First proof of Poincaré Lemma
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I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !
reference-request differential-geometry manifolds homology-cohomology
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add a comment |
$begingroup$
I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !
reference-request differential-geometry manifolds homology-cohomology
$endgroup$
2
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
1
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
2
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52
add a comment |
$begingroup$
I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !
reference-request differential-geometry manifolds homology-cohomology
$endgroup$
I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !
reference-request differential-geometry manifolds homology-cohomology
reference-request differential-geometry manifolds homology-cohomology
asked Nov 7 '13 at 16:30
user90041user90041
1,7791233
1,7791233
2
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
1
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
2
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52
add a comment |
2
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
1
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
2
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52
2
2
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
1
1
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
2
2
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ are exact. To do so we construct a linear operator
$$alpha:Omega^p(mathbb R^n) rightarrow Omega^{p-1}(mathbb R^n) $$
s.t. $$dalpha+alpha d=1.$$
Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define
$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$
where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)
$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$
and we are done. The diff. form $theta:=t(omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* omega$ where $M_t : mathbb R^n to mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p omega(tx) = M_t^* omega (x)$.
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$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
add a comment |
$begingroup$
We describe a linear functional $alpha : Omega^p mathbb R^n to Omega^{p-1} mathbb R^n$. The space alternating $p$-linear functions on $mathbb R^n$ has dimension $n choose p$, and you can write the basis as $dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$ where $1 leq i_1 < i_2 < cdots < i_p leq n$.
If $I = (i_1, i_2, cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.
Given a $p$-form $f dx_I$ with $f: mathbb R^n to mathbb R$ we define $alpha$ linearly, by $alpha (f dx_I) = 0$ if $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ if $I in I_1$. You can think of $alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that
$$ d(alpha(omega)) + alpha(d omega) = omega - pi^*(i^* omega)$$
for every $p$-form $omega$. Here $i : mathbb R^{n-1} to mathbb R^n$ is
the inclusion $i(x_2,cdots,x_n) = (0,x_2,cdots,x_n)$ and $pi : mathbb R^n to mathbb R^{n-1}$ is projection $pi(x_1,x_2,cdots,x_n) = (x_2,cdots,x_n)$.
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.
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$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
add a comment |
$begingroup$
The above calculation in more detail. Define $h_k:Omega^p(mathbb{R}^n)rightarrow Omega^{p-1}(mathbb{R}^n)$ by,
begin{eqnarray}
h_k(omega)(x)=int_0^1t^{p-1}i_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. To see this note that $X^i(t)=frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)&=&int_0^1t^{p-1}Big(d(i_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x
end{eqnarray}
Now we can compute the pullback explicitly,
begin{eqnarray}
[((Phi_s)^*omega)_x]_{k_1...k_p}=frac{partial y^{i_1}}{partial x^{k_1}}...frac{partial y^{i_p}}{partial x^{k_p}}[omega_{Phi_s(x)}]_{i_1...i_p}=e^{ps}[omega_{e^sx}]_{k_1...k_p}
end{eqnarray}
Due to linearity we can restrict to the simple case where $omega(x)=f(x)dx^I$. Therefore,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ are exact. To do so we construct a linear operator
$$alpha:Omega^p(mathbb R^n) rightarrow Omega^{p-1}(mathbb R^n) $$
s.t. $$dalpha+alpha d=1.$$
Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define
$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$
where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)
$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$
and we are done. The diff. form $theta:=t(omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* omega$ where $M_t : mathbb R^n to mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p omega(tx) = M_t^* omega (x)$.
$endgroup$
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
add a comment |
$begingroup$
We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ are exact. To do so we construct a linear operator
$$alpha:Omega^p(mathbb R^n) rightarrow Omega^{p-1}(mathbb R^n) $$
s.t. $$dalpha+alpha d=1.$$
Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define
$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$
where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)
$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$
and we are done. The diff. form $theta:=t(omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* omega$ where $M_t : mathbb R^n to mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p omega(tx) = M_t^* omega (x)$.
$endgroup$
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
add a comment |
$begingroup$
We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ are exact. To do so we construct a linear operator
$$alpha:Omega^p(mathbb R^n) rightarrow Omega^{p-1}(mathbb R^n) $$
s.t. $$dalpha+alpha d=1.$$
Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define
$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$
where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)
$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$
and we are done. The diff. form $theta:=t(omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* omega$ where $M_t : mathbb R^n to mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p omega(tx) = M_t^* omega (x)$.
