Mixing
Space of conformal classes
$begingroup$
Let $M^n$ be a smooth, closed, simply connected manifold.
Question: "How many" distinct conformal classes of metrics does $M$ admit?
For example, when $n=2$, then $M^2=S^2$. The uniformization theorem tells us that any metric on $M^2$ is conformally diffeomorphic to the constant curvature metric on $M^2$. Therefore, in this case, $M^2$ has a unique conformal class up to diffeomorphism.
What is known in higher dimensions, and for specific cases like $M^n=S^n$?
I would appreciate references. Thanks.
differential-geometry riemannian-geometry conformal-geometry
asked May 12 '18 at 20:33
rpfrpf
1,100513
$endgroup$
add a comment |
$begingroup$
Let $M^n$ be a smooth, closed, simply connected manifold.
Question: "How many" distinct conformal classes of metrics does $M$ admit?
For example, when $n=2$, then $M^2=S^2$. The uniformization theorem tells us that any metric on $M^2$ is conformally diffeomorphic to the constant curvature metric on $M^2$. Therefore, in this case, $M^2$ has a unique conformal class up to diffeomorphism.
What is known in higher dimensions, and for specific cases like $M^n=S^n$?
I would appreciate references. Thanks.
differential-geometry riemannian-geometry conformal-geometry
asked May 12 '18 at 20:33
rpfrpf
1,100513
$endgroup$
1
$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25
add a comment |
$begingroup$
Let $M^n$ be a smooth, closed, simply connected manifold.
Question: "How many" distinct conformal classes of metrics does $M$ admit?
For example, when $n=2$, then $M^2=S^2$. The uniformization theorem tells us that any metric on $M^2$ is conformally diffeomorphic to the constant curvature metric on $M^2$. Therefore, in this case, $M^2$ has a unique conformal class up to diffeomorphism.
What is known in higher dimensions, and for specific cases like $M^n=S^n$?
I would appreciate references. Thanks.
differential-geometry riemannian-geometry conformal-geometry
asked May 12 '18 at 20:33
rpfrpf
1,100513
$endgroup$
Let $M^n$ be a smooth, closed, simply connected manifold.
Question: "How many" distinct conformal classes of metrics does $M$ admit?
For example, when $n=2$, then $M^2=S^2$. The uniformization theorem tells us that any metric on $M^2$ is conformally diffeomorphic to the constant curvature metric on $M^2$. Therefore, in this case, $M^2$ has a unique conformal class up to diffeomorphism.
What is known in higher dimensions, and for specific cases like $M^n=S^n$?
I would appreciate references. Thanks.
differential-geometry riemannian-geometry conformal-geometry
differential-geometry riemannian-geometry conformal-geometry
asked May 12 '18 at 20:33
rpfrpf
1,100513
asked May 12 '18 at 20:33
rpfrpf
1,100513
edited Jun 24 '18 at 3:28
rpf
asked May 12 '18 at 20:33
rpfrpf
1,100513
asked May 12 '18 at 20:33
rpfrpf
1,100513
asked May 12 '18 at 20:33
rpfrpf
1,100513
1,100513
1
$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25
add a comment |
1
$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25
1
1
$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The space of conformal classes (up to conformal isomorphism) is infinite-dimensional starting with dimension 3. I will prove this only in 3-dimensional case. I will consider a family of Riemannian metrics on $R^3$ of the form
$$
dx^2 + (1+f) dy^2 + (1+h)dz^2,
$$
where $f=f(z) > 0$ and $h=h(y) > 0$.
Remark. This choice of metrics is not random. It is proven in
D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.
that every Riemannian metric in dimension 3 can be locally conformally transformed to one of the above form with $f=f(x,y,z), h=h(x,y,z)$. I made further assumptions $f=f(z), h=h(y)$ in order to have cleaner computations.
In order to distinguish conformal classes of these metrics I will be using the Cotton tensor (due to its conformal invariance).
