Fundemental group of $S^2$ with equator is identified $z sim z^3$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Van Kampen with complicated attaching mapNeed help understanding statement of Van Kampen's Theorem and using it to compute the fundamental group of Projective PlaneUnion of 2-sphere with line segment in $mathbbR^3$ removing one point homotopy equivalence.A simple fundamental groupComputing fundamental group of the complement to three infinite straight lines, and of complement to $S^1 cup Z $Fundamental group of the sphere with two points identified.Fundamental group of the sphere with n-points identifiedHomotopy group of three spheresGiven a fundamental polygon and an attaching word, is the resulting space “unique?”Calculate the fundamental group of the 2-sphere with 2 disks removed

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Fundemental group of $S^2$ with equator is identified $z sim z^3$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Van Kampen with complicated attaching mapNeed help understanding statement of Van Kampen's Theorem and using it to compute the fundamental group of Projective PlaneUnion of 2-sphere with line segment in $mathbbR^3$ removing one point homotopy equivalence.A simple fundamental groupComputing fundamental group of the complement to three infinite straight lines, and of complement to $S^1 cup Z $Fundamental group of the sphere with two points identified.Fundamental group of the sphere with n-points identifiedHomotopy group of three spheresGiven a fundamental polygon and an attaching word, is the resulting space “unique?”Calculate the fundamental group of the 2-sphere with 2 disks removed










4












$begingroup$


You split a sphere in half at the equator and glue back together the boundaries with the function f:$S^1 rightarrow S^1$ that maps $z$ to $z^3$. What is the fundamental group of this space?



The exact statement of the question:




Cut a sphere $S^2$ through its equator and glue it back using the attaching map f:$S^1 rightarrow S^1$ defined by $f(z) = z^3$ . What is the fundamental group of this space?"




We take $S^1$ as the set of complex numbers with norm 1.



With Van kampen's theorem I compute the fundamental group as the trivial group. However, consider the path $gamma$ on the equator starting at some $x_0$ and moving $2pi/3$ degree around the equator. This path is correspond to a loop in the identification space. I fail to see how this path is homotopic to the trivial path.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
    $endgroup$
    – Tempestas Ludi
    Mar 26 at 19:02










  • $begingroup$
    Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
    $endgroup$
    – Daniel P
    Mar 26 at 19:05






  • 1




    $begingroup$
    Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
    $endgroup$
    – Daniel P
    Mar 26 at 19:13










  • $begingroup$
    Okay, let me write the exact statement in the question. Please see the edited version.
    $endgroup$
    – Baran Zadeoglu
    Mar 26 at 19:24










  • $begingroup$
    Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
    $endgroup$
    – Daniel P
    Mar 26 at 19:27















4












$begingroup$


You split a sphere in half at the equator and glue back together the boundaries with the function f:$S^1 rightarrow S^1$ that maps $z$ to $z^3$. What is the fundamental group of this space?



The exact statement of the question:




Cut a sphere $S^2$ through its equator and glue it back using the attaching map f:$S^1 rightarrow S^1$ defined by $f(z) = z^3$ . What is the fundamental group of this space?"




We take $S^1$ as the set of complex numbers with norm 1.



With Van kampen's theorem I compute the fundamental group as the trivial group. However, consider the path $gamma$ on the equator starting at some $x_0$ and moving $2pi/3$ degree around the equator. This path is correspond to a loop in the identification space. I fail to see how this path is homotopic to the trivial path.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
    $endgroup$
    – Tempestas Ludi
    Mar 26 at 19:02










  • $begingroup$
    Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
    $endgroup$
    – Daniel P
    Mar 26 at 19:05






  • 1




    $begingroup$
    Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
    $endgroup$
    – Daniel P
    Mar 26 at 19:13










  • $begingroup$
    Okay, let me write the exact statement in the question. Please see the edited version.
    $endgroup$
    – Baran Zadeoglu
    Mar 26 at 19:24










  • $begingroup$
    Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
    $endgroup$
    – Daniel P
    Mar 26 at 19:27













4












4








4


1



$begingroup$


You split a sphere in half at the equator and glue back together the boundaries with the function f:$S^1 rightarrow S^1$ that maps $z$ to $z^3$. What is the fundamental group of this space?



