Hjdcti

Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?

Can we say “you can pay when the order gets ready”?

What is the most expensive material in the world that could be used to create Pun-Pun's lute?

Is the claim "Employers won't employ people with no 'social media presence'" realistic?

Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?

"The cow" OR "a cow" OR "cows" in this context

Phrase for the opposite of "foolproof"

As an international instructor, should I openly talk about my accent?

Why did C use the -> operator instead of reusing the . operator?

a sore throat vs a strep throat vs strep throat

I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?

Re-entry to Germany after vacation using blue card

Was there a Viking Exchange as well as a Columbian one?

Can an Area of Effect spell cast outside a Prismatic Wall extend inside it?

Can someone publish a story that happened to you?

What happened to Captain America in Endgame?

What is the philosophical significance of speech acts/implicature?

Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?

A ​Note ​on ​N!

Alignment of various blocks in tikz

Is it idiomatic to construct against `this`

Pre-plastic human skin alternative

Dynamic SOQL query relationship with field visibility for Users

What are the steps to solving this definite integral?

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

4

$begingroup$

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

share|cite|improve this question

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 | 

4

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

$endgroup$

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

share|cite|improve this question

asked 3 hours ago

KarenKaren

1336

asked 3 hours ago

KarenKaren

1336

asked 3 hours ago

KarenKaren

1336

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

add a comment | 

2

$begingroup$

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 $g$.

share|cite|improve this answer

answered 2 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
);



);

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

2

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

answered 2 hours ago

Adam ChalumeauAdam Chalumeau

48010

2

$begingroup$

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 $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

add a comment | 

2

$begingroup$

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 $g$.

share|cite|improve this answer

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 $g$.

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354

answered 2 hours ago

Eric WofseyEric Wofsey

194k14223354