What is the universal cover of SL(2,R)?
$begingroup$
Wikipedia [1] says,
For $n geq 2$, the universal cover of the special linear group SL($n, mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).
What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, mathbb{R}$) have in it?
group-theory
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
$endgroup$
add a comment |
$begingroup$
Wikipedia [1] says,
For $n geq 2$, the universal cover of the special linear group SL($n, mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).
What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, mathbb{R}$) have in it?
group-theory
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
$endgroup$
add a comment |
$begingroup$
Wikipedia [1] says,
For $n geq 2$, the universal cover of the special linear group SL($n, mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).
What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, mathbb{R}$) have in it?
group-theory
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
$endgroup$
Wikipedia [1] says,
For $n geq 2$, the universal cover of the special linear group SL($n, mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).
What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, mathbb{R}$) have in it?
group-theory
group-theory
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
edited May 4 '13 at 14:35
user59083
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
asked May 4 '13 at 14:05
The VeeThe Vee
2,225823
2,225823
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The maximal compact subgroup of $SL(n,mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,mathbb{R}) cong SO(n)times mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$pi_1(SL(n,mathbb{R})) = pi_1(SO(n)) =begin{cases} mathbb{Z}_2, & ngeq 3; \ mathbb{Z}, & n=2.end{cases}$$
This implies that $widetilde{SL}(n,mathbb{R})$ is a two-sheeted cover for $ngeq 3$ and an infinite cyclic cover for $n = 2$.
As for what group it is, well, it's the universal cover of $SL(n,mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.
For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $sqrt{2}$. We just haven't met it before so the only way we know it is because $sqrt{2}sqrt{2} = 2.$ Same here: $widetilde{SL}(n,mathbb{R})$ is the group which is simply connected and covers $SL(n,mathbb{R})$. That is its defining characteristic.
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
$endgroup$
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
add a comment |
$begingroup$
John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $tilde{Sp}(2n,{mathbb R})$ of $Sp(2n,{mathbb R})$ for any $n geq 1$.
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The maximal compact subgroup of $SL(n,mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,mathbb{R}) cong SO(n)times mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$pi_1(SL(n,mathbb{R})) = pi_1(SO(n)) =begin{cases} mathbb{Z}_2, & ngeq 3; \ mathbb{Z}, & n=2.end{cases}$$
This implies that $widetilde{SL}(n,mathbb{R})$ is a two-sheeted cover for $ngeq 3$ and an infinite cyclic cover for $n = 2$.
As for what group it is, well, it's the universal cover of $SL(n,mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.
For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $sqrt{2}$. We just haven't met it before so the only way we know it is because $sqrt{2}sqrt{2} = 2.$ Same here: $widetilde{SL}(n,mathbb{R})$ is the group which is simply connected and covers $SL(n,mathbb{R})$. That is its defining characteristic.
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
$endgroup$
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
add a comment |
$begingroup$
The maximal compact subgroup of $SL(n,mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,mathbb{R}) cong SO(n)times mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$pi_1(SL(n,mathbb{R})) = pi_1(SO(n)) =begin{cases} mathbb{Z}_2, & ngeq 3; \ mathbb{Z}, & n=2.end{cases}$$
This implies that $widetilde{SL}(n,mathbb{R})$ is a two-sheeted cover for $ngeq 3$ and an infinite cyclic cover for $n = 2$.
As for what group it is, well, it's the universal cover of $SL(n,mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.
For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $sqrt{2}$. We just haven't met it before so the only way we know it is because $sqrt{2}sqrt{2} = 2.$ Same here: $widetilde{SL}(n,mathbb{R})$ is the group which is simply connected and covers $SL(n,mathbb{R})$. That is its defining characteristic.
