New answers tagged group-theory
1
Your source is being sloppy with notation. The two-dimensional square lattice has a two-dimensional symmetry point group isomorphic to $C_{4v}$. However, $C_{4v}$ is really only supposed to be used to refer to subgroups of the three-dimensional point group $O(3)$ of orthogonal transformations. The three-dimensional point group of the square lattice is $D_{...
0
The Mathematica application lieART can do decompositions for all classical and exceptional Lie algebras including $SO(N)$. The documentation in https://arxiv.org/pdf/1206.6379.pdf also explains the algorithm used to do the decomposition. The only downside is that it does not work with Young Tableaus for SO(N). This means that you will have to convert your ...
0
This is not a complete answer to this question
and is adapted from my answer to Irrep decompositions for $SO(N)$ tensors for $N>3$ as it also seems to apply here. Indeed you can do decompositions and find the dimension of irreps using young tableau also for SO(n). The young tableaux describe permutation of indices and thus are relevant for all lie algebra'...
2
A Majorana spinor is a spinor for which the Dirac conjugate spinor $\psi^\dagger \gamma^0$ is equal to the Majorana conjugate spinor $\psi^T C$, i.e. $\psi$ satisfies the Majorana condition
$$
(\gamma^0)^T \psi^* = C^T \psi.
$$
Here $T$ stands for transposition, $*$ for (complex) conjugation, $\dagger$ for conjugate transposition, and $C$ is the charge ...
3
I'm not sure this is an answer to your question, but per your comment I will leave it anyway. Consider the Kagome lattice, shown below [ref]. One can see in the imagine that the Kagome lattice has the same translational symmetries of a triangular lattice: each of the red, blue, and green sublattices form a triangular lattice, connected by the Bravais lattice ...
1
There is a simple answer to your question. First, in $\mathrm{Cl3}$ (geometric algebra of 3D Euclidean vector space), the elements of the even part of the algebra are just quaternions.
Second, the simplest representation of orthonormal vectors in $\mathrm{Cl3}$ are the Pauli matrices. In addition, we have a geometric meaning of both the Pauli matrices and ...
3
As stated in other answers, you did in fact get a symmetric matrix, but only for some given values of $\phi$. One way to look at this is noticing you started off with two degrees of freedom ($\theta_1$ and $\theta_2$) but were finishing with three. By then $\textit{imposing}$ that the matrix you obtained is symmetric (as you know it must be), you can ...
1
Just because a matrix isn’t obviously symmetric doesn’t mean it isn’t symmetric. Given $\theta_1$ and $\theta_2$, there is a $\phi$ which makes the final matrix above symmetric. For example, when $\theta_1$ and $\theta_2$ are rapidities corresponding to boosts to speed $0.500c$, $\phi$ is $0.143$. Put in the numbers and see.
The general algebraic solution ...
1
The $(1,0)$ and $(0,1)$ representations are not four-dimensional, but three-dimensional. Together - as the direct sum $(1,0)\oplus (0,1)$ - they form the 2-forms, i.e. antisymmetric tensors, on $\mathbb{R}^4$, as opposed to vectors. The parts of the sum are the self-dual and anti-self-dual parts of the 2-form under the Hodge dual.
So this is, for example, ...
3
Many mathematical texts (and Wikipedia) often assume that Lie groups are finite-dimensional differential manifolds. Yang-Mills theories and Chern-Simons theories are based on such Lie Groups. However, in physics, gauge theories are sometimes governed by (infinite-dimensional) Lie groupoids. See also this related Phys.SE post.
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I think that you are groping for the equation
$$
U(R) \sigma_i U^{-1}(R)= \sigma_j {R^j}_i,
$$
where ${R^j}_i$ is an ${\rm SO}(3)$ rotation matrix, the $\sigma_i$ are the matrix generators of $\mathfrak{su}(2)$ in some representation, and the $U(R)$ are the corresponding ${\rm SU}(2)$ matrices in the same representation.
This is saying that the the ...
0
I reckon that there is a gap in your argument. The Killing vectors of S² are correct, but two questions:
These vectors, in this representation, are they valid to the coordinates that you chose? Change the domain is change the system of coordinates.
The step φ-> βφ is not so clear to me.
I don't know if this object is a 2-sphere. You cut a part.
Even if ...
2
You're on the right track! The first two steps are using Jensen's and upper bounding in terms of the 2-norm. You now want to write out the 2-norm and then compute the Haar integral. After the first step, you should find:
$$
\big\|\rho_A(U) - I_A/|A|\big\|_2^2 = {\rm tr}\big(\rho_A(U)^2\big) - 1/|A|
$$
and then integrate the purity using the 2nd moment ...
5
I would say that it means I have a lagrangian $\mathcal{L}$ that depends on a bunch of fields. I can transform those fields under $G$. They may or may not transform under the same representation $\rho_G$ of some group $G$. The objects in any given representation are not invariant under the group (unless they are in a trivial representation). It's the system ...
1
To add to the discussion what @user1379857 did not regarding "what does the most general proper Lorentz transformation look like written down explicitly", I will provide this as a derivation.
The three rotation matrices (note here that I prefer the Tait-Bryan angles instead of the more "physicist-y" Euler angles because I find them more ...
3
No, the Lorentz group contains all rotations, boosts, and compositions of rotations and boosts. What you have written is the most general form for a boost. However, a pure rotation, like
\begin{equation}
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos \theta & 0 & \sin \theta \\
0 & 0 & 1 & 0 \\
0 & -\sin \theta & 0 &...
8
Your quote already contains the answer: Symmetric, antisymmetric and mixed.
4
They needn't.
People have definitely been interested in "non-invertible symmetries". We typically don't call them "symmetries", i.e., we reserve the name "symmetry" for those operations that are invertible. In this sense, it would be better to say that a symmetry is, by definition, an operation that is invertible; but non-...
answered Jul 7 at 1:46
AccidentalFourierTransform
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4
I will give a "physicist" pictorial explanation, very qualitative. For technical details I agree that Math SE would be a better place.
Imagine the group as a manifold and the associated Lie algebra as the tangent space to the point of the manifold that represents the identity element of the group. This is not such a "crazy thing" and you ...
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