Mixing
De Rham cohomology groups of $mathbb{R}^n$
$begingroup$
I want to show that for each $1 le kle n$ we have $$ H_{dR}^k(mathbb{R}^n)=0 $$
The strategy is to construct for each $k$ a linear map $$h_k:Omega^k(mathbb{R}^n)to Omega^{k-1}(mathbb{R}^n)$$
such that $$ d_{k-1}circ h_k + h_{k+1}circ d_{k}=Id_{Omega^k(mathbb{R}^n)} qquadqquad[1]$$ where $d_k:Omega^{k}(mathbb{R}^n)to Omega^{k+1}(mathbb{R}^n)$ is the exterior derivative.
I have the following definition: for each $omega=omega_{i_1dots i_k}dx^{i_1}wedgedotswedge dx^{i_k}$ in $Omega^k(mathbb{R}^n)$ we define $$h_k(omega)(x)=k left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1} dx^{i_2}wedgedotswedge dx^{i_k}$$
Can you help me to show that this definition satisfies $[1]$?
For example I have $$d_k(w)=sum_j frac{partial(omega_{i_1dots i_k})}{partial x^j} dx^jwedge dx^{i_1}wedgedotswedge dx^{i_k}$$ and
$$ d_{k-1}( h_k (w))(x)=ksum_j frac{partial(left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1})}{partial x^j} dx^jwedge dx^{i_2}wedgedotswedge dx^{i_k}$$ and
$$ h_{k+1}(d_k(omega))(x)=(k+1)left[ int_0^1t^{k} frac{partial(omega_{i_1dots i_k})}{partial x^j} (tx) ;dtright]x^{j}dx^{i_1}wedgedotswedge dx^{i_k}$$
I don't know if my calculations above are correct, and even if they are, I can't go on.
Please give most details you can if you do sofisticated manipulations/calculatons.
multivariable-calculus smooth-manifolds group-cohomology de-rham-cohomology
asked Oct 30 '18 at 17:44
MinatoMinato
554313
$endgroup$
add a comment |
$begingroup$
I want to show that for each $1 le kle n$ we have $$ H_{dR}^k(mathbb{R}^n)=0 $$
The strategy is to construct for each $k$ a linear map $$h_k:Omega^k(mathbb{R}^n)to Omega^{k-1}(mathbb{R}^n)$$
such that $$ d_{k-1}circ h_k + h_{k+1}circ d_{k}=Id_{Omega^k(mathbb{R}^n)} qquadqquad[1]$$ where $d_k:Omega^{k}(mathbb{R}^n)to Omega^{k+1}(mathbb{R}^n)$ is the exterior derivative.
I have the following definition: for each $omega=omega_{i_1dots i_k}dx^{i_1}wedgedotswedge dx^{i_k}$ in $Omega^k(mathbb{R}^n)$ we define $$h_k(omega)(x)=k left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1} dx^{i_2}wedgedotswedge dx^{i_k}$$
Can you help me to show that this definition satisfies $[1]$?
For example I have $$d_k(w)=sum_j frac{partial(omega_{i_1dots i_k})}{partial x^j} dx^jwedge dx^{i_1}wedgedotswedge dx^{i_k}$$ and
$$ d_{k-1}( h_k (w))(x)=ksum_j frac{partial(left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1})}{partial x^j} dx^jwedge dx^{i_2}wedgedotswedge dx^{i_k}$$ and
$$ h_{k+1}(d_k(omega))(x)=(k+1)left[ int_0^1t^{k} frac{partial(omega_{i_1dots i_k})}{partial x^j} (tx) ;dtright]x^{j}dx^{i_1}wedgedotswedge dx^{i_k}$$
I don't know if my calculations above are correct, and even if they are, I can't go on.
Please give most details you can if you do sofisticated manipulations/calculatons.
multivariable-calculus smooth-manifolds group-cohomology de-rham-cohomology
asked Oct 30 '18 at 17:44
MinatoMinato
554313
$endgroup$
$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54
add a comment |
$begingroup$
I want to show that for each $1 le kle n$ we have $$ H_{dR}^k(mathbb{R}^n)=0 $$
The strategy is to construct for each $k$ a linear map $$h_k:Omega^k(mathbb{R}^n)to Omega^{k-1}(mathbb{R}^n)$$
such that $$ d_{k-1}circ h_k + h_{k+1}circ d_{k}=Id_{Omega^k(mathbb{R}^n)} qquadqquad[1]$$ where $d_k:Omega^{k}(mathbb{R}^n)to Omega^{k+1}(mathbb{R}^n)$ is the exterior derivative.
