What is the field of de Rham cohomology vector space over?
$begingroup$
On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below)
It said that all the closed $k$-form and exact $k$-forms on a manifold are both vector space. I think the vector space are both over $ mathbb{R}$.
My question is:
Given a manifold $M$ of dimension $n$ and a coordinate of $M:x_1,x_2cdots,x_n $ I think the $k$-form on $M$ is some thing like $f(p),x_{i_1}wedge x_{i_2} wedge x_{i_3} cdots x_{i_k}$, where $1le{i_1} <{i_2}<cdots {i_k} le n$, $p in M$ . Now it look like the vector space should be over $f(p)$, all the smooth function on $M$, because multiply a real number seem not give us all the $k$-form on $M$. Nevertheless, if $omega$ is a closed form, $f(p)omega$ is not necessary a closed form. It makes me confused.
Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?
vector-spaces de-rham-cohomology
asked Jan 7 at 11:08
RikeijinRikeijin
999
$endgroup$
add a comment |
$begingroup$
On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below)
It said that all the closed $k$-form and exact $k$-forms on a manifold are both vector space. I think the vector space are both over $ mathbb{R}$.
My question is:
Given a manifold $M$ of dimension $n$ and a coordinate of $M:x_1,x_2cdots,x_n $ I think the $k$-form on $M$ is some thing like $f(p),x_{i_1}wedge x_{i_2} wedge x_{i_3} cdots x_{i_k}$, where $1le{i_1} <{i_2}<cdots {i_k} le n$, $p in M$ . Now it look like the vector space should be over $f(p)$, all the smooth function on $M$, because multiply a real number seem not give us all the $k$-form on $M$. Nevertheless, if $omega$ is a closed form, $f(p)omega$ is not necessary a closed form. It makes me confused.
Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?
vector-spaces de-rham-cohomology
asked Jan 7 at 11:08
RikeijinRikeijin
999
$endgroup$
$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43
add a comment |
$begingroup$
On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below)
It said that all the closed $k$-form and exact $k$-forms on a manifold are both vector space. I think the vector space are both over $ mathbb{R}$.
My question is:
Given a manifold $M$ of dimension $n$ and a coordinate of $M:x_1,x_2cdots,x_n $ I think the $k$-form on $M$ is some thing like $f(p),x_{i_1}wedge x_{i_2} wedge x_{i_3} cdots x_{i_k}$, where $1le{i_1} <{i_2}<cdots {i_k} le n$, $p in M$ . Now it look like the vector space should be over $f(p)$, all the smooth function on $M$, because multiply a real number seem not give us all the $k$-form on $M$. Nevertheless, if $omega$ is a closed form, $f(p)omega$ is not necessary a closed form. It makes me confused.
Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?
vector-spaces de-rham-cohomology
asked Jan 7 at 11:08
RikeijinRikeijin
999
$endgroup$
On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below)
It said that all the closed $k$-form and exact $k$-forms on a manifold are both vector space. I think the vector space are both over $ mathbb{R}$.
My question is:
Given a manifold $M$ of dimension $n$ and a coordinate of $M:x_1,x_2cdots,x_n $ I think the $k$-form on $M$ is some thing like $f(p),x_{i_1}wedge x_{i_2} wedge x_{i_3} cdots x_{i_k}$, where $1le{i_1} <{i_2}<cdots {i_k} le n$, $p in M$ . Now it look like the vector space should be over $f(p)$, all the smooth function on $M$, because multiply a real number seem not give us all the $k$-form on $M$. Nevertheless, if $omega$ is a closed form, $f(p)omega$ is not necessary a closed form. It makes me confused.
Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?
vector-spaces de-rham-cohomology
vector-spaces de-rham-cohomology
asked Jan 7 at 11:08
RikeijinRikeijin
999
asked Jan 7 at 11:08
RikeijinRikeijin
999
edited Jan 7 at 15:58
Rikeijin
asked Jan 7 at 11:08
RikeijinRikeijin
999
asked Jan 7 at 11:08
RikeijinRikeijin
999
asked Jan 7 at 11:08
RikeijinRikeijin
999
999
$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43
add a comment |
$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43
$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43
add a comment |
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$begingroup$
Yes, they are vector spaces over $mathbf R$. Smooth functions do not form a field.
$endgroup$
– Asal Beag Dubh
Jan 7 at 16:03
$begingroup$
To make sense of calling them vector space, it must be $mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed.
$endgroup$
– Rikeijin
Jan 7 at 16:43