive bounded linear functional is tracial when its value on a product is unchanged by reversing the two factors. When normalized by $\tau (1)=1$, it is a tracial state. A vector functional $\tau (a)=⟨a\Omega ,\Omega ⟩$ is faithful if the vector is separating, since $\tau (a^{\ast }a)=\Vert a\Omega {\Vert }^2$. Its trace identity makes the associated Tomita operator isometric.

Positivity of $f((x+ty{)}^{\ast }(x+ty))$ for every complex $t$ proves the displayed Cauchy-Schwarz inequality, including zero diagonal cases. Replacing $x,y$ by their adjoints gives $|f(x{y}^{\ast }){|}^2\le f(x{x}^{\ast })f(y{y}^{\ast })$. These two forms cannot be mixed for a general positive functional on a C-star algebra. For the state evaluating the lower-right entry of $2\times 2$ matrices and $x=y=E_{12}$, $f(x^{\ast }y)=1$ but $f(x{x}^{\ast })=f(y{y}^{\ast })=0$, disproving the mixed form.

Define positivity spectrally: a self-adjoint element has nonnegative spectrum. First, addition preserves it. For $p,q\ge 0$ and $s=\Vert p\Vert +\Vert q\Vert$, functional calculus gives $\Vert s1-p-q\Vert \le s$, forcing every real spectral value of $p+q$ to be nonnegative. The cone is proper because a self-adjoint element with spectrum contained in $\{0\}$ has norm zero.

For $h=a^{\ast }a$, let $h_{-}=\max (-h,0)$ by the continuous functional calculus, and set $c=a{h}_{-}^{1/2}$. Then $c^{\ast }c=-h_{-}^2$ is nonpositive. The nonzero spectra of products in opposite orders imply that $c{c}^{\ast }$ is nonpositive too. Write $c=u+iv$ with $u,v$ self-adjoint. Then $c^{\ast }c+c{c}^{\ast }=2(u^2+v^2)$ is positive, since self-adjoint squares have nonnegative spectrum. It is also nonpositive, hence zero. It follows that $c^{\ast }c=-c{c}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $c=0$, so $h_{-}^2=0$ and $h_{-}=0$. Thus $h\ge 0$, without assuming positivity of adjoint products in an earlier step.

For a positive element $a$, its positive square root in a C-star algebra gives $c^{\ast }ac=(a^{1/2}c{)}^{\ast }(a^{1/2}c)$, which is positive. Applying this to $b-a$ shows that $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. The map need not preserve multiplication, but it preserves order; if $c$ is invertible, the inverse congruence also preserves order.

Here strictly positive means positive and invertible. Set $k=a^{-1/2}b{a}^{-1/2}\ge 1$ using congruence preserves positivity in a C-star algebra. The continuous functional calculus gives $k^{-1}\le 1$, since $1-t^{-1}\ge 0$ for $t\ge 1$. Then $a^{-1}-b^{-1}=a^{-1/2}(1-k^{-1})a^{-1/2}\ge 0$. The proof works without commutativity. If $1\le a$, functional calculus also gives $a^{-1}\le 1$.

In $M_2(\mathbb{C})$ take $a=\mathrm{diag}(1,2)$ and $b=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$. Then $a-1\ge 0$ and $b-a=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ge 0$, but $b^2-a^2=\left(\begin{array}{cc}4& 5\\ 5& 6\end{array}\right)$ has determinant $-1$ and a negative eigenvalue. In contrast, for commuting positive elements $b^2-a^2=(b-a)(b+a)$ is positive in the commutative function algebra supplied by continuous functional calculus. The obstruction is noncommutativity, not a failure of the scalar inequality.

The positive elements of a C-star algebra form a norm-closed proper convex cone. For $p,q\ge 0$ and $s=\Vert p\Vert +\Vert q\Vert$, continuous functional calculus gives $\Vert s1-p-q\Vert \le s$, forcing the real spectrum of $p+q$ to be nonnegative. The same bound with one fixed upper norm bound proves norm closedness under limits; scaling by nonnegative real scalars preserves positivity. If both $h$ and $-h$ are positive, their spectra lie in $\{0\}$, and spectral radius norm equality for normal elements forces $h=0$. This cone defines the order $a\le b$ precisely when $b-a\in A_{+}$.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

An antipodal simplicial map from a regular antipodal triangulation of a sphere $S^k$ to $F^n$ has an odd number of positive alternating $k$-simplices. For each $k$-simplex, the number of positive alternating facets is odd exactly when the simplex is positive or negative alternating; otherwise it is even, including collapsed images. Count these incidences in one hemisphere: interior facets contribute twice, boundary facets once. Antipodal pairing then gives $p_k\equiv p_{k-1}$, starting with one positive vertex on $S^0$. In particular, no such map exists when $k>n$.

