Computing de Rham Cohomology
I'm stuck on the following problem.
Let $X=S^{n}setminus A$, where $A$ is the union of $kgeq 1$ disks $D_{k}$. Use the Mayer-Vietoris sequence to compute the de Rham cohomology $H_{mathrm{dR}}^{*}(X)$.
For $k=1$, I see that $S^{n}setminus D_{1}$ is diffeomorphic to $mathbb{R}^{n}$. Since $mathbb{R}^{n}$ is connected, we know that $H_{mathrm{dR}}^{0}(mathbb{R}^{n})congmathbb{R}$, so $H_{mathrm{dR}}^{0}(S^{n}setminus D_{1})congmathbb{R}$. However, I have no idea of show to handle the general case (i.e. $ngeq 0$ and $k> 1$) to find $H^{n}_{mathrm{dR}}(S^{n}setminus A)$. I know that I need to decompose the spaces to apply Mayer-Vietoris, but I can't seem to figure out anything past this. Any help is appreciated.
algebraic-topology differential-topology smooth-manifolds de-rham-cohomology
asked Nov 26 at 0:38
user608571
1959
add a comment |
I'm stuck on the following problem.
Let $X=S^{n}setminus A$, where $A$ is the union of $kgeq 1$ disks $D_{k}$. Use the Mayer-Vietoris sequence to compute the de Rham cohomology $H_{mathrm{dR}}^{*}(X)$.
For $k=1$, I see that $S^{n}setminus D_{1}$ is diffeomorphic to $mathbb{R}^{n}$. Since $mathbb{R}^{n}$ is connected, we know that $H_{mathrm{dR}}^{0}(mathbb{R}^{n})congmathbb{R}$, so $H_{mathrm{dR}}^{0}(S^{n}setminus D_{1})congmathbb{R}$. However, I have no idea of show to handle the general case (i.e. $ngeq 0$ and $k> 1$) to find $H^{n}_{mathrm{dR}}(S^{n}setminus A)$. I know that I need to decompose the spaces to apply Mayer-Vietoris, but I can't seem to figure out anything past this. Any help is appreciated.
algebraic-topology differential-topology smooth-manifolds de-rham-cohomology
asked Nov 26 at 0:38
user608571
1959
Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36
add a comment |
I'm stuck on the following problem.
Let $X=S^{n}setminus A$, where $A$ is the union of $kgeq 1$ disks $D_{k}$. Use the Mayer-Vietoris sequence to compute the de Rham cohomology $H_{mathrm{dR}}^{*}(X)$.
For $k=1$, I see that $S^{n}setminus D_{1}$ is diffeomorphic to $mathbb{R}^{n}$. Since $mathbb{R}^{n}$ is connected, we know that $H_{mathrm{dR}}^{0}(mathbb{R}^{n})congmathbb{R}$, so $H_{mathrm{dR}}^{0}(S^{n}setminus D_{1})congmathbb{R}$. However, I have no idea of show to handle the general case (i.e. $ngeq 0$ and $k> 1$) to find $H^{n}_{mathrm{dR}}(S^{n}setminus A)$. I know that I need to decompose the spaces to apply Mayer-Vietoris, but I can't seem to figure out anything past this. Any help is appreciated.
algebraic-topology differential-topology smooth-manifolds de-rham-cohomology
asked Nov 26 at 0:38
user608571
1959
I'm stuck on the following problem.
Let $X=S^{n}setminus A$, where $A$ is the union of $kgeq 1$ disks $D_{k}$. Use the Mayer-Vietoris sequence to compute the de Rham cohomology $H_{mathrm{dR}}^{*}(X)$.
For $k=1$, I see that $S^{n}setminus D_{1}$ is diffeomorphic to $mathbb{R}^{n}$. Since $mathbb{R}^{n}$ is connected, we know that $H_{mathrm{dR}}^{0}(mathbb{R}^{n})congmathbb{R}$, so $H_{mathrm{dR}}^{0}(S^{n}setminus D_{1})congmathbb{R}$. However, I have no idea of show to handle the general case (i.e. $ngeq 0$ and $k> 1$) to find $H^{n}_{mathrm{dR}}(S^{n}setminus A)$. I know that I need to decompose the spaces to apply Mayer-Vietoris, but I can't seem to figure out anything past this. Any help is appreciated.
algebraic-topology differential-topology smooth-manifolds de-rham-cohomology
algebraic-topology differential-topology smooth-manifolds de-rham-cohomology
asked Nov 26 at 0:38
user608571
1959
asked Nov 26 at 0:38
user608571
1959
asked Nov 26 at 0:38
user608571
1959
asked Nov 26 at 0:38
user608571
1959
asked Nov 26 at 0:38
user608571
1959
1959
Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36
add a comment |
Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36
Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36
add a comment |
1 Answer
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Hint: You have the $k=1$ case. By the same reasoning, for $k>1$ $X$ is diffeomorphic to $mathbb R^n$ with $k-1$ disks removed. Use Mayer-Vietoris and induction on $k$ to compute the cohomology of such spaces.
answered Nov 26 at 1:02
Aweygan
13.4k21441
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Hint: You have the $k=1$ case. By the same reasoning, for $k>1$ $X$ is diffeomorphic to $mathbb R^n$ with $k-1$ disks removed. Use Mayer-Vietoris and induction on $k$ to compute the cohomology of such spaces.
answered Nov 26 at 1:02
Aweygan
13.4k21441
add a comment |
Hint: You have the $k=1$ case. By the same reasoning, for $k>1$ $X$ is diffeomorphic to $mathbb R^n$ with $k-1$ disks removed. Use Mayer-Vietoris and induction on $k$ to compute the cohomology of such spaces.
answered Nov 26 at 1:02
Aweygan
13.4k21441
add a comment |
Hint: You have the $k=1$ case. By the same reasoning, for $k>1$ $X$ is diffeomorphic to $mathbb R^n$ with $k-1$ disks removed. Use Mayer-Vietoris and induction on $k$ to compute the cohomology of such spaces.
answered Nov 26 at 1:02
Aweygan
13.4k21441
Hint: You have the $k=1$ case. By the same reasoning, for $k>1$ $X$ is diffeomorphic to $mathbb R^n$ with $k-1$ disks removed. Use Mayer-Vietoris and induction on $k$ to compute the cohomology of such spaces.
answered Nov 26 at 1:02
Aweygan
13.4k21441
edited Nov 26 at 1:13
answered Nov 26 at 1:02
Aweygan
13.4k21441
answered Nov 26 at 1:02
Aweygan
13.4k21441
answered Nov 26 at 1:02
Aweygan
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13.4k21441
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Hint: instead of trying to decompose $X$, try to decompose $S^n$ (for which the homology is known) into two pieces, one of which is $X$.
– hunter
Nov 26 at 0:56
@hunter I hadn't thought of it from that perspective. So would I decompose $S^{n}$ as $X=S^{n}setminus A$ and $A$?
– user608571
Nov 26 at 1:01
yes. You have to thicken them a bit so there's an overlap.
– hunter
Nov 26 at 1:36