New answers tagged group-theory
0
Group theory does play an important role in general relativity, and I'm aware of three different types of relevant symmetries:
First, there are the physical symmetries of specific solutions to the field equations, formalized by Killing fields, the generators of one-parameter groups of local isometries.
Second, there's general covariance. Mathematically, ...
4
Symmetry is just as important in General Relativity as it is in Special Relativity.
In SR the symmetry is the Poincare Group which is the group of mappings of space-time to itself that preserves the metric formula between events given by $d^2 = \Delta{x}^2+\Delta{y}^2 + \Delta{z}^2 - c^2\Delta{t}^2$ The Poincare group is an extension of the Lorentz group ...
2
To be honest, I have a hard time interpreting geometrically what's going on here when using quaternions or biquaternions or anything else. All the algebra of rotations in 4d is adequately handled by a geometric algebra, with the elements of that algebra having clear geometric interpretations. The mathematics is similar to quaternions, but differs in some ...
1
This may not be intuitive at first but I think it is valuable in understanding the relationship between rotation matrices and angular velocities. Also, I know it does not direction answer the question, but I sense there is confusion in the OP and this might help.
So given the rotation matrices $E_1$ and $E_2$ for two connected rigid bodies how do be ...
1
You are mixing up different things. The first is a rotation transformation. Such a transformation is linear and can therefore be written as a matrix.
$$\bf\vec x'=A\vec x $$
Now, angular velocity, is the velocity of a physical rotation.
$$\vec\omega=\frac{\mbox d\vec\theta}{\mbox dt}$$
This theta is the angular displacement, or the angle of rotation. This ...
8
As you point out, the Minkowski metric $\eta = \mathrm{diag}(-1,+1, \dots, +1)$ in $d+1$ dimensions possesses a global Lorentz symmetry. A highbrow way of saying this in terms of differential geometry is that the (global) isometry group of the metric is the Lorentz group. Well actually, translations are also isometries of Minkowski, so the full isometry ...
2
The problem is that your coordinates aren't well defined at $\theta=0$ and $\phi=\pi/2$. Note in particular that
$$
U|_{(0,\frac{\pi}{2},\gamma)} = \begin{pmatrix}1&0\\0&1\end{pmatrix}
$$
for any value of $\gamma$. A simpler choice is
$$
\tilde{U} =
\begin{pmatrix}
x+iy & z+iw \\
-z+iw & x-iy
\end{pmatrix},
$$
with
$$
x = \sqrt{1 - y^2 - z^2 ...
-1
I don't see what you're missing.
$$\frac{\partial e^{ \pm i \gamma}}{\partial \gamma} = \pm i$$ which gives you the third generator, doesn't it?
2
There is no necessity to start from the Gaussian measure on $\mathbb{C}^{2N}$. Any $U(N)$ invariant measure would result the same $U(N)$ invariant measure of the Grassmannian. However, this is the standard choice for two reasons.
1) When GL(2) is factorized out, the resulting measure on GL(2) is the "Ginibre measure" , Please see for example: ...
3
I) First note that there is a group action $\rho: GL(2,\mathbb{C})\times u(2) \to u(2)$ given by
$$\tag{A} g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger},
\qquad g\in GL(2,\mathbb{C}),\qquad\sigma\in u(2). $$
In detail, the Lie group
$$\tag{B} GL(2,\mathbb{C})~:=~ \{ g\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \det(g)\neq 0\}$$
acts ...
3
User twistor59 has addressed the part regarding the "generator" terminology, but let me give a bit more detail on the second part of the question. I'm going to restrict the discussion to matrix Lie groups for simplicity.
Some background.
Given a Lie group $G$ with Lie algebra $\mathfrak g$, there exist two mappings $\mathrm{Ad}$ and $\mathrm{ad}$, both ...
3
If you have a basis for the Lie algebra, you can talk of these basis vectors as being "generators for the Lie group". This is true in the sense that, by using the exponential map on linear combinations of them, you generate (at least locally) a copy of the Lie group. So they're sort of "primitive infinitesimal elements" that you can use to build the local ...
7
I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC.
Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then
$$ [t_a,t_b] = C_{ab}{}^c t_c$$
defines the structure constants. If you define
$$ g_{ab} = C_{ad}{}^e C_{be}{}^d,$$
then this ...
7
I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$.
Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$.
II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create ...
6
The different definitions you mentioned are NOT definitions. In fact, what you are describing are different representations of the Lorentz Algebra. Representation theory plays a very important role in physics.
As far as the Lie algebra are concerned, the generators $L_{\mu\nu}$ are simply some operators with some defined commutation properties.
The choices ...
3
No, elements of $Spin(n)$ don't obey the Clifford algebra. Instead, it's the gamma matrices that obey it. And no, the commutator of the $Spin(n)$ Lie algebra isn't the commutators of the elements of the group but elements of the Lie algebra.
Now positively.
The spinor representation is the representation on which the generators $J_{ij}$ (the basis of the ...
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