Special Linear Group - why positive determinant?
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What is the reason for restricting the Special Linear Group $SL_n(mathbb{C})$ to $det(A)=+1$? What would be the practical consequence of including $det(A)=-1$?
Someone else answered before with some hints, that it will not be a subgroup if negative determinant, but he removed his answer. Anyway, in the comments to that answer, I wrote the following, for $A in SL$, invertibility requirement gives $pm 1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1}) = frac{det(A)}{det(A)}$. This leads to a contradiction for negative determinant, since $-det(A) = det(A) implies det(A) = 0$. This means that $SL_n(mathbb{C)}$ is only a subgroup when restricting to positive determinants. Is this correct?
group-theory linear-groups
asked Nov 24 at 14:12
bjorn
1104
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up vote
1
down vote
favorite
What is the reason for restricting the Special Linear Group $SL_n(mathbb{C})$ to $det(A)=+1$? What would be the practical consequence of including $det(A)=-1$?
Someone else answered before with some hints, that it will not be a subgroup if negative determinant, but he removed his answer. Anyway, in the comments to that answer, I wrote the following, for $A in SL$, invertibility requirement gives $pm 1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1}) = frac{det(A)}{det(A)}$. This leads to a contradiction for negative determinant, since $-det(A) = det(A) implies det(A) = 0$. This means that $SL_n(mathbb{C)}$ is only a subgroup when restricting to positive determinants. Is this correct?
group-theory linear-groups
asked Nov 24 at 14:12
bjorn
1104
By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
4
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
2
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the reason for restricting the Special Linear Group $SL_n(mathbb{C})$ to $det(A)=+1$? What would be the practical consequence of including $det(A)=-1$?
Someone else answered before with some hints, that it will not be a subgroup if negative determinant, but he removed his answer. Anyway, in the comments to that answer, I wrote the following, for $A in SL$, invertibility requirement gives $pm 1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1}) = frac{det(A)}{det(A)}$. This leads to a contradiction for negative determinant, since $-det(A) = det(A) implies det(A) = 0$. This means that $SL_n(mathbb{C)}$ is only a subgroup when restricting to positive determinants. Is this correct?
group-theory linear-groups
asked Nov 24 at 14:12
bjorn
1104
What is the reason for restricting the Special Linear Group $SL_n(mathbb{C})$ to $det(A)=+1$? What would be the practical consequence of including $det(A)=-1$?
Someone else answered before with some hints, that it will not be a subgroup if negative determinant, but he removed his answer. Anyway, in the comments to that answer, I wrote the following, for $A in SL$, invertibility requirement gives $pm 1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1}) = frac{det(A)}{det(A)}$. This leads to a contradiction for negative determinant, since $-det(A) = det(A) implies det(A) = 0$. This means that $SL_n(mathbb{C)}$ is only a subgroup when restricting to positive determinants. Is this correct?
group-theory linear-groups
group-theory linear-groups
asked Nov 24 at 14:12
bjorn
1104
asked Nov 24 at 14:12
bjorn
1104
edited Nov 24 at 16:18
asked Nov 24 at 14:12
bjorn
1104
asked Nov 24 at 14:12
bjorn
1104
asked Nov 24 at 14:12
bjorn
1104
1104
By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
4
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
2
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19
add a comment |
By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
4
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
2
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19
By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
4
4
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
2
2
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19
add a comment |
1 Answer
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Your claim at the bottom is not correct. The orthogonal group $O_n(Bbb C) := {M in GL_n(Bbb C) mid det(M) = pm 1}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(Bbb C) = {M in GL_n(Bbb C) mid |det(M)| = 1}$.
Your calculation goes wrong when you state $pm 1 = det(I)$. But $det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your claim at the bottom is not correct. The orthogonal group $O_n(Bbb C) := {M in GL_n(Bbb C) mid det(M) = pm 1}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(Bbb C) = {M in GL_n(Bbb C) mid |det(M)| = 1}$.
Your calculation goes wrong when you state $pm 1 = det(I)$. But $det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
add a comment |
up vote
1
down vote
accepted
Your claim at the bottom is not correct. The orthogonal group $O_n(Bbb C) := {M in GL_n(Bbb C) mid det(M) = pm 1}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(Bbb C) = {M in GL_n(Bbb C) mid |det(M)| = 1}$.
Your calculation goes wrong when you state $pm 1 = det(I)$. But $det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your claim at the bottom is not correct. The orthogonal group $O_n(Bbb C) := {M in GL_n(Bbb C) mid det(M) = pm 1}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(Bbb C) = {M in GL_n(Bbb C) mid |det(M)| = 1}$.
Your calculation goes wrong when you state $pm 1 = det(I)$. But $det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
Your claim at the bottom is not correct. The orthogonal group $O_n(Bbb C) := {M in GL_n(Bbb C) mid det(M) = pm 1}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(Bbb C) = {M in GL_n(Bbb C) mid |det(M)| = 1}$.
Your calculation goes wrong when you state $pm 1 = det(I)$. But $det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
answered Nov 24 at 23:10
Paul Sinclair
18.9k21440
18.9k21440
add a comment |
add a comment |
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By the way, what makes you want to include $-1$ and not something else?
– the_fox
Nov 24 at 14:56
Because we want matrices in this group to preserve the volume, no? Meaning absolute value of determinant should equal 1.
– bjorn
Nov 24 at 14:58
4
You can take ${rm det}(A) = pm 1$. The resulting group is sometimes denoted by ${rm SL}^pm_n({mathbb C})$. I am not sure if there are any practical consequences of doing this!
– Derek Holt
Nov 24 at 15:55
2
It's no longer connected. That seems significant.
– C Monsour
Nov 24 at 21:19