Understanding the homotopy operator for de Rham Cohomology
$begingroup$
This is in John Lee's Smooth Manifold 2nd Edition, pg 444
For any smooth manifold $M$, there exists a linear map
$$ h:Omega^p(M times I ) rightarrow Omega^{p-1}(M)$$
such that $$ h(dw)+d(hw) = i_1^*w - i^*_0w$$
The proof begins goes as follows. This is the same one here.
Let $s$ be the standard coordinate on $Bbb R$ and let $S$ be the vector field on $M times Bbb R$ given by $S_{(q,s)} = (0, partial/partial s|_s)$. Let $w$ be a smooth $p$-form on $M times I$. Define $hw in Omega^{p-1}(M)$ by
$$ hw = int_0^1 i^*_t (S lrcorner w) , dt. $$
Specifically, given $q in M$, this means,
$$ (hw)_q = int_0^1 i^*_tBig( (S lrcorner w )_{(q,t)} Big) , dt $$
The notation $S lrcorner w$ is interior multiplication. Otherwise denoted by $iota_Sw$.
My question is, what does this integral even mean?
At each point $q in M$, the integrand is a function of $t$ with values in the vector space $wedge^{p-1} (T^*q M)$. How do we integrate over this?
My thoughts on the way to see this:
We work locally, in nhood $U_q$. For $r in U_q$ our integrand is given by
$$ sum a_I(r,t) dx^I$$
summing over indices $I$ increasing $p-1$ length subset of $0$ to $n$. Our new $p-1$ form is given by
$$ sum int a_I(r,t) , dt , d x^I $$
There is a problem however, that we have to show this is independent of the choice of coordinate.
differential-geometry algebraic-topology homotopy-theory de-rham-cohomology
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
$endgroup$
add a comment |
$begingroup$
This is in John Lee's Smooth Manifold 2nd Edition, pg 444
For any smooth manifold $M$, there exists a linear map
$$ h:Omega^p(M times I ) rightarrow Omega^{p-1}(M)$$
such that $$ h(dw)+d(hw) = i_1^*w - i^*_0w$$
The proof begins goes as follows. This is the same one here.
Let $s$ be the standard coordinate on $Bbb R$ and let $S$ be the vector field on $M times Bbb R$ given by $S_{(q,s)} = (0, partial/partial s|_s)$. Let $w$ be a smooth $p$-form on $M times I$. Define $hw in Omega^{p-1}(M)$ by
$$ hw = int_0^1 i^*_t (S lrcorner w) , dt. $$
Specifically, given $q in M$, this means,
$$ (hw)_q = int_0^1 i^*_tBig( (S lrcorner w )_{(q,t)} Big) , dt $$
The notation $S lrcorner w$ is interior multiplication. Otherwise denoted by $iota_Sw$.
My question is, what does this integral even mean?
At each point $q in M$, the integrand is a function of $t$ with values in the vector space $wedge^{p-1} (T^*q M)$. How do we integrate over this?
My thoughts on the way to see this:
We work locally, in nhood $U_q$. For $r in U_q$ our integrand is given by
$$ sum a_I(r,t) dx^I$$
summing over indices $I$ increasing $p-1$ length subset of $0$ to $n$. Our new $p-1$ form is given by
$$ sum int a_I(r,t) , dt , d x^I $$
There is a problem however, that we have to show this is independent of the choice of coordinate.
differential-geometry algebraic-topology homotopy-theory de-rham-cohomology
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
$endgroup$
3
$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
1
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08
add a comment |
$begingroup$
This is in John Lee's Smooth Manifold 2nd Edition, pg 444
For any smooth manifold $M$, there exists a linear map
$$ h:Omega^p(M times I ) rightarrow Omega^{p-1}(M)$$
such that $$ h(dw)+d(hw) = i_1^*w - i^*_0w$$
The proof begins goes as follows. This is the same one here.
Let $s$ be the standard coordinate on $Bbb R$ and let $S$ be the vector field on $M times Bbb R$ given by $S_{(q,s)} = (0, partial/partial s|_s)$. Let $w$ be a smooth $p$-form on $M times I$. Define $hw in Omega^{p-1}(M)$ by
$$ hw = int_0^1 i^*_t (S lrcorner w) , dt. $$
Specifically, given $q in M$, this means,
$$ (hw)_q = int_0^1 i^*_tBig( (S lrcorner w )_{(q,t)} Big) , dt $$
The notation $S lrcorner w$ is interior multiplication. Otherwise denoted by $iota_Sw$.
My question is, what does this integral even mean?
At each point $q in M$, the integrand is a function of $t$ with values in the vector space $wedge^{p-1} (T^*q M)$. How do we integrate over this?
My thoughts on the way to see this:
We work locally, in nhood $U_q$. For $r in U_q$ our integrand is given by
$$ sum a_I(r,t) dx^I$$
summing over indices $I$ increasing $p-1$ length subset of $0$ to $n$. Our new $p-1$ form is given by
$$ sum int a_I(r,t) , dt , d x^I $$
There is a problem however, that we have to show this is independent of the choice of coordinate.
differential-geometry algebraic-topology homotopy-theory de-rham-cohomology
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
$endgroup$
This is in John Lee's Smooth Manifold 2nd Edition, pg 444
For any smooth manifold $M$, there exists a linear map
$$ h:Omega^p(M times I ) rightarrow Omega^{p-1}(M)$$
such that $$ h(dw)+d(hw) = i_1^*w - i^*_0w$$
The proof begins goes as follows. This is the same one here.
Let $s$ be the standard coordinate on $Bbb R$ and let $S$ be the vector field on $M times Bbb R$ given by $S_{(q,s)} = (0, partial/partial s|_s)$. Let $w$ be a smooth $p$-form on $M times I$. Define $hw in Omega^{p-1}(M)$ by
$$ hw = int_0^1 i^*_t (S lrcorner w) , dt. $$
Specifically, given $q in M$, this means,
$$ (hw)_q = int_0^1 i^*_tBig( (S lrcorner w )_{(q,t)} Big) , dt $$
The notation $S lrcorner w$ is interior multiplication. Otherwise denoted by $iota_Sw$.
My question is, what does this integral even mean?
At each point $q in M$, the integrand is a function of $t$ with values in the vector space $wedge^{p-1} (T^*q M)$. How do we integrate over this?
My thoughts on the way to see this:
We work locally, in nhood $U_q$. For $r in U_q$ our integrand is given by
$$ sum a_I(r,t) dx^I$$
summing over indices $I$ increasing $p-1$ length subset of $0$ to $n$. Our new $p-1$ form is given by
$$ sum int a_I(r,t) , dt , d x^I $$
There is a problem however, that we have to show this is independent of the choice of coordinate.
differential-geometry algebraic-topology homotopy-theory de-rham-cohomology
differential-geometry algebraic-topology homotopy-theory de-rham-cohomology
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
asked Dec 25 '18 at 8:06
CL.CL.
2,2692925
2,2692925
3
$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
1
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08
add a comment |
3
$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
1
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08
3
3
$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
1
1
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08
add a comment |
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$begingroup$
So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $gamma:[0,1]to V$ one can define the integral $int_0^1gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:40
1
$begingroup$
I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral.
$endgroup$
– Amitai Yuval
Dec 25 '18 at 8:47
$begingroup$
You can also define $h$ via $int_M h(alpha)wedgebeta=int_{Mtimes I} alphawedge p^*beta$ where $p:Mtimes Ito M$ is the projection and $beta$ is any compactly supported form on $M$
$endgroup$
– user8268
Dec 25 '18 at 9:08