Dimension of $SU(2)$ from a quadratic condition
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The way I understand it, the dimension of a Lie group is the (minimal) number of real parameters required to label a single element of the group. I know that $SU(2)$ is a 3-dimensional group, but the argumentation I've read in some textbooks/references is sloppy. Here's how it usually goes.
Consider $hin SU(2)$ a $2times2$ complex matrix. Since $h^dagger=h^-1$, we can parametrize $h$ this way:
$$h=beginpmatrix a & b \ -b^* & a^* endpmatrix.$$
Let $a = x+iy$ and $b=z+iw$ with $x,y,z,win mathbbR$. The condition $det h = 1$ implies that:
$$x^2+y^2+z^2+w^2 =1.$$
Most references conclude here by saying that there are 4 real parameters that must satisfy the previous constraint, so the dimension is $4-1=3$.
My problem is that, using this parametrization, setting $x,y,z$ for instance is usually not sufficient to indicate a single matrix of $SU(2)$. Since the condition above is quadratic, then $w$ could take 2 values: $w=pm sqrt1-x^2-y^2-z^2$.
I know there are other parametrizations that don't include a 4th dependent variable, such as $h= expi alpha^i sigma_i$ (where $sigma_i$ are the Pauli matrices) or
$$ h = beginpmatrix e^ialphacostheta & -e^-ibetasintheta \ e^ibetasintheta & e^-ialphacostheta endpmatrix.$$
Using these, it is much clearer that $SU(2)$ is 3-dimensional. But is it always necessary to find such a parametrization (without any dependent variable) to know the dimension of a Lie group? As I said, the proofs I read didn't care to go this far. They seemed to assume that any condition, even in a quadratic form as above, is sufficient to decrease the dimension of the group. Is this true or the authors of the proofs were just being sloppy?
My hypothesis is that perhaps they should have mentioned that the above condition is special because we recognize it as the equation of a 3-sphere (on which any point can be labeled using 3 angles), but that in general the dimension is not so trivial to find.
group-theory lie-groups
asked Aug 7 at 17:07
Jasmeru
101
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The way I understand it, the dimension of a Lie group is the (minimal) number of real parameters required to label a single element of the group. I know that $SU(2)$ is a 3-dimensional group, but the argumentation I've read in some textbooks/references is sloppy. Here's how it usually goes.
Consider $hin SU(2)$ a $2times2$ complex matrix. Since $h^dagger=h^-1$, we can parametrize $h$ this way:
$$h=beginpmatrix a & b \ -b^* & a^* endpmatrix.$$
Let $a = x+iy$ and $b=z+iw$ with $x,y,z,win mathbbR$. The condition $det h = 1$ implies that:
$$x^2+y^2+z^2+w^2 =1.$$
Most references conclude here by saying that there are 4 real parameters that must satisfy the previous constraint, so the dimension is $4-1=3$.
My problem is that, using this parametrization, setting $x,y,z$ for instance is usually not sufficient to indicate a single matrix of $SU(2)$. Since the condition above is quadratic, then $w$ could take 2 values: $w=pm sqrt1-x^2-y^2-z^2$.
I know there are other parametrizations that don't include a 4th dependent variable, such as $h= expi alpha^i sigma_i$ (where $sigma_i$ are the Pauli matrices) or
$$ h = beginpmatrix e^ialphacostheta & -e^-ibetasintheta \ e^ibetasintheta & e^-ialphacostheta endpmatrix.$$
Using these, it is much clearer that $SU(2)$ is 3-dimensional. But is it always necessary to find such a parametrization (without any dependent variable) to know the dimension of a Lie group? As I said, the proofs I read didn't care to go this far. They seemed to assume that any condition, even in a quadratic form as above, is sufficient to decrease the dimension of the group. Is this true or the authors of the proofs were just being sloppy?
My hypothesis is that perhaps they should have mentioned that the above condition is special because we recognize it as the equation of a 3-sphere (on which any point can be labeled using 3 angles), but that in general the dimension is not so trivial to find.
group-theory lie-groups
asked Aug 7 at 17:07
Jasmeru
101
While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The way I understand it, the dimension of a Lie group is the (minimal) number of real parameters required to label a single element of the group. I know that $SU(2)$ is a 3-dimensional group, but the argumentation I've read in some textbooks/references is sloppy. Here's how it usually goes.
Consider $hin SU(2)$ a $2times2$ complex matrix. Since $h^dagger=h^-1$, we can parametrize $h$ this way:
$$h=beginpmatrix a & b \ -b^* & a^* endpmatrix.$$
Let $a = x+iy$ and $b=z+iw$ with $x,y,z,win mathbbR$. The condition $det h = 1$ implies that:
$$x^2+y^2+z^2+w^2 =1.$$
Most references conclude here by saying that there are 4 real parameters that must satisfy the previous constraint, so the dimension is $4-1=3$.