The exact statement of the question:




Cut a sphere $S^2$ through its equator and glue it back using the attaching map f:$S^1 rightarrow S^1$ defined by $f(z) = z^3$ . What is the fundamental group of this space?"




We take $S^1$ as the set of complex numbers with norm 1.



With Van kampen's theorem I compute the fundamental group as the trivial group. However, consider the path $gamma$ on the equator starting at some $x_0$ and moving $2pi/3$ degree around the equator. This path is correspond to a loop in the identification space. I fail to see how this path is homotopic to the trivial path.










share|cite|improve this question











$endgroup$




You split a sphere in half at the equator and glue back together the boundaries with the function f:$S^1 rightarrow S^1$ that maps $z$ to $z^3$. What is the fundamental group of this space?



The exact statement of the question:




Cut a sphere $S^2$ through its equator and glue it back using the attaching map f:$S^1 rightarrow S^1$ defined by $f(z) = z^3$ . What is the fundamental group of this space?"




We take $S^1$ as the set of complex numbers with norm 1.



With Van kampen's theorem I compute the fundamental group as the trivial group. However, consider the path $gamma$ on the equator starting at some $x_0$ and moving $2pi/3$ degree around the equator. This path is correspond to a loop in the identification space. I fail to see how this path is homotopic to the trivial path.







algebraic-topology homotopy-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 26 at 19:28







Baran Zadeoglu

















asked Mar 26 at 18:23









Baran ZadeogluBaran Zadeoglu

546




546







  • 1




    $begingroup$
    By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
    $endgroup$
    – Tempestas Ludi
    Mar 26 at 19:02










  • $begingroup$
    Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
    $endgroup$
    – Daniel P
    Mar 26 at 19:05






  • 1




    $begingroup$
    Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
    $endgroup$
    – Daniel P
    Mar 26 at 19:13










  • $begingroup$
    Okay, let me write the exact statement in the question. Please see the edited version.
    $endgroup$
    – Baran Zadeoglu
    Mar 26 at 19:24










  • $begingroup$
    Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
    $endgroup$
    – Daniel P
    Mar 26 at 19:27












  • 1




    $begingroup$
    By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
    $endgroup$
    – Tempestas Ludi
    Mar 26 at 19:02










  • $begingroup$
    Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
    $endgroup$
    – Daniel P
    Mar 26 at 19:05






  • 1




    $begingroup$
    Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
    $endgroup$
    – Daniel P
    Mar 26 at 19:13










  • $begingroup$
    Okay, let me write the exact statement in the question. Please see the edited version.
    $endgroup$
    – Baran Zadeoglu
    Mar 26 at 19:24










  • $begingroup$
    Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
    $endgroup$
    – Daniel P
    Mar 26 at 19:27







1




1




$begingroup$
By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
$endgroup$
– Tempestas Ludi
Mar 26 at 19:02




$begingroup$
By "glueing $z$ to $z^3$", you view them as complex numbers with $|z| = 1$?
$endgroup$
– Tempestas Ludi
Mar 26 at 19:02












$begingroup$
Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
$endgroup$
– Daniel P
Mar 26 at 19:05




$begingroup$
Please include more detail. What $S^1$ do you exactly rotate? Do you rotate the entire upper hemisphere with it using the $z rightarrow z^3$ function and then paste it back? How exactly does your example path look like? Since $pi_1(S^2) = 1$, I don't see how your example is different from a regular $S^2$.
$endgroup$
– Daniel P
Mar 26 at 19:05




1




1




$begingroup$
Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
$endgroup$
– Daniel P
Mar 26 at 19:13




