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}$${\ displaystyle (k1)}$${\ displaystyle \ nu}$${\ displaystyle \ omega = \ mathrm {d} \ nu}$${\ displaystyle \ nu}$${\ displaystyle \ omega.}$
The potential form is not clearly defined, but only "except for recalibration" (see below).
Because of this , every exact differential form is also closed. The Poincaré lemma specifies conditions under which the opposite statement also applies. In the proof, there is also a generalization of the lemma: An exact part can be split off from every differential form “by construction”.
${\ displaystyle \ mathrm {d} \ circ \ mathrm {d} \ equiv 0}$
statement
The Poincaré lemma says that every closed differential form defined on a starshaped open set is exact.
${\ displaystyle U \ subseteq \ mathbb {R} ^ {d}}$
The statement can also be formulated more abstractly as follows: For a starshaped open set , the th De Rham cohomology vanishes for all :
${\ displaystyle U \ subseteq \ mathbb {R} ^ {d}}$${\ displaystyle k}$${\ displaystyle k> 0}$
 ${\ displaystyle \ mathrm {H} _ {\ mathrm {dR}} ^ {k} (U) = 0}$
In the threedimensional special case, the Poincaré lemma, translated into the language of vector analysis , says that an eddyfree vector field defined on a simply connected area as a gradient of a potential field ( ), a sourcefree vector field on a convex area through rotation of a vector potential ( ), and a scalar field density ("source density") can be represented as the divergence of a vector field ( ).
${\ displaystyle \ Phi (\ mathbf {r})}$${\ displaystyle k = 1}$ ${\ displaystyle {\ vec {A}} (\ mathbf {r}, t)}$${\ displaystyle k = 2}$${\ displaystyle k = 3}$
Proof (constructive)
Be the point around which is starshaped. 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 (k1)}$${\ displaystyle k}$${\ displaystyle \ textstyle \ omega ^ {k} = \ sum \ omega _ {I} {\ rm {d}} x_ {I}}$${\ displaystyle (k1)}$${\ displaystyle P ^ {k1} (\ omega ^ {k})}$

${\ displaystyle P ^ {k1} (\ omega ^ {k}) (x) \,: = \, \ sum _ {i_ {1} <\ cdots <i_ {k}} \ sum _ {\ alpha = 1} ^ {k} ( 1) ^ {\ alpha 1} {\ Big (} \ int _ {0} ^ {1} t ^ {k1} \ omega _ {i_ {1} \ cdots i_ {k}} (x_ {0} + tx) dt {\ Big)} x ^ {i _ {\ alpha}} {\ rm {d}} x ^ {i_ {1}} \ wedge \ cdots \ wedge { \ widehat {{\ rm {d}} x ^ {i _ {\ alpha}}}} \ wedge \ cdots \ wedge {\ rm {d}} x ^ {i_ {k}}}$.
(The roof symbol in the th column on the righthand 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} ^ {k1} (} \ omega ^ {k} \ mathrm {)} \ ,,}$${\ displaystyle \ omega ^ {k}}$
Because of the prerequisite and because of this, this initially applies without restricting the generality even without the foremost of the righthand side, namely because the requirement only considers the form at the zero point, so that, as with the total differential, a function from up to socalled Calibration transformations (see below) can also be deduced.
${\ displaystyle {\ rm {d}} \ omega ^ {k} \ equiv 0}$${\ 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 ^ {k1},}$${\ displaystyle \ eta ^ {k1} \,: = \, \ mathrm {P} ^ {k1} (\ omega ^ {k}) \ ,.}$
The given identity also generalizes Poincaré's lemma by breaking down any differential form into an inexact (“anholonomic”) and an exact (“holonomic”) part (the bracketed names correspond to the socalled constraining forces in analytical mechanics ). At the same time, it corresponds to the decomposition of any vector field into a vortex part and a source part.
${\ displaystyle \ omega}$
In the language of homological algebra , a contracting homotopy , e.g. B. contracted to the central point of the starshaped area considered here .
${\ displaystyle P}$
Recalibration
What is so defined is not the only form whose external differential is. All others differ from one another by the differential of a form: If and are two such forms, then there exists a form such that it holds.
${\ displaystyle \ eta ^ {k1}}$${\ displaystyle (k1)}$${\ displaystyle {\ omega} ^ {k}}$${\ displaystyle (k2)}$${\ displaystyle \ eta _ {2} ^ {k1}}$${\ displaystyle \ eta _ {1} ^ {k1}}$${\ displaystyle (k1)}$${\ displaystyle (k2)}$${\ displaystyle \ xi ^ {k2}}$${\ displaystyle \ eta _ {2} ^ {k1} = \ eta _ {1} ^ {k1} + \ mathrm {d} \ xi ^ {k2}}$
The addition is also known as calibration transformation or recalibration of .
${\ displaystyle + \, \ mathrm {d} \ xi ^ {k2}}$${\ displaystyle \ eta _ {1} ^ {k1}}$
Application in electrodynamics
The case of a magnetic field generated by a stationary current is known from electrodynamics , with the socalled vector potential . This case corresponds , the starshaped area being the . The vector of the current density is and corresponds to the current shape.
The same applies to the magnetic field : it corresponds to the magnetic flux shape and can be derived from the vector potential:, or . The vector potential corresponds to the potential form.
The closeness of the magnetic flux form corresponds to the absence of sources of the magnetic field ${\ displaystyle {\ vec {A}} (\ mathbf {r}) \ ,.}$${\ 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 socalled magnetic field constant .
${\ displaystyle \ mu _ {0}}$
In this equation u. a. remarkable that it fully corresponds to a wellknown 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 socalled "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 rotationfree. 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 9783528372521 .
 John M. Lee: Introduction to Smooth Manifolds (= Graduate Texts in Mathematics 218). SpringerVerlag, New York NY et al. 2003, ISBN 0387954481 .
 C. Truesdell: Solid Mechanics II in S. Flügge (Ed.): Handbook of Physics , Volume VIa / 2. SpringerVerlag, 1972, ISBN 3540055355 , ISBN 0387055355 .