$endgroup$
We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ are exact. To do so we construct a linear operator
$$alpha:Omega^p(mathbb R^n) rightarrow Omega^{p-1}(mathbb R^n) $$
s.t. $$dalpha+alpha d=1.$$
Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define
$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$
where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)
$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$
and we are done. The diff. form $theta:=t(omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* omega$ where $M_t : mathbb R^n to mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p omega(tx) = M_t^* omega (x)$.
edited Mar 4 '15 at 6:02
Ryan Budney
19.9k35496
19.9k35496
answered Nov 11 '13 at 21:28
AvitusAvitus
11.7k11841
11.7k11841
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
add a comment |
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
$begingroup$
I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality.
$endgroup$
– Lucky
Feb 4 '18 at 17:41
add a comment |
$begingroup$
We describe a linear functional $alpha : Omega^p mathbb R^n to Omega^{p-1} mathbb R^n$. The space alternating $p$-linear functions on $mathbb R^n$ has dimension $n choose p$, and you can write the basis as $dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$ where $1 leq i_1 < i_2 < cdots < i_p leq n$.
If $I = (i_1, i_2, cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.
Given a $p$-form $f dx_I$ with $f: mathbb R^n to mathbb R$ we define $alpha$ linearly, by $alpha (f dx_I) = 0$ if $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ if $I in I_1$. You can think of $alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that
$$ d(alpha(omega)) + alpha(d omega) = omega - pi^*(i^* omega)$$
for every $p$-form $omega$. Here $i : mathbb R^{n-1} to mathbb R^n$ is
the inclusion $i(x_2,cdots,x_n) = (0,x_2,cdots,x_n)$ and $pi : mathbb R^n to mathbb R^{n-1}$ is projection $pi(x_1,x_2,cdots,x_n) = (x_2,cdots,x_n)$.
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.
$endgroup$
$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
add a comment |
$begingroup$
We describe a linear functional $alpha : Omega^p mathbb R^n to Omega^{p-1} mathbb R^n$. The space alternating $p$-linear functions on $mathbb R^n$ has dimension $n choose p$, and you can write the basis as $dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$ where $1 leq i_1 < i_2 < cdots < i_p leq n$.
If $I = (i_1, i_2, cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.
Given a $p$-form $f dx_I$ with $f: mathbb R^n to mathbb R$ we define $alpha$ linearly, by $alpha (f dx_I) = 0$ if $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ if $I in I_1$. You can think of $alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that
$$ d(alpha(omega)) + alpha(d omega) = omega - pi^*(i^* omega)$$
for every $p$-form $omega$. Here $i : mathbb R^{n-1} to mathbb R^n$ is
the inclusion $i(x_2,cdots,x_n) = (0,x_2,cdots,x_n)$ and $pi : mathbb R^n to mathbb R^{n-1}$ is projection $pi(x_1,x_2,cdots,x_n) = (x_2,cdots,x_n)$.
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.
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$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
add a comment |
$begingroup$
We describe a linear functional $alpha : Omega^p mathbb R^n to Omega^{p-1} mathbb R^n$. The space alternating $p$-linear functions on $mathbb R^n$ has dimension $n choose p$, and you can write the basis as $dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$ where $1 leq i_1 < i_2 < cdots < i_p leq n$.
If $I = (i_1, i_2, cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.
Given a $p$-form $f dx_I$ with $f: mathbb R^n to mathbb R$ we define $alpha$ linearly, by $alpha (f dx_I) = 0$ if $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ if $I in I_1$. You can think of $alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that
$$ d(alpha(omega)) + alpha(d omega) = omega - pi^*(i^* omega)$$
for every $p$-form $omega$. Here $i : mathbb R^{n-1} to mathbb R^n$ is
the inclusion $i(x_2,cdots,x_n) = (0,x_2,cdots,x_n)$ and $pi : mathbb R^n to mathbb R^{n-1}$ is projection $pi(x_1,x_2,cdots,x_n) = (x_2,cdots,x_n)$.
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.
$endgroup$
We describe a linear functional $alpha : Omega^p mathbb R^n to Omega^{p-1} mathbb R^n$. The space alternating $p$-linear functions on $mathbb R^n$ has dimension $n choose p$, and you can write the basis as $dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$ where $1 leq i_1 < i_2 < cdots < i_p leq n$.
If $I = (i_1, i_2, cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} wedge dx_{i_2} wedge cdots wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.
Given a $p$-form $f dx_I$ with $f: mathbb R^n to mathbb R$ we define $alpha$ linearly, by $alpha (f dx_I) = 0$ if $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ if $I in I_1$. You can think of $alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that
$$ d(alpha(omega)) + alpha(d omega) = omega - pi^*(i^* omega)$$
for every $p$-form $omega$. Here $i : mathbb R^{n-1} to mathbb R^n$ is
the inclusion $i(x_2,cdots,x_n) = (0,x_2,cdots,x_n)$ and $pi : mathbb R^n to mathbb R^{n-1}$ is projection $pi(x_1,x_2,cdots,x_n) = (x_2,cdots,x_n)$.