A direct computation of the Cotton tensor of metrics from this family (cf. Appendix C here for a general computation of the Cotton tensor; such explicit computations are also done in Eisenhart's "Riemannian Geometry" which is a bit dated and, thus, hard to read) yields:
$$
C= frac{1}{4} left[ begin{array}{ccc}
0&f'''& - h'''\
f''' & 0 & 0\
-h''' & 0 & 0
end{array}right]
$$
Assuming that $f''', h'''$ are nowhere vanishing, every conformal isomorphism of two metrics from this family has the form
$$
Phi: (x,y,z)mapsto (phi(x), psi(y), eta(z)),
$$
which implies linearity of $Phi$ (since I am assuming that $f$ is independent of $x, y$ and $h$ is independent of $x, z$). From this, we see that the space of conformal isomorphism classes of metrics in this family is infinite-dimensional. The construction is purely local, hence, can be done on any 3-dimensional manifold.
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
$endgroup$
add a comment |
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$begingroup$
The space of conformal classes (up to conformal isomorphism) is infinite-dimensional starting with dimension 3. I will prove this only in 3-dimensional case. I will consider a family of Riemannian metrics on $R^3$ of the form
$$
dx^2 + (1+f) dy^2 + (1+h)dz^2,
$$
where $f=f(z) > 0$ and $h=h(y) > 0$.
Remark. This choice of metrics is not random. It is proven in
D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.
that every Riemannian metric in dimension 3 can be locally conformally transformed to one of the above form with $f=f(x,y,z), h=h(x,y,z)$. I made further assumptions $f=f(z), h=h(y)$ in order to have cleaner computations.
In order to distinguish conformal classes of these metrics I will be using the Cotton tensor (due to its conformal invariance).
A direct computation of the Cotton tensor of metrics from this family (cf. Appendix C here for a general computation of the Cotton tensor; such explicit computations are also done in Eisenhart's "Riemannian Geometry" which is a bit dated and, thus, hard to read) yields:
$$
C= frac{1}{4} left[ begin{array}{ccc}
0&f'''& - h'''\
f''' & 0 & 0\
-h''' & 0 & 0
end{array}right]
$$
Assuming that $f''', h'''$ are nowhere vanishing, every conformal isomorphism of two metrics from this family has the form
$$
Phi: (x,y,z)mapsto (phi(x), psi(y), eta(z)),
$$
which implies linearity of $Phi$ (since I am assuming that $f$ is independent of $x, y$ and $h$ is independent of $x, z$). From this, we see that the space of conformal isomorphism classes of metrics in this family is infinite-dimensional. The construction is purely local, hence, can be done on any 3-dimensional manifold.
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
$endgroup$
add a comment |
$begingroup$
The space of conformal classes (up to conformal isomorphism) is infinite-dimensional starting with dimension 3. I will prove this only in 3-dimensional case. I will consider a family of Riemannian metrics on $R^3$ of the form
$$
dx^2 + (1+f) dy^2 + (1+h)dz^2,
$$
where $f=f(z) > 0$ and $h=h(y) > 0$.
Remark. This choice of metrics is not random. It is proven in
D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.
that every Riemannian metric in dimension 3 can be locally conformally transformed to one of the above form with $f=f(x,y,z), h=h(x,y,z)$. I made further assumptions $f=f(z), h=h(y)$ in order to have cleaner computations.
In order to distinguish conformal classes of these metrics I will be using the Cotton tensor (due to its conformal invariance).
A direct computation of the Cotton tensor of metrics from this family (cf. Appendix C here for a general computation of the Cotton tensor; such explicit computations are also done in Eisenhart's "Riemannian Geometry" which is a bit dated and, thus, hard to read) yields:
$$
C= frac{1}{4} left[ begin{array}{ccc}
0&f'''& - h'''\
f''' & 0 & 0\
-h''' & 0 & 0
end{array}right]
$$
Assuming that $f''', h'''$ are nowhere vanishing, every conformal isomorphism of two metrics from this family has the form
$$
Phi: (x,y,z)mapsto (phi(x), psi(y), eta(z)),
$$
which implies linearity of $Phi$ (since I am assuming that $f$ is independent of $x, y$ and $h$ is independent of $x, z$). From this, we see that the space of conformal isomorphism classes of metrics in this family is infinite-dimensional. The construction is purely local, hence, can be done on any 3-dimensional manifold.
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
$endgroup$
add a comment |
$begingroup$
The space of conformal classes (up to conformal isomorphism) is infinite-dimensional starting with dimension 3. I will prove this only in 3-dimensional case. I will consider a family of Riemannian metrics on $R^3$ of the form
$$
dx^2 + (1+f) dy^2 + (1+h)dz^2,
$$
where $f=f(z) > 0$ and $h=h(y) > 0$.
