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The free group $F(S)$ on a set $S$ consists of reduced words in letters $s^{\pm 1}$, with multiplication followed by cancellation. Every map from $S$ to a group extends uniquely to a homomorphism from $F(S)$.

A null geodesic congruence has an affinely parametrized null tangent $k^a$ satisfying $k^a{k}_a=0$ and $k^b{\nabla }_b{k}^a=0$.

A Yang-Mills instanton is a finite-action Euclidean gauge field with self-dual or anti-self-dual curvature. On ${\mathbb{R}}^4$, its asymptotic pure gauge defines a map $S_{\infty }^3\to SU(2)$ whose degree is the instanton number.

It is a gauge-field instanton solving the Euclidean Yang-Mills equations, with additional finite-action and duality conditions.

The Abelian Higgs model couples a complex scalar field to a $U(1)$ gauge field and permits magnetic vortex solitons in two spatial dimensions.

The AdS/CFT correspondence identifies a conformal field theory with quantum gravity on an asymptotically anti-de Sitter spacetime. Boundary sources couple to CFT operators and equal boundary values of bulk fields.

A dominated strategy can be replaced by another strategy that performs at least as well against every opponent action.

In a zero-sum game, the players' payoffs sum to zero at every pure strategy strategy profile. For two players this means $u_2=-u_1$: maximizing the column player's payoff is minimizing the row player's. A single payoff matrix $A$ therefore specifies a finite two-player matrix game, and independent mixed strategies $x,y$ give expected payoffs $x^{\mathsf{T}}Ay$ and $-x^{\mathsf{T}}Ay$.

Matching pennies and the usual rock paper scissors are two-player zero-sum games. The asymmetric coordination example is not: its total payoff is $5$ when choices match and $0$ otherwise. Applying a two-player minimax theorem to general bimatrix games would discard this possibility of shared gains. For a two-player zero-sum game, however, optimal mixed strategies guarantee the value of a zero-sum game, and pairing any two opposing optimal strategies gives a Nash equilibrium.

A bimatrix game is a finite two-player game specified by one payoff matrix for each player.

A mixed strategy is a probability distribution over a player's pure strategies. For a finite game with $m_i$ such strategies, the possible mixed strategies form the probability simplexA pure strategy is included as a mixed strategy placing all probability on one choice. The support of a mixed strategy consists of the choices with positive probability. In a sequential game, sampling a complete pure strategy once is the definition; it is not automatically equivalent to fresh randomization at every move.

In a bimatrix game, write $A$ for the row player's payoff matrix and $B$ for the column player's, and write $x,y$ for their probability vectors. Independent randomization gives expected payoffsFor example, in matching pennies, choosing heads with probabilities $p=3/4$ and $q=1/4$ gives joint probabilities $3/16,9/16,1/16,3/16$ for $(H,H),(H,T),(T,H),(T,T)$. The row player's expected payoff is $(3-9-1+3)/16=-1/4$, and the column player's is $1/4$.

Randomization can prevent an opponent exploiting a predictable choice. It does not mean every choice is equally good against every opponent: only best responses belong in the support of a mixed strategy at a Nash equilibrium. A shared random signal can instead produce a lottery over joint action profiles whose joint probabilities do not factor into independent mixed strategies.

For offspring mean $\mu >1$ and variance ${\sigma }^2$, Taylor expansion at one givesfor the extinction probability $d$.

For a Galton-Watson process with one ancestor and offspring probability generating function $F$, the branching extinction probability $q$ is the smallest fixed point of $F$ in $[0,1]$. A direct way to find it is to compute extinction by generation $n$:Zero is an absorbing state, so extinction by generation $n$ is exactly the event $Z_n=0$. ^[ref]

For $F(s)=1/4+s/4+s^2/2$, the equation isBoth $1/2$ and $1$ are fixed points, but iteration from zero selects $q=1/2$. The first values are $q_1=1/4$, $q_2=11/32$, $q_3=809/2048\approx 0.395020$, and $q_6\approx 0.462186$. Iteration started at $1$ would stay at $1$ forever and would give the wrong extinction answer for this offspring distribution.

The vertical steps in the figure apply the offspring probability generating function; horizontal steps return to the diagonal to start the next iteration. This gives a useful numerical calculation even when solving $F(s)=s$ explicitly is difficult. For the displayed law, every finite iteration is a lower bound for the eventual branching extinction probability; at generation six, about $0.037814$ additional extinction probability remains beyond the current empty families.

Why the smallest fixed point? The ancestor's children's descendant families are independent; given $k$ children, all are empty $n$ generations later with probability $q_n^k$. Averaging gives $q_{n+1}={\sum }_k{p}_k{q}_n^k=F(q_n)$. The increasing extinction events give the limit $q$, and continuity of the probability generating function yields $F(q)=q$. For any other fixed point $r\in [0,1]$, $q_0\le r$ and monotonicity of $F$ imply $q_n\le r$ at every step; hence $q\le r$. ^[ref]

Let $x_1,x_2,x_3$ be the roots of the separable polynomial $T^3+aT+b$ in characteristic two. The cyclic sumsare the distinct roots ofEven root permutations fix $r,s$, while odd permutations interchange them. The cubic Galois group is therefore contained in $A_3$ exactly when this quadratic splits over the base field.

An irreducible cubic over a field of characteristic other than two has Galois group $A_3$ when its discriminant is a square and $S_3$ otherwise.

A polynomial is solvable by radicals when its roots lie in a tower obtained by adjoining successive roots; equivalently, its Galois group is solvable.

For cubic roots ${\alpha }_i$ and a cube root of unity $\omega$, the Fourier combinations $u={\alpha }_1+\omega {\alpha }_2+{\omega }^2{\alpha }_3$ and $v$ have cubes satisfying a quadratic resolvent.

For a monic polynomial with roots $r_i$, its discriminant is ${\prod }_{i<j}(r_i-r_j{)}^2$ and vanishes exactly for a repeated root.

For a finite Galois extension, subgroups correspond in reverse order to intermediate fields by taking fixed fields and field-fixing subgroups.

If a finite group $G$ acts faithfully by automorphisms on a field $L$, then $L/L^G$ is Galois, has degree $|G|$, and has Galois group $G$.

This theorem constructs a Galois extension from a finite group of field automorphisms; it is not an alternative name for the extension itself.

For a group $G$ of automorphisms of a field $L$, the fixed field isFor one automorphism $\sigma$, it is also written $L^{\sigma }$.

A finite real extension $K/\mathbb{Q}$ is constructible when it is contained in a towerEquivalently, every element of $K$ can be obtained from rational numbers by field operations and successive square roots.