$begingroup$
Nevermind, I get it now. Just had to think about it. So all points on the equator are equivalent to their cube, when viewed as a $|z| = 1$ complex number.
$endgroup$
– Daniel P
Mar 26 at 19:13












$begingroup$
Okay, let me write the exact statement in the question. Please see the edited version.
$endgroup$
– Baran Zadeoglu
Mar 26 at 19:24




$begingroup$
Okay, let me write the exact statement in the question. Please see the edited version.
$endgroup$
– Baran Zadeoglu
Mar 26 at 19:24












$begingroup$
Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
$endgroup$
– Daniel P
Mar 26 at 19:27




$begingroup$
Please also include what path you exactly mean by "starting at some $x_0$ and moving $2pi/3$ degree around the equator". Is this a loop? Where exactly does it go?
$endgroup$
– Daniel P
Mar 26 at 19:27










2 Answers
2






active

oldest

votes


















3












$begingroup$

Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z in partial N$ with $z^3 in partial S$. Then the path $c(t) = exp(frac2pi i3 t), t in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = exp(2pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.



    So, this does mean that a path on the first equator which goes around an arc of $2pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).



    (If you instead had a space which is the quotient of $S^2$ by the equivalence relation $zsim e^2pi i/3z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z in partial N$ with $z^3 in partial S$. Then the path $c(t) = exp(frac2pi i3 t), t in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = exp(2pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.






      share|cite|improve this answer











      $endgroup$

















        3












        $begingroup$

        Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z in partial N$ with $z^3 in partial S$. Then the path $c(t) = exp(frac2pi i3 t), t in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = exp(2pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.






        share|cite|improve this answer











        $endgroup$















          3












          3








          3





          $begingroup$

          Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z in partial N$ with $z^3 in partial S$. Then the path $c(t) = exp(frac2pi i3 t), t in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = exp(2pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.






          share|cite|improve this answer











          $endgroup$



          Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z in partial N$ with $z^3 in partial S$. Then the path $c(t) = exp(frac2pi i3 t), t in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = exp(2pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 26 at 19:46

























          answered Mar 26 at 19:33









          John HughesJohn Hughes

          65.5k24293




          65.5k24293





















              2












              $begingroup$

              I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.



              So, this does mean that a path on the first equator which goes around an arc of $2pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).



              (If you instead had a space which is the quotient of $S^2$ by the equivalence relation $zsim e^2pi i/3z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.



                So, this does mean that a path on the first equator which goes around an arc of $2pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).



                (If you instead had a space which is the quotient of $S^2$ by the equivalence relation $zsim e^2pi i/3z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.



                  So, this does mean that a path on the first equator which goes around an arc of $2pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).



                  (If you instead had a space which is the quotient of $S^2$ by the equivalence relation $zsim e^2pi i/3z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)






                  share|cite|improve this answer









                  $endgroup$



                  I think you are getting confused because the space you describe (which I'll call $X$) really has two different "equators". Note that $X$ is not simply the quotient of $S^2$ by an equivalence relation on the equator. Instead, $X$ is two hemispheres attached together where $z$ in the equator of one hemisphere is identified with $z^3$ in the equator of the other hemisphere.



                  So, this does mean that a path on the first equator which goes around an arc of $2pi/3$ is a loop, since its endpoints become identified when you glue it to the second hemisphere. There is no nullhomotopy of this loop inside the first hemisphere. But when you look at this loop as living in the second hemisphere instead, it's just an ordinary loop that goes around the equator once. So, it is nullhomotopic inside the second hemisphere, which really is just homeomorphic to an ordinary disk (unlike the first hemisphere whose equator has gotten glued together).



                  (If you instead had a space which is the quotient of $S^2$ by the equivalence relation $zsim e^2pi i/3z$ on the equator, then it would not be simply connected and indeed the loop going a third of the way around the equator would not be nullhomotopic.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 26 at 21:11









                  Eric WofseyEric Wofsey

                  193k14221352




                  193k14221352



























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                      Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye