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Classification of closed 3-manifolds with finite first homology group?
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I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$.
Since $H_1(M)$ is the abelization of the fundamental group $pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group.
It is known that each closed 3-manifold with finite homotopy group $Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_sim$ of the 3-sphere, endowed with a free action of the group $Gamma$).
Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite?
at.algebraic-topology gt.geometric-topology 3-manifolds manifolds homology
asked 2 hours ago
Taras Banakh
13.9k2881
 |Â
show 4 more comments
up vote
1
down vote
favorite
I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$.
Since $H_1(M)$ is the abelization of the fundamental group $pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group.
It is known that each closed 3-manifold with finite homotopy group $Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_sim$ of the 3-sphere, endowed with a free action of the group $Gamma$).
Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite?
at.algebraic-topology gt.geometric-topology 3-manifolds manifolds homology
asked 2 hours ago
Taras Banakh
13.9k2881
1
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
1
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
2
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$.
Since $H_1(M)$ is the abelization of the fundamental group $pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group.
It is known that each closed 3-manifold with finite homotopy group $Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_sim$ of the 3-sphere, endowed with a free action of the group $Gamma$).
Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite?
at.algebraic-topology gt.geometric-topology 3-manifolds manifolds homology
asked 2 hours ago
Taras Banakh
13.9k2881
I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$.
Since $H_1(M)$ is the abelization of the fundamental group $pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group.
It is known that each closed 3-manifold with finite homotopy group $Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_sim$ of the 3-sphere, endowed with a free action of the group $Gamma$).
Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite?
at.algebraic-topology gt.geometric-topology 3-manifolds manifolds homology
at.algebraic-topology gt.geometric-topology 3-manifolds manifolds homology
asked 2 hours ago
Taras Banakh
13.9k2881
asked 2 hours ago
Taras Banakh
13.9k2881
edited 2 hours ago
asked 2 hours ago
Taras Banakh
13.9k2881
asked 2 hours ago
Taras Banakh
13.9k2881
asked 2 hours ago
Taras Banakh
13.9k2881
13.9k2881
1
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
1
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
2
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago
 |Â
show 4 more comments
1
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
1
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
2
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago
1
1
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
1
1
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
2
2
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago
 |Â
show 4 more comments
1 Answer
1
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up vote
5
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The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
- N. Dunfield, W. Thurston, The virtual Haken conjecture: experiments and examples https://mathscinet.ams.org/mathscinet-getitem?mr=1988291
- J. Brock, N. Dunfield, Injectivity radii of hyperbolic integer homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=3318758
- N. Dunfield, W. Thurston, Finite covers of random 3-manifolds https://mathscinet.ams.org/mathscinet-getitem?mr=2257389
- F. Calegari, N. Dunfield, Automorphic forms and rational homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=2224458
answered 1 hour ago
Jean Raimbault
1,483717
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
- N. Dunfield, W. Thurston, The virtual Haken conjecture: experiments and examples https://mathscinet.ams.org/mathscinet-getitem?mr=1988291
- J. Brock, N. Dunfield, Injectivity radii of hyperbolic integer homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=3318758
- N. Dunfield, W. Thurston, Finite covers of random 3-manifolds https://mathscinet.ams.org/mathscinet-getitem?mr=2257389
- F. Calegari, N. Dunfield, Automorphic forms and rational homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=2224458
answered 1 hour ago
Jean Raimbault
1,483717
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
 |Â
show 3 more comments
up vote
5
down vote
The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
- N. Dunfield, W. Thurston, The virtual Haken conjecture: experiments and examples https://mathscinet.ams.org/mathscinet-getitem?mr=1988291
- J. Brock, N. Dunfield, Injectivity radii of hyperbolic integer homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=3318758
- N. Dunfield, W. Thurston, Finite covers of random 3-manifolds https://mathscinet.ams.org/mathscinet-getitem?mr=2257389
- F. Calegari, N. Dunfield, Automorphic forms and rational homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=2224458
answered 1 hour ago
Jean Raimbault
1,483717
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
 |Â
show 3 more comments
up vote
5
down vote
up vote
5
down vote
The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
- N. Dunfield, W. Thurston, The virtual Haken conjecture: experiments and examples https://mathscinet.ams.org/mathscinet-getitem?mr=1988291
- J. Brock, N. Dunfield, Injectivity radii of hyperbolic integer homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=3318758
- N. Dunfield, W. Thurston, Finite covers of random 3-manifolds https://mathscinet.ams.org/mathscinet-getitem?mr=2257389
- F. Calegari, N. Dunfield, Automorphic forms and rational homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=2224458
answered 1 hour ago
Jean Raimbault
1,483717
The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.
For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:
- N. Dunfield, W. Thurston, The virtual Haken conjecture: experiments and examples https://mathscinet.ams.org/mathscinet-getitem?mr=1988291
- J. Brock, N. Dunfield, Injectivity radii of hyperbolic integer homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=3318758
- N. Dunfield, W. Thurston, Finite covers of random 3-manifolds https://mathscinet.ams.org/mathscinet-getitem?mr=2257389
- F. Calegari, N. Dunfield, Automorphic forms and rational homology 3-spheres https://mathscinet.ams.org/mathscinet-getitem?mr=2224458
answered 1 hour ago
Jean Raimbault
1,483717
answered 1 hour ago
Jean Raimbault
1,483717
answered 1 hour ago
Jean Raimbault
1,483717
answered 1 hour ago
Jean Raimbault
1,483717
1,483717
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
 |Â
show 3 more comments
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)?
– Taras Banakh
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.)
– YCor
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason?
– Taras Banakh
1 hour ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
@YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true?
– Taras Banakh
54 mins ago
 |Â
show 3 more comments
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1
You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/Gamma$ with $G$ the universal covering of $mathrmSL_2(mathbfR)$ and $Gamma$ a cocompact lattice.
– YCor
2 hours ago
@YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group).
– Taras Banakh
2 hours ago
1
@TarasBanakh the answer of your current question is already in my previous comment.
– YCor
2 hours ago
@YCor, But $SL_2(R)$ has fundamental group isomorphic to $mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/Gamma$?
– Taras Banakh
2 hours ago
2
The fundamental group in my example is $Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups.
– YCor
1 hour ago