Closed orientable 2-manifold
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Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?
I want a prove with triangulation.
geometry differential-geometry smooth-manifolds
asked Nov 21 at 22:44
hussein
456
add a comment |
up vote
0
down vote
favorite
Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?
I want a prove with triangulation.
geometry differential-geometry smooth-manifolds
asked Nov 21 at 22:44
hussein
456
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?
I want a prove with triangulation.
geometry differential-geometry smooth-manifolds
asked Nov 21 at 22:44
hussein
456
Suppose that $M$ closed orientable 2-manifold and $chi (M) = 2$.
Why is $M$ a sphere?
I want a prove with triangulation.
geometry differential-geometry smooth-manifolds
geometry differential-geometry smooth-manifolds
asked Nov 21 at 22:44
hussein
456
asked Nov 21 at 22:44
hussein
456
edited Nov 21 at 23:04
asked Nov 21 at 22:44
hussein
456
asked Nov 21 at 22:44
hussein
456
asked Nov 21 at 22:44
hussein
456
456
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32
add a comment |
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32
2
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32
add a comment |
1 Answer
1
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votes
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1
down vote
accepted
By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.
answered Nov 23 at 0:35
abstract
256212
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.
answered Nov 23 at 0:35
abstract
256212
add a comment |
up vote
1
down vote
accepted
By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.
answered Nov 23 at 0:35
abstract
256212
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.
answered Nov 23 at 0:35
abstract
256212
By classification of surfaces, two closed orientable surfaces are homeomorphic if and only if their Euler characteristics agree. To show that $chi(S^2) = 2$, consider a vertex with a loop to itself (1 edge), then glue two discs (2 faces) with disjoint interiors along the loop.
answered Nov 23 at 0:35
abstract
256212
answered Nov 23 at 0:35
abstract
256212
answered Nov 23 at 0:35
abstract
256212
answered Nov 23 at 0:35
abstract
256212
256212
add a comment |
add a comment |
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2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 at 22:55
@Nick your answer has one mistake. definition of genus is $ g := frac{- chi (M) + 2}{2}$ and your answer has cyclic.
– hussein
Nov 21 at 23:32