My problem is that, using this parametrization, setting $x,y,z$ for instance is usually not sufficient to indicate a single matrix of $SU(2)$. Since the condition above is quadratic, then $w$ could take 2 values: $w=pm sqrt1-x^2-y^2-z^2$.
I know there are other parametrizations that don't include a 4th dependent variable, such as $h= expi alpha^i sigma_i$ (where $sigma_i$ are the Pauli matrices) or
$$ h = beginpmatrix e^ialphacostheta & -e^-ibetasintheta \ e^ibetasintheta & e^-ialphacostheta endpmatrix.$$
Using these, it is much clearer that $SU(2)$ is 3-dimensional. But is it always necessary to find such a parametrization (without any dependent variable) to know the dimension of a Lie group? As I said, the proofs I read didn't care to go this far. They seemed to assume that any condition, even in a quadratic form as above, is sufficient to decrease the dimension of the group. Is this true or the authors of the proofs were just being sloppy?
My hypothesis is that perhaps they should have mentioned that the above condition is special because we recognize it as the equation of a 3-sphere (on which any point can be labeled using 3 angles), but that in general the dimension is not so trivial to find.
group-theory lie-groups
asked Aug 7 at 17:07
Jasmeru
101
The way I understand it, the dimension of a Lie group is the (minimal) number of real parameters required to label a single element of the group. I know that $SU(2)$ is a 3-dimensional group, but the argumentation I've read in some textbooks/references is sloppy. Here's how it usually goes.
Consider $hin SU(2)$ a $2times2$ complex matrix. Since $h^dagger=h^-1$, we can parametrize $h$ this way:
$$h=beginpmatrix a & b \ -b^* & a^* endpmatrix.$$
Let $a = x+iy$ and $b=z+iw$ with $x,y,z,win mathbbR$. The condition $det h = 1$ implies that:
$$x^2+y^2+z^2+w^2 =1.$$
Most references conclude here by saying that there are 4 real parameters that must satisfy the previous constraint, so the dimension is $4-1=3$.
My problem is that, using this parametrization, setting $x,y,z$ for instance is usually not sufficient to indicate a single matrix of $SU(2)$. Since the condition above is quadratic, then $w$ could take 2 values: $w=pm sqrt1-x^2-y^2-z^2$.
I know there are other parametrizations that don't include a 4th dependent variable, such as $h= expi alpha^i sigma_i$ (where $sigma_i$ are the Pauli matrices) or
$$ h = beginpmatrix e^ialphacostheta & -e^-ibetasintheta \ e^ibetasintheta & e^-ialphacostheta endpmatrix.$$
Using these, it is much clearer that $SU(2)$ is 3-dimensional. But is it always necessary to find such a parametrization (without any dependent variable) to know the dimension of a Lie group? As I said, the proofs I read didn't care to go this far. They seemed to assume that any condition, even in a quadratic form as above, is sufficient to decrease the dimension of the group. Is this true or the authors of the proofs were just being sloppy?
My hypothesis is that perhaps they should have mentioned that the above condition is special because we recognize it as the equation of a 3-sphere (on which any point can be labeled using 3 angles), but that in general the dimension is not so trivial to find.
group-theory lie-groups
asked Aug 7 at 17:07
Jasmeru
101
asked Aug 7 at 17:07
Jasmeru
101
asked Aug 7 at 17:07
Jasmeru
101
asked Aug 7 at 17:07
Jasmeru
101
101
While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35
add a comment |Â
While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35
While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35
add a comment |Â
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While I think thinking about dimension this way builds good intuition, I don’t think it’s rigorous enough to get much out of it formally.
– Randall
Aug 7 at 17:23
Do you know what a manifold is and how its dimension is defined?
– joriki
Aug 7 at 17:41
@joriki I just started to read on them, I don't find it quite intuitive yet. I've worked with Lie groups/algebra in my physics courses, but re-reading my notes more carefully it seems the definitions were not exactly rigorous. A better definition for the dimension of a Lie group would be the dimension of its manifold? So the correct argumentation regarding the above condition would be : "The parameters form a 3-manifold $S^3$, so its 3-dimensional" rather than "We have 4 values and one constraint, so 3 independent variable (as in linear algebra)"?
– Jasmeru
Aug 7 at 18:29
Yes, that argument works -- more from first principles, you could say that the group is everywhere locally homeomorphic to $mathbb R^3$. Both of your parametrizations, including the one with the fourth discrete binary parameter, demonstrate this.
– joriki
Aug 7 at 18:35