Homogeneous Riemann Surfaces
$begingroup$
A Riemann surface $X$ is a connected complex manifold of complex dimension one. A homogeneous space is a manifold with a transitive smooth action of a Lie group. I guess there must be a classification of those Riemann surfaces which are also homogeneous spaces . . . however, I can't seem to find it.
dg.differential-geometry complex-geometry lie-groups riemann-surfaces kahler-manifolds
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
$endgroup$
add a comment |
$begingroup$
A Riemann surface $X$ is a connected complex manifold of complex dimension one. A homogeneous space is a manifold with a transitive smooth action of a Lie group. I guess there must be a classification of those Riemann surfaces which are also homogeneous spaces . . . however, I can't seem to find it.
dg.differential-geometry complex-geometry lie-groups riemann-surfaces kahler-manifolds
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
$endgroup$
3
$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago
add a comment |
$begingroup$
A Riemann surface $X$ is a connected complex manifold of complex dimension one. A homogeneous space is a manifold with a transitive smooth action of a Lie group. I guess there must be a classification of those Riemann surfaces which are also homogeneous spaces . . . however, I can't seem to find it.
dg.differential-geometry complex-geometry lie-groups riemann-surfaces kahler-manifolds
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
$endgroup$
A Riemann surface $X$ is a connected complex manifold of complex dimension one. A homogeneous space is a manifold with a transitive smooth action of a Lie group. I guess there must be a classification of those Riemann surfaces which are also homogeneous spaces . . . however, I can't seem to find it.
dg.differential-geometry complex-geometry lie-groups riemann-surfaces kahler-manifolds
dg.differential-geometry complex-geometry lie-groups riemann-surfaces kahler-manifolds
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
asked 7 hours ago
Pierre DuboisPierre Dubois
1275
1275
3
$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago
add a comment |
3
$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago
3
3
$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First a purely topological fact. Here surface means a Hausdorff, connected nonempty metrizable space locally homeomorphic to the plane. Topological Lie group means Hausdorff topological group locally homeomorphic to some Euclidean space.
(a) The only surfaces that admit a transitive continuous action of a topological Lie group are (up to homeomorphism): $mathbf{R}^2$ (plane), $mathbf{S}^2$ (sphere), $mathbf{R}^2/mathbf{Z}$ (cylinder), $mathbf{T}^2$ (torus) for the orientable ones, and $mathbf{P}^2$ (projective plane) as non-orientable one.
Indeed, these are precisely the surfaces with abelian fundamental group. It is easy to check that they are all homogeneous (indeed, in a smooth way). For the converse, let us check that the being homogeneous implies abelian $pi_1$. Indeed, write $S=G/H$ with $G$ a topological Lie group; we can suppose that $G$ is simply connected. Then we have $1to pi_1(G/H)to pi_0(H)to 1$. Here $pi_0(H)=H/H^0$ is a discrete subgroup in $G/H^0$, which is a connected topological Lie group of dimension $le 2$. This implies that it is abelian (all discrete subgroups of the connected real affine group are abelian). So $pi_1(G/H)$ is abelian.
(b) It remains to classify all Riemann surfaces with these few allowed topologies. For the plane, we get $mathbf{C}$, $mathbf{H}^2$; both are homogeneous (under $mathbf{C}$ itself for the first, under either $mathrm{PSL}(2,mathbf{R})$ or its upper triangular subgroup for the second. For the sphere, we get $mathbf{P}^1(mathbf{C})$ which is homogeneous under $mathrm{PGL}(2,mathbf{C})$ or its subgroup $mathrm{PSU}(2)$.
For the cylinder, we have the genuine quotient $mathbf{C}/mathbf{Z}$ which is homogeneous as a complex Lie group. We also have the quotients of the hyperbolic plane, which can be realized as annuli (for $0le r<1$, the set $C_r$ of complex numbers whose radius is in $mathopen]r,1mathclose[$). The automorphism group of the Riemann surface $C_r$ does not act transitively on $C_r$ and hence it cannot be homogeneous as a Riemann surface.
Finally, tori correspond to elliptic curves $mathbf{C}/Lambda$, which are homogeneous as complex Lie groups.
answered 6 hours ago
YCorYCor
27.5k481133
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
First a purely topological fact. Here surface means a Hausdorff, connected nonempty metrizable space locally homeomorphic to the plane. Topological Lie group means Hausdorff topological group locally homeomorphic to some Euclidean space.
(a) The only surfaces that admit a transitive continuous action of a topological Lie group are (up to homeomorphism): $mathbf{R}^2$ (plane), $mathbf{S}^2$ (sphere), $mathbf{R}^2/mathbf{Z}$ (cylinder), $mathbf{T}^2$ (torus) for the orientable ones, and $mathbf{P}^2$ (projective plane) as non-orientable one.
