by Jip
Tags: differential, geometry
 P: 11 Hi, I read in Padmanabhan's book that $\nabla_a J^a=0$ implies that there exists an antisymetric tensor P such that $J^a= \nabla_b P^{ba}$. What's the name of the theorem? Any reference? Thanks
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 Quote by jedishrfu having trouble reading your latex rendering. Can you fix it?
$\nabla_a J^a=0$ implies that there exists an antisymetric tensor $P$ such that $J^a= \nabla_b P^{ba}$

HW Helper
P: 11,915

 Quote by Jip Hi, I read in Padmanabhan's book that \nabla_a J^a=0 implies that there exists an antisymetric tensor P such that J^a= nabla_b P^{ba}. What's the name of the theorem? Any reference? Thanks
It should be the Poincaré's lemma for the codifferential. You might check Nakahara's text.
C. Spirit
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P: 5,585
 Quote by Jip Hi, I read in Padmanabhan's book that \nabla_a J^a=0 implies that there exists an antisymetric tensor P such that J^a= nabla_b P^{ba}. What's the name of the theorem? Any reference? Thanks
Check out appendix B in Wald's "General Relativity" and also problem 5 in chapter 4. It is essentially the converse of the poincare lemma. The lemma itself comes out of a combination of differential and algebraic topology; for its proof you would need to consult a proper text on differentiable manifolds.
 Sci Advisor HW Helper P: 11,915 It's not the converse. It's the direct lemma. http://en.wikipedia.org/wiki/Poincar...ar.C3.A9_lemma Because d2 = 0, any exact form is automatically closed. The question of whether every closed form is exact depends on the topology of the domain of interest. On a contractible domain, every closed form is exact by the Poincaré lemma. More general questions of this kind on an arbitrary differentiable manifold are the subject of de Rham cohomology, that allows one to obtain purely topological information using differential methods.
 C. Spirit Sci Advisor Thanks P: 5,585 The result $\partial _{a }v^{a} = 0 \Rightarrow \exists P^{ab} = -P^{ba}:v^{a}=\partial _{b}P^{ab}$ is gotten by applying a consequence of the poincare lemma. The result needed by the OP comes from the fact that if $d\alpha = 0$ then locally $\exists \beta :\alpha = d\beta$. Maybe converse wasn't the word to use here if that's what you are saying dexter; Wald does use that word but Lee doesn't so I don't know what to say other than that Wald might not be using it in the logical sense but rather in an informal/literal sense of the word (EDIT: Lee proves it and Wald doesn't - for the OP's interest)
P: 2,059
 Quote by WannabeNewton Check out appendix B in Wald's "General Relativity" and also problem 5 in chapter 4. It is essentially the converse of the poincare lemma. The lemma itself comes out of a combination of differential and algebraic topology; for its proof you would need to consult a proper text on differentiable manifolds.
It seems to me that the converse is just a generalization of the various 3D vector equations:
• $\vec{\nabla} \times (\vec{\nabla} \Phi) = 0$
$\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0$

The converses in the 3D case are:
• If $\vec{F}$ is a vector field such that $\vec{\nabla} \times \vec{F}= 0$, then $\vec{F} = \nabla \Phi$ for some scalar field $\Phi$.
• If $\vec{F}$ is a vector field such that $\vec{\nabla} \cdot \vec{F}= 0$, then $\vec{F} = \nabla \times \vec{A}$ for some vector field $\vec{A}$.

I think that these cases follow from Gauss' theorem and
Mentor
P: 18,231
 Quote by stevendaryl It seems to me that the converse is just a generalization of the various 3D vector equations:$\vec{\nabla} \times (\vec{\nabla} \Phi) = 0$ $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{A}) = 0$ The converses in the 3D case are: If $\vec{F}$ is a vector field such that $\vec{\nabla} \times \vec{F}= 0$, then $\vec{F} = \nabla \Phi$ for some scalar field $\Phi$. If $\vec{F}$ is a vector field such that $\vec{\nabla} \cdot \vec{F}= 0$, then $\vec{F} = \nabla \times \vec{A}$ for some vector field $\vec{A}$. I think that these cases follow from Gauss' theorem and
Yeah, the converse is just that $d^2=0$, where d is the exterior derivative. This is of course crucial for establishing the De Rham cohomology.
 Sci Advisor P: 897 Some weeks ago there was a question about this also here. You can check Tomas Ortin's book "Gravity and Strings" on this theorem, it probably states its name.
 Sci Advisor HW Helper P: 11,915 I couldn't find it in Ortin (excellent book, btw).
HW Helper
P: 11,915
 Quote by WannabeNewton Check out appendix B in Wald's "General Relativity" and also problem 5 in chapter 4. It is essentially the converse of the poincare lemma. The lemma itself comes out of a combination of differential and algebraic topology; for its proof you would need to consult a proper text on differentiable manifolds.
I don't use Wald for a mathematical reference. He's picked the 'Poincaré lemma' part from Flanders's text on Differential Forms*. But merely stating that the exterior differential is nilpotent to the second order is not an interesting/difficult result, but rather

<Poincare lemma. Let U be an open ball in E and let A be a differential
form of degree >= 1 on U such that dA = 0. Then there exists a differential
form B on U such that dB = A.>

This is the mathematical standard result picked up from <Serge Lang, Differential Manifolds, Springer Verlag, 1985>.

Let's go to <Spivak, Calculus on Manifolds, Addison-Wesley> Page 94:

<4-11 Theorem (Poincare Lemma). If $A\subset \mathbb{R}^{n}$ is an open
set star-shaped with respect to 0, then every closed form on A
is exact.>

*From Flanders's text, his first words from this preface to the first (1963) edition (quoted by Wald).

<Last spring the author gave a series of lectures on exterior differential
forms to a group of faculty members and graduate students from the Purdue
Engineering Schools. The material that was covered in these lectures is
presented here in an expanded version. The book is aimed primarily at
engineers and physical scientists in the hope of making available to them new
tools of very great power in modern mathematics
.>

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