I have the following definition: for each $omega=omega_{i_1dots i_k}dx^{i_1}wedgedotswedge dx^{i_k}$ in $Omega^k(mathbb{R}^n)$ we define $$h_k(omega)(x)=k left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1} dx^{i_2}wedgedotswedge dx^{i_k}$$
Can you help me to show that this definition satisfies $[1]$?
For example I have $$d_k(w)=sum_j frac{partial(omega_{i_1dots i_k})}{partial x^j} dx^jwedge dx^{i_1}wedgedotswedge dx^{i_k}$$ and
$$ d_{k-1}( h_k (w))(x)=ksum_j frac{partial(left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1})}{partial x^j} dx^jwedge dx^{i_2}wedgedotswedge dx^{i_k}$$ and
$$ h_{k+1}(d_k(omega))(x)=(k+1)left[ int_0^1t^{k} frac{partial(omega_{i_1dots i_k})}{partial x^j} (tx) ;dtright]x^{j}dx^{i_1}wedgedotswedge dx^{i_k}$$
I don't know if my calculations above are correct, and even if they are, I can't go on.
Please give most details you can if you do sofisticated manipulations/calculatons.
multivariable-calculus smooth-manifolds group-cohomology de-rham-cohomology
asked Oct 30 '18 at 17:44
MinatoMinato
554313
$endgroup$
I want to show that for each $1 le kle n$ we have $$ H_{dR}^k(mathbb{R}^n)=0 $$
The strategy is to construct for each $k$ a linear map $$h_k:Omega^k(mathbb{R}^n)to Omega^{k-1}(mathbb{R}^n)$$
such that $$ d_{k-1}circ h_k + h_{k+1}circ d_{k}=Id_{Omega^k(mathbb{R}^n)} qquadqquad[1]$$ where $d_k:Omega^{k}(mathbb{R}^n)to Omega^{k+1}(mathbb{R}^n)$ is the exterior derivative.
I have the following definition: for each $omega=omega_{i_1dots i_k}dx^{i_1}wedgedotswedge dx^{i_k}$ in $Omega^k(mathbb{R}^n)$ we define $$h_k(omega)(x)=k left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1} dx^{i_2}wedgedotswedge dx^{i_k}$$
Can you help me to show that this definition satisfies $[1]$?
For example I have $$d_k(w)=sum_j frac{partial(omega_{i_1dots i_k})}{partial x^j} dx^jwedge dx^{i_1}wedgedotswedge dx^{i_k}$$ and
$$ d_{k-1}( h_k (w))(x)=ksum_j frac{partial(left[ int_0^1t^{k-1} omega_{i_1dots i_k}(tx) ;dtright]x^{i1})}{partial x^j} dx^jwedge dx^{i_2}wedgedotswedge dx^{i_k}$$ and
$$ h_{k+1}(d_k(omega))(x)=(k+1)left[ int_0^1t^{k} frac{partial(omega_{i_1dots i_k})}{partial x^j} (tx) ;dtright]x^{j}dx^{i_1}wedgedotswedge dx^{i_k}$$
I don't know if my calculations above are correct, and even if they are, I can't go on.
Please give most details you can if you do sofisticated manipulations/calculatons.
multivariable-calculus smooth-manifolds group-cohomology de-rham-cohomology
multivariable-calculus smooth-manifolds group-cohomology de-rham-cohomology
asked Oct 30 '18 at 17:44
MinatoMinato
554313
asked Oct 30 '18 at 17:44
MinatoMinato
554313
edited Jan 22 at 15:58
Minato
asked Oct 30 '18 at 17:44
MinatoMinato
554313
asked Oct 30 '18 at 17:44
MinatoMinato
554313
asked Oct 30 '18 at 17:44
MinatoMinato
554313
554313
$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54
add a comment |
$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54
$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54
add a comment |
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$begingroup$
This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:47
$begingroup$
@MatthewLeingang This is homework I gave myself, since I'm self-studing this subject.
$endgroup$
– Minato
Oct 30 '18 at 17:49
$begingroup$
look en.wikipedia.org/wiki/…
$endgroup$
– Tsemo Aristide
Oct 30 '18 at 17:51
$begingroup$
@Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on.
$endgroup$
– Matthew Leingang
Oct 30 '18 at 17:52
$begingroup$
@TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula.
$endgroup$
– Minato
Oct 30 '18 at 17:54