# Poincaré lemma

The Poincaré lemma is a mathematical theorem and was named after the French mathematician Henri Poincaré .

## Exact and closed differential forms

• A differential form of degree is called closed if applies. The outer derivation denotes .${\ displaystyle \ omega}$${\ displaystyle k}$${\ displaystyle \ mathrm {d} \ omega = 0}$${\ displaystyle \ mathrm {d}}$
• A differential form of degree is called exact if there is a differential form such that it holds. The form is called a potential form of${\ displaystyle \ omega}$${\ displaystyle k}$${\ displaystyle (k-1)}$${\ displaystyle \ nu}$${\ displaystyle \ omega = \ mathrm {d} \ nu}$${\ displaystyle \ nu}$${\ displaystyle \ omega.}$

The potential form is not clearly defined, but only "except for re-calibration" (see below).

Because of this , every exact differential form is also closed. The Poincaré lemma specifies conditions under which the opposite statement also applies. In the proof, there is also a generalization of the lemma: An exact part can be split off from every differential form “by construction”. ${\ displaystyle \ mathrm {d} \ circ \ mathrm {d} \ equiv 0}$

## statement

The Poincaré lemma says that every closed differential form defined on a star-shaped open set is exact. ${\ displaystyle U \ subseteq \ mathbb {R} ^ {d}}$

The statement can also be formulated more abstractly as follows: For a star-shaped open set , the -th De Rham cohomology vanishes for all : ${\ displaystyle U \ subseteq \ mathbb {R} ^ {d}}$${\ displaystyle k}$${\ displaystyle k> 0}$

${\ displaystyle \ mathrm {H} _ {\ mathrm {dR}} ^ {k} (U) = 0}$

In the three-dimensional special case, the Poincaré lemma, translated into the language of vector analysis , says that an eddy-free vector field defined on a simply connected area as a gradient of a potential field ( ), a source-free vector field on a convex area through rotation of a vector potential ( ), and a scalar field density ("source density") can be represented as the divergence of a vector field ( ). ${\ displaystyle \ Phi (\ mathbf {r})}$${\ displaystyle k = 1}$ ${\ displaystyle {\ vec {A}} (\ mathbf {r}, t)}$${\ displaystyle k = 2}$${\ displaystyle k = 3}$

## Proof (constructive)

Be the point around which is star-shaped. The Poincaré Lemma gives an explicit form, and although having the formula: any form unity can not necessary provided that a form assign, from which the desired potential shape results in unity: This associated shape can by itself define the following figure: ${\ displaystyle x_ {0} \ in U}$${\ displaystyle U \ subset \ mathbb {R} ^ {n}}$${\ displaystyle (k-1)}$${\ displaystyle k}$${\ displaystyle \ textstyle \ omega ^ {k} = \ sum \ omega _ {I} {\ rm {d}} x_ {I}}$${\ displaystyle (k-1)}$${\ displaystyle P ^ {k-1} (\ omega ^ {k})}$

${\ displaystyle P ^ {k-1} (\ omega ^ {k}) (x) \,: = \, \ sum _ {i_ {1} <\ cdots .

(The roof symbol in the -th column on the right-hand side means that the corresponding differential is omitted.) ${\ displaystyle i _ {\ alpha}}$

Now one shows directly that the following identity applies: what formally corresponds to the product rule of differentiation and splits the properties represented by into two parts, of which the second has the property sought. ${\ displaystyle \ omega ^ {k} \ equiv \ mathrm {P} ^ {k} ({\ rm {d}} \ omega ^ {k}) + {\ rm {d}} {\ mathrm {P} ^ {k-1} (} \ omega ^ {k} \ mathrm {)} \ ,,}$${\ displaystyle \ omega ^ {k}}$

