Mixing
Embedding $S^1timescdotstimes S^1$ into $mathbbR^k+1$
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I'm currently attempting to come up with an explicit parametrization of the $k$-torus into $mathbbR^k+1$. I've been following the answer here but I'm struggling to prove injectivity.
Let $e_1,ldots, e_k+1$ be the standard basis in $mathbbR^k+1$ and $varepsilon<1$. So according to the link, I should start with an element $v_1$ of length 1 in span$(e_1,e_2)$, i.e.
$$
v_1(theta_1)=cos(theta_1)e_1+sin(theta_2)e_2
$$
Then from there I should choose an element $v_2$ of length $varepsilon$ in span$(v_1,e_3)$, i.e.
$$
v_2(theta_1,theta_2)=varepsiloncos(theta_2)v_1(theta_1)+varepsilonsin(theta_2)e_3
$$
and in general (for $1leq jleq k$), set
$$
v_j(theta_1,ldots,theta_j)=varepsilon^j-1cos(theta_j)v_j-1(theta_1,ldots ,theta_j)+varepsilon^j-1sin(theta_j)e_j+1.
$$
From this, the map
$$
Phi:(e^itheta_1,ldots,e^itheta_k)mapstosum_j=1^kv_j(theta_1,ldots,theta_j)
$$
is supposed to be the necessary embedding. This makes sense intuitively, but I can't seem to prove injectivity.
Things Tried
The only case I can make any progress is in the case $k=3$. I've tried doing induction and immitating this argument in the inductive step, but it doesn't seem to work.
If $k=3$, then $$Phi(e^itheta_1,e^itheta_1)=Phi(e^itheta_1',e^itheta_1')$$
implies
$$
(1+varepsiloncos(theta_2))v_1(theta_1)+varepsilonsin(theta_2)e_3=(1+varepsiloncos(theta_2'))v_1(theta_1')+varepsilonsin(theta_2')e_3.
$$
In this case the vectors being added together are orthogonal, so taking the norm-square of both sides gives
$$
(1+varepsiloncos(theta_2))^2+varepsilon^2sin(theta_2)=(1+varepsiloncos(theta_2'))^2+varepsilon^2sin(theta_2'),
$$
which after simplifying turns into
$$
2varepsiloncos(theta_2)=2varepsiloncos(theta_2),
$$
so $cos(theta_2)=cos(theta_2')$. This combined with the immediate fact that $sin(theta_2)=sin(theta_2')$ shows that $theta_2=theta_2'$ (in $[0,2pi)$). The rest follows from the fact that $v_1$ is injective.
Any help is greatly appreciated. Thanks in advance.
general-topology differential-geometry
asked Jul 20 at 3:05
Blake
1,086516
 |Â
show 5 more comments
up vote
5
down vote
favorite
I'm currently attempting to come up with an explicit parametrization of the $k$-torus into $mathbbR^k+1$. I've been following the answer here but I'm struggling to prove injectivity.
Let $e_1,ldots, e_k+1$ be the standard basis in $mathbbR^k+1$ and $varepsilon<1$. So according to the link, I should start with an element $v_1$ of length 1 in span$(e_1,e_2)$, i.e.
$$
v_1(theta_1)=cos(theta_1)e_1+sin(theta_2)e_2
$$
Then from there I should choose an element $v_2$ of length $varepsilon$ in span$(v_1,e_3)$, i.e.
$$
v_2(theta_1,theta_2)=varepsiloncos(theta_2)v_1(theta_1)+varepsilonsin(theta_2)e_3
$$
and in general (for $1leq jleq k$), set
$$
v_j(theta_1,ldots,theta_j)=varepsilon^j-1cos(theta_j)v_j-1(theta_1,ldots ,theta_j)+varepsilon^j-1sin(theta_j)e_j+1.
$$
From this, the map
$$
Phi:(e^itheta_1,ldots,e^itheta_k)mapstosum_j=1^kv_j(theta_1,ldots,theta_j)
$$
is supposed to be the necessary embedding. This makes sense intuitively, but I can't seem to prove injectivity.