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.
answered Mar 25 '15 at 7:46
Ryan BudneyRyan Budney
19.9k35496
19.9k35496
$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
add a comment |
$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
$begingroup$
This looks very similar to Bott&Tu's proof!
$endgroup$
– Bombyx mori
May 22 '16 at 5:13
add a comment |
$begingroup$
The above calculation in more detail. Define $h_k:Omega^p(mathbb{R}^n)rightarrow Omega^{p-1}(mathbb{R}^n)$ by,
begin{eqnarray}
h_k(omega)(x)=int_0^1t^{p-1}i_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. To see this note that $X^i(t)=frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)&=&int_0^1t^{p-1}Big(d(i_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x
end{eqnarray}
Now we can compute the pullback explicitly,
begin{eqnarray}
[((Phi_s)^*omega)_x]_{k_1...k_p}=frac{partial y^{i_1}}{partial x^{k_1}}...frac{partial y^{i_p}}{partial x^{k_p}}[omega_{Phi_s(x)}]_{i_1...i_p}=e^{ps}[omega_{e^sx}]_{k_1...k_p}
end{eqnarray}
Due to linearity we can restrict to the simple case where $omega(x)=f(x)dx^I$. Therefore,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.
$endgroup$
add a comment |
$begingroup$
The above calculation in more detail. Define $h_k:Omega^p(mathbb{R}^n)rightarrow Omega^{p-1}(mathbb{R}^n)$ by,
begin{eqnarray}
h_k(omega)(x)=int_0^1t^{p-1}i_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. To see this note that $X^i(t)=frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)&=&int_0^1t^{p-1}Big(d(i_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x
end{eqnarray}
Now we can compute the pullback explicitly,
begin{eqnarray}
[((Phi_s)^*omega)_x]_{k_1...k_p}=frac{partial y^{i_1}}{partial x^{k_1}}...frac{partial y^{i_p}}{partial x^{k_p}}[omega_{Phi_s(x)}]_{i_1...i_p}=e^{ps}[omega_{e^sx}]_{k_1...k_p}
end{eqnarray}
Due to linearity we can restrict to the simple case where $omega(x)=f(x)dx^I$. Therefore,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.
$endgroup$
add a comment |
$begingroup$
The above calculation in more detail. Define $h_k:Omega^p(mathbb{R}^n)rightarrow Omega^{p-1}(mathbb{R}^n)$ by,
begin{eqnarray}
h_k(omega)(x)=int_0^1t^{p-1}i_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. To see this note that $X^i(t)=frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)&=&int_0^1t^{p-1}Big(d(i_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x
end{eqnarray}
Now we can compute the pullback explicitly,
begin{eqnarray}
[((Phi_s)^*omega)_x]_{k_1...k_p}=frac{partial y^{i_1}}{partial x^{k_1}}...frac{partial y^{i_p}}{partial x^{k_p}}[omega_{Phi_s(x)}]_{i_1...i_p}=e^{ps}[omega_{e^sx}]_{k_1...k_p}
end{eqnarray}
Due to linearity we can restrict to the simple case where $omega(x)=f(x)dx^I$. Therefore,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.
$endgroup$
The above calculation in more detail. Define $h_k:Omega^p(mathbb{R}^n)rightarrow Omega^{p-1}(mathbb{R}^n)$ by,
begin{eqnarray}
h_k(omega)(x)=int_0^1t^{p-1}i_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. To see this note that $X^i(t)=frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)&=&int_0^1t^{p-1}Big(d(i_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x
end{eqnarray}
Now we can compute the pullback explicitly,
begin{eqnarray}
[((Phi_s)^*omega)_x]_{k_1...k_p}=frac{partial y^{i_1}}{partial x^{k_1}}...frac{partial y^{i_p}}{partial x^{k_p}}[omega_{Phi_s(x)}]_{i_1...i_p}=e^{ps}[omega_{e^sx}]_{k_1...k_p}
end{eqnarray}
Due to linearity we can restrict to the simple case where $omega(x)=f(x)dx^I$. Therefore,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.
edited Dec 15 '18 at 14:16
answered Dec 15 '18 at 3:49
Sam CollieSam Collie
507
507
add a comment |
add a comment |
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2
$begingroup$
I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $omega$ is closed, then $dtomega=omega$ and $theta:=tomega$ gives the solution (i.e. proves that $omega$ is exact). Would you like it in an answer?
$endgroup$
– Avitus
Nov 10 '13 at 21:03
1
$begingroup$
@Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference.
$endgroup$
– user90041
Nov 11 '13 at 2:06
$begingroup$
@Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well.
$endgroup$
– user90041
Nov 11 '13 at 2:13
2
$begingroup$
According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language.
$endgroup$
– Ben
Nov 11 '13 at 22:52