derivations of smooth functions are vector fields


Differential geometry

synthetic differential geometry


from point-set topology to differentiable manifolds

geometry of physics: coordinate systems, smooth spaces, manifolds, smooth homotopy types, supergeometry



smooth space


The magic algebraic facts




tangent cohesion

differential cohesion

graded differential cohesion

id id fermionic bosonic bosonic Rh rheonomic reduced infinitesimal infinitesimal & étale cohesive ʃ discrete discrete continuous * \array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& Rh & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& ʃ &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }


Lie theory, ∞-Lie theory

differential equations, variational calculus

Chern-Weil theory, ∞-Chern-Weil theory

Cartan geometry (super, higher)



On a smooth manifold XX the smooth tangent vector fields vΓ X(TX)v \in \Gamma_X(T X) are geometrically defined as smoothly varying collections of smooth paths γ x: 1X\gamma_x \colon \mathbb{R}^1 \to X through every point xXx \in X (i.e. γ(0)=x\gamma(0) = x), with two such paths regarded as equivalent if their first derivative v xT xXv_x \in T_x X at xx, seen in any chart, coincides.

By pre-composing a smooth function f:X 1f \colon X \longrightarrow \mathbb{R}^1 which such a path, we obtain a function fγ x: 1 1f\circ \gamma_x \;\colon\; \mathbb{R}^1 \to \mathbb{R}^1 from the real line to itself. Therefore its derivative

(D vf)(x)d(fγ) 0 (D_v f)(x) \coloneqq d (f \circ \gamma)_0

may be identified with a real number (measuring the rate of change of ff along γ\gamma at xx to first order). Since γ x\gamma_x, or rather its derivative v xT xXv_x \in T_x X, is assumed to depend smoothly on xx, this defines a new smooth function

D vf:X 1. D_v f \;\colon\; X \longrightarrow \mathbb{R}^1 \,.

Due to the product rule of differentiation, this assignment fD vff \mapsto D_v f is such that for f,gC (X)f,g \in C^\infty(X) two smooth functions, with pointwise product function fgf \cdot g then

D v(fg)=D v(f)g+fD v(g). D_v(f \cdot g) \;=\; D_v(f) \cdot g + f \cdot D_v(g) \,.

This means that D v:C (C)C (X)D_v \;\colon\; C^\infty(C) \to C^\infty(X) is a derivation on the algebra of smooth functions.

Remarkably, all derivations on C (X)C^\infty(X) arise this way, and for a unique vector field vΓ X(TX)v \in \Gamma_X(T X):

{derivations on algebra of smooth functions onX}{smooth tangent vector fields onX} \left\{ \array{ \text{derivations on} \\ \text{algebra of smooth functions} \\ \text{on}\, X } \right\} \;\simeq\; \left\{ \array{ \text{smooth} \\ \text{tangent vector fields} \\ \text{on}\, X } \right\}

This is prop. 1 below. What makes this work is the Hadamard lemma, see the proof below for details.

Notice that the concept of derivations is purely a concept of algebra, with no input from the topology and differential geometry that goes into the definition and construction of smooth manifolds and their tangent bundles. Therefore the identification of smooth vector fields with derivations is an algebraic incarnation of an aspect of differential geometry, an identification comparable two two other such phenomena:

These statements mean that differential geometry has more in common with algebraic geometry than is manifest from the traditional definitions. This serves as the bases for the definition of formal smooth manifolds and the theory of synthetic differential geometry. For more exposition of this relation see at geometry of physics -- supergeometry.

Beware however that related algebraic properties familiar from affine schemes may break in differential geometry: For the Kähler differential forms of C (X)C^\infty(X) to come out as the expected smooth differential forms one needs to refine the plain \mathbb{R}-commutative algebra C nfty(X)C^\nfty(X) to the structure of a smooth algebra. See at Kähler differential forms for discussion of this issue.



Let XX be a smooth manifold.


