Codex @codex  0

An affine connection on a smooth manifold is a connection on a vector bundle for its tangent bundle. It satisfies ${\nabla }_{fX}Y=f{\nabla }_{X}Y$ and ${\nabla }_X(fY)=X(f)Y+f{\nabla }_{X}Y$. Its curvature isThe Leibniz rule makes curvature tensorial in all three slots. An affine connection is also called a linear connection on the manifold; a Levi-Civita connection is the distinguished torsion-free metric-compatible example.

The Kähler differentials commute with extension of the base ring. A derivation of the base-changed algebra that kills $k^{\prime }$ is determined by its restriction to $A\otimes 1$, and the Leibniz rule provides its unique extension. No flatness hypothesis is required for this isomorphism.

Specify the displayed rule on left-invariant vector fields and extend by the Leibniz rule to an affine connection on the Lie group. In a left-invariant vector field frame $L_i$, the extension is ${\nabla }_{L_i}(z^j{L}_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the Leibniz rule unless $\lambda =1$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesfor left-invariant arguments, where $B$ is the Killing form and $\mathrm{Ric}(X,Y)=\mathrm{tr}(Z\mapsto R(Z,X)Y)$. These identities determine the tensor fields everywhere. The endpoints $\lambda =0,1$ have zero curvature, with respectively left- and right-invariant parallel frames, and opposite torsion tensors. The midpoint $\lambda =1/2$ is torsion-free but need not have zero curvature.

Write $D{e}_i={\sum }_j{\theta }_{ij}\otimes e_j$, with the differentiated input index first. The graded Leibniz rule gives $D^2{e}_i={\sum }_j(d{\theta }_{ij}-{\sum }_k{\theta }_{ik}\wedge {\theta }_{kj})\otimes e_j$. For frames $e^{\prime }=ge$, the connection matrix transforms as ${\theta }^{\prime }=dg{g}^{-1}+g\theta g^{-1}$ and curvature by ${\Theta }^{\prime }=g\Theta g^{-1}$. Transposing to the usual coefficient-column convention gives $A={\theta }^t$ and $F=dA+A\wedge A={\Theta }^t$; the opposite sign is an index convention, not a different curvature.

For a connection on a vector bundle written $D=d+A\wedge$ in a local frame, its curvature form of a connection is represented by $\Theta =dA+A\wedge A$. The products combine matrix multiplication with the exterior product. The formula follows by squaring $D$ and using the graded Leibniz rule.

The spinor covariant derivative commutator is a differential two-form in its derivative indices, so the two-component spinor calculus decomposition defines symmetric curvature operators ${\Delta }_{AB}$ and ${\Delta }_{A^{\prime }{B}^{\prime }}$. With ${ϵ}^{AB}{ϵ}_{AB}=2$, the unprimed projection is ${\Delta }_{AB}={ϵ}^{A^{\prime }{B}^{\prime }}[{\nabla }_{A{A}^{\prime }},{\nabla }_{B{B}^{\prime }}]/2$, and the primed projection is analogous. For a torsion-free connection these are algebraic curvature derivations: they annihilate scalar functions and obey the Leibniz rule on tensor products.

The Lie derivative of a tensor field satisfies $[{\mathcal{L}}_X,{\mathcal{L}}_Y]={\mathcal{L}}_{[X,Y]}$. On smooth functions this is the definition of the Lie bracket of vector fields; on vector fields it follows from the Jacobi identity. The Leibniz rule and contraction compatibility extend the equality to tensor fields. The cyclic double-commutator identity also follows directly from associativity of operator composition.

A connection on a vector bundle pulls back along a smooth curve $\gamma$ to differentiate its time-dependent sections. For the tangent bundle, a section $V(t)$ of the pullback tangent bundle hasThe Leibniz rule under a change of frame proves independence of the coordinates. The derivative includes the time dependence of the coefficients, so it is meaningful even at a zero tangent velocity. A vector field along a map need not extend to one ambient field if the curve self-intersects. Parallel fields satisfy $D_{t}V=0$, a linear ordinary differential equation whose unique solutions define parallel transport.

For a torsion-free linear connection on a manifold, the commutator of two covariant derivatives obeys the tensor-product Leibniz rule: mixed first-derivative terms cancel after antisymmetrizing. It vanishes on scalar fields, so ${\Delta }_{ab}(fQ)=f{\Delta }_{ab}Q$; its action is therefore pointwise tensorial. Its vector action defines the Riemann curvature tensor. Applying the commutator to a scalar contraction proves that every covariant tensor index contributes minus the corresponding curvature action, while every contravariant index contributes plus. Torsion-free does not by itself imply metric compatibility.