Because of the prerequisite and because of this, this initially applies without restricting the generality even without the foremost of the right-hand side, namely because the requirement only considers the form at the zero point, so that, as with the total differential, a function from up to so-called Calibration transformations (see below) can also be deduced. ${\ displaystyle {\ rm {d}} \ omega ^ {k} \ equiv 0}$${\ displaystyle \ mathrm {d} \ circ \ mathrm {d} = 0}$${\ displaystyle 0 \ equiv {\ mathrm {d}} P ^ {k} ({\ mathrm {d} {\ omega} ^ {k}} \ to 0).}$${\ displaystyle \ mathrm {d}}$${\ displaystyle \ mathrm {d} \ omega ^ {k} \ to 0}$${\ displaystyle \ mathrm {d} P ^ {k}}$${\ displaystyle \ mathrm {d} P ^ {k} = 0}$${\ displaystyle \ mathrm {P} ^ {k} = 0}$

This leaves only the last term of the above identity, and the required statement follows: with${\ displaystyle \ omega ^ {k} \ equiv \ mathrm {d} \ eta ^ {k-1},}$${\ displaystyle \ eta ^ {k-1} \,: = \, \ mathrm {P} ^ {k-1} (\ omega ^ {k}) \ ,.}$

The given identity also generalizes Poincaré's lemma by breaking down any differential form into an inexact (“anholonomic”) and an exact (“holonomic”) part (the bracketed names correspond to the so-called constraining forces in analytical mechanics ). At the same time, it corresponds to the decomposition of any vector field into a vortex part and a source part. ${\ displaystyle \ omega}$

In the language of homological algebra , a contracting homotopy , e.g. B. contracted to the central point of the star-shaped area considered here . ${\ displaystyle P}$

## Re-calibration

What is so defined is not the only form whose external differential is. All others differ from one another by the differential of a -form: If and are two such -forms, then there exists a -form such that it holds. ${\ displaystyle \ eta ^ {k-1}}$${\ displaystyle (k-1)}$${\ displaystyle {\ omega} ^ {k}}$${\ displaystyle (k-2)}$${\ displaystyle \ eta _ {2} ^ {k-1}}$${\ displaystyle \ eta _ {1} ^ {k-1}}$${\ displaystyle (k-1)}$${\ displaystyle (k-2)}$${\ displaystyle \ xi ^ {k-2}}$${\ displaystyle \ eta _ {2} ^ {k-1} = \ eta _ {1} ^ {k-1} + \ mathrm {d} \ xi ^ {k-2}}$

The addition is also known as calibration transformation or re-calibration of . ${\ displaystyle + \, \ mathrm {d} \ xi ^ {k-2}}$${\ displaystyle \ eta _ {1} ^ {k-1}}$

## Application in electrodynamics

The case of a magnetic field generated by a stationary current is known from electrodynamics , with the so-called vector potential . This case corresponds , the star-shaped area being the . The vector of the current density is and corresponds to the current shape. The same applies to the magnetic field : it corresponds to the magnetic flux shape and can be derived from the vector potential:, or . The vector potential corresponds to the potential form. The closeness of the magnetic flux form corresponds to the absence of sources of the magnetic field ${\ displaystyle {\ vec {A}} (\ mathbf {r}) \ ,.}$${\ displaystyle k = 2}$${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle {\ vec {j}}}$${\ displaystyle \ mathbf {I}: = j_ {1} (x, y, z) {\ rm {d}} x_ {2} \ wedge {\ rm {d}} x_ {3} + j_ {2} (x, y, z) {\ rm {d}} x_ {3} \ wedge {\ rm {d}} x_ {1} + j_ {3} (x, y, z) {\ rm {d}} x_ {1} \ wedge {\ rm {d}} x_ {2} \ ,.}$${\ displaystyle {\ vec {B}}}$ ${\ displaystyle \ Phi _ {B}: = B_ {1} {\ rm {d}} x_ {2} \ wedge {\ rm {d}} x_ {3} + \ dots}$${\ displaystyle \ textstyle {\ vec {B}} = \ operatorname {rot} {\ vec {A}} = \ left ({\ tfrac {\ partial A_ {3}} {\ partial x_ {2}}} - {\ tfrac {\ partial A_ {2}} {\ partial x_ {3}}}, {\ tfrac {\ partial A_ {1}} {\ partial x_ {3}}} - {\ tfrac {\ partial A_ { 3}} {\ partial x_ {1}}}, {\ tfrac {\ partial A_ {2}} {\ partial x_ {1}}} - {\ tfrac {\ partial A_ {1}} {\ partial x_ { 2}}} \ right) ^ {t}}$${\ displaystyle \ Phi _ {B} = {\ rm {d}} \ mathbf {A}}$${\ displaystyle {\ vec {A}}}$${\ displaystyle \ mathbf {A}: = A_ {1} {\ rm {d}} x_ {1} + A_ {2} {\ rm {d}} x_ {2} + A_ {3} {\ rm { d}} x_ {3} \ ,.}$${\ displaystyle (\ operatorname {div} {\ vec {B}} \ equiv 0 \,).}$

