An aldohexose is a six-carbon aldose: its open-chain reference form has an aldehyde group at C1. Glucose is an example.^[ref]

The unmodified open-chain skeleton has four stereocenters, C2–C5, giving $2^4=16$ stereoisomers, or eight enantiomer pairs. Fixing the C5 hydroxy group on the right leaves $2^3=8$ D forms; D-glucose's right/left/right/right pattern selects one. An aldohexose label therefore specifies less than a particular sugar name, and a molecular formula specifies less still. The existing glucose article checks its chemical atom inventory; monosaccharide ring-chain interconversion follows those same numbered carbon positions into a ring.

An aldopentose is a five-carbon aldose: its open-chain reference form has an aldehyde group at C1. Ribose is an example; deoxyribose is its 2-deoxy derivative.^[ref]

For the unmodified open-chain aldopentose skeleton, count one $\mathrm{CHO}$, three $\mathrm{CHOH}$ and one $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group: ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5$. The three interior carbon positions, C2–C4, are stereocenters. Each allows two configurations, giving $2^3=8$ stereoisomers, grouped into four enantiomer pairs. The unlike chain ends prevent a meso compound identification. D-ribose selects the right/right/right hydroxy group pattern in the Fischer projection; the word aldopentose alone does not select that pattern. Compare the existing ribose and deoxyribose diagrams for the separate question of which oxygen is present.

The associator $a$ and unitors of a monoidal category satisfyFor the first identity, postcompose the pentagon on $X,Y,I,Z$ by $1_X\otimes (1_Y\otimes {\lambda }_Z)$, then use the triangle twice and cancel $a_{X,Y,Z}$. This gives the first equality after tensoring with $1_Z$; set $Z=I$ and cancel using the natural invertible right unitor. Reversing tensor order proves the second identity. Naturality of the left unitor at ${\lambda }_X$ gives ${\lambda }_{I\otimes X}=1_I\otimes {\lambda }_X$; comparing the second identity with the triangle on $I,I,X$ gives $({\lambda }_I\otimes 1_X)=({\rho }_I\otimes 1_X)$ and hence the last identity. These are consequences of the axioms, not extra axioms imported from the monoidal coherence theorem.

For an associator $a$ in a monoidal category, the two paths from $((W\otimes X)\otimes Y)\otimes Z$ to $W\otimes (X\otimes (Y\otimes Z))$ agree:This five-vertex diagram is the basic compatibility for reassociating four tensor factors. The monoidal coherence theorem extends it to arbitrary finite tensor expressions.

The associator and unitors of a monoidal category satisfyThus the two ways to remove a unit between two tensor factors agree. Together with the pentagon identity for a monoidal category this implies all unit identities derived from the monoidal pentagon and triangle and the monoidal coherence theorem.

Cartesian product with singleton unit and disjoint union with empty unit each make the Category of sets monoidal. Associators and unitors simply retag or rebracket the same elements, so the pentagon and triangle identities hold elementwise. These structures differ even up to monoidal equivalence: any category equivalence preserves initial and terminal objects, so it cannot identify the empty unit with the singleton unit.

For a linear ordinary differential equation with holomorphic coefficients on a punctured disc, analytic continuation of a basis gives an invertible monodromy matrix. A nonzero eigenvector with eigenvalue $\lambda =e^{2\pi i\sigma }$ selects a solution $w$ for which $z^{-\sigma }w$ is single-valued and holomorphic. Its Laurent series gives $w=z^{\sigma }{\sum }_{n\in \mathbb{Z}}{c}_n{z}^n$. No regular-singular hypothesis is needed; infinitely many negative powers may occur.

An unramified formal fundamental matrix $\hat{Y}=\hat{G}(\lambda ){\lambda }^{\Theta }{e}^{Q(\lambda )}$ acquires the right factor $e^{2\pi i\Theta }$ on a positive turn of the logarithm branch. This is formal monodromy. It can be nontrivial even when the true equation has an entire fundamental matrix and trivial actual monodromy, because Stokes matrices compensate for it.

A parameter deformation preserves generalized monodromy data, including Stokes matrices and formal monodromy at irregular points. If normalized sectorial matrices have parameter-independent jumps, $B=Y_z{Y}^{-1}$ agrees across overlaps. Its finite singularities and growth at infinity determine it by complex analytic arguments. Compatibility of $Y_{\lambda }=AY$ and $Y_z=BY$ is precisely the displayed zero-curvature identity, obtained by differentiating both equations and subtracting.

