Find Levi-Civita Connection in Hyperbolic Space
$begingroup$
With:
$$
mathbb{H}^n=left{ (x_0,x_1,dots, x_n)in mathbb{R}^{n+1}: ; x_0^2=1+x_1^2+cdots +x_n^2,; x_0>0right}.
$$
and the form
$$
langlelangle(u_0,u_1,dots, u_n),(v_0,v_1,dots,v_n)ranglerangle=-u_0v_0+u_1v_1+cdots+u_nv_n,
$$
I have to prove that
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle P,
$$
where $overline{nabla}$ is the connection in $mathbb{R}^{n+1}$ and $P(p)=p$.
I have tried what follows:
I know that the normal vector to a point $p=(p_0,p_1,dots,p_n)inmathbb{H}^n$ is $(-p_0,p_1,dots,p_n)$ and also that ($langle, rangle$ is the usual product):
$$
nabla_X Y =overline{nabla}_X Y -langle overline{nabla}_XY,Nrangle N.
$$
Since $langle overline{nabla}_XY,Nrangle=X(langle Y,Nrangle)-langle Y,overline{nabla}_X Nrangle=-langle Y,overline{nabla}_X Nrangle$. Now $overline{nabla}_X N=sum X(N_i)frac{partial}{partial x_i}=X'$, where $X'$ is $X$, but with a change of sign in the first coordinate. Now $langle Y,overline{nabla}_X Nrangle=langle Y,X'rangle=langle langle Y,X ranglerangle$.
So I get
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle N,
$$
which is the same as above but with a change in a sign. Where is my mistake?
geometry differential-geometry riemannian-geometry
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
$endgroup$
add a comment |
$begingroup$
With:
$$
mathbb{H}^n=left{ (x_0,x_1,dots, x_n)in mathbb{R}^{n+1}: ; x_0^2=1+x_1^2+cdots +x_n^2,; x_0>0right}.
$$
and the form
$$
langlelangle(u_0,u_1,dots, u_n),(v_0,v_1,dots,v_n)ranglerangle=-u_0v_0+u_1v_1+cdots+u_nv_n,
$$
I have to prove that
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle P,
$$
where $overline{nabla}$ is the connection in $mathbb{R}^{n+1}$ and $P(p)=p$.
I have tried what follows:
I know that the normal vector to a point $p=(p_0,p_1,dots,p_n)inmathbb{H}^n$ is $(-p_0,p_1,dots,p_n)$ and also that ($langle, rangle$ is the usual product):
$$
nabla_X Y =overline{nabla}_X Y -langle overline{nabla}_XY,Nrangle N.
$$
Since $langle overline{nabla}_XY,Nrangle=X(langle Y,Nrangle)-langle Y,overline{nabla}_X Nrangle=-langle Y,overline{nabla}_X Nrangle$. Now $overline{nabla}_X N=sum X(N_i)frac{partial}{partial x_i}=X'$, where $X'$ is $X$, but with a change of sign in the first coordinate. Now $langle Y,overline{nabla}_X Nrangle=langle Y,X'rangle=langle langle Y,X ranglerangle$.
So I get
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle N,
$$
which is the same as above but with a change in a sign. Where is my mistake?
geometry differential-geometry riemannian-geometry
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
$endgroup$
$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39
add a comment |
$begingroup$
With:
$$
mathbb{H}^n=left{ (x_0,x_1,dots, x_n)in mathbb{R}^{n+1}: ; x_0^2=1+x_1^2+cdots +x_n^2,; x_0>0right}.
$$
and the form
$$
langlelangle(u_0,u_1,dots, u_n),(v_0,v_1,dots,v_n)ranglerangle=-u_0v_0+u_1v_1+cdots+u_nv_n,
$$
I have to prove that
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle P,
$$
where $overline{nabla}$ is the connection in $mathbb{R}^{n+1}$ and $P(p)=p$.
I have tried what follows:
I know that the normal vector to a point $p=(p_0,p_1,dots,p_n)inmathbb{H}^n$ is $(-p_0,p_1,dots,p_n)$ and also that ($langle, rangle$ is the usual product):
$$
nabla_X Y =overline{nabla}_X Y -langle overline{nabla}_XY,Nrangle N.
$$
Since $langle overline{nabla}_XY,Nrangle=X(langle Y,Nrangle)-langle Y,overline{nabla}_X Nrangle=-langle Y,overline{nabla}_X Nrangle$. Now $overline{nabla}_X N=sum X(N_i)frac{partial}{partial x_i}=X'$, where $X'$ is $X$, but with a change of sign in the first coordinate. Now $langle Y,overline{nabla}_X Nrangle=langle Y,X'rangle=langle langle Y,X ranglerangle$.
So I get
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle N,
$$
which is the same as above but with a change in a sign. Where is my mistake?
geometry differential-geometry riemannian-geometry
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
$endgroup$
With:
$$
mathbb{H}^n=left{ (x_0,x_1,dots, x_n)in mathbb{R}^{n+1}: ; x_0^2=1+x_1^2+cdots +x_n^2,; x_0>0right}.
$$
and the form
$$
langlelangle(u_0,u_1,dots, u_n),(v_0,v_1,dots,v_n)ranglerangle=-u_0v_0+u_1v_1+cdots+u_nv_n,
$$
I have to prove that
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle P,
$$
where $overline{nabla}$ is the connection in $mathbb{R}^{n+1}$ and $P(p)=p$.
I have tried what follows:
I know that the normal vector to a point $p=(p_0,p_1,dots,p_n)inmathbb{H}^n$ is $(-p_0,p_1,dots,p_n)$ and also that ($langle, rangle$ is the usual product):
$$
nabla_X Y =overline{nabla}_X Y -langle overline{nabla}_XY,Nrangle N.
$$
Since $langle overline{nabla}_XY,Nrangle=X(langle Y,Nrangle)-langle Y,overline{nabla}_X Nrangle=-langle Y,overline{nabla}_X Nrangle$. Now $overline{nabla}_X N=sum X(N_i)frac{partial}{partial x_i}=X'$, where $X'$ is $X$, but with a change of sign in the first coordinate. Now $langle Y,overline{nabla}_X Nrangle=langle Y,X'rangle=langle langle Y,X ranglerangle$.
So I get
$$
nabla_X Y =overline{nabla}_X Y -langlelangle X,Yranglerangle N,
$$
which is the same as above but with a change in a sign. Where is my mistake?
geometry differential-geometry riemannian-geometry
geometry differential-geometry riemannian-geometry
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
edited Jan 7 at 8:35
davidivadful
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
asked Jan 6 at 19:41
davidivadfuldavidivadful
13610
13610
$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39
add a comment |
$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39
$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39
add a comment |
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$begingroup$
I don't see any sign difference. Can you proofread? Also, you have a typo in the definition of $langlelangle u,vranglerangle$.
$endgroup$
– Ted Shifrin
Jan 7 at 0:49
$begingroup$
The difference is that in one equation appears $P$ and in other $N$. And $p=(p_0, p_1,dots,p_n$ while $N=(-p_0,p_1,dots,p_n)$.
$endgroup$
– davidivadful
Jan 7 at 8:39