Indeed, these are precisely the surfaces with abelian fundamental group. It is easy to check that they are all homogeneous (indeed, in a smooth way). For the converse, let us check that the being homogeneous implies abelian $pi_1$. Indeed, write $S=G/H$ with $G$ a topological Lie group; we can suppose that $G$ is simply connected. Then we have $1to pi_1(G/H)to pi_0(H)to 1$. Here $pi_0(H)=H/H^0$ is a discrete subgroup in $G/H^0$, which is a connected topological Lie group of dimension $le 2$. This implies that it is abelian (all discrete subgroups of the connected real affine group are abelian). So $pi_1(G/H)$ is abelian.
(b) It remains to classify all Riemann surfaces with these few allowed topologies. For the plane, we get $mathbf{C}$, $mathbf{H}^2$; both are homogeneous (under $mathbf{C}$ itself for the first, under either $mathrm{PSL}(2,mathbf{R})$ or its upper triangular subgroup for the second. For the sphere, we get $mathbf{P}^1(mathbf{C})$ which is homogeneous under $mathrm{PGL}(2,mathbf{C})$ or its subgroup $mathrm{PSU}(2)$.
For the cylinder, we have the genuine quotient $mathbf{C}/mathbf{Z}$ which is homogeneous as a complex Lie group. We also have the quotients of the hyperbolic plane, which can be realized as annuli (for $0le r<1$, the set $C_r$ of complex numbers whose radius is in $mathopen]r,1mathclose[$). The automorphism group of the Riemann surface $C_r$ does not act transitively on $C_r$ and hence it cannot be homogeneous as a Riemann surface.
Finally, tori correspond to elliptic curves $mathbf{C}/Lambda$, which are homogeneous as complex Lie groups.
answered 6 hours ago
YCorYCor
27.5k481133
$endgroup$
add a comment |
$begingroup$
First a purely topological fact. Here surface means a Hausdorff, connected nonempty metrizable space locally homeomorphic to the plane. Topological Lie group means Hausdorff topological group locally homeomorphic to some Euclidean space.
(a) The only surfaces that admit a transitive continuous action of a topological Lie group are (up to homeomorphism): $mathbf{R}^2$ (plane), $mathbf{S}^2$ (sphere), $mathbf{R}^2/mathbf{Z}$ (cylinder), $mathbf{T}^2$ (torus) for the orientable ones, and $mathbf{P}^2$ (projective plane) as non-orientable one.
Indeed, these are precisely the surfaces with abelian fundamental group. It is easy to check that they are all homogeneous (indeed, in a smooth way). For the converse, let us check that the being homogeneous implies abelian $pi_1$. Indeed, write $S=G/H$ with $G$ a topological Lie group; we can suppose that $G$ is simply connected. Then we have $1to pi_1(G/H)to pi_0(H)to 1$. Here $pi_0(H)=H/H^0$ is a discrete subgroup in $G/H^0$, which is a connected topological Lie group of dimension $le 2$. This implies that it is abelian (all discrete subgroups of the connected real affine group are abelian). So $pi_1(G/H)$ is abelian.
(b) It remains to classify all Riemann surfaces with these few allowed topologies. For the plane, we get $mathbf{C}$, $mathbf{H}^2$; both are homogeneous (under $mathbf{C}$ itself for the first, under either $mathrm{PSL}(2,mathbf{R})$ or its upper triangular subgroup for the second. For the sphere, we get $mathbf{P}^1(mathbf{C})$ which is homogeneous under $mathrm{PGL}(2,mathbf{C})$ or its subgroup $mathrm{PSU}(2)$.
For the cylinder, we have the genuine quotient $mathbf{C}/mathbf{Z}$ which is homogeneous as a complex Lie group. We also have the quotients of the hyperbolic plane, which can be realized as annuli (for $0le r<1$, the set $C_r$ of complex numbers whose radius is in $mathopen]r,1mathclose[$). The automorphism group of the Riemann surface $C_r$ does not act transitively on $C_r$ and hence it cannot be homogeneous as a Riemann surface.
Finally, tori correspond to elliptic curves $mathbf{C}/Lambda$, which are homogeneous as complex Lie groups.
answered 6 hours ago
YCorYCor
27.5k481133
$endgroup$
add a comment |
$begingroup$
First a purely topological fact. Here surface means a Hausdorff, connected nonempty metrizable space locally homeomorphic to the plane. Topological Lie group means Hausdorff topological group locally homeomorphic to some Euclidean space.
(a) The only surfaces that admit a transitive continuous action of a topological Lie group are (up to homeomorphism): $mathbf{R}^2$ (plane), $mathbf{S}^2$ (sphere), $mathbf{R}^2/mathbf{Z}$ (cylinder), $mathbf{T}^2$ (torus) for the orientable ones, and $mathbf{P}^2$ (projective plane) as non-orientable one.