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
$endgroup$
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
add a comment |
$begingroup$
The maximal compact subgroup of $SL(n,mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,mathbb{R}) cong SO(n)times mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$pi_1(SL(n,mathbb{R})) = pi_1(SO(n)) =begin{cases} mathbb{Z}_2, & ngeq 3; \ mathbb{Z}, & n=2.end{cases}$$
This implies that $widetilde{SL}(n,mathbb{R})$ is a two-sheeted cover for $ngeq 3$ and an infinite cyclic cover for $n = 2$.
As for what group it is, well, it's the universal cover of $SL(n,mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.
For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $sqrt{2}$. We just haven't met it before so the only way we know it is because $sqrt{2}sqrt{2} = 2.$ Same here: $widetilde{SL}(n,mathbb{R})$ is the group which is simply connected and covers $SL(n,mathbb{R})$. That is its defining characteristic.
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
$endgroup$
The maximal compact subgroup of $SL(n,mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,mathbb{R}) cong SO(n)times mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$pi_1(SL(n,mathbb{R})) = pi_1(SO(n)) =begin{cases} mathbb{Z}_2, & ngeq 3; \ mathbb{Z}, & n=2.end{cases}$$
This implies that $widetilde{SL}(n,mathbb{R})$ is a two-sheeted cover for $ngeq 3$ and an infinite cyclic cover for $n = 2$.
As for what group it is, well, it's the universal cover of $SL(n,mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.
For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $sqrt{2}$. We just haven't met it before so the only way we know it is because $sqrt{2}sqrt{2} = 2.$ Same here: $widetilde{SL}(n,mathbb{R})$ is the group which is simply connected and covers $SL(n,mathbb{R})$. That is its defining characteristic.
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
edited May 4 '13 at 14:51
Stefan Hansen
20.9k73765
20.9k73765
answered May 4 '13 at 14:29
NealNeal
23.8k23886
answered May 4 '13 at 14:29
NealNeal
23.8k23886
answered May 4 '13 at 14:29
NealNeal
23.8k23886
23.8k23886
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
add a comment |
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
$begingroup$
In the case $n=2$ I agree that there is no extra nomenclature available; $tilde{SL}(2,mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,mathbb{R})$".
$endgroup$
– Lee Mosher
May 4 '13 at 14:37
5
5
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
However, when $n ge 3$, we have an isomorphism $tilde {SL}(n,mathbb{R}) approx tilde{SO}(n,mathbb{R}) times mathbb{E}^k$, the group $tilde{SO}(n,mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics.
$endgroup$
– Lee Mosher
May 4 '13 at 14:40
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
$begingroup$
Thanks! I think $tilde{SL}(2,mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics.
$endgroup$
– The Vee
May 4 '13 at 15:23
1
1
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
@LeeMosher Do you mean that $widetilde{SL}(n,mathbb{R})$ is isomorphic as a group to $Spin(n)times mathbb{E}^k$?
$endgroup$
– Neal
May 5 '13 at 1:21
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
$begingroup$
I agree that its a new object un same sense that $sqrt 2$ is a new number, but that doesn't mean there isn't explicit construction as a "matrix group" for example the reduced heissenber group is not a matrix group but it has the usual matrix multipkication.
$endgroup$
– k76u4vkweek547v7
Feb 1 '16 at 21:40
add a comment |
$begingroup$
John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $tilde{Sp}(2n,{mathbb R})$ of $Sp(2n,{mathbb R})$ for any $n geq 1$.
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
$endgroup$
add a comment |
$begingroup$
John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $tilde{Sp}(2n,{mathbb R})$ of $Sp(2n,{mathbb R})$ for any $n geq 1$.
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
$endgroup$
add a comment |
$begingroup$
John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $tilde{Sp}(2n,{mathbb R})$ of $Sp(2n,{mathbb R})$ for any $n geq 1$.
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
$endgroup$
John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $tilde{Sp}(2n,{mathbb R})$ of $Sp(2n,{mathbb R})$ for any $n geq 1$.
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
answered Aug 5 '14 at 9:05
Andrew RanickiAndrew Ranicki
30214
30214
add a comment |
add a comment |
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