Mixing
Generators of Lie groups in physics
$begingroup$
I asked this question in physics SE here. But I was not satisfied.
As we solving something related to symmetry transformations, we need Lie groups. Also Lie algebra is very important to generate those transformations.
E.g. the generators of $SO(3)$:
$$begin{pmatrix}0&i&0\-i&0&0\0&0&0end{pmatrix},
begin{pmatrix}0&0&i\0&0&0\-i&0&0end{pmatrix},
begin{pmatrix}0&0&0\0&0&i\0&-i&0end{pmatrix}.
$$
Why there is an i in the front.. The thing it is not so wrong if we just consider the Lie algebra as a vector space. But when we use the Lie bracket we will also get those real ones.. But the Lie algebra is three real dimensional.. Now, it is 6.
lie-groups lie-algebras mathematical-physics
asked Jan 21 at 15:55
CO2CO2
1679
$endgroup$
|
show 1 more comment
$begingroup$
I asked this question in physics SE here. But I was not satisfied.
As we solving something related to symmetry transformations, we need Lie groups. Also Lie algebra is very important to generate those transformations.
E.g. the generators of $SO(3)$:
$$begin{pmatrix}0&i&0\-i&0&0\0&0&0end{pmatrix},
begin{pmatrix}0&0&i\0&0&0\-i&0&0end{pmatrix},
begin{pmatrix}0&0&0\0&0&i\0&-i&0end{pmatrix}.
$$
Why there is an i in the front.. The thing it is not so wrong if we just consider the Lie algebra as a vector space. But when we use the Lie bracket we will also get those real ones.. But the Lie algebra is three real dimensional.. Now, it is 6.
lie-groups lie-algebras mathematical-physics
asked Jan 21 at 15:55
CO2CO2
1679
$endgroup$
4
$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
1
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
2
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04
|
show 1 more comment
$begingroup$
I asked this question in physics SE here. But I was not satisfied.
As we solving something related to symmetry transformations, we need Lie groups. Also Lie algebra is very important to generate those transformations.
E.g. the generators of $SO(3)$:
$$begin{pmatrix}0&i&0\-i&0&0\0&0&0end{pmatrix},
begin{pmatrix}0&0&i\0&0&0\-i&0&0end{pmatrix},
begin{pmatrix}0&0&0\0&0&i\0&-i&0end{pmatrix}.
$$
Why there is an i in the front.. The thing it is not so wrong if we just consider the Lie algebra as a vector space. But when we use the Lie bracket we will also get those real ones.. But the Lie algebra is three real dimensional.. Now, it is 6.
lie-groups lie-algebras mathematical-physics
asked Jan 21 at 15:55
CO2CO2
1679
$endgroup$
I asked this question in physics SE here. But I was not satisfied.
As we solving something related to symmetry transformations, we need Lie groups. Also Lie algebra is very important to generate those transformations.
E.g. the generators of $SO(3)$:
$$begin{pmatrix}0&i&0\-i&0&0\0&0&0end{pmatrix},
begin{pmatrix}0&0&i\0&0&0\-i&0&0end{pmatrix},
begin{pmatrix}0&0&0\0&0&i\0&-i&0end{pmatrix}.
$$
Why there is an i in the front.. The thing it is not so wrong if we just consider the Lie algebra as a vector space. But when we use the Lie bracket we will also get those real ones.. But the Lie algebra is three real dimensional.. Now, it is 6.
lie-groups lie-algebras mathematical-physics
lie-groups lie-algebras mathematical-physics
asked Jan 21 at 15:55
CO2CO2
1679
asked Jan 21 at 15:55
CO2CO2
1679
edited Jan 21 at 18:04
CO2
asked Jan 21 at 15:55
CO2CO2
1679
asked Jan 21 at 15:55
CO2CO2
1679
asked Jan 21 at 15:55
CO2CO2
1679
1679
4
$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
1
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
2
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04
|
show 1 more comment
4
$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
1
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
2
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04
4
4
$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
1
1
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
2
2
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Just a reminder of facts concerning physics usage.
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators.
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
$endgroup$
add a comment |
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$begingroup$
Just a reminder of facts concerning physics usage.
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators.
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
$endgroup$
add a comment |
$begingroup$
Just a reminder of facts concerning physics usage.
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators.
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
$endgroup$
add a comment |
$begingroup$
Just a reminder of facts concerning physics usage.
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators.
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
$endgroup$
Just a reminder of facts concerning physics usage.
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators.
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
answered Jan 21 at 20:49
Cosmas ZachosCosmas Zachos
1,777522
1,777522
add a comment |
add a comment |
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$begingroup$
No, $mathfrak{so}(3)$ is a real $3$-dimensional vector space. Simply forget about the $i$ in front. The last matrix not skew-symmetric, as it should be.
$endgroup$
– Dietrich Burde
Jan 21 at 15:58
1
$begingroup$
I thought $SO(3)$ was the Lie group, not the Lie algebra, which I'd write as $mathfrak{so}(3)$. Also, I know physicists love their $i$s, but I don't.
$endgroup$
– Lord Shark the Unknown
Jan 21 at 15:58
2
$begingroup$
that $i$ is just a part of a physics convention, coming from quantum mechanics. The correct math definition has no $i$'s, so just delete them.
$endgroup$
– user8268
Jan 21 at 15:59
$begingroup$
@LordSharktheUnknown So true.. I gave the link to where I asked in physics SE. They were still persisting on it... mathematically, it is simply wrong...
$endgroup$
– CO2
Jan 21 at 18:01
$begingroup$
@DietrichBurde Well, in physics, they call the three matrices generators (or infinitesimal generators), just simply elements form the Lie algebra of a Lie group when we $exp$ them we get the elements in the Lie group. For me it does not make sense... Because generators of a (Lie) group in algebra are a subset of the group and span the whole group... Last one was a typo.
$endgroup$
– CO2
Jan 21 at 18:04