Things Tried
The only case I can make any progress is in the case $k=3$. I've tried doing induction and immitating this argument in the inductive step, but it doesn't seem to work.
If $k=3$, then $$Phi(e^itheta_1,e^itheta_1)=Phi(e^itheta_1',e^itheta_1')$$
implies
$$
(1+varepsiloncos(theta_2))v_1(theta_1)+varepsilonsin(theta_2)e_3=(1+varepsiloncos(theta_2'))v_1(theta_1')+varepsilonsin(theta_2')e_3.
$$
In this case the vectors being added together are orthogonal, so taking the norm-square of both sides gives
$$
(1+varepsiloncos(theta_2))^2+varepsilon^2sin(theta_2)=(1+varepsiloncos(theta_2'))^2+varepsilon^2sin(theta_2'),
$$
which after simplifying turns into
$$
2varepsiloncos(theta_2)=2varepsiloncos(theta_2),
$$
so $cos(theta_2)=cos(theta_2')$. This combined with the immediate fact that $sin(theta_2)=sin(theta_2')$ shows that $theta_2=theta_2'$ (in $[0,2pi)$). The rest follows from the fact that $v_1$ is injective.
Any help is greatly appreciated. Thanks in advance.
general-topology differential-geometry
asked Jul 20 at 3:05
Blake
1,086516
Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
1
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30
 |Â
show 5 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm currently attempting to come up with an explicit parametrization of the $k$-torus into $mathbbR^k+1$. I've been following the answer here but I'm struggling to prove injectivity.
Let $e_1,ldots, e_k+1$ be the standard basis in $mathbbR^k+1$ and $varepsilon<1$. So according to the link, I should start with an element $v_1$ of length 1 in span$(e_1,e_2)$, i.e.
$$
v_1(theta_1)=cos(theta_1)e_1+sin(theta_2)e_2
$$
Then from there I should choose an element $v_2$ of length $varepsilon$ in span$(v_1,e_3)$, i.e.
$$
v_2(theta_1,theta_2)=varepsiloncos(theta_2)v_1(theta_1)+varepsilonsin(theta_2)e_3
$$
and in general (for $1leq jleq k$), set
$$
v_j(theta_1,ldots,theta_j)=varepsilon^j-1cos(theta_j)v_j-1(theta_1,ldots ,theta_j)+varepsilon^j-1sin(theta_j)e_j+1.
$$
From this, the map
$$
Phi:(e^itheta_1,ldots,e^itheta_k)mapstosum_j=1^kv_j(theta_1,ldots,theta_j)
$$
is supposed to be the necessary embedding. This makes sense intuitively, but I can't seem to prove injectivity.
Things Tried
The only case I can make any progress is in the case $k=3$. I've tried doing induction and immitating this argument in the inductive step, but it doesn't seem to work.
If $k=3$, then $$Phi(e^itheta_1,e^itheta_1)=Phi(e^itheta_1',e^itheta_1')$$
implies
$$
(1+varepsiloncos(theta_2))v_1(theta_1)+varepsilonsin(theta_2)e_3=(1+varepsiloncos(theta_2'))v_1(theta_1')+varepsilonsin(theta_2')e_3.
$$
In this case the vectors being added together are orthogonal, so taking the norm-square of both sides gives
$$
(1+varepsiloncos(theta_2))^2+varepsilon^2sin(theta_2)=(1+varepsiloncos(theta_2'))^2+varepsilon^2sin(theta_2'),
$$
which after simplifying turns into
$$
2varepsiloncos(theta_2)=2varepsiloncos(theta_2),
$$
so $cos(theta_2)=cos(theta_2')$. This combined with the immediate fact that $sin(theta_2)=sin(theta_2')$ shows that $theta_2=theta_2'$ (in $[0,2pi)$). The rest follows from the fact that $v_1$ is injective.
