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A one-parameter point transformation has infinitesimal generatorIts flow recovers the finite transformations.

The Lie derivative of a connection is the tensorAn isometry preserves the Levi-Civita connection, so ${\mathcal{L}}_K\nabla =0$ for a Killing vector field.

A metric connection satisfies metric compatibility, ${\nabla }_{\alpha }{g}_{\mu \nu }=0$. This allows the metric tensor to commute with covariant derivatives and index raising or lowering.

Let $p$ be an odd prime and let $a,b,c$ be integers with $p\nmid a$. The quadratic congruencehas a solution exactly when its discriminant $\Delta =b^2-4ac$ is zero or a quadratic residue modulo $p$.

Indeed, completing the square gives the equivalent equationMultiplication by $2a$ and translation by $b$ give a bijection of the residue classes, because $2a$ is invertible modulo $p$. Each square root $r$ of $\Delta$ therefore gives exactly one solution $X\equiv (r-b)(2a{)}^{-1}(\mathrm{mod}p)$.

Consequently the number of distinct solutions is $1+(\Delta /p)$, using the Legendre symbol: one for zero discriminant, two for a nonzero quadratic residue, and none for a quadratic nonresidue. The hypotheses matter: if $p\mid a$ the equation can become linear, while at $p=2$ division by $2a$ is unavailable.

For a prime number $p\equiv 7(\mathrm{mod}8)$, let $L$ be the sum of the least positive quadratic residues below $p/2$. ThenFor example, at $p=23$ the relevant quadratic residues are $1,2,3,4,6,8,9$, and their sum is $33=(23^2-1)/16$.

To prove the identity, put $m=(p-1)/2$, let $S$ be the sum of all least positive nonzero quadratic residues, and let $b$ count those above $p/2$. The second supplementary law for quadratic reciprocity gives $(2/p)=1$, so doubling permutes these residue classes. Reducing each doubled representative modulo $p$ subtracts $p$ exactly $b$ times. Hence $S=2S-pb$, or $S=pb$.

Since $p\equiv 3(\mathrm{mod}4)$, the first supplementary law for quadratic reciprocity says $-1$ is a quadratic nonresidue. Exactly one of $j,p-j$ is therefore a quadratic residue for each $1\le j\le m$. The lower-half choices sum to $L$; the upper-half choices sum to $pb-(1+\cdots +m-L)$. ConsequentlySubstituting $S=pb$ gives $L=m(m+1)/4=(p^2-1)/16$. Both congruence conditions have a role: one makes doubling a permutation, and the other makes each opposite pair contain exactly one quadratic residue.

For an odd prime $p$, putThen $T=0$ if $p\equiv 1(\mathrm{mod}4)$, and $T\equiv 0(\mathrm{mod}p)$ for every $p>3$.

For the first statement, the first supplementary law for quadratic reciprocity gives $\chi (-1)=1$, where $\chi$ is the Legendre symbol. Pairing $a$ with $p-a$ givesThe same pairing in the complete Legendre-symbol sum shows that this half sum is zero, proving the exact equality.

For the second statement, multiplication by $2$ permutes the nonzero residue classes modulo $p$. Reducing the weighted sum modulo $p$ and using multiplicativity of the Legendre symbol therefore gives $T\equiv 2\chi (2)T(\mathrm{mod}p)$. The coefficient $2\chi (2)-1$ is either $1$ or $-3$, and hence is invertible for $p>3$. Thus $T\equiv 0(\mathrm{mod}p)$. The restriction is necessary: at $p=3$, $T=1-2=-1$.

For an odd prime $p$,For $p\nmid a$, write $a=g^m$ using a primitive root $g$; then the left side is $(-1{)}^m$, which is one exactly when $m$ is even and $a$ is a square.

For an odd prime $p$ and $p\nmid a$, Euler's criterion identifies the Legendre symbol by modular exponentiation:

Thus a quadratic residue gives one and a quadratic nonresidue gives minus one, represented by $p-1$ if a nonnegative remainder is used. The formula also holds when $p\mid a$, because $(p-1)/2>0$ and both sides are zero modulo $p$.^[ref]

For example, modulo seven the exponent is three: $2^3\equiv 1$ confirms that two is a quadratic residue, while $3^3\equiv 6\equiv -1$ confirms that three is a quadratic nonresidue. Modulo eleven the exponent is five, and $2^5=32\equiv 10\equiv -1$ confirms that two is a quadratic nonresidue. The test answers whether a square root exists; it does not itself return the roots $3,4$ of $x^2\equiv 2(\mathrm{mod}7)$.

A short proof needs only Fermat's little theorem and the root bound for a polynomial. Put $m=(p-1)/2$. If $a=x^2\ne 0$ in the prime field, then $a^m=x^{p-1}=1$. There are already $m$ such quadratic residues. The nonzero polynomial $T^m-1$ has polynomial degree $m$, so it has no other polynomial roots. Finally, for every nonzero $a$, Fermat's little theorem gives $(a^m{)}^2=a^{p-1}=1$. Factoring $(a^m-1)(a^m+1)=0$ in the field shows $a^m\in \{1,-1\}$; every quadratic nonresidue must take the remaining value $-1$. This proves the criterion without first finding a primitive root.

For a larger original example, decide whether seven is a quadratic residue modulo the prime number $101$. Exponentiation by squaring computes the required power without listing fifty squares:Therefore $(7/101)=-1$ and $x^2\equiv 7(\mathrm{mod}101)$ has no solution. Each intermediate product is reduced modulo $101$, as in modular exponentiation. The advantage is that exponentiation by squaring uses a number of multiplications proportional to the binary digit length of the exponent rather than testing every potential square root.

