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Finite-dimensional faithful unitary representations of SL(2,Z)
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Does $SL(2,mathbbZ)$ have a finite-dimensional faithful unitary representation? No such representation exists for $SL(2,mathbbR)$, but I don't see a reason why one shouldn't exist for $SL(2,mathbbZ)$.
gr.group-theory rt.representation-theory unitary-representations
asked Aug 24 at 19:08
DanielHarlow
1127
add a comment |Â
up vote
9
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Does $SL(2,mathbbZ)$ have a finite-dimensional faithful unitary representation? No such representation exists for $SL(2,mathbbR)$, but I don't see a reason why one shouldn't exist for $SL(2,mathbbZ)$.
gr.group-theory rt.representation-theory unitary-representations
asked Aug 24 at 19:08
DanielHarlow
1127
1
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
4
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
4
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Does $SL(2,mathbbZ)$ have a finite-dimensional faithful unitary representation? No such representation exists for $SL(2,mathbbR)$, but I don't see a reason why one shouldn't exist for $SL(2,mathbbZ)$.
gr.group-theory rt.representation-theory unitary-representations
asked Aug 24 at 19:08
DanielHarlow
1127
Does $SL(2,mathbbZ)$ have a finite-dimensional faithful unitary representation? No such representation exists for $SL(2,mathbbR)$, but I don't see a reason why one shouldn't exist for $SL(2,mathbbZ)$.
gr.group-theory rt.representation-theory unitary-representations
asked Aug 24 at 19:08
DanielHarlow
1127
asked Aug 24 at 19:08
DanielHarlow
1127
asked Aug 24 at 19:08
DanielHarlow
1127
asked Aug 24 at 19:08
DanielHarlow
1127
1127
1
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
4
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
4
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21
add a comment |Â
1
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
4
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
4
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21
1
1
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
4
4
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
4
4
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
17
down vote
accepted
Here a non-explicit proof of the existence of a faithful representation of $mathrmSL_2(mathbfZ)$ in $mathrmSU(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $mathrmSL_2(mathbfZ)$.
[The basic idea is that if all representations in $mathrmSU(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $mathrmSL_2$. We need to use the particular form of the presentation of $mathrmSL_2(mathbfZ)$, since the argument will not carry over representations of $mathrmSL_3(mathbfZ)$ in $mathrmSU(3)$.]
Let $P_t$ be the set of $2times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $mathrmSL_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $mathrmSL_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $mathrmSL_2(K)$.
For every $(g,h)in P_0times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$mathrmSL_2(mathbfZ)=langle u,vmid u^4=v^6=[u^2,v]=[u,v^3]=1rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6)
into $mathrmSL_2(K)$ can be naturally identified to $(P_0times P_1)(K)$. Note that $P_0times P_1$ is irreducible.
Write $P_t^sharp=P_t(mathbfC)capmathrmSU(2)$. Then using that $mathrmSU(2)$ is Zariski-dense in $mathrmSL_2(mathbfC)$ and describing $P_t(mathbfC)$ as a conjugacy class, one deduces that $P_t^sharp$ is Zariski-dense in $P_t(mathbfC)$. So $P_0^sharptimes P_1^sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0times P_1)(mathbfC)$.
For every given nontrivial element $w$ in $mathrmSL_2(mathbfZ)$ the set of representations vanishing on $w$ is a proper subvariety of $P_0times P_1$, hence has dimension $le 3$. By Zariski density of $P_0^sharptimes P_1^sharp$, we deduce that its intersection with $P_0^sharptimes P_1^sharp$ is a proper Zariski closed subset (because $mathrmSL_2(mathbfZ)$ admits one faithful representation into $mathrmSL_2(mathbfC)$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^sharptimes P_1^sharp$ defining a faithful representation.
answered Aug 24 at 22:52
YCor
24.6k273116
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
Here a non-explicit proof of the existence of a faithful representation of $mathrmSL_2(mathbfZ)$ in $mathrmSU(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $mathrmSL_2(mathbfZ)$.
[The basic idea is that if all representations in $mathrmSU(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $mathrmSL_2$. We need to use the particular form of the presentation of $mathrmSL_2(mathbfZ)$, since the argument will not carry over representations of $mathrmSL_3(mathbfZ)$ in $mathrmSU(3)$.]
