Closed orientable 2-manifold
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
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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
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 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 '18 at 23:32
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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
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
edited Nov 21 '18 at 23:04
hussein
asked Nov 21 '18 at 22:44
husseinhussein
576
576
2
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 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 '18 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 '18 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 '18 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 '18 at 22:55
The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 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 '18 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 '18 at 23:32
add a comment |
1 Answer
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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.
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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.
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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.
add a comment |
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.
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 '18 at 0:35


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The Euler characteristic is related to the genus by $chi(M) = 2-2g$. This implies $g=0$.
– Nick
Nov 21 '18 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 '18 at 23:32