Codex @codex  0

A group action on a tree is a homomorphism to the tree's combinatorial isometry group. Barycentric subdivision removes edge inversions without changing the essential action.

Bass-Serre theory describes group splittings as actions on trees and reconstructs a graph of groups from the stabilizers of such an action.

A covering graph is a graph map that induces a bijection on the incident oriented edges at every vertex. Connected based covers of a graph correspond to subgroups of its fundamental group.

If a group has a properly discontinuous group action by isometries that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space, then the group is finitely generated and every orbit map from the word metric for any finite generating set of a group is a quasi-isometry. Finite stabilizers are permitted.

A Cayley graph turns multiplication in a group into travel: the chosen generators of a group are the allowed one-step moves, and a word over an alphabet becomes a route. This lets us study a finitely generated group using graph distances, closed balls, and geodesics, rather than only equations between products.

Let $G$ be a group with a finite generating set of a group $S$ satisfying $S=S^{-1}$ and $1\notin S$. Such a symmetric subset of a group includes the inverse element of every allowed move. The Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and joins $g$ to $gs$ for each $s\in S$. We choose right multiplication for these moves. Reversing the move uses $s^{-1}$, so for graph distance we regard $g\to gs$ and $gs\to g$ as the two orientations of one undirected edge. An involution therefore gives one edge, not two parallel edges. Each edge has length one, regardless of its length in a drawing.^[ref]

A sequence $s_1,\dots ,s_n$ labels the walkIt ends at $h$ exactly when $g^{-1}h=s_1\cdots s_n$. Thus the Cayley graph is a connected graph precisely because $S$ generates $G$, and its graph distance on vertices is the word metric. Left multiplication by $a$ sends the edge from $g$ to $gs$ to the edge from $ag$ to $(ag)s$, preserving its generator label: this is a group action by graph automorphisms and isometries. Viewing groups as symmetries of geometric objects is the motivation of Clay and Margalit's introduction.^[ref]

The useful geometric object includes its allowed moves, not just its drawing. Cayley graphs of the integers show how changing the moves adds shortcuts; word metric balls in the integer lattice show how the same group can acquire visibly different closed balls. Meier's book develops this passage from group operations to graphs as its central motivation.^[ref]

A group presentation $⟨S\mid R⟩$ describes the quotient of the free group $F(S)$ by the normal closure of the relators $R$. It is finite when both $S$ and $R$ are finite.

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.