Let $P_t$ be the set of $2times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $mathrmSL_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $mathrmSL_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $mathrmSL_2(K)$.
For every $(g,h)in P_0times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$mathrmSL_2(mathbfZ)=langle u,vmid u^4=v^6=[u^2,v]=[u,v^3]=1rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6)
into $mathrmSL_2(K)$ can be naturally identified to $(P_0times P_1)(K)$. Note that $P_0times P_1$ is irreducible.
Write $P_t^sharp=P_t(mathbfC)capmathrmSU(2)$. Then using that $mathrmSU(2)$ is Zariski-dense in $mathrmSL_2(mathbfC)$ and describing $P_t(mathbfC)$ as a conjugacy class, one deduces that $P_t^sharp$ is Zariski-dense in $P_t(mathbfC)$. So $P_0^sharptimes P_1^sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0times P_1)(mathbfC)$.
For every given nontrivial element $w$ in $mathrmSL_2(mathbfZ)$ the set of representations vanishing on $w$ is a proper subvariety of $P_0times P_1$, hence has dimension $le 3$. By Zariski density of $P_0^sharptimes P_1^sharp$, we deduce that its intersection with $P_0^sharptimes P_1^sharp$ is a proper Zariski closed subset (because $mathrmSL_2(mathbfZ)$ admits one faithful representation into $mathrmSL_2(mathbfC)$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^sharptimes P_1^sharp$ defining a faithful representation.
answered Aug 24 at 22:52
YCor
24.6k273116
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
add a comment |Â
up vote
17
down vote
accepted
Here a non-explicit proof of the existence of a faithful representation of $mathrmSL_2(mathbfZ)$ in $mathrmSU(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $mathrmSL_2(mathbfZ)$.
[The basic idea is that if all representations in $mathrmSU(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $mathrmSL_2$. We need to use the particular form of the presentation of $mathrmSL_2(mathbfZ)$, since the argument will not carry over representations of $mathrmSL_3(mathbfZ)$ in $mathrmSU(3)$.]
Let $P_t$ be the set of $2times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $mathrmSL_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $mathrmSL_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $mathrmSL_2(K)$.
For every $(g,h)in P_0times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$mathrmSL_2(mathbfZ)=langle u,vmid u^4=v^6=[u^2,v]=[u,v^3]=1rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6)
into $mathrmSL_2(K)$ can be naturally identified to $(P_0times P_1)(K)$. Note that $P_0times P_1$ is irreducible.
Write $P_t^sharp=P_t(mathbfC)capmathrmSU(2)$. Then using that $mathrmSU(2)$ is Zariski-dense in $mathrmSL_2(mathbfC)$ and describing $P_t(mathbfC)$ as a conjugacy class, one deduces that $P_t^sharp$ is Zariski-dense in $P_t(mathbfC)$. So $P_0^sharptimes P_1^sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0times P_1)(mathbfC)$.
For every given nontrivial element $w$ in $mathrmSL_2(mathbfZ)$ the set of representations vanishing on $w$ is a proper subvariety of $P_0times P_1$, hence has dimension $le 3$. By Zariski density of $P_0^sharptimes P_1^sharp$, we deduce that its intersection with $P_0^sharptimes P_1^sharp$ is a proper Zariski closed subset (because $mathrmSL_2(mathbfZ)$ admits one faithful representation into $mathrmSL_2(mathbfC)$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^sharptimes P_1^sharp$ defining a faithful representation.
answered Aug 24 at 22:52
YCor
24.6k273116
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
Here a non-explicit proof of the existence of a faithful representation of $mathrmSL_2(mathbfZ)$ in $mathrmSU(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $mathrmSL_2(mathbfZ)$.
[The basic idea is that if all representations in $mathrmSU(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $mathrmSL_2$. We need to use the particular form of the presentation of $mathrmSL_2(mathbfZ)$, since the argument will not carry over representations of $mathrmSL_3(mathbfZ)$ in $mathrmSU(3)$.]
Let $P_t$ be the set of $2times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $mathrmSL_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $mathrmSL_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $mathrmSL_2(K)$.
For every $(g,h)in P_0times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$mathrmSL_2(mathbfZ)=langle u,vmid u^4=v^6=[u^2,v]=[u,v^3]=1rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6)
into $mathrmSL_2(K)$ can be naturally identified to $(P_0times P_1)(K)$. Note that $P_0times P_1$ is irreducible.