Any help is greatly appreciated. Thanks in advance.
general-topology differential-geometry
asked Jul 20 at 3:05
Blake
1,086516
I'm currently attempting to come up with an explicit parametrization of the $k$-torus into $mathbbR^k+1$. I've been following the answer here but I'm struggling to prove injectivity.
Let $e_1,ldots, e_k+1$ be the standard basis in $mathbbR^k+1$ and $varepsilon<1$. So according to the link, I should start with an element $v_1$ of length 1 in span$(e_1,e_2)$, i.e.
$$
v_1(theta_1)=cos(theta_1)e_1+sin(theta_2)e_2
$$
Then from there I should choose an element $v_2$ of length $varepsilon$ in span$(v_1,e_3)$, i.e.
$$
v_2(theta_1,theta_2)=varepsiloncos(theta_2)v_1(theta_1)+varepsilonsin(theta_2)e_3
$$
and in general (for $1leq jleq k$), set
$$
v_j(theta_1,ldots,theta_j)=varepsilon^j-1cos(theta_j)v_j-1(theta_1,ldots ,theta_j)+varepsilon^j-1sin(theta_j)e_j+1.
$$
From this, the map
$$
Phi:(e^itheta_1,ldots,e^itheta_k)mapstosum_j=1^kv_j(theta_1,ldots,theta_j)
$$
is supposed to be the necessary embedding. This makes sense intuitively, but I can't seem to prove injectivity.
Things Tried
The only case I can make any progress is in the case $k=3$. I've tried doing induction and immitating this argument in the inductive step, but it doesn't seem to work.
If $k=3$, then $$Phi(e^itheta_1,e^itheta_1)=Phi(e^itheta_1',e^itheta_1')$$
implies
$$
(1+varepsiloncos(theta_2))v_1(theta_1)+varepsilonsin(theta_2)e_3=(1+varepsiloncos(theta_2'))v_1(theta_1')+varepsilonsin(theta_2')e_3.
$$
In this case the vectors being added together are orthogonal, so taking the norm-square of both sides gives
$$
(1+varepsiloncos(theta_2))^2+varepsilon^2sin(theta_2)=(1+varepsiloncos(theta_2'))^2+varepsilon^2sin(theta_2'),
$$
which after simplifying turns into
$$
2varepsiloncos(theta_2)=2varepsiloncos(theta_2),
$$
so $cos(theta_2)=cos(theta_2')$. This combined with the immediate fact that $sin(theta_2)=sin(theta_2')$ shows that $theta_2=theta_2'$ (in $[0,2pi)$). The rest follows from the fact that $v_1$ is injective.
Any help is greatly appreciated. Thanks in advance.
general-topology differential-geometry
asked Jul 20 at 3:05
Blake
1,086516
edited Jul 20 at 3:20
asked Jul 20 at 3:05
Blake
1,086516
asked Jul 20 at 3:05
Blake
1,086516
asked Jul 20 at 3:05
Blake
1,086516
1,086516
Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
1
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30
 |Â
show 5 more comments
Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
1
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30
Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
1
1
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $Bbb R^n+1$. We want to use this
as a basis for one of $T^n+1=T^ntimes S^1$ into $Bbb R^n+2$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_n+1>0$. Then let the embedding of $T^n$
be given by functions
$pmapsto (phi(p),psi(p))$ where $phi(p)inBbb R^n$ and $psi(p)in(0,infty)$. Now the embedding of $T^n+1$ into $Bbb R^n+2$ by
$$(p,e^it)mapsto(phi(p),psi(p)cos t,psi(p)sin t).$$
As $psi(p)>0$ the last two coordinates determine $psi(p)$ and $e^it$.
