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Applying the curvature commutator to the parallel metric gives $R_{\mu \alpha \beta \gamma }=-R_{\alpha \mu \beta \gamma }$.

The Weyl tensor is the completely trace-free part of the Riemann curvature tensor. In four dimensions,

For a torsion-free connection, $R^{\mu }{}_{\alpha \beta \gamma }+R^{\mu }{}_{\beta \gamma \alpha }+R^{\mu }{}_{\gamma \alpha \beta }=0$.

For a torsion-free connection, $U_{\mu }={\partial }_{\mu }\varphi$ satisfies ${\nabla }_{\alpha }{U}_{\beta }={\nabla }_{\beta }{U}_{\alpha }$.

For a family of affinely parametrized geodesics with tangent $U$ and connecting field $J$, the Jacobi equation in the convention used here isIt describes the relative acceleration produced by spacetime curvature.

Here $J$ describes an infinitesimal separation and the covariant derivatives compare it in a frame transported along the reference geodesic. The Riemann curvature tensor thus becomes something measured with neighboring trajectories. The Newtonian version follows by differentiating gravitational acceleration with respect to position. In the exterior of a spherical mass, let ${\xi }_r$ be radial separation and ${\xi }_{\perp }$ separation in either transverse direction. Newton's law of universal gravitation givesThe upper particle falls less strongly than the lower one, stretching their radial separation; side-by-side particles accelerate toward converging radial directions, compressing their transverse separation. See Oregon State's derivation of tidal acceleration.

For a spherical Earth model with $r=R_{\oplus }=6.371\times 10^6\mathrm{m}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $GM=g{R}_{\oplus }^2$, a one-metre radial separation gives ${\ddot{\xi }}_r=2g/R_{\oplus }=3.07958\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-2}$; a one-metre transverse separation gives $-1.53979\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-2}$. Initially comoving particles therefore change separation after one second by approximately $\frac{1}{2}\ddot{\xi }{t}^2$: $+1.54$ micrometres radially or $-0.770$ micrometres transversely. This short-time calculation holds the reference radius fixed to leading order. Its beauty is that common falling motion drops out, leaving a measurable stretching pattern that reveals curvature.

For a rank-two covariant tensor, the derivative commutator contributes one curvature action on each index:

When a residue theorem argument encloses increasingly many poles, the resulting infinite sum needs a specified grouping and convergence justification. Symmetric summation means taking poles in matched groups, such as $n$ and $-n$, before sending the contour size to infinity.

The meromorphic function $\pi \csc (\pi z)$ has a simple pole of residue $(-1{)}^n$ at each integer $n$. Its partial-fraction expansion isThe paired terms are $O(n^{-2})$ uniformly on each compact set avoiding the poles, so the paired series converges absolutely and locally uniformly. The individual one-sided terms are only $O(n^{-1})$; one cannot infer absolute convergence from the unpaired expression.^[ref]

This example separates two mechanisms: symmetry can improve the decay of a pair, while alternating signs can permit conditional convergence. Neither mechanism licenses arbitrary regrouping of an unproved infinite residue sum.

Symmetry can relate residues, but the sign depends on the symmetry. If a meromorphic function is even, $f(-z)=f(z)$, and $a\ne 0$, thenIndeed the substitution $z=-a+w$ replaces the local variable at $a$ by $-w$, reversing the coefficient of $w^{-1}$ in the Laurent series. A contour containing both poles with the same winding number therefore receives cancelling residue contributions.

For example, $f(z)=1/(z^2+1)$ has residues $1/(2i)$ and $-1/(2i)$ at $i$ and $-i$. An upper-half-plane contour contains only the first pole, so those residues do not cancel in that contour. For an odd meromorphic function, the residues at opposite nonzero poles are instead equal: $z/(z^2+1)$ has residue$1/2$ at both poles. One must check both the local symmetry and which poles the contour actually encloses.

Suppose a contour follows the real axis, closes in the upper half-plane, and avoids each simple real pole $a_j$ by a small semicircle above it. Each indentation is clockwise and tends toIf the large arc vanishes and there are no enclosed poles away from the real axis, the Cauchy principal value is therefore

Ifwith $g$ holomorphic at $a$, then

An indecomposable representation is nonzero and cannot be expressed as a direct sum of two nonzero invariant subspaces. It is an indecomposable module over the associated group algebra or representing algebra.

A group representation $\rho :G\to GL(V)$ factors through a finite quotient group exactly when its kernel of a group homomorphism has finite subgroup index, equivalently when its image of a function is a finite group. If a specific quotient group $G/N$ is prescribed, factorization through it is equivalent to $N\subseteq \ker \rho$.

For $G=\mathbb{Z}$, the one-dimensional group representation $n\mapsto {\lambda }^n$ factors through a finite quotient group exactly when $\lambda$ is a root of unity. Taking $\lambda =2$ gives a one-dimensional group representation with infinite image. Finite dimension therefore does not imply factorization through a finite quotient group.

A one-dimensional character is a character of a representation arising from a one-dimensional representation, equivalently a homomorphism from a group to the multiplicative group of its scalar field.

The infinite dihedral group is $⟨r,t\mid t^2=1,tr{t}^{-1}=r^{-1}⟩$.

An irreducible group representation is nonzero and has no nonzero proper invariant subspace. These are the pieces that cannot be decomposed further by taking a smaller symmetry-preserved vector space.

The real two-dimensional rotation through $120$ degrees illustrates the role of the coefficient field. A one-dimensional real invariant subspace would supply a real eigenvector, but the characteristic polynomial ${\lambda }^2+\lambda +1$ has no real root. The representation is therefore irreducible over the real numbers. Over the complex numbers, that polynomial has distinct roots of unity $e^{2\pi i/3}$ and $e^{-2\pi i/3}$, giving two invariant eigenlines. Irreducibility over the real numbers does not imply absolute irreducibility.

For a subgroup $H\le G$ and a left group algebra $k[H]$-module $V$, its induced representation is ${\mathrm{Ind}}_H^{G}V=k[G]{\otimes }_{k[H]}V$, with $G$ acting by left multiplication on the first tensor factor. If $H$ has finite index of a subgroup and $V$ is a finite-dimensional vector space, the dimension of the induced representation is $[G:H]\dim V$.

Inducing the trivial representation gives the permutation representation on cosets $G/H$. Frobenius reciprocity characterizes the construction: intertwining operators from the induced representation to $W$ correspond to $H$-intertwining operators from $V$ to the restriction of a representation $W$. Thus induction of representations extends a smaller group action while imposing only the identifications forced by that subgroup.

For a representation on $V$, the dual action on $V^{\ast }$ is $(g\phi )(v)=\phi (g^{-1}v)$.

An intertwining operator between representations $\rho$ and $\sigma$ is a linear equivariant map $T$ satisfying $T\rho (g)=\sigma (g)T$ for every group element $g$.

For a finite $G$-set $X$, the character of $\mathbb{C}[X]$ at $g$ is $|X^g|$. Its invariant dimension is the number of $G$-orbits.

The character of a finite-dimensional representation $\rho$ is the class function ${\chi }_{\rho }(g)=\mathrm{tr}(\rho (g))$. Isomorphic representations have the same character.