Remark. This choice of metrics is not random. It is proven in
D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.
that every Riemannian metric in dimension 3 can be locally conformally transformed to one of the above form with $f=f(x,y,z), h=h(x,y,z)$. I made further assumptions $f=f(z), h=h(y)$ in order to have cleaner computations.
In order to distinguish conformal classes of these metrics I will be using the Cotton tensor (due to its conformal invariance).
A direct computation of the Cotton tensor of metrics from this family (cf. Appendix C here for a general computation of the Cotton tensor; such explicit computations are also done in Eisenhart's "Riemannian Geometry" which is a bit dated and, thus, hard to read) yields:
$$
C= frac{1}{4} left[ begin{array}{ccc}
0&f'''& - h'''\
f''' & 0 & 0\
-h''' & 0 & 0
end{array}right]
$$
Assuming that $f''', h'''$ are nowhere vanishing, every conformal isomorphism of two metrics from this family has the form
$$
Phi: (x,y,z)mapsto (phi(x), psi(y), eta(z)),
$$
which implies linearity of $Phi$ (since I am assuming that $f$ is independent of $x, y$ and $h$ is independent of $x, z$). From this, we see that the space of conformal isomorphism classes of metrics in this family is infinite-dimensional. The construction is purely local, hence, can be done on any 3-dimensional manifold.
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
$endgroup$
The space of conformal classes (up to conformal isomorphism) is infinite-dimensional starting with dimension 3. I will prove this only in 3-dimensional case. I will consider a family of Riemannian metrics on $R^3$ of the form
$$
dx^2 + (1+f) dy^2 + (1+h)dz^2,
$$
where $f=f(z) > 0$ and $h=h(y) > 0$.
Remark. This choice of metrics is not random. It is proven in
D. DeTurck, D. Yang, Existence of elastic deformations with prescribed principal strains and triply orthogonal systems. Duke Math. J. 51 (1984), no. 2, 243–260.
that every Riemannian metric in dimension 3 can be locally conformally transformed to one of the above form with $f=f(x,y,z), h=h(x,y,z)$. I made further assumptions $f=f(z), h=h(y)$ in order to have cleaner computations.
In order to distinguish conformal classes of these metrics I will be using the Cotton tensor (due to its conformal invariance).
A direct computation of the Cotton tensor of metrics from this family (cf. Appendix C here for a general computation of the Cotton tensor; such explicit computations are also done in Eisenhart's "Riemannian Geometry" which is a bit dated and, thus, hard to read) yields:
$$
C= frac{1}{4} left[ begin{array}{ccc}
0&f'''& - h'''\
f''' & 0 & 0\
-h''' & 0 & 0
end{array}right]
$$
Assuming that $f''', h'''$ are nowhere vanishing, every conformal isomorphism of two metrics from this family has the form
$$
Phi: (x,y,z)mapsto (phi(x), psi(y), eta(z)),
$$
which implies linearity of $Phi$ (since I am assuming that $f$ is independent of $x, y$ and $h$ is independent of $x, z$). From this, we see that the space of conformal isomorphism classes of metrics in this family is infinite-dimensional. The construction is purely local, hence, can be done on any 3-dimensional manifold.
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
answered Jan 18 at 23:25
Moishe CohenMoishe Cohen
47.4k343108
47.4k343108
add a comment |
add a comment |
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$begingroup$
It's not true that every surface has a unique conformal class. The uniformization theorem says that any metric on $M$ is conformal to a constant-curvature metric, but unless $M$ is a sphere or a real projective plane, the constant-curvature metric is not unique, and neither is the conformal structure.
$endgroup$
– Jack Lee
May 13 '18 at 0:23
$begingroup$
As I wrote in the question, I am assuming that $M^n$ is closed and simply connected. A closed simply connected surface is a sphere, so for $n=2$, $M^2=S^2$ and the conformal class is unique, as you point out.
$endgroup$
– rpf
May 13 '18 at 1:08
$begingroup$
Sorry, I missed the phrase “simply connected.”
$endgroup$
– Jack Lee
May 13 '18 at 1:25