A morphism of monads on one category is a natural transformation $\alpha :S\Rightarrow T$ with $\alpha {\eta }^S={\eta }^T$ and $\alpha {\mu }^S={\mu }^T(T\alpha )(\alpha S)$. These equations preserve the unit and multiplication. A terminal endofunctor in a full composition-closed subcategory containing the identity receives a unique such morphism from every monad whose functor belongs to that subcategory.

In a category with equalizers, the equalizer of ${\eta }_{TA},T{\eta }_A$ defines a functor $R$ by ${\beta }_{B}Rf=Tf{\beta }_A$. It has unit $\alpha$ and multiplication $\theta$ characterized byCancelling the monomorphisms ${\beta }_A$ reduces its unit and associativity laws to those of $T$, so $\beta :R\Rightarrow T$ is a monad morphism. The split equalizers associated with a monad show that ${\alpha }_{TA}$ is invertible. If every $T{\beta }_A$ is monic, then every $T{\alpha }_A$ is invertible, the naturality square of $\beta$ at ${\beta }_A$ is a pullback in a category, and $R$ is an idempotent monad. This monicity hypothesis is essential to this last argument; arbitrary functors need not preserve monomorphisms.

The unit repeats an element and multiplication selects the two outer diagonal entries. Both unit laws hold directly, and the associativity law selects the same two diagonal entries of an eight-entry array. Its algebras for a monad are precisely rectangular bands, including the empty algebra.

For a uniform disc of radius $a$ and mass $M$, annuli of radius $r$ and thickness $dr$ have mass $(M/\pi a^2)2\pi rdr$. Their axial moment of inertia sums to $I={\int }_0^a{r}^{2}dm=M{a}^2/2$.

A uniform planar annulus of outer radius $a$, inner radius $b<a$ and mass $M$ has surface density $M/[\pi (a^2-b^2)]$. Integrating $r^2$ over its mass gives the axial moment of inertia $I=M(a^2+b^2)/2$. Subtracting the inner disc from a complete disc gives the same result, provided their surface densities match.

// Target: electromagnetism.bigb

If a fixed-axis wheel of radius $R$ drives a fixed-axis roller of radius $S$ without slip, their angular-speed magnitudes satisfy $\sigma =(R/S)\omega$. Their kinetic energy is $\frac{1}{2}[I+J(R/S{)}^2]{\omega }^2$, so the displayed reflected moment of inertia governs the wheel coordinate. A roller resisting torque of magnitude $\gamma {\sigma }^2$ consumes power $\gamma {\sigma }^3$ and therefore reflects to wheel torque $\gamma (R/S{)}^3{\omega }^2$. The squared ratio belongs to inertia; the cubed ratio belongs to this quadratic resisting torque.

For a uniform solid cylinder of radius $a$, length $L$ and mass density $\rho$, a coaxial shell carries mass $dm=2\pi \rho Lrdr$. The axial moment of inertia is $I={\int }_0^a{r}^{2}dm=\pi \rho L{a}^4/2=m{a}^2/2$. The axial moment of inertia is independent of the length once mass and radius are fixed. A thin cylindrical shell instead has $I=m{a}^2$, demonstrating why the uniform solid assumption matters.

The centroid of a uniform solid hemisphere of radius $a$ lies $3a/8$ from its flat face. About a diameter of the underlying sphere parallel to that face, its moment of inertia is $2m{a}^2/5$. The parallel axis theorem therefore gives $I_G=2m{a}^2/5-m(3a/8{)}^2=83m{a}^2/320$ about the parallel centroidal axis. A thin hemispherical shell has a different centroid and inertia.

For a toroidally compactified heterotic string, the torus metric, antisymmetric tensor and gauge Wilson lines determine a positive $d$-plane in a charge space of signature $(d,d+16)$, modulo lattice identifications. The continuous dimension is $d(d+16)$; the coupling is an additional modulus. The three-torus case has fifty-seven shape and Wilson-line parameters.

For stable compact Riemann surfaces with $n$ distinct labelled marked points, where $2g-2+n>0$, the displayed complex dimension counts independent conformal shapes and positions after conformal automorphisms are removed. On the sphere it is $n-3$, since three points fix a Möbius transformation. On a torus it is $1+n-1=n$: one shape modulus and $n$ positions, less one translation. For genus $g\ge 2$, the unmarked shape dimension is $3g-3$ and continuous conformal automorphisms vanish. The shape dimension also follows from the Riemann-Roch theorem applied to holomorphic quadratic differentials.

The socle of a module is the sum of all its simple submodules, equivalently its largest semisimple module.