- #1
teddd
- 62
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Hi everybody!
Ok, so from a few days I've begun a group theory class, and i have to say i love the subject.
In particular i happened to like Lie groups, but there are things that are not cristal clear to me, hope you'll help to figure'em out!First of all, Lie groups are continuous group, so their set of elements form a topological group.
Now, the definition of the Lie group (at least the one given to me) says:
Take any neighbourhood of the set V of the group element which contains the unit element.
Every elements g of the group G in the neighbour V can be identified with some parameters [itex](\alpha_1,\alpha_2...\alpha_n)[/itex] ,called canonical coordinates, in a 1-1 corrispondence,
and if g1,g2 and g=g1g2 lie in V then [itex]g=g(\alpha'_1,\alpha'_2...\alpha'_n)[/itex] where the [itex]\alpha'_k[/itex] are analytic function of the parameters of g1g2.
This is an n-dimensional Lie group.Now, the questions!
1)
Is it right to say that since we're restricting to a certain neighbour (the one which contains the unit element) we're actually ignoring (or puttin aside, choose the term you prefer) the global topology of the group? I mean, if we only work in a fixed neighbour we are always working in a topological space of an n-dimensional ball!
2)
Take the group of the rotation in two dimension.
IN the realization which employs the unimodular complex number, i.e. [itex]\mathbb{T}=\{z\epsilon\mathbb{C},zz^*=1\}[/itex] the group elements have the form [itex]g(\theta)=e^{i\theta}[/itex].
Restricting to [0,2∏] the parameter you need to identify an element is just the angle theta, so that this Lie group is a 1-dimensional group (and thus will have just one generator).
But now I ask: i do not have to restrict to a neighbour of the unit element to make that work, i just can take all of the set [0,2∏] becaouse within that the identification parameters-group IS 1-1, so effectively i can take V=[0,2∏], and so I can see the global topological structure (which is a circle minus a point that is, if we start from 0, ∏----wait a minute is that the reason?? so that I'm working on a line and not a circle becaouse i cannot close the latter (hope I've explained...))
How is that??
3)This one comes from an explicit example.
Take SU(2), the group of all the matrix with determinant=+1, which certainly IS a Lie group.
An element g of such a group can be written as
[tex]
g=\left(
\begin{array}{cc}
v^*&-u^*\\
u&v
\end{array}
\right)
[/tex]
with [itex]uu^*+vv^*=1[/itex].
Now my teacher's note, after writing that matrix, says that this is clearly a 3-dimensional group.
How do you see that?? At first sight I'm pushed to say that a generic element g is determined only by u and v!
Then he (the teacher) writes down the generators, which are 3 (proportional to the Pauli marices, you know this better than me) making clear that this is a 3 dimensional LIe group.
But what are then the 3 parameters to identify one element of the group, and why u and v alone are not enough?
There is a strange precisation too: it's said that the 3 euler angles are NOT the canonical coordinates. But why? what special features do the canonical coordinates have to have?4)
Finally, the last question!
It regrds the exponential representation.
Now, why on Earth do i have to take only those element close to the unit one??
Take the unimodualr complex group in question 2.
Now, if the element of such group is called [itex]\phi[/itex], the exponential map is just [itex]T(\phi)=e^{i\phi}[/itex], and that is valid in all the set of elements on which there is a 1-1 corrispondence between [itex]\phi[/itex] and the representation chosen!I thing that, for now, that's all!
Thanks for your time!
Ok, so from a few days I've begun a group theory class, and i have to say i love the subject.
In particular i happened to like Lie groups, but there are things that are not cristal clear to me, hope you'll help to figure'em out!First of all, Lie groups are continuous group, so their set of elements form a topological group.
Now, the definition of the Lie group (at least the one given to me) says:
Take any neighbourhood of the set V of the group element which contains the unit element.
Every elements g of the group G in the neighbour V can be identified with some parameters [itex](\alpha_1,\alpha_2...\alpha_n)[/itex] ,called canonical coordinates, in a 1-1 corrispondence,
and if g1,g2 and g=g1g2 lie in V then [itex]g=g(\alpha'_1,\alpha'_2...\alpha'_n)[/itex] where the [itex]\alpha'_k[/itex] are analytic function of the parameters of g1g2.
This is an n-dimensional Lie group.Now, the questions!
1)
Is it right to say that since we're restricting to a certain neighbour (the one which contains the unit element) we're actually ignoring (or puttin aside, choose the term you prefer) the global topology of the group? I mean, if we only work in a fixed neighbour we are always working in a topological space of an n-dimensional ball!
2)
Take the group of the rotation in two dimension.
IN the realization which employs the unimodular complex number, i.e. [itex]\mathbb{T}=\{z\epsilon\mathbb{C},zz^*=1\}[/itex] the group elements have the form [itex]g(\theta)=e^{i\theta}[/itex].
Restricting to [0,2∏] the parameter you need to identify an element is just the angle theta, so that this Lie group is a 1-dimensional group (and thus will have just one generator).
But now I ask: i do not have to restrict to a neighbour of the unit element to make that work, i just can take all of the set [0,2∏] becaouse within that the identification parameters-group IS 1-1, so effectively i can take V=[0,2∏], and so I can see the global topological structure (which is a circle minus a point that is, if we start from 0, ∏----wait a minute is that the reason?? so that I'm working on a line and not a circle becaouse i cannot close the latter (hope I've explained...))
How is that??
3)This one comes from an explicit example.
Take SU(2), the group of all the matrix with determinant=+1, which certainly IS a Lie group.
An element g of such a group can be written as
[tex]
g=\left(
\begin{array}{cc}
v^*&-u^*\\
u&v
\end{array}
\right)
[/tex]
with [itex]uu^*+vv^*=1[/itex].
Now my teacher's note, after writing that matrix, says that this is clearly a 3-dimensional group.
How do you see that?? At first sight I'm pushed to say that a generic element g is determined only by u and v!
Then he (the teacher) writes down the generators, which are 3 (proportional to the Pauli marices, you know this better than me) making clear that this is a 3 dimensional LIe group.
But what are then the 3 parameters to identify one element of the group, and why u and v alone are not enough?
There is a strange precisation too: it's said that the 3 euler angles are NOT the canonical coordinates. But why? what special features do the canonical coordinates have to have?4)
Finally, the last question!
It regrds the exponential representation.
Now, why on Earth do i have to take only those element close to the unit one??
Take the unimodualr complex group in question 2.
Now, if the element of such group is called [itex]\phi[/itex], the exponential map is just [itex]T(\phi)=e^{i\phi}[/itex], and that is valid in all the set of elements on which there is a 1-1 corrispondence between [itex]\phi[/itex] and the representation chosen!I thing that, for now, that's all!
Thanks for your time!