Indeed, these are precisely the surfaces with abelian fundamental group. It is easy to check that they are all homogeneous (indeed, in a smooth way). For the converse, let us check that the being homogeneous implies abelian $pi_1$. Indeed, write $S=G/H$ with $G$ a topological Lie group; we can suppose that $G$ is simply connected. Then we have $1to pi_1(G/H)to pi_0(H)to 1$. Here $pi_0(H)=H/H^0$ is a discrete subgroup in $G/H^0$, which is a connected topological Lie group of dimension $le 2$. This implies that it is abelian (all discrete subgroups of the connected real affine group are abelian). So $pi_1(G/H)$ is abelian.
(b) It remains to classify all Riemann surfaces with these few allowed topologies. For the plane, we get $mathbf{C}$, $mathbf{H}^2$; both are homogeneous (under $mathbf{C}$ itself for the first, under either $mathrm{PSL}(2,mathbf{R})$ or its upper triangular subgroup for the second. For the sphere, we get $mathbf{P}^1(mathbf{C})$ which is homogeneous under $mathrm{PGL}(2,mathbf{C})$ or its subgroup $mathrm{PSU}(2)$.
For the cylinder, we have the genuine quotient $mathbf{C}/mathbf{Z}$ which is homogeneous as a complex Lie group. We also have the quotients of the hyperbolic plane, which can be realized as annuli (for $0le r<1$, the set $C_r$ of complex numbers whose radius is in $mathopen]r,1mathclose[$). The automorphism group of the Riemann surface $C_r$ does not act transitively on $C_r$ and hence it cannot be homogeneous as a Riemann surface.
Finally, tori correspond to elliptic curves $mathbf{C}/Lambda$, which are homogeneous as complex Lie groups.
answered 6 hours ago
YCorYCor
27.5k481133
$endgroup$
First a purely topological fact. Here surface means a Hausdorff, connected nonempty metrizable space locally homeomorphic to the plane. Topological Lie group means Hausdorff topological group locally homeomorphic to some Euclidean space.
(a) The only surfaces that admit a transitive continuous action of a topological Lie group are (up to homeomorphism): $mathbf{R}^2$ (plane), $mathbf{S}^2$ (sphere), $mathbf{R}^2/mathbf{Z}$ (cylinder), $mathbf{T}^2$ (torus) for the orientable ones, and $mathbf{P}^2$ (projective plane) as non-orientable one.
Indeed, these are precisely the surfaces with abelian fundamental group. It is easy to check that they are all homogeneous (indeed, in a smooth way). For the converse, let us check that the being homogeneous implies abelian $pi_1$. Indeed, write $S=G/H$ with $G$ a topological Lie group; we can suppose that $G$ is simply connected. Then we have $1to pi_1(G/H)to pi_0(H)to 1$. Here $pi_0(H)=H/H^0$ is a discrete subgroup in $G/H^0$, which is a connected topological Lie group of dimension $le 2$. This implies that it is abelian (all discrete subgroups of the connected real affine group are abelian). So $pi_1(G/H)$ is abelian.
(b) It remains to classify all Riemann surfaces with these few allowed topologies. For the plane, we get $mathbf{C}$, $mathbf{H}^2$; both are homogeneous (under $mathbf{C}$ itself for the first, under either $mathrm{PSL}(2,mathbf{R})$ or its upper triangular subgroup for the second. For the sphere, we get $mathbf{P}^1(mathbf{C})$ which is homogeneous under $mathrm{PGL}(2,mathbf{C})$ or its subgroup $mathrm{PSU}(2)$.
For the cylinder, we have the genuine quotient $mathbf{C}/mathbf{Z}$ which is homogeneous as a complex Lie group. We also have the quotients of the hyperbolic plane, which can be realized as annuli (for $0le r<1$, the set $C_r$ of complex numbers whose radius is in $mathopen]r,1mathclose[$). The automorphism group of the Riemann surface $C_r$ does not act transitively on $C_r$ and hence it cannot be homogeneous as a Riemann surface.
Finally, tori correspond to elliptic curves $mathbf{C}/Lambda$, which are homogeneous as complex Lie groups.
answered 6 hours ago
YCorYCor
27.5k481133
answered 6 hours ago
YCorYCor
27.5k481133
answered 6 hours ago
YCorYCor
27.5k481133
answered 6 hours ago
YCorYCor
27.5k481133
27.5k481133
add a comment |
add a comment |
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$begingroup$
So you don't require the action to preserve the holomorphic structure? Eventually I think it doesn't matter at least in the closed case (where homogeneous $Leftrightarrow$ genus $le 1$). In the non-closed case, it matters: the quotient of the hyperbolic plane by a loxodromic isometry is homogeneous as smooth manifold, but not as Riemann surface.
$endgroup$
– YCor
7 hours ago
$begingroup$
What if I include compact? This is what I am really interested in.
$endgroup$
– Pierre Dubois
6 hours ago
$begingroup$
You're not answering my question. In any case, I posted an answer covering both cases.
$endgroup$
– YCor
6 hours ago
$begingroup$
Ok, so I guess yes, I would like that the action preserves the holomorphic structure
$endgroup$
– Pierre Dubois
4 hours ago