Show that the $Delta$-complex obtained from $Delta^3$ by performing edge identifications deformation retracts...
$begingroup$
I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?
algebraic-topology simplicial-complex
edited Nov 6 '18 at 9:51
Batominovski
1
asked Jan 18 '12 at 15:07
0986709867
294414
$endgroup$
add a comment |
$begingroup$
I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?
algebraic-topology simplicial-complex
edited Nov 6 '18 at 9:51
Batominovski
1
asked Jan 18 '12 at 15:07
0986709867
294414
$endgroup$
add a comment |
$begingroup$
I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?
algebraic-topology simplicial-complex
edited Nov 6 '18 at 9:51
Batominovski
1
asked Jan 18 '12 at 15:07
0986709867
294414
$endgroup$
I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?
algebraic-topology simplicial-complex
algebraic-topology simplicial-complex
edited Nov 6 '18 at 9:51
Batominovski
1
asked Jan 18 '12 at 15:07
0986709867
294414
edited Nov 6 '18 at 9:51
Batominovski
1
asked Jan 18 '12 at 15:07
0986709867
294414
edited Nov 6 '18 at 9:51
Batominovski
1
edited Nov 6 '18 at 9:51
Batominovski
1
edited Nov 6 '18 at 9:51
Batominovski
1
1
asked Jan 18 '12 at 15:07
0986709867
294414
asked Jan 18 '12 at 15:07
0986709867
294414
asked Jan 18 '12 at 15:07
0986709867
294414
294414
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.
http://oi40.tinypic.com/2u6jzlu.jpg sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
$endgroup$
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
add a comment |
$begingroup$
The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.
http://oi40.tinypic.com/2u6jzlu.jpg sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
$endgroup$
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
add a comment |
$begingroup$
flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.
http://oi40.tinypic.com/2u6jzlu.jpg sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
$endgroup$
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
add a comment |
$begingroup$
flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.
http://oi40.tinypic.com/2u6jzlu.jpg sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
$endgroup$
flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.
http://oi40.tinypic.com/2u6jzlu.jpg sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
edited Jun 4 '16 at 0:13
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
answered Jan 18 '12 at 16:09
yoyoyoyo
6,5521626
6,5521626
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
add a comment |
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
$endgroup$
– NKS
Jan 18 '12 at 16:50
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
$endgroup$
– 09867
Jan 18 '12 at 16:52
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@NKS yes you do have to cut it up, my bad
$endgroup$
– yoyo
Jan 18 '12 at 19:12
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
$begingroup$
@yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
$endgroup$
– yoyostein
Jun 1 '16 at 9:28
add a comment |
$begingroup$
The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
$endgroup$
add a comment |
$begingroup$
The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
$endgroup$
add a comment |
$begingroup$
The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
$endgroup$
The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
edited Jul 3 '17 at 10:39
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
answered Jun 29 '17 at 14:22
Ka HoKa Ho
62
62
add a comment |
add a comment |
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