Using the Coulomb calibration or as appropriate, the following then applies to i = 1,2,3 ${\ displaystyle \ operatorname {div} {\ vec {A}} {\ stackrel {!} {=}} 0}$${\ displaystyle \ operatorname {div} {\ vec {j}} {\ stackrel {!} {=}} 0}$

${\ displaystyle A_ {i} ({\ vec {r}}) = \ int {\ frac {\ mu _ {0} j_ {i} ({\ vec {r}} ^ {\, '}) \, \, dx_ {1} 'dx_ {2}' dx_ {3} '} {4 \ pi | {\ vec {r}} - {\ vec {r}} ^ {\,'} |}} \ ,, }$

there is a natural constant , the so-called magnetic field constant . ${\ displaystyle \ mu _ {0}}$

In this equation u. a. remarkable that it fully corresponds to a well-known formula for the electric field , the Coulomb potential of a given charge distribution with density . At this point it is already assumed that ${\ displaystyle {\ vec {E}}}$ ${\ displaystyle \, \ phi (x_ {1}, x_ {2}, x_ {3})}$${\ displaystyle \ rho (x_ {1}, x_ {2}, x_ {3})}$

• ${\ displaystyle {\ vec {E}}}$and or${\ displaystyle {\ vec {B}}}$
• ${\ displaystyle \ rho}$and as well${\ displaystyle {\ vec {j}}}$
• ${\ displaystyle \, \ phi}$ and ${\ displaystyle {\ vec {A}}}$

can be summarized and that the relativistic invariance of Maxwell's electrodynamics results from it, see also electrodynamics .

If you give up the condition of stationarity , the time argument must be added to the space coordinates on the left side of the above equation , while on the right side the so-called "retarded time" must be added. As before, it is integrated using the three spatial coordinates . After all, the speed of light is in a vacuum . ${\ displaystyle A_ {i}}$${\ displaystyle t}$${\ displaystyle j_ {i} '}$${\ displaystyle t ': = t - {\ tfrac {| {\ vec {r}} - {\ vec {r}} ^ {\,'} |} {c}}}$${\ displaystyle {\ vec {r}} ^ {\, '}}$${\ displaystyle c}$

## Application in continuum mechanics

In continuum mechanics , the lemma is applied to tensors . B. is needed for the establishment of the compatibility conditions . The starting point is the lemma in the formulation:

 ${\ displaystyle \ operatorname {rot} {\ vec {u}} = {\ hat {e}} _ {k} \ times {\ frac {\ partial {\ vec {u}}} {\ partial x_ {k} }} = {\ vec {0}} \ quad \ rightarrow \ quad \ exists \ varphi \ colon {\ vec {u}} = \ operatorname {grad} \ varphi}$ (I)

The operator "grad" forms the gradient , the vectors are the standard basis of the Cartesian coordinate system with coordinates and Einstein's summation convention was applied, according to which indices occurring twice in a product, here k, are to be summed from one to three, which is also the case in The following should be practiced. ${\ displaystyle {\ hat {e}} _ {1,2,3}}$${\ displaystyle x_ {1,2,3}}$

Let us now be given a tensor field whose row vectors are combined with the dyadic product “ ” to form the tensor. The rotation of the transposed tensor vanishes ${\ displaystyle \ mathbf {T} = {\ hat {e}} _ {i} \ otimes {\ vec {t}} _ {i}}$${\ displaystyle {\ vec {t}} _ {1,2,3}}$${\ displaystyle \ otimes}$

${\ displaystyle \ operatorname {rot} (\ mathbf {T} ^ {\ top}): = {\ hat {e}} _ {k} \ times {\ frac {\ partial} {\ partial x_ {k}} } ({\ vec {t}} _ {i} \ otimes {\ hat {e}} _ {i}) = \ left ({\ hat {e}} _ {k} \ times {\ frac {\ partial {\ vec {t}} _ {i}} {\ partial x_ {k}}} \ right) \ otimes {\ hat {e}} _ {i} = \ mathbf {0} \ quad \ rightarrow \ quad { \ hat {e}} _ {k} \ times {\ frac {\ partial {\ vec {t}} _ {i}} {\ partial x_ {k}}} = {\ vec {0}} \ ,, \ quad i = 1,2,3 \ ,,}$

so that every row vector is rotation-free. Then there is a scalar field for every row vector , the gradient of which it is: ${\ displaystyle u_ {i}}$

${\ displaystyle {\ vec {t}} _ {i} = \ operatorname {grad} u_ {i} \ quad \ rightarrow \ quad \ mathbf {T} = {\ hat {e}} _ {i} \ otimes { \ vec {t}} _ {i} = {\ hat {e}} _ {i} \ otimes \ operatorname {grad} u_ {i} = \ operatorname {grad} {\ vec {u}} \ ,,}$

because the gradient of the vector is formed according to: ${\ displaystyle {\ vec {u}}: = u_ {i} {\ hat {e}} _ {i}}$

${\ displaystyle \ operatorname {grad} {\ vec {u}}: = {\ frac {\ partial u_ {i}} {\ partial x_ {k}}} {\ hat {e}} _ {i} \ otimes {\ hat {e}} _ {k} = {\ hat {e}} _ {i} \ otimes {\ frac {\ partial u_ {i}} {\ partial x_ {k}}} {\ hat {e }} _ {k} = {\ hat {e}} _ {i} \ otimes \ operatorname {grad} u_ {i} \ ,.}$

Thus the second form of the lemma applies:

 ${\ displaystyle \ operatorname {rot} (\ mathbf {T} ^ {\ top}) = \ mathbf {0} \ quad \ rightarrow \ quad \ exists {\ vec {u}} \ colon \ mathbf {T} = \ operatorname {grad} {\ vec {u}}}$ (II)

If the trace of the tensor also disappears, then the vector field is free of divergence:

${\ displaystyle \ operatorname {Sp} (\ mathbf {T}) = \ operatorname {Sp} ({\ hat {e}} _ {i} \ otimes \ operatorname {grad} u_ {i}) = {\ hat { e}} _ {i} \ cdot {\ frac {\ partial u_ {i}} {\ partial x_ {k}}} {\ hat {e}} _ {k} = {\ frac {\ partial u_ {i }} {\ partial x_ {i}}} = \ operatorname {div} {\ vec {u}} = 0 \ ,.}$

In this case, the Kronecker delta δ ij is used to calculate :