Write $P_t^sharp=P_t(mathbfC)capmathrmSU(2)$. Then using that $mathrmSU(2)$ is Zariski-dense in $mathrmSL_2(mathbfC)$ and describing $P_t(mathbfC)$ as a conjugacy class, one deduces that $P_t^sharp$ is Zariski-dense in $P_t(mathbfC)$. So $P_0^sharptimes P_1^sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0times P_1)(mathbfC)$.
For every given nontrivial element $w$ in $mathrmSL_2(mathbfZ)$ the set of representations vanishing on $w$ is a proper subvariety of $P_0times P_1$, hence has dimension $le 3$. By Zariski density of $P_0^sharptimes P_1^sharp$, we deduce that its intersection with $P_0^sharptimes P_1^sharp$ is a proper Zariski closed subset (because $mathrmSL_2(mathbfZ)$ admits one faithful representation into $mathrmSL_2(mathbfC)$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^sharptimes P_1^sharp$ defining a faithful representation.
answered Aug 24 at 22:52
YCor
24.6k273116
Here a non-explicit proof of the existence of a faithful representation of $mathrmSL_2(mathbfZ)$ in $mathrmSU(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $mathrmSL_2(mathbfZ)$.
[The basic idea is that if all representations in $mathrmSU(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $mathrmSL_2$. We need to use the particular form of the presentation of $mathrmSL_2(mathbfZ)$, since the argument will not carry over representations of $mathrmSL_3(mathbfZ)$ in $mathrmSU(3)$.]
Let $P_t$ be the set of $2times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $mathrmSL_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $mathrmSL_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $mathrmSL_2(K)$.
For every $(g,h)in P_0times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$mathrmSL_2(mathbfZ)=langle u,vmid u^4=v^6=[u^2,v]=[u,v^3]=1rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6)
into $mathrmSL_2(K)$ can be naturally identified to $(P_0times P_1)(K)$. Note that $P_0times P_1$ is irreducible.
Write $P_t^sharp=P_t(mathbfC)capmathrmSU(2)$. Then using that $mathrmSU(2)$ is Zariski-dense in $mathrmSL_2(mathbfC)$ and describing $P_t(mathbfC)$ as a conjugacy class, one deduces that $P_t^sharp$ is Zariski-dense in $P_t(mathbfC)$. So $P_0^sharptimes P_1^sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0times P_1)(mathbfC)$.
For every given nontrivial element $w$ in $mathrmSL_2(mathbfZ)$ the set of representations vanishing on $w$ is a proper subvariety of $P_0times P_1$, hence has dimension $le 3$. By Zariski density of $P_0^sharptimes P_1^sharp$, we deduce that its intersection with $P_0^sharptimes P_1^sharp$ is a proper Zariski closed subset (because $mathrmSL_2(mathbfZ)$ admits one faithful representation into $mathrmSL_2(mathbfC)$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^sharptimes P_1^sharp$ defining a faithful representation.
answered Aug 24 at 22:52
YCor
24.6k273116
answered Aug 24 at 22:52
YCor
24.6k273116
answered Aug 24 at 22:52
YCor
24.6k273116
answered Aug 24 at 22:52
YCor
24.6k273116
24.6k273116
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
add a comment |Â
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
5
5
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group...
– Uri Bader
Aug 25 at 8:54
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
@UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24-dimensional representation that has finite kernel, hence kernel of order $le 2$ (I guess 2). Using a 1-dimensional representation (through $C_12$ that's faithful on $-I_2$, we eventually get a faithful 25-dimensional unitary representation.
– YCor
Aug 25 at 9:25
add a comment |Â
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1
The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finite-dimensional unitary representation of $SL(3,mathbfZ)$.
– YCor
Aug 24 at 19:31
Thanks. Do you mind saying something about how such representations are constructed or providing a reference?
– DanielHarlow
Aug 24 at 19:38
4
$SL(2,Z)$ is the amalgam $<u,v,z|u^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,-i)$ (so $z=-1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^-1)$, where $t=exp(2ipi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam.
– YCor
Aug 24 at 19:42
4
@YCor's suggestion works - one can prove a faithful action of $SL(2,Z)$ on the Bruhat-Tits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that low-dimensional representations are classified in this paper: mathscinet.ams.org/mathscinet-getitem?mr=1815266
– Ian Agol
Aug 24 at 20:21