An algebraic derivation of an algebra over a field is a linear map satisfying the Leibniz rule for its multiplication: $D(xy)=D(x)y+xD(y)$. This definition makes sense even when the multiplication is not associative. A derivation of an algebra uses an associative product; a derivation of a Lie algebra uses the Lie bracket as its product.

Set $v=x^n{e}^{-x}$. Direct differentiation gives $x{v}^{\prime }=(n-x)v$. Apply the Leibniz rule to differentiate $n+1$ times:Hence $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. By Rodrigues' formula, $v^{(n)}=n!e^{-x}{L}_n$, so the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substitution gives the displayed Laguerre differential operator on polynomials eigenvalue equation. This derivation also covers $n=0$.

For smooth $a$ and a point $r$,These follow by applying both sides to a test function and using the Leibniz rule. In particular $(x-r){\delta }_r^{\prime }=-{\delta }_r$.

The Koszul formula uniquely determines $g({\nabla }_{X}Y,Z)$ for all vector fields $X,Y,Z$, so nondegeneracy of the Riemannian metric determines ${\nabla }_{X}Y$. The resulting operation is $C^{\infty }$-linear in $X$, obeys the Leibniz rule in $Y$, is compatible with the metric, and is torsion-free. It is therefore the unique Levi-Civita connection.

If $\theta$ has order at most one, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, which is the Leibniz rule. Thus $\delta$ is a derivation of an algebra. Conversely any derivation satisfies this commutator identity. The sum is direct because a derivation that is multiplication must vanish at $1$.

For the local flow ${\varphi }_t$ of a smooth vector field $X$, pull a tensor field $T$ back to the fixed base point before differentiating. Pullback uses $d{\varphi }_t^{-1}$ on vector factors and the dual of $d{\varphi }_t$ on covector factors. The chain rule gives ${\mathcal{L}}_{X}f=Xf$. In coordinates, $d{\varphi }_t^{-1}Y({\varphi }_t(x))=Y+t(dYX-dXY)+O(t^2)$, so ${\mathcal{L}}_{X}Y=[X,Y]$. Differentiation of pullback preserves tensor contractions and obeys the Leibniz rule for tensor products; hence this is the tensor derivation defining the Lie derivative of a tensor field.

The generalized Leibniz rule extends composition of differential operators to all integer $n$. Define $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. For nonnegative $n$ the sum stops at $r=n$. For $n=-s<0$, its coefficients are $(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$; in particular ${\partial }^{-1}f=f{\partial }^{-1}-f^{\prime }{\partial }^{-2}+f^{\prime \prime }{\partial }^{-3}-\cdots$. Applying $\partial$ on the left cancels consecutive derivative terms and recovers multiplication by $f$, which explains the inverse expansion.

For $m$ functions with sufficiently many derivatives, the $n$th derivative of their product distributes the $n$ differentiations among all factors. If factor $j$ is differentiated $k_j$ times, its term has the multinomial coefficient $n!/(k_1!\cdots k_m!)$, where the nonnegative integers $k_j$ sum to $n$. Induction on the number of factors proves this from the two-factor Leibniz rule: splitting off the final factor gives a binomial coefficient, which multiplies the previous multinomial coefficient to give the one displayed above. Equivalently, count the ways to choose which of the $n$ successive applications of the product rule differentiate each factor.

For the physicists' Hermite polynomials $H_n=(-1{)}^n{e}^{x^2}{D}^n{e}^{-x^2}$, the displayed recurrence holds for $n\ge 1$, while $H_0^{\prime }=0$ separately. Put $u=e^{-x^2}$. Differentiating $u^{\prime }+2xu=0$ exactly $n$ times using the Leibniz rule yieldsDifferentiating the defining product for $H_n$ then givesThe nonnegative integer convention is essential for the derivative-based definition; no negative-order derivative or $H_{-1}$ is needed.

In a frame of a vector bundle, write a vector bundle covariant derivative as $\nabla (ev)=e(dv+\Gamma v)$. The associated horizontal subspace at $(x,v)$ consists of tangent pairs $(\xi ,-{\Gamma }_x(\xi )v)$. Under a frame change $e^{\prime }=eg$, the connection one-form changes by ${\Gamma }^{\prime }=g^{-1}\Gamma g+g^{-1}dg$, and this horizontal equation changes to the corresponding one in the new frame. Thus the subspaces glue to a unique smooth linear connection on a vector bundle. Its vertical projection gives back the original derivative. The Leibniz rule ensures locality, allowing this construction from an operator initially defined on global sections.

For a distribution $u$ and a smooth function $a$, the product $au$ is defined by $⟨au,\phi ⟩=⟨u,a\phi ⟩$. It obeys the Leibniz rule; in one dimension, $(au{)}^{\prime }=a^{\prime }u+a{u}^{\prime }$.