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An oriented volume retains the sign of a determinant; reversing the orientation changes that sign.

The scalar-QED term $e^2{A}_{\mu }{A}^{\mu }{\varphi }^{\ast }\varphi$ produces a local two-photon two-scalar vertex, conventionally called the seagull vertex.

Characteristic crossing occurs when the map from an initial point to its position at time $t$ loses injectivity. For $u_t+f(u{)}_x=0$, its Jacobian is $1+t{f}^{\prime \prime }(u_0)u_0^{\prime }$ in the autonomous case, so a negative value of $f^{\prime \prime }(u_0)u_0^{\prime }$ causes finite-time crossing.

The Inviscid Burgers equation is the scalar conservation lawFor a classical solution, the chain rule makes this equivalent to $u_t+u{u}_x=0$, a nonlinear transport equation with characteristic speed $u$. The conservative form remains meaningful for a discontinuous weak solution, whereas the product $u{u}_x$ cannot simply be evaluated pointwise at a jump. Its discontinuities obey the Rankine-Hugoniot condition $s=(u_L+u_R)/2$; this condition alone permits inadmissible expansion jumps. For bounded initial conditions on the infinite line, imposing the Kruzhkov entropy inequalities and the initial trace selects a unique entropy solution. For step data, incoming characteristic curves select $u_L>s>u_R$, while $u_L<u_R$ produces a rarefaction wave.^[ref]

For $k\gg 1$, the low-concentration saddle node of the saturating autocatalytic switch has

For a pattern of winding number $w$,whenever the relevant signatures are defined by nonsingular forms. The usual averaged convention extends the identity through roots.

The $(p,q)$-cable of $K$ uses the $(p,q)$ torus knot as a pattern in a tubular neighborhood of $K$.

A pattern is a knot $P$ inside a solid torus. Embedding that torus as a tubular neighborhood of $K$ produces the satellite $P(K)$.

If $f$ is holomorphic near the closed unit disc, $|f|\ge 1$ on its boundary, and $|f(z_0)|<1$ at one interior point, then $f$ maps the open unit disc onto a set containing the open unit disc. Rouché's theorem first shows that $f$ has a zero and then that every $f-w$ with $|w|<1$ has one.

A nonconstant holomorphic function on a domain maps open sets to open sets. Around any point, isolate its zero relative to the value there and use Rouché's theorem to show that every sufficiently nearby value has a preimage.

A locally uniform limit of nonvanishing holomorphic functions on a connected open set is either nonvanishing or identically zero. More generally, isolated zeros persist nearby with their multiplicity.

The torque transmitted across any concentric sphere isIt approaches $8\pi \mu a^3\mathbf{\Omega }$ when $a\ll b$, and for a thin gap $h=b-a\ll a$ it approaches

Conservative binary mass transfer moves material between two stars while retaining total binary mass and, in the usual approximation, orbital angular momentum. For a circular orbit with negligible spin angular momentum,At fixed $M$ and $J$, the separation obeys $a(M_d{M}_a{)}^2=\mathrm{constant}$. When an initially more massive donor star loses material, the product of component masses first increases and the orbit shrinks. After equal masses, continued transfer decreases that product and widens the orbit.

The period-product invariant for conservative mass transfer follows using Kepler's third law. Matter escaping the system, external torques or appreciable exchange with stellar spins changes the assumptions, so the simple invariant need not apply to nonconservative binary mass transfer.

Every nonzero ideal of the ring of integers of a number field factors uniquely as a finite product of nonzero prime ideals. Thus each $x\in K^{\times }$ has integer valuations ${\mathrm{ord}}_{\mathfrak{p}}(x)$, almost all zero, determined by the factorization of its principal ideal.

Every ring homomorphism $\varphi :R\to S$ induces an isomorphism

For an element $a$ of a commutative $R$-algebra, evaluation on a polynomial ring is the ring homomorphism

For a ring homomorphism $f:R\to A$ and an ideal $J\subseteq A$, its contraction is $J^c=f^{-1}(J)$.

For a ring homomorphism $f:R\to A$, the extension $I^e$ of a two-sided ideal $I\subseteq R$ is the smallest two-sided ideal of $A$ containing $f(I)$. Thus its elements are finite sums ${\sum }_j{a}_{j}f(i_j)b_j$ with $i_j\in I$ and $a_j,b_j\in A$.

When $A$ is commutative, the formula becomes $I^e=IA=Af(I)$. For example, extending $(2)$ from $\mathbb{Z}$ to $\mathbb{Z}[i]$ gives the multiples of two in the Gaussian integers; extending it to $\mathbb{Q}$ gives the whole field because two is invertible there. Extension can therefore enlarge an ideal substantially even when the original ring map is injective.

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

For a smooth scalar function $u$ on a compact Riemannian manifold $M$, its Dirichlet energy isThe Riemannian gradient measures local variation and the Riemannian volume weights its contribution. With source function $f$, consider $J(u)=E(u)-{\int }_{M}fud{\mathrm{vol}}_g$. On a manifold without boundary, or for variations vanishing at the boundary, the first variation in direction $v$ isHere ${\Delta }_g={\mathrm{div}}_g{\nabla }_g$ is the Laplace-Beltrami operator. A critical function therefore solves the Poisson equation $-{\Delta }_{g}u=f$.

On a closed manifold, adding a constant leaves $E$ unchanged. If ${\int }_{M}fd{\mathrm{vol}}_g\ne 0$, shifting that constant makes $J$ unbounded below. Thus zero integral of the source is a necessary compatibility condition for this unconstrained closed-manifold problem.