New answers tagged group-theory
0
votes
Does Cauchy stress tensor act on $SO(3)$?
Cauchy’s theorem, under standard physical laws of continuous bodies proves that the stress vector $s$ at $p$ referred to the normal unit $n$ has the form
$$s(p,n)_k = \sigma(p)_{kj} n_j.$$
Above, I ...
1
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Vector/axial current and vector/axial transformation
They are not just related, they are identical, joined at the hip. I'll illustrate by the original legendary 2-flavor chiral model, with the chiral group $\text{SU}(2)_L \times \text{SU}(2)_R$. This ...
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Holonomy group of Schwarzschild spacetime, other interesting examples?
Firstly, it is important to note that the holonomy groups are non-local in nature, even in trivial topology.
Consider, for example, a Lorentzian manifold $(\mathcal M,g)$ such that the Riemann ...
0
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Can we do better than "a spinor is something that transforms like a spinor"?
What you are asking about are what are called "coordinate free descriptions". A tensor can be defined transformatively or to use the correct term, covariantly, once we define a basis in the ...
1
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Projective representations and reduction of half-integer spin representations under $C_{\infty v}$
I have a partial answer to some of the questions posed.
First off, by analyzing the characters of the action of group $G \equiv \{R^z_\phi : \phi \in [0,4\pi)\} \cup \{\Sigma^z_\phi : \phi \in [0,4\pi)...
2
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On JD Jackson's derivation of Matrix Representations of Lorentz Tranformations
The "sines and cosines" comes from the algebraic properties of the generator.
The Lie algebra of the rotation group $SO(3)$ is given (as can be checked by differentiating the $SO(3)$ ...
1
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Accepted
Group of time translations for finite-dimensional quantum systems and incommensurable eigenvalues
Not a group theory person, but here are my thoughts on the problem.
For general $e_1$ and $e_2$, when is the function $U\left(t\right)$ one-to-one?
If $U\left(t_{1}\right)=U\left(t_{2}\right)$ then $\...
0
votes
Physical intuition of spin connection and spinor bundles?
A spinor bundle is defined via the associated bundle construction. Lets first make a couple of preliminary definitions:
A rotation bundle is defined as a principal bundle whose gauge structure group ...
6
votes
Showing that a generator exponentiates to a $\mathbb{R}$ group
From now on $\{\cdot,\cdot\}$ denotes the Lie commutator of smooth vector fields on a smooth manifold.
As far as I understand, you have a vector field $X$ on $TM$, where $M$ is the Schwarzschild ...
1
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How can one says that a particle IS a representation of some group?
I think when someone says “a particle IS a representation of some group” in the context of quantum field theory it would always be more precise to say, “at any point in space time, the state space of ...
1
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How can one says that a particle IS a representation of some group?
But how else do you define a particle? Something that has a definite position or momentum, definite mass, and definite spin? Just like, erm, a representation of the Poincare group?
(By the way, a ...
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How can one says that a particle IS a representation of some group?
Saying that particles are irreducible representation of Poincare group is like saying that energy is a number, not a full story. Also note that a particle is a tensor field.
1
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Accepted
Cartesian tensor transformation and representations of $SO(3)$
The question is very unclear so maybe what the OP means is that any matrix
$$
M=\pmatrix{M_{xx}&M_{xy}&M_{xz}\\ M_{yx}&M_{yy}&M_{yz}\\ M_{zx}&M_{zy}&M_zz}
$$ can be written in ...
1
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Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation
The $(1,0)\oplus (0,1)$ representation is the $2$-covariant anti-symmetric tensor representation $$F = \frac{1}{2}F_{\mu\nu}dx^\mu\wedge dx^\nu.$$
Moreover, the $F_{\mu\nu}$ are already the components ...
0
votes
Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation
After some calculations I found what I wanted. We can take the change-of-basis matrix $P$ to be
\begin{gather}
P = \frac{1}{2}
\begin{bmatrix}
1 & 0 &0 &-i & 0 & 0 \\
0 & 1 &...
3
votes
Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation
The $(1,0)$ representation is $F_{(\alpha\beta)}$ and the $(0,1)$ representation is $F_{({\dot \alpha}{\dot \beta})}$. The field strength in spacetime is then given by
$$
F_{\mu\nu} = (\sigma_{\mu\nu})...
1
vote
Accepted
Group actions confusion
This should be familiar to you from the first example of generators of transformations we usually encounter: The bracket between position $q$ and momentum $p$ is constant as $\{q,p\} = 1$, and ...
6
votes
Accepted
Why is the gauge field $A_\mu$ real for $\mathrm{U}(N)$ symmetry?
The first point is that the Lie group $G=U(N)$ [which consists of unitary $N\times N$ matrices] is a real Lie group, which is perhaps best explained as that the tangent spaces, or equivalently, the ...
3
votes
Why is the gauge field $A_\mu$ real for $\mathrm{U}(N)$ symmetry?
First, I question the importance of this in practice. Because if you have an action where the $A_\mu^a$ are initially real and you decide that you don't like this, you can always change this by ...
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