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $Bbb R^n+1$. We want to use this
as a basis for one of $T^n+1=T^ntimes S^1$ into $Bbb R^n+2$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_n+1>0$. Then let the embedding of $T^n$
be given by functions
$pmapsto (phi(p),psi(p))$ where $phi(p)inBbb R^n$ and $psi(p)in(0,infty)$. Now the embedding of $T^n+1$ into $Bbb R^n+2$ by
$$(p,e^it)mapsto(phi(p),psi(p)cos t,psi(p)sin t).$$
As $psi(p)>0$ the last two coordinates determine $psi(p)$ and $e^it$.
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
add a comment |Â
up vote
2
down vote
accepted
Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $Bbb R^n+1$. We want to use this
as a basis for one of $T^n+1=T^ntimes S^1$ into $Bbb R^n+2$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_n+1>0$. Then let the embedding of $T^n$
be given by functions
$pmapsto (phi(p),psi(p))$ where $phi(p)inBbb R^n$ and $psi(p)in(0,infty)$. Now the embedding of $T^n+1$ into $Bbb R^n+2$ by
$$(p,e^it)mapsto(phi(p),psi(p)cos t,psi(p)sin t).$$
As $psi(p)>0$ the last two coordinates determine $psi(p)$ and $e^it$.
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $Bbb R^n+1$. We want to use this
as a basis for one of $T^n+1=T^ntimes S^1$ into $Bbb R^n+2$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_n+1>0$. Then let the embedding of $T^n$
be given by functions
$pmapsto (phi(p),psi(p))$ where $phi(p)inBbb R^n$ and $psi(p)in(0,infty)$. Now the embedding of $T^n+1$ into $Bbb R^n+2$ by
$$(p,e^it)mapsto(phi(p),psi(p)cos t,psi(p)sin t).$$
As $psi(p)>0$ the last two coordinates determine $psi(p)$ and $e^it$.
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
Here's a description of an embedding, which can be elaborated into an
explicit formula if you like.
Take an embedding of $T^n$ into $Bbb R^n+1$. We want to use this
as a basis for one of $T^n+1=T^ntimes S^1$ into $Bbb R^n+2$.
Translate your $T^n$ so that it lies within the half-plane
defined by $x_n+1>0$. Then let the embedding of $T^n$
be given by functions
$pmapsto (phi(p),psi(p))$ where $phi(p)inBbb R^n$ and $psi(p)in(0,infty)$. Now the embedding of $T^n+1$ into $Bbb R^n+2$ by
$$(p,e^it)mapsto(phi(p),psi(p)cos t,psi(p)sin t).$$
As $psi(p)>0$ the last two coordinates determine $psi(p)$ and $e^it$.
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
answered Jul 20 at 5:33
Lord Shark the Unknown
85.3k950111
85.3k950111
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
add a comment |Â
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
I like this a lot. Thank you.
– Blake
Jul 20 at 6:30
add a comment |Â
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Can you give some more background? Is there a reason you want $k+1$ space instead of just some Euclidean space? $Bbb R^2k$ is particularly pleasant for this problem, for example.
– Adam Hughes
Jul 20 at 3:23
@AdamHughes I probably didn't make it very clear. I'm attempting to embed the $k$-dimensional torus into $(k+1)$-dimensional Euclidean space, for any $k$. I'm pretty sure $k+1$ is the smallest dimension where this is possible. Does this clear up the confusion?
– Blake
Jul 20 at 3:26
Right, I got what you were asking, I was just wondering why you cared about it being the smallest $k$, mostly because there is a ridiculously easy embedding if you're willing to use a few more dimensions.
– Adam Hughes
Jul 20 at 3:52
Do you care about an explicit embedding, or just an existence proof?
– Steve D
Jul 20 at 5:03
1
@SteveD: I think this is an equivalent formulation. I guess, in fact, that it holds for any product of hypersurfaces in Euclidean space. No need to restrict to spheres.
– Ted Shifrin
Jul 22 at 1:30