  1. C (X)C^\infty(X) for its real algebra of smooth functions X 1X \to \mathbb{R}^1 (under pointwise multiplication);

  2. Γ X(TX)\Gamma_X(T X) for the real vector space of smooth tangent vector fields on XX, hence the space of smooth sections of the tangent bundle TXT X of XX, regarded as a smooth vector bundle.

1, Der(C (X))Der(C^\infty(X)) for the real vector space of \mathbb{R}-linear derivations of the \mathbb{R}-algebra C (X)C^\infty(X)


  1. The operation of differentiation with respect to a vector field is a derivation of C (X)C^\infty(X).

  2. Every derivation on C (X)C^\infty(X) arises this way, for a unique vector field.

Hence there is an isomorphism

D ():Γ X(TX)Der(C (X)). D_{(-)} \;\colon\; \Gamma_X(T X) \stackrel{\simeq}{\longrightarrow} Der(C^\infty(X)) \,.

First to discuss that vector fields induce derivations. Let vΓ X(TX)v \in \Gamma_X(T X) be a smooth tangent vector field.

This means first of all that for each xXx \in X there is a smooth function γ x: 1X\gamma_x \;\colon\; \mathbb{R}^1 \longrightarrow X such that

  1. γ x(0)=x\gamma_x(0) = x

  2. dγ x0=v xT xXd \gamma_x_0 = v_x \in T_x X.

Then for f:X 1f \colon X \to \mathbb{R}^1 a smooth function, define a new function

D v(f):X 1 D_v(f) \;\colon\; X \longrightarrow \mathbb{R}^1


(D v(f))(x)d(fγ x) 0=ddt(f(γ x(t)))| t=0. (D_v(f))(x) \coloneqq d (f \circ \gamma_x)_0 = \frac{d}{d t} (f(\gamma_x(t)))\vert_{t = 0} \,.

By the linearity of differentation we have for cc \in \mathbb{R} that D v(cf)=cD v(f)D_v(c \cdot f) = c \cdot D_v(f). Moreover, by the product rule for differentiation we have for f,gC (X)f,g \in C^\infty(X) two smooth functions, that

D v(fg) =ddt((fg)(γ(t)))| t=0 =ddt(f(γ x(t))g(γ x(t)))| t=0 =(ddtf(γ x(t))| t=0)g(γ x(0))+f(γ x(0))(ddtg(γ(t))| t=0) =(D v(f)g+fD v(g))(x). \begin{aligned} D_v( f \cdot g) & = \frac{d}{d t}( (f \cdot g)(\gamma(t)) )\vert_{t = 0} \\ & = \frac{d}{ d t}( f(\gamma_x(t)) \cdot g(\gamma_x(t)) )\vert_{t = 0} \\ & = \left( \frac{d}{d t} f(\gamma_x(t))\vert_{t = 0} \right) \cdot g(\gamma_x(0)) + f(\gamma_x(0)) \cdot \left( \frac{d}{d t} g(\gamma(t))\vert_{t = 0} \right) \\ & = (D_v(f) \cdot g + f \cdot D_v(g))(x) \end{aligned} \,.

Therefore it only remains to check that the function D v(f)D_v(f) is a smooth function in the first place.


{ nϕ iU iX} \left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X \right\}

be an atlas that exhibits the smooth structure of XX, hence an open cover by subsets equipped with a diffeomorphism ϕ i\phi_i from a Cartesian space n\mathbb{R}^n with its standard smooth structure.

By definition of the tangent bundle TXT X, its restriction to any chart of this atlas is given by a diffeomorphism ψ i\psi_i as in the following diagram

n× n ψ i TX| U i ϕ ipr 1 p| U i U i. \array{ \mathbb{R}^n \times \mathbb{R}^n && \underoverset{\simeq}{\psi_i}{\longrightarrow} && T X\vert_{U_i} \\ & {}_{\phi_i \circ \mathllap{pr_1}}\searrow && \swarrow_{p\vert_{U_i}} \\ && U_i } \,.

For D v(f):X 1D_v(f) \colon X \to \mathbb{R}^1 to be continuous and smooth, it is sufficient to see that its restrictions

D v(f)ψ i D_v(f) \circ \psi_i

are continuous and smooth, for all iIi \in I. (For continuity this follows by this example and for differentiability this is clear from the fact that derivatives at any point may be computed in an arbitrary open neighbourhood).