Effective porosity is the pore-volume fraction accessible to connected transport under the specified measurement or model. It excludes isolated pore volume and may also exclude poorly exchanging regions on the transport timescale, so $0\le {\varphi }_e\le \varphi$ for total porosity $\varphi$. In a simple mobile-volume model the mean pore velocity is $q/{\varphi }_e$; the assumptions behind that model matter when interpreting tracer arrival times. A sample with total porosity $0.25$ but mobile pore fraction $0.20$ and Darcy flux $2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ has mobile mean pore velocity $1.0\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-1}$. Woessner and Poeter discuss the role of connectivity and slowly exchanging pores.^[ref]

For a nonnegative random constant individual hazard $\Lambda$, write $M_k(t)=\mathbb{E}[{\Lambda }^k{e}^{-\Lambda t}]$. The population survivor function is $M_0$ and its hazard is $M_1/M_0$. Under the integrability needed to differentiate, $M_k^{\prime }=-M_{k+1}$, soThe conditional distribution is weighted by $e^{-\Lambda t}$. This proves decreasing population hazard without any decrease in individual hazards and explains survival selection in a heterogeneous population.

For monotone missing data, let $R_{ij}$ indicate observation and let ${\rho }_{ij}$ be the product of the conditional retention probabilities through visit $j$. Under sequential missing at random, conditioning on the complete response history makes these retention probabilities depend only on the history already observed at each decision. Therefore $\mathbb{E}(R_{ij}/{\rho }_{ij}\mid Y_i,X_i)=1$. Weighting a working-independence generalized estimating equation by $R_{ij}/{\rho }_{ij}$ restores its mean-zero property. This needs correct retention probabilities, positivity and regularity; uncertainty must account for estimating the weights.

For simply connected $Y$, rational Hurewicz homomorphism identifies ${\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q}$ with the primitive elements of a Hopf algebra of $H_{\ast }(\Omega Y;\mathbb{Q})$ and induces an isomorphism $U({\pi }_{\ast }(\Omega Y)\otimes \mathbb{Q})\cong H_{\ast }(\Omega Y;\mathbb{Q})$ of graded Hopf algebras. The bracket on homotopy is the Samelson product. The underlying algebra theorem says that a connected graded cocommutative Hopf algebra over characteristic zero is the universal enveloping algebra of its primitive Lie algebra. These statements let a free tensor algebra of loop-space homology detect a free graded Lie algebra of homotopy.

The character group of a topological group that is an abelian group consists of its continuous group homomorphisms to the circle group, with pointwise multiplication. For a finite group, all characters are continuous and their values are roots of unity, so this agrees with the character group of a finite abelian group.

For a locally compact Hausdorff Abelian group $G$, its Pontryagin dual is the group of continuous unitary characters, with pointwise multiplication and the compact-open topology. Its topology agrees with the Gelfand topology of the L1 convolution algebra. It is again a locally compact Hausdorff Abelian group.

// Target: analysis.bigb

For a gas-pressure-supported polytrope of index five with uniform mean molecular weight, $T/T_c=\theta$. A specific stellar energy-generation rate $ϵ={ϵ}_0\rho T^q$ therefore integrates to $4\pi {ϵ}_0{\alpha }^3{\rho }_c^2{T}_c^q{\int }_0^{\infty }{\xi }^2{\theta }^{10+q}d\xi$. Substituting $u=\xi /\sqrt{3+{\xi }^2}$ gives the displayed formula, convergent for $q>-7$. At $q=11$ its dimensionless coefficient is approximately $1.02942$. The scale factor is a volume, ${\alpha }^3$, rather than an area.

A level-one modular form with rational Fourier coefficients is a weighted homogeneous polynomial with rational coefficients in the normalized Eisenstein series $E_4,E_6$. Subtract its constant coefficient times a same-weight monomial. The remainder is a cusp form; divide it by ${\Delta }_0=(E_4^3-E_6^2)/1728=q+O(q^2)$. The valence formula for the modular group shows that ${\Delta }_0$ has no interior zeros, so this quotient is holomorphic and has rational coefficients. Induction on weight proves the assertion. Euler's Bernoulli number formula for even zeta values then gives the corresponding rational weighted polynomial in the unnormalized lattice Eisenstein sums $G_4,G_6$.

Let $P_w(E_4,E_6)$ be weighted homogeneous of modular weight $w$, and suppose ${\sum }_w{P}_w(z)=0$ as a holomorphic function. Apply the modular group matrices $\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At a fixed point $z$ in the complex upper half-plane, this gives ${\sum }_w(nz+1{)}^w{P}_w(z)=0$ for every integer $n$. A polynomial with infinitely many roots vanishes identically, so every $P_w(z)=0$. Within one modular weight, the monomials reduce to powers of $E_4^3/E_6^2$, after taking out a common factor. This ratio is nonconstant, as its Fourier expansion begins $1+1728q+\cdots$, so its powers are linearly independent. This proves algebraic independence even when the generators are regarded as analytic functions rather than elements of a formal graded ring.

In characteristic zero, averaging over a compact group, or over a Zariski-dense compact form of a complex reductive group, gives a degree-preserving projection onto the polynomial invariant ring. Its module property is $\mathcal{R}(af)=a\mathcal{R}(f)$ for invariant $a$. This property turns finite ideal generation in the ambient polynomial ring into finite algebra generation of the invariant subring.

For a finite-dimensional rational representation of a complex reductive algebraic group, choose finitely many homogeneous positive-degree invariants generating the ideal generated by all positive-degree invariants in the ambient polynomial ring. Such a finite set exists because a polynomial ring is Noetherian. Apply the Reynolds operator on a polynomial invariant ring to an expression for any homogeneous invariant in that ideal. Its coefficients become invariants of smaller degree, so induction proves that the chosen set generates the entire invariant ring as an algebra.