For a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, a state $a$ is absorbing exactly when $P_{aa}=1$. Nonnegativity and the row sum then force $P_{aj}=0$ for $j\ne a$: once entered, $a$ is never left. Equivalently, $\{a\}$ is a closed communicating class. A larger closed communicating class may keep moving internally; closure means confinement to a set, whereas absorption at a state means remaining at one point.

An absorbing state is useful for recording a terminal outcome: after a toy routine finishes, keep its outcome in the state space so that the same transition matrix still describes subsequent steps. Calling a finite Markov chain absorbing usually also requires that some absorbing state be accessible from every state. Merely containing an absorbing state is insufficient: another closed communicating class could prevent absorption. See Grinstead and Snell, Introduction to Probability, section 11.2 for this convention.

Consider a constructed routine with unfinished modes $u,v$ and terminal outcomes $A,B$. All the following transition probabilities are chosen for illustration, not estimated from observations. In state order $(u,v,A,B)$, takeThe communicating class $\{u,v\}$ is open, while $\{A\}$ and $\{B\}$ are closed communicating classes consisting of absorbing states. Both terminal outcomes are accessible from both unfinished modes, sometimes by an indirect path.

Let $T=\inf \{n\ge 0:X_n\in \{A,B\}\}$ be the terminal hitting time. Write $h_i={\mathbb{P}}_i(X_T=A)$ for the absorption probability and $t_i={\mathbb{E}}_{i}T$ for the expected hitting time, with the subscript denoting the starting state. First-step analysis givesHere $h_A=1$, $h_B=0$, and $t_A=t_B=0$. Solving these two pairs of linear equations yieldsStarting at $u$ therefore ends at $A$ with probability $0.6$ and takes $3.2$ steps on average to reach either terminal state. The terminal hitting time counts the transition into the outcome, but no later self-transitions. Outcome and duration answer different questions: from $v$, outcome $A$ is less likely, yet termination is faster on average.

The matrix power $P^2$ gives the probability distribution after two steps from $u$ asThus ${\mathbb{P}}_u(T>2)=1/2$, and the probability of having already reached $A$ is $3/8$, still below its eventual absorption probability $3/5$. From either unfinished mode, each next step has terminal probability at least $1/4$. Consequently ${\mathbb{P}}_i(T>n)\le (3/4{)}^n$, proving termination almost surely and finite expected hitting time directly. The Fundamental matrix of a finite absorbing Markov chain packages both calculations and counts visits to unfinished modes.

List the primes as $p_0,p_1,\dots$ and form the mapping telescope of degree-$p_i$ maps $S^n\to S^n$. Its reduced homology vanishes outside degree $n$, while

Let $Q_n$ be the number left immediately after departure $n$, and let $A_n$ be the arrivals during the next service. ThenConditional on service duration $S$, $A_n$ is Poisson with parameter $\lambda S$, so

Let $X\subseteq {\mathbb{R}}^n$ be Borel. If $L^q(X)\subseteq L^p(X)$ for $1\le p<q\le \infty$, the closed graph theorem makes the inclusion continuous. Applying its norm bound to indicators of $X\cap B(0,R)$ shows that their measures are uniformly bounded, hence $\mu (X)<\infty$.

If $f_n\to f$ in $L^p$ for $1\le p<\infty$, choose a subsequence with $\Vert f_{n_k}-f{\Vert }_p^p\le 2^{-k}$. Tonelli's theorem makes ${\sum }_k|f_{n_k}-f{|}^p$ finite almost everywhere, so $f_{n_k}\to f$ almost everywhere. For $p=\infty$, norm convergence itself gives almost-everywhere convergence after removing one common null set.

The algebra $T_n(k)$ consists of the lower triangular $n$ by $n$ matrices over $k$. Its Jacobson radical $N$ consists of the strictly lower triangular matrices, andThe $n$ simple left modules $S_r$ are one-dimensional, with a matrix acting through its $r$th diagonal entry. For the left regular module,

For finite-length modules $U,V$ and every nonnegative integer $i$,

Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

Kernel linear discriminant analysis performs regularized linear discriminant analysis in a Reproducing kernel Hilbert space, using kernel evaluations to obtain nonlinear decision boundaries in the original input space.

The Laplace contour-integral method seeks a solution in the formReplacing multiplication by $z$ with differentiation of $e^{zt}$ in $t$ and integrating by parts turns the differential equation for $w$ into a first-order equation for $f$. The contour $\gamma$ must make the resulting endpoint term vanish; finite endpoints are often branch points of $f$, while infinite ends must enter sectors where the exponential decays.

The past light cone of a spacetime event $O$ is the past-directed part of its light cone. In Minkowski spacetime, its spacetime events satisfy $|\Delta \mathbf{x}|=-c\Delta t$ with $\Delta t<0$. They are the possible emission events of light rays arriving at $O$. The interior also contains events that can send slower signals to $O$. The retarded potential of a moving point electric charge, given by the Liénard–Wiechert potential, uses the emission event where the source's world line intersects this boundary.

A connected Lie group is simple when its Lie algebra is nonabelian and simple. Some conventions additionally quotient out or exclude a discrete center; the infinitesimal gauge fields depend only on the Lie algebra.

A matrix Lie group is a subgroup of a general linear group that is also a Lie group for its matrix topology. Near the identity, the matrix exponential and matrix logarithm give inverse coordinate charts whenever they are restricted to sufficiently small neighborhoods in the group and its tangent space.

A Lie group action is a smooth group action of a Lie group on a smooth manifold. Differentiating the action gives an infinitesimal action of its Lie algebra by vector fields.

The direct sum $\mathfrak{g}\oplus \mathfrak{h}$ has componentwise bracketThe two summands are commuting ideals.

A Lie algebra isomorphism is a bijective linear map $\phi :\mathfrak{g}\to \mathfrak{h}$ satisfying $\phi ([x,y])=[\phi (x),\phi (y)]$. Its inverse automatically preserves the Lie bracket.

For a Matrix Lie group $G$, its tangent space $\mathfrak{g}=T_{I}G$ is closed under the matrix commutator. Indeed, for $X,Y\in \mathfrak{g}$, the group commutatorlies in $G$, and the mixed derivative at $(0,0)$ of its matrix logarithm is $XY-YX$. Thus $[X,Y]=XY-YX$ belongs to $\mathfrak{g}$.

A Lie algebra generator is an element of a chosen spanning or generating set for a Lie algebra. In a matrix representation, an infinitesimal group transformation has the form $1+i{\alpha }_a{t}^a+O({\alpha }^2)$ in terms of generator matrices $t^a$.

A compound Poisson process has the formwhere $N$ is a Poisson process and the independent identically distributed marks $Y_n$ are independent of $N$. It is a Lévy process with a finite Lévy measure.

The Lévy measure $\nu$ of a Lévy process records the intensity of its jumps: for every measurable set $A$ bounded away from zero, $\nu (A)$ is the expected number per unit time of jumps whose sizes lie in $A$. It satisfies