Mixing
Weyl curvature of hypersurface in $mathbb{R}^{n+1}$
$begingroup$
Let $M^nsubset mathbb{R}^{n+1}$ be an embedded hypersurface and let $g$ be the metric induced on $M$ by the flat metric on $mathbb{R}^{n+1}$.
Q: What is a formula for the Weyl tensor $W$ of $g$ in terms of the second fundamental form $A$ of $M$? More precisely, what is $|W|^2$?
For example, the scalar curvature $S$ of $g$ can be written as
$$
S=H^2-|A|^2=mathrm{trace}(A)^2-mathrm{trace}(A^2)=sum_{1leq i_1<i_2leq n}lambda_{i_1}lambda_{i_2},
$$
where $H$ is the mean curvature (trace of $A$) and $lambda_1,dots,lambda_nin mathbb{R}$ are the principal curvatures of $M$, i.e. the eigenvalues of $A$. Note that $S$ is thus an elementary symmetric polynomial in $lambda_1,dots,lambda_n$.
Is $|W|^2$ also given by such a polynomial? I guess this is probably not true as $|W|^2$ is zero for any 2-manifold (all surfaces are locally conformally flat).
differential-geometry riemannian-geometry smooth-manifolds
asked Jan 24 at 2:43
srpsrp
2728
$endgroup$
add a comment |
$begingroup$
Let $M^nsubset mathbb{R}^{n+1}$ be an embedded hypersurface and let $g$ be the metric induced on $M$ by the flat metric on $mathbb{R}^{n+1}$.
Q: What is a formula for the Weyl tensor $W$ of $g$ in terms of the second fundamental form $A$ of $M$? More precisely, what is $|W|^2$?
For example, the scalar curvature $S$ of $g$ can be written as
$$
S=H^2-|A|^2=mathrm{trace}(A)^2-mathrm{trace}(A^2)=sum_{1leq i_1<i_2leq n}lambda_{i_1}lambda_{i_2},
$$
where $H$ is the mean curvature (trace of $A$) and $lambda_1,dots,lambda_nin mathbb{R}$ are the principal curvatures of $M$, i.e. the eigenvalues of $A$. Note that $S$ is thus an elementary symmetric polynomial in $lambda_1,dots,lambda_n$.
Is $|W|^2$ also given by such a polynomial? I guess this is probably not true as $|W|^2$ is zero for any 2-manifold (all surfaces are locally conformally flat).
differential-geometry riemannian-geometry smooth-manifolds
asked Jan 24 at 2:43
srpsrp
2728
$endgroup$
add a comment |
$begingroup$
Let $M^nsubset mathbb{R}^{n+1}$ be an embedded hypersurface and let $g$ be the metric induced on $M$ by the flat metric on $mathbb{R}^{n+1}$.
Q: What is a formula for the Weyl tensor $W$ of $g$ in terms of the second fundamental form $A$ of $M$? More precisely, what is $|W|^2$?
For example, the scalar curvature $S$ of $g$ can be written as
$$
S=H^2-|A|^2=mathrm{trace}(A)^2-mathrm{trace}(A^2)=sum_{1leq i_1<i_2leq n}lambda_{i_1}lambda_{i_2},
$$
where $H$ is the mean curvature (trace of $A$) and $lambda_1,dots,lambda_nin mathbb{R}$ are the principal curvatures of $M$, i.e. the eigenvalues of $A$. Note that $S$ is thus an elementary symmetric polynomial in $lambda_1,dots,lambda_n$.
Is $|W|^2$ also given by such a polynomial? I guess this is probably not true as $|W|^2$ is zero for any 2-manifold (all surfaces are locally conformally flat).
differential-geometry riemannian-geometry smooth-manifolds
asked Jan 24 at 2:43
srpsrp
2728
$endgroup$
Let $M^nsubset mathbb{R}^{n+1}$ be an embedded hypersurface and let $g$ be the metric induced on $M$ by the flat metric on $mathbb{R}^{n+1}$.
Q: What is a formula for the Weyl tensor $W$ of $g$ in terms of the second fundamental form $A$ of $M$? More precisely, what is $|W|^2$?
For example, the scalar curvature $S$ of $g$ can be written as
$$
S=H^2-|A|^2=mathrm{trace}(A)^2-mathrm{trace}(A^2)=sum_{1leq i_1<i_2leq n}lambda_{i_1}lambda_{i_2},
$$
where $H$ is the mean curvature (trace of $A$) and $lambda_1,dots,lambda_nin mathbb{R}$ are the principal curvatures of $M$, i.e. the eigenvalues of $A$. Note that $S$ is thus an elementary symmetric polynomial in $lambda_1,dots,lambda_n$.