With respect to this chart the tangent vector field vΓ X(TX)v \in \Gamma_X(T X) is given by a smooth function

v ipr 2ψ i 1(v| U i):U i n v_i \;\coloneqq\; pr_2 \circ \psi_i^{-1}(v \vert_{U_i}) \;\colon\; U_i \longrightarrow \mathbb{R}^n

By the identification of tangent vectors on n\mathbb{R}^n with elements of n\mathbb{R}^n, these tangent vectors are represented by straight smooth curves (this remark):

γ x,i:tϕ i 1(x)+tv i(x). \gamma_{x, i} \;\colon\; t \mapsto \phi_i^{-1}(x) + t \cdot v_i(x) \,.

and for f ifϕ i: n 1f_i \coloneqq f \circ \phi_i \;\colon\; \mathbb{R}^n \to \mathbb{R}^1 the restriction of the given smooth function, we have, by the chain rule

ddt(f i(γ x,i(t)))| t=0= k=1 n(v i) k(x)f i(y)y k| y=x. \frac{d}{d t} ( f_i(\gamma_{x,i}(t)) )\vert_{t = 0} = \sum_{k = 1}^n (v_i)^k(x) \cdot \frac{\partial f_i(y)}{\partial y^k}\vert_{y = x} \,.

(Here and in the following we are writing () k(-)^k for the kkth component of an element of n\mathbb{R}^n.)

This is the value of a function at xx with is the sum over the product of two smooth functions, namely the component functions (v i) k(v_i)^k of the smooth section in the given chart, and the derivatives of f if_i (which are functions by the assumption that ff is smooth).

This shows that D v(f)D_v(f) is a smooth function.

Now for the converse. Let DDer(C (X))D \in Der(\C^\infty(X)) be a derivation, we need to show that it arises on each chart ϕ i\phi_i from smooth vector field v iv_i as above.

The Hadamard lemma gives that every smooth function f i: n 1f_i \colon \mathbb{R}^n \to \mathbb{R}^1 may be written for each point x nx \in \mathbb{R}^n as

f(y)=f(yx)+ k=0 n(y kx k)g x,k(y) f(y) = f(y-x) + \sum_{k = 0}^n (y^k - x^k) \cdot g_{x,k}(y)

for smooth functions

{g kC ( n)} k{1,,n} \{g_k \in C^\infty(\mathbb{R}^n)\}_{k \in \{1,\cdots, n\}}


g x,k(x)=f(y)y k| y=x. g_{x,k}(x) = \frac{\partial f(y)}{\partial y^k}\vert_{y = x} \,.

Now by the derivation property we have

D(f)(x) =D(f(0)+ k=1 n(() kx k)g x,k()) = k=1 n((D(() kx k))g x,k()+(() kx k)Dg x,k())(x) = k=1 n(v i) k(x)f(y)y k| y=x. \begin{aligned} D(f)(x) & = D \left( f(0) + \sum_{k = 1}^n ((-)^k - x^k) \cdot g_{x,k}(-) \right) \\ & = \sum_{k = 1}^n \big( {(D( (-)^k - x^k ))} \cdot g_{x,k}(-) + ((-)^k - x^k) \cdot D g_{x,k}(-) \big)(x) \\ & = \sum_{k = 1}^n (v_i)^k(x) \cdot \frac{\partial f(y)}{\partial y^k}\vert_{y = x} \end{aligned} \,.

Here in the last step we defined

(v i) k(x)D(() k)(x) (v_i)^k(x) \coloneqq D( (-)^k )(x)

to be the value at xx of the function which the derivation produces when applied to the kkth coordinate function (which is a smooth function of xx by definition of DD), and we used the above property of the g x,kg_{x,k} as well as that () kx k(-)^k - x^k vanishes at xx.

This shows that the derivation DD comes, in the chart ϕ i\phi_i, from the vector field v iv_i as above.



Revised on July 5, 2017 11:36:46 by Urs Schreiber (