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Why we use (0,1,0) for free group proof in Banach-Tarski paradox
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Why we use point $(0,1,0)$ as a starting point when we prove that rotations with $theta = arccos(frac13)$ form free group? Why not $(fracsqrt 22, fracsqrt 22, 0)$ or any other not on one of the rotations axes?
If $rho: mathbfR^3tomathbfR^3$ is an expression in $mathcal G$ with length $n$ in reduced form, then $rho(0,1,0)$ is of the following form, where $a,b, $ and $c$ are intergers:
$$rho(0,1,0) = frac13^nleft(asqrt 2, b, csqrt 2right).$$
geometry
edited Jul 28 at 15:27
amWhy
189k25219431
asked Jul 28 at 15:12
Yola
742721
add a comment |Â
up vote
0
down vote
favorite
Why we use point $(0,1,0)$ as a starting point when we prove that rotations with $theta = arccos(frac13)$ form free group? Why not $(fracsqrt 22, fracsqrt 22, 0)$ or any other not on one of the rotations axes?
If $rho: mathbfR^3tomathbfR^3$ is an expression in $mathcal G$ with length $n$ in reduced form, then $rho(0,1,0)$ is of the following form, where $a,b, $ and $c$ are intergers:
$$rho(0,1,0) = frac13^nleft(asqrt 2, b, csqrt 2right).$$
geometry
edited Jul 28 at 15:27
amWhy
189k25219431
asked Jul 28 at 15:12
Yola
742721
why not use (0,1,0)?
– amWhy
Jul 28 at 15:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why we use point $(0,1,0)$ as a starting point when we prove that rotations with $theta = arccos(frac13)$ form free group? Why not $(fracsqrt 22, fracsqrt 22, 0)$ or any other not on one of the rotations axes?
If $rho: mathbfR^3tomathbfR^3$ is an expression in $mathcal G$ with length $n$ in reduced form, then $rho(0,1,0)$ is of the following form, where $a,b, $ and $c$ are intergers:
$$rho(0,1,0) = frac13^nleft(asqrt 2, b, csqrt 2right).$$
geometry
edited Jul 28 at 15:27
amWhy
189k25219431
asked Jul 28 at 15:12
Yola
742721
Why we use point $(0,1,0)$ as a starting point when we prove that rotations with $theta = arccos(frac13)$ form free group? Why not $(fracsqrt 22, fracsqrt 22, 0)$ or any other not on one of the rotations axes?
If $rho: mathbfR^3tomathbfR^3$ is an expression in $mathcal G$ with length $n$ in reduced form, then $rho(0,1,0)$ is of the following form, where $a,b, $ and $c$ are intergers:
$$rho(0,1,0) = frac13^nleft(asqrt 2, b, csqrt 2right).$$
geometry
edited Jul 28 at 15:27
amWhy
189k25219431
asked Jul 28 at 15:12
Yola
742721
edited Jul 28 at 15:27
amWhy
189k25219431
edited Jul 28 at 15:27
amWhy
189k25219431
edited Jul 28 at 15:27
amWhy
189k25219431
189k25219431
asked Jul 28 at 15:12
Yola
742721
asked Jul 28 at 15:12
Yola
742721
asked Jul 28 at 15:12
Yola
742721
742721
why not use (0,1,0)?
– amWhy
Jul 28 at 15:24
add a comment |Â
why not use (0,1,0)?
– amWhy
Jul 28 at 15:24
why not use (0,1,0)?
– amWhy
Jul 28 at 15:24
why not use (0,1,0)?
– amWhy
Jul 28 at 15:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
Having proven it for one point we know it is true for all the other points not on the axis because it is a rigid rotation. We use $(0,1,0)$ because it makes the proof easy. It is easy to multiply by $1$ and $0$. It is similar to some statements "prove such and such about all triangles." If the property is maintained by affine transformations you can prove it about one triangle and be done. Often using a triangle like $(0,0),(1,0),(0,1)$ makes the calculations easier.
answered Jul 28 at 15:33
Ross Millikan
275k21185351
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Having proven it for one point we know it is true for all the other points not on the axis because it is a rigid rotation. We use $(0,1,0)$ because it makes the proof easy. It is easy to multiply by $1$ and $0$. It is similar to some statements "prove such and such about all triangles." If the property is maintained by affine transformations you can prove it about one triangle and be done. Often using a triangle like $(0,0),(1,0),(0,1)$ makes the calculations easier.
answered Jul 28 at 15:33
Ross Millikan
275k21185351
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
add a comment |Â
up vote
1
down vote
accepted
Having proven it for one point we know it is true for all the other points not on the axis because it is a rigid rotation. We use $(0,1,0)$ because it makes the proof easy. It is easy to multiply by $1$ and $0$. It is similar to some statements "prove such and such about all triangles." If the property is maintained by affine transformations you can prove it about one triangle and be done. Often using a triangle like $(0,0),(1,0),(0,1)$ makes the calculations easier.
answered Jul 28 at 15:33
Ross Millikan
275k21185351
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Having proven it for one point we know it is true for all the other points not on the axis because it is a rigid rotation. We use $(0,1,0)$ because it makes the proof easy. It is easy to multiply by $1$ and $0$. It is similar to some statements "prove such and such about all triangles." If the property is maintained by affine transformations you can prove it about one triangle and be done. Often using a triangle like $(0,0),(1,0),(0,1)$ makes the calculations easier.
answered Jul 28 at 15:33
Ross Millikan
275k21185351
Having proven it for one point we know it is true for all the other points not on the axis because it is a rigid rotation. We use $(0,1,0)$ because it makes the proof easy. It is easy to multiply by $1$ and $0$. It is similar to some statements "prove such and such about all triangles." If the property is maintained by affine transformations you can prove it about one triangle and be done. Often using a triangle like $(0,0),(1,0),(0,1)$ makes the calculations easier.
answered Jul 28 at 15:33
Ross Millikan
275k21185351
answered Jul 28 at 15:33
Ross Millikan
275k21185351
answered Jul 28 at 15:33
Ross Millikan
275k21185351
answered Jul 28 at 15:33
Ross Millikan
275k21185351
275k21185351
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
add a comment |Â
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
I think for this to be true we would need to be sure that we can get to all the other points not on the axis starting at $(0,1,0)$.
– Yola
Jul 28 at 15:41
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
You know, I think that this is true not because all the points not on the axes are equivalent in this sense, but because that is our setup. We start with $(0,1,0)$ and two axes $x$ and $z$. So coordinate system is per Caley graph.
– Yola
Jul 28 at 17:59
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
No, starting at $(0,1,0)$ you can only get to its orbit. which for a free group on two generators is only countably many points. The important thing to the proof is that no word applied to $(0,1,0)$ except the empty word brings us back to $(0,1,0)$. Using this starting point makes that easier.
– Ross Millikan
Jul 28 at 19:33
add a comment |Â
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why not use (0,1,0)?
– amWhy
Jul 28 at 15:24