{\ displaystyle {\ begin {aligned} \ operatorname {rot} ({\ vec {u}} \ times \ mathbf {I}) = & {\ hat {e}} _ {k} \ times {\ frac {\ partial} {\ partial x_ {k}}} [u_ {i} {\ hat {e}} _ {i} \ times ({\ hat {e}} _ {j} \ otimes {\ hat {e}} _ {j})] = {\ frac {\ partial u_ {i}} {\ partial x_ {k}}} [{\ hat {e}} _ {k} \ times ({\ hat {e}} _ {i} \ times {\ hat {e}} _ {j})] \ otimes {\ hat {e}} _ {j} = {\ frac {\ partial u_ {i}} {\ partial x_ {k} }} (\ delta _ {jk} {\ hat {e}} _ {i} - \ delta _ {ik} {\ hat {e}} _ {j}) \ otimes {\ hat {e}} _ { j} \\ = & {\ frac {\ partial u_ {i}} {\ partial x_ {j}}} {\ hat {e}} _ {i} \ otimes {\ hat {e}} _ {j} - {\ frac {\ partial u_ {i}} {\ partial x_ {i}}} {\ hat {e}} _ {j} \ otimes {\ hat {e}} _ {j} = \ operatorname {grad } {\ vec {u}} \ end {aligned}}}

and the tensor is skew symmetric : ${\ displaystyle \ mathbf {W}: = {\ vec {u}} \ times \ mathbf {I}}$

${\ displaystyle ({\ vec {u}} \ times \ mathbf {I}) ^ {\ top} = (u_ {i} {\ hat {e}} _ {i} \ times {\ hat {e}} _ {j} \ otimes {\ hat {e}} _ {j}) ^ {\ top} = \ epsilon _ {ijk} u_ {i} {\ hat {e}} _ {j} \ otimes {\ hat {e}} _ {k} = - u_ {i} {\ hat {e}} _ {i} \ times {\ hat {e}} _ {k} \ otimes {\ hat {e}} _ {k } = - {\ vec {u}} \ times \ mathbf {I} \ ,.}$

Inside is the permutation symbol . The third form of the lemma follows: ${\ displaystyle \ epsilon _ {ijk} = ({\ hat {e}} _ {i} \ times {\ hat {e}} _ {j}) \ cdot {\ hat {e}} _ {k}}$

 ${\ displaystyle \ operatorname {rot} (\ mathbf {T} ^ {\ top}) = \ mathbf {0} \ quad {\ text {and}} \ quad \ operatorname {Sp} (\ mathbf {T}) = 0 \ quad \ rightarrow \ quad \ exists \ mathbf {W} \ colon \ mathbf {T} = \ operatorname {red} \ mathbf {W} \ quad {\ text {with}} \ quad \ mathbf {W} = - \ mathbf {W} ^ {\ top}}$ (III)

In the literature, a rotation operator is also used, which directly forms the rotation of the row vectors:

${\ displaystyle \ operatorname {\ tilde {red}} \ mathbf {T} = \ operatorname {red} (\ mathbf {T} ^ {\ top}) \ ,.}$

With this operator:

${\ displaystyle {\ begin {array} {rrcll} {\ textsf {II}}: & \ operatorname {\ tilde {red}} \ mathbf {T} = \ mathbf {0} & \ rightarrow & \ exists {\ vec {u}} \ colon & \ mathbf {T} = \ operatorname {grad} {\ vec {u}} \\ {\ textsf {III}}: & \ operatorname {\ tilde {red}} \ mathbf {T} = \ mathbf {0} \ quad {\ text {and}} \ quad \ operatorname {Sp} (\ mathbf {T}) = 0 & \ rightarrow & \ exists \ mathbf {V} \ colon & \ mathbf {T} = \ operatorname {\ tilde {rot}} \ mathbf {V} \ quad {\ text {with}} \ quad \ mathbf {V} ^ {\ top} = - \ mathbf {V} \ end {array}}}$

## literature

• Otto Forster : Analysis. Volume 3: Integral calculus in R n with applications. 4th edition. Vieweg + Teubner, Braunschweig et al. 2007, ISBN 978-3-528-37252-1 .
• John M. Lee: Introduction to Smooth Manifolds (= Graduate Texts in Mathematics 218). Springer-Verlag, New York NY et al. 2003, ISBN 0-387-95448-1 .
• C. Truesdell: Solid Mechanics II in S. Flügge (Ed.): Handbook of Physics , Volume VIa / 2. Springer-Verlag, 1972, ISBN 3-540-05535-5 , ISBN 0-387-05535-5 .