Classification of surfacesconnected sum of torus with projective planeWhy can all surfaces with boundary be realized in $mathbbR^3$?2-cell embeddings of graphs in surfaces and Euler formulaAbout zeros of vector fields in compact surfacesClassification of orientable non-closed surfacesClosed, orientable surface whose genus is very hard to find intuitivelyNonorientable surfaces: genus or demigenus?Surface has Euler characteristic 2 iff equal to sphereClassification of surfaces theoremClassification of 2-dim topological manifold (not necessarily second countable)Connected sum of two non homeomorphic surfaces
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Classification of surfaces
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Classification of surfaces
connected sum of torus with projective planeWhy can all surfaces with boundary be realized in $mathbbR^3$?2-cell embeddings of graphs in surfaces and Euler formulaAbout zeros of vector fields in compact surfacesClassification of orientable non-closed surfacesClosed, orientable surface whose genus is very hard to find intuitivelyNonorientable surfaces: genus or demigenus?Surface has Euler characteristic 2 iff equal to sphereClassification of surfaces theoremClassification of 2-dim topological manifold (not necessarily second countable)Connected sum of two non homeomorphic surfaces
$begingroup$
The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.
From there, it is easy to compute $chi(M)=2-2g-b-c$.
Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.
I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?
manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces
asked 3 hours ago
KarenKaren
1336
$endgroup$
add a comment |
$begingroup$
The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.
From there, it is easy to compute $chi(M)=2-2g-b-c$.
Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.
I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?
manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces
asked 3 hours ago
KarenKaren
1336
$endgroup$
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
2 hours ago
add a comment |
$begingroup$
The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.
From there, it is easy to compute $chi(M)=2-2g-b-c$.
Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.
I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?
manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces
asked 3 hours ago
KarenKaren
1336
$endgroup$
The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2# (#_gT^2)# (#_b D^2)# (#_c mathbbRP^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.
From there, it is easy to compute $chi(M)=2-2g-b-c$.
Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.
I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $chi(M)$, and orientability only tells us if $c=0$ or $cgeq 1$. Can someone help me, please?
manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces
manifolds surfaces orientation manifolds-with-boundary non-orientable-surfaces
asked 3 hours ago
KarenKaren
1336
asked 3 hours ago
KarenKaren
1336
asked 3 hours ago
KarenKaren
1336
asked 3 hours ago
KarenKaren
1336
asked 3 hours ago
KarenKaren
1336
1336
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
2 hours ago
add a comment |
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
2 hours ago
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
2 hours ago
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
2 hours ago
add a comment |
2 Answers
2
active
oldest
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$begingroup$
If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
$endgroup$
add a comment |
$begingroup$
Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.
Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
$endgroup$
add a comment |
$begingroup$
If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
$endgroup$
add a comment |
$begingroup$
If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
$endgroup$
If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 2 hours ago
Adam ChalumeauAdam Chalumeau
48010
48010
add a comment |
add a comment |
$begingroup$
Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.
Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
$begingroup$
Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.
Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
$begingroup$
Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.
Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
Since $mathbbRP^2# mathbbRP^2#mathbbRP^2cong mathbbRP^2 # T^2$, $c$ and $g$ are not uniquely determined: if $cgeq 3$, you can subtract $2$ from $c$ and add $1$ to $g$ and get the same surface, or if $c,ggeq 1$, you can subtract $1$ from $g$ and add $2$ to $c$.
Note, though, that the first operation can always be used to get a connected sum presentation where $cleq 2$. If you impose the additional restriction that $cleq 2$, then $c$ and $g$ can be uniquely determined and can be calculated from the data you mention. If the surface is orientable, then $c=0$ and then you can just solve for $g$. If the surface is not orientable, then you can determine whether $c=1$ or $c=2$ since $c$ must have the same parity as $chi(M)+b$. Once $c$ is determined, you can solve for $g$.
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 2 hours ago
Eric WofseyEric Wofsey
194k14223354
194k14223354
add a comment |
add a comment |
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This can help you math.stackexchange.com/q/358724/654562
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– dcolazin
2 hours ago