Is $|W|^2$ also given by such a polynomial? I guess this is probably not true as $|W|^2$ is zero for any 2-manifold (all surfaces are locally conformally flat).
differential-geometry riemannian-geometry smooth-manifolds
differential-geometry riemannian-geometry smooth-manifolds
asked Jan 24 at 2:43
srpsrp
2728
asked Jan 24 at 2:43
srpsrp
2728
edited Jan 24 at 15:35
srp
asked Jan 24 at 2:43
srpsrp
2728
asked Jan 24 at 2:43
srpsrp
2728
asked Jan 24 at 2:43
srpsrp
2728
2728
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $Msubseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-frac{1}{n-2}(Ric-frac{R}{n}g)wedge g-frac{R}{2n(n-1)}gwedge g$
where $wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame ${Z_1,dots, Z_n}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(lambda_1,dots, lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
add a comment |
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$begingroup$
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $Msubseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-frac{1}{n-2}(Ric-frac{R}{n}g)wedge g-frac{R}{2n(n-1)}gwedge g$
where $wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame ${Z_1,dots, Z_n}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(lambda_1,dots, lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
add a comment |
$begingroup$
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $Msubseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-frac{1}{n-2}(Ric-frac{R}{n}g)wedge g-frac{R}{2n(n-1)}gwedge g$
where $wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame ${Z_1,dots, Z_n}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(lambda_1,dots, lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
add a comment |
$begingroup$
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $Msubseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-frac{1}{n-2}(Ric-frac{R}{n}g)wedge g-frac{R}{2n(n-1)}gwedge g$
where $wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame ${Z_1,dots, Z_n}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(lambda_1,dots, lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
$endgroup$
You can observe that by Gauss Equation and Codazzi-Mainardi equation you can write the Riemann tensor and the Ricci tensor in function of the second fundamental form of your hypersurface that is in $mathbb{R}^{n+1}$, so the Riemann and Ricci tensor of the ambient are zero.
The Weyl tensor is defined in function of the induced metric, the Riemann tensor and the Ricci tensor of the hypersurface so it is true, it is possible write the Weyl tensor in function of the second fundamental form.
We know that Gauss equation of an Hypersurface $Msubseteq N$ is
$R^N(X,Y,Z,W)=R^M(X,Y,Z,W)+h(X,W)h(Y,Z)-h(X,Z)h(Y,W)$
where $h$ is the second fundamental form of $M$.
In local coordinates we have that
$R^N_{ijkl}=R^M_{ijkl}+h_{il}h_{jk}-h_{ik}h_{jl}$
In our case $R^N=0$ so
$R^M_{ijkl}= h_{ik}h_{jl}-h_{il}h_{jk}$
If we trace the equation we can write also Ricci Tensor in function of the second fundamental form:
$R^M_{jl}=Hh_{jl}-g^{ik} h_{il}h_{jk}$
where $H$ is the mean curvature of $M$.
If we trace twice Gauss equation we get a relationship between the scalar curvature $R$ of $M$ and the second fundamental form:
$R=H^2-|h|^2$
We consider the (0,4)-Weyl tensor of $M$
$W=R^M-frac{1}{n-2}(Ric-frac{R}{n}g)wedge g-frac{R}{2n(n-1)}gwedge g$
where $wedge$ is the Kulkarni-Nomizou product.
In local coordinate we have that
$W_{ijkl}=R^M_{ijkl}+frac{1}{n-2}(R^M_{il}g_{jk}-R^M_{ik}g_{il}+R^M_{jk}g_{il}-R^M_{jl}g_{ik})+$
$+frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})$
So you have all elements to write Weyl tensor in function of the second fundamental form of $M$ (and in function of the metric $g$).
You can observe also that the (0,2)-Tensor $h$ is symmetric so there exists a local orthonormal frame ${Z_1,dots, Z_n}$ such that the matrix $A$ associated with $h$ with respect that frame is $diag(lambda_1,dots, lambda_n)$ and th matrix $G$ associated with the metric $g$ with respect that frame will be the identity $I$. In this case you can write all our formula in an easier way than before.
If it is not clear I can explain more precisely.
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
edited Jan 24 at 17:43
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
answered Jan 24 at 15:59
Federico FalluccaFederico Fallucca
2,270210
2,270210
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
add a comment |
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Thank you for your answer. I am aware that the second fundamental form of a hypersurface completely determines its intrinsic curvature. My question is what the precise formula of the Weyl curvature is in terms of the second fundamental form.
$endgroup$
– srp
Jan 24 at 16:54
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
$begingroup$
Oook one moment that I write it
$endgroup$
– Federico Fallucca
Jan 24 at 17:03
add a comment |
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