Riemannian metrics with discrete isometry group

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Does anybody knows an example of a Riemannian metric on a manifold (other than the compact nonpositive curvature case) whose isometry group is discrete? How about such a Riemannian metric with postive curvature?

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asked Jul 24 at 22:01

Mehdi Rafie-rad

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Does anybody knows an example of a Riemannian metric on a manifold (other than the compact nonpositive curvature case) whose isometry group is discrete? How about such a Riemannian metric with postive curvature?

differential-geometry

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asked Jul 24 at 22:01

Mehdi Rafie-rad

161

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Does anybody knows an example of a Riemannian metric on a manifold (other than the compact nonpositive curvature case) whose isometry group is discrete? How about such a Riemannian metric with postive curvature?

differential-geometry

share|cite|improve this question

asked Jul 24 at 22:01

Mehdi Rafie-rad

161

Does anybody knows an example of a Riemannian metric on a manifold (other than the compact nonpositive curvature case) whose isometry group is discrete? How about such a Riemannian metric with postive curvature?

differential-geometry

share|cite|improve this question

asked Jul 24 at 22:01

Mehdi Rafie-rad

161

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 22:01

Mehdi Rafie-rad

161

asked Jul 24 at 22:01

Mehdi Rafie-rad

161

asked Jul 24 at 22:01

Mehdi Rafie-rad

161

161

add a comment | 

add a comment | 

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A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $x,y,z = A$. $Y = X setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y to Y$, you get $g = i circ f : Y to S^2$, by inclusion $i : Y to S^2$. Next, since $S^2$ is complete and $i circ f$ is an isometry, this extends continuously to a map $h : S^2 to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y in S^2$ so that $d( h(x) , h(y) ) not = d(x,y)$. But we can take $x_n, y_n in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n to infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)

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answered Jul 24 at 22:43

Lorenzo

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A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $x,y,z = A$. $Y = X setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y to Y$, you get $g = i circ f : Y to S^2$, by inclusion $i : Y to S^2$. Next, since $S^2$ is complete and $i circ f$ is an isometry, this extends continuously to a map $h : S^2 to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y in S^2$ so that $d( h(x) , h(y) ) not = d(x,y)$. But we can take $x_n, y_n in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n to infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)

share|cite|improve this answer

answered Jul 24 at 22:43

Lorenzo

11.5k31537

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up vote
1
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A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $x,y,z = A$. $Y = X setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y to Y$, you get $g = i circ f : Y to S^2$, by inclusion $i : Y to S^2$. Next, since $S^2$ is complete and $i circ f$ is an isometry, this extends continuously to a map $h : S^2 to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y in S^2$ so that $d( h(x) , h(y) ) not = d(x,y)$. But we can take $x_n, y_n in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n to infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)

share|cite|improve this answer

answered Jul 24 at 22:43

Lorenzo

11.5k31537

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up vote
1
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up vote
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down vote

A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $x,y,z = A$. $Y = X setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y to Y$, you get $g = i circ f : Y to S^2$, by inclusion $i : Y to S^2$. Next, since $S^2$ is complete and $i circ f$ is an isometry, this extends continuously to a map $h : S^2 to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y in S^2$ so that $d( h(x) , h(y) ) not = d(x,y)$. But we can take $x_n, y_n in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n to infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)

share|cite|improve this answer

answered Jul 24 at 22:43

Lorenzo

11.5k31537

A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $x,y,z = A$. $Y = X setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y to Y$, you get $g = i circ f : Y to S^2$, by inclusion $i : Y to S^2$. Next, since $S^2$ is complete and $i circ f$ is an isometry, this extends continuously to a map $h : S^2 to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y in S^2$ so that $d( h(x) , h(y) ) not = d(x,y)$. But we can take $x_n, y_n in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n to infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)

share|cite|improve this answer

answered Jul 24 at 22:43

Lorenzo

11.5k31537

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 22:43

Lorenzo

11.5k31537

answered Jul 24 at 22:43

Lorenzo

11.5k31537

answered Jul 24 at 22:43

Lorenzo

11.5k31537

11.5k31537

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