Mixing
Sheafification of constant real sheaf on smooth manifold and sheaf of smooth functions
0
$begingroup$
I have a question that has been motivated by this one.
Since $H^k_{dR}(M)cong hat{H}^k(M;mathbb{R}_M)$ I was wondering if the constant sheaf $mathbb{R}_M$ was isomorphic to the sheaf of $C^{infty}$ functions $f:Mto mathbb{R}$. From the definition of the constant sheaf I think it's clear that it "contains" the sheaf of $C^{infty}$ functions since the stalks determine germs of these functions, but it looks like there can be many more functions, since one can consider just $C^1$ functions with those germs for example.
If they are not isomorphic, is $H^k_{dR}(M)cong hat{H}(M;C^{infty})$?
differential-geometry algebraic-geometry sheaf-theory sheaf-cohomology de-rham-cohomology
asked Jan 11 at 19:17
JaviJavi
2,6442826
$endgroup$
|
show 1 more comment
0
$begingroup$
I have a question that has been motivated by this one.
Since $H^k_{dR}(M)cong hat{H}^k(M;mathbb{R}_M)$ I was wondering if the constant sheaf $mathbb{R}_M$ was isomorphic to the sheaf of $C^{infty}$ functions $f:Mto mathbb{R}$. From the definition of the constant sheaf I think it's clear that it "contains" the sheaf of $C^{infty}$ functions since the stalks determine germs of these functions, but it looks like there can be many more functions, since one can consider just $C^1$ functions with those germs for example.
If they are not isomorphic, is $H^k_{dR}(M)cong hat{H}(M;C^{infty})$?
differential-geometry algebraic-geometry sheaf-theory sheaf-cohomology de-rham-cohomology
asked Jan 11 at 19:17
JaviJavi
2,6442826
$endgroup$
3
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
2
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
1
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
1
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40
|
show 1 more comment
0
0
0
$begingroup$
I have a question that has been motivated by this one.
Since $H^k_{dR}(M)cong hat{H}^k(M;mathbb{R}_M)$ I was wondering if the constant sheaf $mathbb{R}_M$ was isomorphic to the sheaf of $C^{infty}$ functions $f:Mto mathbb{R}$. From the definition of the constant sheaf I think it's clear that it "contains" the sheaf of $C^{infty}$ functions since the stalks determine germs of these functions, but it looks like there can be many more functions, since one can consider just $C^1$ functions with those germs for example.
If they are not isomorphic, is $H^k_{dR}(M)cong hat{H}(M;C^{infty})$?
differential-geometry algebraic-geometry sheaf-theory sheaf-cohomology de-rham-cohomology
asked Jan 11 at 19:17
JaviJavi
2,6442826
$endgroup$
I have a question that has been motivated by this one.
Since $H^k_{dR}(M)cong hat{H}^k(M;mathbb{R}_M)$ I was wondering if the constant sheaf $mathbb{R}_M$ was isomorphic to the sheaf of $C^{infty}$ functions $f:Mto mathbb{R}$. From the definition of the constant sheaf I think it's clear that it "contains" the sheaf of $C^{infty}$ functions since the stalks determine germs of these functions, but it looks like there can be many more functions, since one can consider just $C^1$ functions with those germs for example.
If they are not isomorphic, is $H^k_{dR}(M)cong hat{H}(M;C^{infty})$?
differential-geometry algebraic-geometry sheaf-theory sheaf-cohomology de-rham-cohomology
differential-geometry algebraic-geometry sheaf-theory sheaf-cohomology de-rham-cohomology
asked Jan 11 at 19:17
JaviJavi
2,6442826
asked Jan 11 at 19:17
JaviJavi
2,6442826
asked Jan 11 at 19:17
JaviJavi
2,6442826
asked Jan 11 at 19:17
JaviJavi
2,6442826
asked Jan 11 at 19:17
JaviJavi
2,6442826
2,6442826
3
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
2
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
1
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
1
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40
|
show 1 more comment
3
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
2
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
1
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
1
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40
3
3
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
2
2
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
1
1
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
1
1
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40
|
show 1 more comment
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3
$begingroup$
The sheaf of constant functions is not isomorphic to the sheaf of all smooth functions. There are much more smooth functions than constant ones... So its not clear at all that the constant sheaf contains the sheaf of $C^infty$ function, this is exactly the opposite !
$endgroup$
– Roland
Jan 11 at 19:24
2
$begingroup$
This is not true that $H^k_{dR}(M)simeq check{H}(M,C^{infty})$. In fact the group $check{H}(M,C^{infty})$ is zero for $k>0$ since $C^infty$ is a fine sheaf.
$endgroup$
– Roland
Jan 11 at 19:35
$begingroup$
@Roland I don't clearly what you say. The constant sheaf is the sheafification of the constant presheaf, and therefore the sections consist of functions $s:Utomathbb{R}$ satisfying the locallity property. Why must these functions be constant?
$endgroup$
– Javi
Jan 11 at 19:37
1
$begingroup$
They are not constant but locally constant. I should have been more precise.
$endgroup$
– Roland
Jan 11 at 19:38
1
$begingroup$
As stated in the comments of @Roland the constant sheaf $mathbb R$ is the sheaf of locally constant functions. The problem in the line of argument in the question is that the stalks of the costant presheaf $mathbb R$ are the set $mathbb R$ endowed with the dicrete topology and not with the standard topology. And of course, a real valued function is continuous for the discrete topoolgy if and only if it is locally constant.
$endgroup$
– Andreas Cap
Jan 12 at 10:40