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
Mistake in years of experience in resume?
Initiative: Do I lose my attack/action if my target moves or dies before my turn in combat?
Check if a string is entirely made of the same substring
What term is being referred to with "reflected-sound-of-underground-spirits"?
Dynamic SOQL query relationship with field visibility for Users
What happens to Mjolnir (Thor's hammer) at the end of Endgame?
What does ゆーか mean?
How to write a column outside the braces in a matrix?
Don’t seats that recline flat defeat the purpose of having seatbelts?
Do I have an "anti-research" personality?
How to limit Drive Letters Windows assigns to new removable USB drives
Could the terminal length of components like resistors be reduced?
Implications of cigar-shaped bodies having rings?
What happened to Captain America in Endgame?
bldc motor, esc and battery draw, nominal vs peak
Rivers without rain
Was there a shared-world project before "Thieves World"?
Who was the lone kid in the line of people at the lake at the end of Avengers: Endgame?
Like totally amazing interchangeable sister outfits II: The Revenge
How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?
Critique of timeline aesthetic
How to have a sharp product image?
Alignment of various blocks in tikz
Why do games have consumables?
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 4 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 4 hours ago
KarenKaren
1336
$endgroup$
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 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 4 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 4 hours ago
KarenKaren
1336
asked 4 hours ago
KarenKaren
1336
asked 4 hours ago
KarenKaren
1336
asked 4 hours ago
KarenKaren
1336
asked 4 hours ago
KarenKaren
1336
1336
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 hours ago
add a comment |
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 hours ago
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 hours ago
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 3 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 3 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203841%2fclassification-of-surfaces%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 3 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 3 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 3 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 3 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 3 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 3 hours ago
Adam ChalumeauAdam Chalumeau
48010
answered 3 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 3 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 3 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 3 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 3 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
194k14223354
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203841%2fclassification-of-surfaces%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This can help you math.stackexchange.com/q/358724/654562
$endgroup$
– dcolazin
3 hours ago