Calculation in proof of Poincare's Lemma

In summary, the (0, p) tensor A_{j_{1}...j_{p}} is defined in a star shaped region of point P with coordinates x^1 = ... = x^n = 0. During the proof of Poincare's lemma, the book uses the chain rule to show that \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l}\frac{\partial(tx^l)}{\partial x^j} = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l
  • #1
oblixps
38
0
[tex] A_{j_{1}...j_{p}} [/tex] is a (0, p) tensor defined in a star shaped region of some point P where the coordinates [tex] x^1 = ... = x^n = 0 [/tex].

in the course of proving Poincare's lemma my book does the following: [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l}\frac{\partial(tx^l)}{\partial x^j} = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} t\delta^{l}_{j} = t\frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} [/tex].

what I'm confused about is why didn't the book do [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial (tx^l)}\frac{\partial(tx^l)}{\partial x^j} [/tex].

what happened to that t in the "denominator" of the first fraction in the chain rule?
 
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  • #2
i was looking around on google and i ran across this related result in some lecture slides.

letting g(t) = f(tx, ty) and using the chain rule:

[tex] g'(t) = (\frac{\partial f}{\partial x})(tx, ty) * x + (\frac{\partial f}{\partial y})(tx, ty) * y [/tex].

once again i am confused on why they wrote [tex] \frac{\partial f}{\partial x} [/tex] instead of [tex] \frac{\partial f}{\partial (tx)} [/tex].
 
  • #3
At the risk of getting an infraction from one of our moderators, I'm going to bump this thread as the OP has waited a bit and posted more info yet hasn't been helped. If possible I'd help myself but alas, this problem is out of my league.
 
  • #4
Hi oblixps, :)

oblixps said:
[tex] A_{j_{1}...j_{p}} [/tex] is a (0, p) tensor defined in a star shaped region of some point P where the coordinates [tex] x^1 = ... = x^n = 0 [/tex].

in the course of proving Poincare's lemma my book does the following: [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l}\frac{\partial(tx^l)}{\partial x^j} = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} t\delta^{l}_{j} = t\frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} [/tex].

what I'm confused about is why didn't the book do [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial (tx^l)}\frac{\partial(tx^l)}{\partial x^j} [/tex].

what happened to that t in the "denominator" of the first fraction in the chain rule?

Can you please tell me what your book is...

oblixps said:
i was looking around on google and i ran across this related result in some lecture slides.

letting g(t) = f(tx, ty) and using the chain rule:

[tex] g'(t) = (\frac{\partial f}{\partial x})(tx, ty) * x + (\frac{\partial f}{\partial y})(tx, ty) * y [/tex].

once again i am confused on why they wrote [tex] \frac{\partial f}{\partial x} [/tex] instead of [tex] \frac{\partial f}{\partial (tx)} [/tex].

... and the web-link where you found the above statement.

Kind Regards,
Sudharaka.
 
  • #5
thanks for the reply.

the book I'm using is "Tensors, Differential Forms, and Variational Principles" by Lovelock and Rund. Just in case you have access to a copy, it should be on page 143.

and the lecture slides I referred to are from: http://www.math.upenn.edu/~ryblair/Math 600/papers/Lec1.pdf

the result i posted is on slide 22 and is what starts off one of the proofs.
 

FAQ: Calculation in proof of Poincare's Lemma

What is Poincare's Lemma?

Poincare's Lemma is a fundamental theorem in mathematics that states that under certain conditions, a differential form on a smooth manifold can be locally expressed as the exterior derivative of another form. It is an important tool in the study of differential geometry and topology.

What is the significance of Poincare's Lemma in mathematics?

Poincare's Lemma has numerous applications in mathematics, including in the fields of differential equations, differential topology, and algebraic topology. It is also used in the development of the de Rham cohomology theory, which is a powerful tool for studying the topology of manifolds.

How is Poincare's Lemma related to the concept of closed and exact forms?

Poincare's Lemma states that a differential form is closed if and only if it is exact. This means that a closed form can always be locally expressed as the exterior derivative of another form, and an exact form can always be locally integrated to obtain a potential function.

What are the conditions required for Poincare's Lemma to hold?

The main condition for Poincare's Lemma to hold is that the manifold must be contractible, meaning that any closed loop on the manifold can be continuously shrunk to a point. Additionally, the form must be defined on a smooth manifold and must satisfy certain smoothness and integrability conditions.

How is Poincare's Lemma used in the proof of the de Rham theorem?

The de Rham theorem is a generalization of Poincare's Lemma, stating that on a contractible manifold, any closed form is exact. Poincare's Lemma is used in the proof of this theorem by showing that the existence of a closed form implies the existence of a potential function, which in turn implies the exactness of the form.

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