- #1
Jason Bennett
- 48
- 3
- Homework Statement
- various questions
- Relevant Equations
- lie theory
1) How do we determine a Lie group's global properties when the manifold that it represents is not immediately obvious?
Allow me to give the definitions I am working with.
A Lie group G is connected iff \forall g_1, g_2 \in G there exists a continuous curve connecting the two, i.e. there exists only one connected component.
A Lie group G is simply connected (if all closed curves on the manifold picture of G) can be contracted to a point.
A Lie group is compact if there are no elements infinitely far away fro the others.
The Lie group U(1) is quite easily identified as a circle in its manifold picture. This is connected, not simply connected, and compact.
However, SO(3) can (apparently) be viewed as manifold as such: a filled sphere of fixed radius, with antipodes identified. Once that has been realized, determining the global properties are straight forward. Getting there is another question, and I believe related to answering...
2) How do we algorithmically determine the parameter space of a Lie group – thus seeing it as a manifold?
For instance, for SU(2), we can write the matrix elements as complex, or decomposed with reals + i(reals), and use the det = 1 condition to determine that the Lie group is 3-deimensional. The step from 3-dimensional to a 3-sphere is not clear to me.
3) Taylor expansion question in the context of Lie algebra elements:
Consider some n-dimensional Lie group whose elements depend on a set of parameters \alpha =(\alpha_1 ... \alpha_n) such that g(0) = e with e as the identity, and that had a d-dimensional representation D(\alpha)=D(g(<br /> \alpha), such that D(0)=\mathbb{1}_{d \times d}. Then in some small neighborhood of \mathbb{1}, we can expand D(\alpha) as,
D(d\alpha) = \mathbb{1} + i d \alpha_i X^i, where X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}
I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:
Please see image [1] below.
Using this terminalology on the Lie case:
<br /> \begin{eqnarray}<br /> D(0+d\alpha) &=& D(0) + \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha <br /> \\<br /> &=& \mathbb{1} + (i) (-i) \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha <br /> \\<br /> &=& \mathbb{1} + (i) X^i d\alpha<br /> \end{eqnarray} <br />
is this correct? Also, why is the "taking the derivative at \alpha=0 important? And can you please point me towards a place to learn these types of Taylor expansions?
Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of (1+a)^x as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
4) Likely an Einstein summation confusion.
Consider Lorentz transformation's defined in the following matter:
Please see image [2] below.
I aim to consider the product L^0{}_0(\Lambda_1\Lambda_2). Consider the following notation L^\mu{}_\nu(\Lambda_i) = L_i{}^\mu{}_\nu. How then, does L^0{}_0(\Lambda_1\Lambda_2) = L_1{}^0{}_\mu L_2{}^\mu{}_0?
5) To right the J and K generators of the Lorentz group in a compact way, one can write (M^{lm})^j{}_k=i (g^<br /> {lj}g^m{}_k - g^{mj}g^l{}_k) where on the left hand side, it is helpful to think of l and m as indices/labels, and the j and k as rows/columns for the whole matrix.
(Apparently) One can write this in an operator representation as M^{\mu\nu} = i(x^\mu \partial^\nu-x^\nu\partial^\mu).
a) where does this come from?
b) why and how is it used? What is it operating on, x^\mu?
c) how does \partial^\nu x^\sigma= \frac{\partial}{\partial x_\nu} x^\sigma equal g^{\nu\sigma}?
[1]: https://i.stack.imgur.com/yAXum.png
[2]: https://i.stack.imgur.com/uPsLc.png
Allow me to give the definitions I am working with.
A Lie group G is connected iff \forall g_1, g_2 \in G there exists a continuous curve connecting the two, i.e. there exists only one connected component.
A Lie group G is simply connected (if all closed curves on the manifold picture of G) can be contracted to a point.
A Lie group is compact if there are no elements infinitely far away fro the others.
The Lie group U(1) is quite easily identified as a circle in its manifold picture. This is connected, not simply connected, and compact.
However, SO(3) can (apparently) be viewed as manifold as such: a filled sphere of fixed radius, with antipodes identified. Once that has been realized, determining the global properties are straight forward. Getting there is another question, and I believe related to answering...
2) How do we algorithmically determine the parameter space of a Lie group – thus seeing it as a manifold?
For instance, for SU(2), we can write the matrix elements as complex, or decomposed with reals + i(reals), and use the det = 1 condition to determine that the Lie group is 3-deimensional. The step from 3-dimensional to a 3-sphere is not clear to me.
3) Taylor expansion question in the context of Lie algebra elements:
Consider some n-dimensional Lie group whose elements depend on a set of parameters \alpha =(\alpha_1 ... \alpha_n) such that g(0) = e with e as the identity, and that had a d-dimensional representation D(\alpha)=D(g(<br /> \alpha), such that D(0)=\mathbb{1}_{d \times d}. Then in some small neighborhood of \mathbb{1}, we can expand D(\alpha) as,
D(d\alpha) = \mathbb{1} + i d \alpha_i X^i, where X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}
I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:
Please see image [1] below.
Using this terminalology on the Lie case:
<br /> \begin{eqnarray}<br /> D(0+d\alpha) &=& D(0) + \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha <br /> \\<br /> &=& \mathbb{1} + (i) (-i) \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha <br /> \\<br /> &=& \mathbb{1} + (i) X^i d\alpha<br /> \end{eqnarray} <br />
is this correct? Also, why is the "taking the derivative at \alpha=0 important? And can you please point me towards a place to learn these types of Taylor expansions?
Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of (1+a)^x as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
4) Likely an Einstein summation confusion.
Consider Lorentz transformation's defined in the following matter:
Please see image [2] below.
I aim to consider the product L^0{}_0(\Lambda_1\Lambda_2). Consider the following notation L^\mu{}_\nu(\Lambda_i) = L_i{}^\mu{}_\nu. How then, does L^0{}_0(\Lambda_1\Lambda_2) = L_1{}^0{}_\mu L_2{}^\mu{}_0?
5) To right the J and K generators of the Lorentz group in a compact way, one can write (M^{lm})^j{}_k=i (g^<br /> {lj}g^m{}_k - g^{mj}g^l{}_k) where on the left hand side, it is helpful to think of l and m as indices/labels, and the j and k as rows/columns for the whole matrix.
(Apparently) One can write this in an operator representation as M^{\mu\nu} = i(x^\mu \partial^\nu-x^\nu\partial^\mu).
a) where does this come from?
b) why and how is it used? What is it operating on, x^\mu?
c) how does \partial^\nu x^\sigma= \frac{\partial}{\partial x_\nu} x^\sigma equal g^{\nu\sigma}?
[1]: https://i.stack.imgur.com/yAXum.png
[2]: https://i.stack.imgur.com/uPsLc.png