The odd prime hypothesis is essential. For the composite modulus nine, $2^{(9-1)/2}=2^4\equiv 7(\mathrm{mod}9)$, which is neither one nor minus one. Although four is a nonzero square modulo nine, $4^4\equiv 4(\mathrm{mod}9)$, so even the square-to-one implication fails. The Jacobi symbol and Euler pseudoprime discuss different questions for odd composite moduli; they do not make this a universal square test.

Let $f:M\to M$ be a smooth self-map of a closed manifold, and suppose every fixed point $x$ is nondegenerate, meaning that $I-D{f}_x$ is an invertible linear map. The fixed points are isolated and, by compactness, finite in number. Their local fixed point indices giveThe determinant is computed in any local coordinate basis; changing that basis conjugates the linear map and does not change its sign. Near $x$, the displacement map is $v\mapsto (I-D{f}_x)v$ to first order, explaining its local mapping degree. Summing these local degrees gives the global Lefschetz number.

Each summand is $+1$ or $-1$, so there are at least $|L(f)|$ fixed points. The nondegeneracy condition is essential for this determinant formula; degenerate isolated fixed points require their actual local fixed point indices. A manifold with boundary requires boundary hypotheses and is not included in the displayed statement.

For a bounded open $U\subset {\mathbb{R}}^n$, $m^2>0$, and $f\in L^2(U)$, there is a unique $u\in H_0^1(U)$ such thatfor every $v\in H_0^1(U)$.

Let $U\subset {\mathbb{R}}^d$ be a bounded connected Lipschitz domain, and let a measurable boundary part ${\Gamma }_D$ have positive surface measure. Impose a homogeneous Dirichlet boundary condition there and a homogeneous Neumann boundary condition on the remainder. For $f\in L^2(U)$, seek $u$ in the closed Sobolev space subspacesatisfyingThe Trace operator makes the boundary restriction meaningful. The Poincare inequality with a partial Dirichlet boundary bounds the full $H^1$ norm by the gradient norm on $V$. Thus the left side is a bounded coercive bilinear form, and the Lax-Milgram theorem gives a unique weak solution. The same proof allows any bounded linear functional on $V$ as the right side.

The Neumann condition is natural: integration by parts leaves the boundary term on the non-Dirichlet portion, and requiring the identity for all $v\in V$ makes that term vanish. If ${\Gamma }_D$ is empty, constants remain in the kernel of a linear map; for homogeneous Neumann data, compatibility requires ${\int }_{U}f=0$, and uniqueness requires a normalization such as zero mean. Positive Dirichlet surface measure is what removes this ambiguity.

For $\ddot{x}=-U^{\prime }(x)-\mu \dot{x}$ with $\mu >0$,If $U$ is radially unbounded, every trajectory is bounded, and LaSalle's principle reduces its omega-limit set to invariant points with zero velocity.

The Heaviside step function switches from zero to one at its threshold. With threshold zero,The value at zero is a convention, commonly zero, one half or one; changing this single value changes no ordinary integral. The delayed signal $H(t-a)f(t-a)$ is zero before time $a$ and then follows the original signal from its beginning. This is the switch used by the Laplace transform shift theorem.

As a distribution, its derivative is the Dirac delta function: $H^{\prime }=\delta$. Indeed, against a smooth compactly supported test function $\phi$, one has $⟨H^{\prime },\phi ⟩=-{\int }_0^{\infty }{\phi }^{\prime }(t)dt=\phi (0)$. Ordinary pointwise differentiation away from zero would miss this concentrated jump.

The delayed unit step has

After $t=zs$, the phase in the integral for ${\Gamma }^{\prime }(z)$ becomes $\log s-s$, with maximum $-1$ and second derivative $-1$ at $s=1$. Hence

A Laplacian eigenfunction satisfies $-\Delta u=\lambda u$ together with specified boundary conditions.

In three-dimensional spherical coordinates $(r,\theta ,\varphi )$, where $\theta$ is colatitude, the Laplacian in the convention $\Delta ={\partial }_x^2+{\partial }_y^2+{\partial }_z^2$ isThe volume element $r^2\sin \theta drd\theta d\varphi$ and the coordinate scale factors give this formula by taking the divergence of the gradient. It separates a radial Laplacian from an angular operator divided by $r^2$.

For a radial function $u=f(r)$, it reduces to $f^{\prime \prime }+2f^{\prime }/r$. Consequently $1/r$ is a harmonic function for $r>0$, while its singularity at the origin supplies a Dirac delta distribution. If $Y$ is a spherical harmonic with angular eigenvalue $-\ell (\ell +1)$ in this sign convention, substituting $u=R(r)Y$ into the Laplace equation gives $r^2{R}^{\prime \prime }+2r{R}^{\prime }-\ell (\ell +1)R=0$, with radial powers $r^{\ell },r^{-\ell -1}$. Coordinate singularities at the axis and origin require regularity in Cartesian coordinates.

For radial $u(r)$ in $n$ dimensions, $\Delta u=u\prime \prime +(n-1)u\prime /r$.

Laplace's equation isIts solutions are harmonic functions.

If a twice differentiable real function has a local maximum at an interior point of a Riemannian manifold, its Hessian matrix is negative semidefinite there and hence its Laplacian is nonpositive.