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The maximal spectrum $\mathrm{MaxSpec}R$ is the set of maximal ideals of a commutative ring $R$.

The residue field at a maximal ideal $\mathfrak{m}$ is the quotient field $\kappa (\mathfrak{m})=R/\mathfrak{m}$. At a prime ideal $\mathfrak{p}$, the residue field is the fraction field of $R/\mathfrak{p}$.

After neglecting the decaying mode, the linear growth between conformal times ${\tau }_1$ and ${\tau }_2$ is

In cosmic time during matter domination, $a(t)\propto t^{2/3}$ and the pressureless density contrast obeysIts growing and decaying solutions are $t^{2/3}$ and $t^{-1}$.

Forthe derivative rule givesand hence $Y(s)=(sI-A{)}^{-1}(y_0+G(s))$ wherever the transforms converge and $sI-A$ is invertible.

If $A$ is an $m$-by-$n$ matrix and $B$ is an $n$-by-$m$ matrix over the same field, both products are square matrices andThe products can even have different sizes: for $A=(12)$ and $B=(34{)}^T$, $AB=(11)$ and $BA=\left(\begin{array}{cc}3& 6\\ 4& 8\end{array}\right)$, but both have trace $11$.^[ref]

Grouping all but one factor proves that $\mathrm{tr}(A_1{A}_2\cdots A_k)$ is unchanged by a cyclic permutation whenever the cyclic products are defined and square. In particular, every commutator has trace zero, since $\mathrm{tr}(AB-BA)=0$ by linearity.

Cyclic rotation does not allow arbitrary reordering. Forone has $ABC=A$, so $\mathrm{tr}(ABC)=1$, but $AC=0$, so $\mathrm{tr}(ACB)=0$. Rotating $ABC$ to $BCA$ or $CAB$ preserves its trace; swapping just $B$ and $C$ need not.

An MPS parent Hamiltonian is a sum of local projectors onto the orthogonal complement of the tensor's local support. The MPS is a frustration-free ground state.

An MPS tensor is injective after blocking when products of its physical matrices span the full virtual matrix algebra.

Caliper matching rejects candidate pairs whose distance exceeds a tolerance.

Covariate balance means treatment groups have similar pretreatment covariate distributions.

If a $k$-regular graph on $n$ vertices has a maximum matching of size $m$, its unmatched vertices are independent. Counting their incident edges gives $k(n-2m)\le 2(k-1)m$, and hence $m\ge kn/(4k-2)$.

For a finite graph $G$, the matching number isIt is the cardinality of a maximum cardinality matching. The vertex capacity constraint gives $0\le \nu (G)\le \lfloor |V(G)|/2\rfloor$. The worked path graph examples have $\nu (P_4)=2$ and $\nu (P_6)=3$. A perfect matching reaches $|V(G)|/2$; the empty graph has matching number zero.

A maximal matching is a matching in a graph to which no extra edge can be added while retaining all current selected edges. Equivalently, no edge joins two unmatched vertices in a matching; those vertices form an independent set. Every maximum cardinality matching is a maximal matching, but the converse fails.

Take the path graph $P_4$ with vertices $1,2,3,4$ and edges $(1,2),(2,3),(3,4)$. A greedy algorithm that first selects the middle edge produces $M=\{(2,3)\}$. Both remaining edges touch a matched vertex, so this size-one matching in a graph is maximal. Yet $N=\{(1,2),(3,4)\}$ has two edges. No matching in a graph on four vertices has more than two edges, so $N$ is a maximum cardinality matching. The two remaining free vertices of $M$, namely $1$ and $4$, are not adjacent; that does not prevent a longer augmenting path in a matching from joining them.

Maximality is useful as a quick stopping rule and a modest guarantee. For any maximal matching $M$ and maximum cardinality matching $N$ in a finite simple graph, every edge of $N$ touches an endpoint of an edge of $M$. Assign it to one such edge of $M$. Since $N$ has disjoint endpoints, each edge of $M$ receives at most two assignments, proving $|N|\le 2|M|$. The P₄ example reaches equality. Thus maximality guarantees at least half the largest cardinality, but says nothing comparable about arbitrary edge benefits. Its selected endpoints also form a vertex cover, as explained under vertex cover.

A perfect matching is a matching in a graph that covers every vertex exactly once. Its edges have no shared endpoints, and every vertex is an endpoint of one selected edge. It is useful when a collection must be divided into disjoint pairs with no element left over.

In a bipartite graph with equally many vertices on its two sides, a perfect matching can assign every person one allowed job and every job one person. The example under permanent of a matrix has three people and three jobs, with exactly three valid assignments. The blue edges below show those three perfect matchings; a crossing in the drawing is not an extra vertex.

Every selected edge covers two vertices, so a graph with an odd integer number of vertices has no perfect matching. An even integer number is necessary but does not guarantee existence: an isolated vertex cannot be covered. A perfect matching also need not be unique. For example, the complete bipartite graph $K_{2,2}$ has two perfect matchings, corresponding to the two ways of assigning two people to two jobs.

Counting all perfect matchings is a different question from finding one. For a zero-one biadjacency matrix of a balanced bipartite graph, the count is its permanent of a matrix. The nonzero determinant criterion for a perfect matching supplies an existence test, but cancellation means that a zero determinant does not rule one out.

The finite-horizon Snell envelope of rewards $Z_n$ is defined backward by $V_N=Z_N$ and $V_n=\max \{Z_n,\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]\}$. It is the least supermartingale dominating $Z$.

In discrete time, a process $A_n$ is predictable or previsible when $A_n$ is ${\mathcal{F}}_{n-1}$-measurable for every $n\ge 1$. In continuous time, predictability means measurability with respect to the predictable sigma-algebra.

A random time $\tau$ is a stopping time when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$.

An integrable adapted process $(V_n)$ is a supermartingale when $\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]\le V_n$.

If $(M_n)$ is a martingale and $U_N[a,b]$ counts its completed upcrossings of $[a,b]$ by time $N$, thenEquivalent conventions may add an initial endpoint term.

If every increment ${\xi }_n=M_n-M_{n-1}$ of a discrete martingale takes values in $\{-1,1\}$, thenIterated conditioning shows that the increments are independent symmetric signs, so $M_n-M_0$ is a simple symmetric random walk.