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suspension of the torus
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The suspension of $S^n$ is $S^n+1$. What about the suspension of $T^n$? Is it just $S^2times T^n-1$? I'm having a hard time visualizing even the first non-trivial case $Sigma T^2$.
general-topology
asked Jul 30 at 10:15
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836
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up vote
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down vote
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The suspension of $S^n$ is $S^n+1$. What about the suspension of $T^n$? Is it just $S^2times T^n-1$? I'm having a hard time visualizing even the first non-trivial case $Sigma T^2$.
general-topology
asked Jul 30 at 10:15
aaa
836
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The suspension of $S^n$ is $S^n+1$. What about the suspension of $T^n$? Is it just $S^2times T^n-1$? I'm having a hard time visualizing even the first non-trivial case $Sigma T^2$.
general-topology
asked Jul 30 at 10:15
aaa
836
The suspension of $S^n$ is $S^n+1$. What about the suspension of $T^n$? Is it just $S^2times T^n-1$? I'm having a hard time visualizing even the first non-trivial case $Sigma T^2$.
general-topology
asked Jul 30 at 10:15
aaa
836
asked Jul 30 at 10:15
aaa
836
asked Jul 30 at 10:15
aaa
836
asked Jul 30 at 10:15
aaa
836
836
add a comment |Â
add a comment |Â
1 Answer
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The cartesian product of two manifolds is a manifold, so $S^2 times T^n-1$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $tildeH_n(X)=tildeH_n+1(Sigma X)$ where $Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.)
A good way to imagine the suspension $Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron.
Now identify the edges. Of course, you need to identify them all the way up and down.
(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres.
answered Jul 30 at 10:40
Alon Amit
10.2k3765
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The cartesian product of two manifolds is a manifold, so $S^2 times T^n-1$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $tildeH_n(X)=tildeH_n+1(Sigma X)$ where $Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.)
A good way to imagine the suspension $Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron.
Now identify the edges. Of course, you need to identify them all the way up and down.
(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres.
answered Jul 30 at 10:40
Alon Amit
10.2k3765
add a comment |Â
up vote
3
down vote
accepted
The cartesian product of two manifolds is a manifold, so $S^2 times T^n-1$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $tildeH_n(X)=tildeH_n+1(Sigma X)$ where $Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.)
A good way to imagine the suspension $Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron.
Now identify the edges. Of course, you need to identify them all the way up and down.
(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres.
answered Jul 30 at 10:40
Alon Amit
10.2k3765
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The cartesian product of two manifolds is a manifold, so $S^2 times T^n-1$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $tildeH_n(X)=tildeH_n+1(Sigma X)$ where $Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.)
A good way to imagine the suspension $Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron.
Now identify the edges. Of course, you need to identify them all the way up and down.
(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres.
answered Jul 30 at 10:40
Alon Amit
10.2k3765
The cartesian product of two manifolds is a manifold, so $S^2 times T^n-1$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $tildeH_n(X)=tildeH_n+1(Sigma X)$ where $Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.)
A good way to imagine the suspension $Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron.
Now identify the edges. Of course, you need to identify them all the way up and down.
(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres.
answered Jul 30 at 10:40
Alon Amit
10.2k3765
answered Jul 30 at 10:40
Alon Amit
10.2k3765
answered Jul 30 at 10:40
Alon Amit
10.2k3765
answered Jul 30 at 10:40
Alon Amit
10.2k3765
10.2k3765
add a comment |Â
add a comment |Â
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