# Tag Info

## Hot answers tagged group-theory

6

There is a new book called Physics From Symmetry which is written specifically for physicists and includes a long, very illustrative introduction to group theory. I especially liked that here concepts like representation or Lie algebra aren't only defined, but motivated and explained in terms that physicists understand. Plus no concepts are introduced which ...

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Comments to the question (v2): Recall that the Lorentz group $G=O(3,1)$ has 4 connected components $$G~=~G_0 ~\cup~ P\cdot G_0 ~\cup~ T\cdot G_0 ~\cup~ PT\cdot G_0.$$ Here the connected components $G_0$ that contains the identity is the restricted Lorentz group $G_0=SO^+(3,1)$. It is straightforward to see that $G_0$ is a normal subgroup of $G$. ...

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If I understand your question correctly, you have two systems $A$ and $B$, on both of which a symmetry group $G$ acts, so we have mappings $$G\times A\to A,\ \ \ (g,a)\mapsto ga$$ and $$G\times B\to B,\ \ \ (g,b)\mapsto gb$$ If the composed system is $A\times B$, then it symmetry group obviously includes $G\times G$ by $(g,h)(a,b) = (ga, hb)$. We have an ...

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Physics from Symmetry is a book that explains all group theoretical concepts needed to understand the foundations of QFT in great detail and is written specifically for physicists. It's not very technical, but it's great if you want to understand quickly what concepts are really important for modern physics and why. For example, it explains why things ...

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A single transformation is not strictly speaking discrete, but a group of transformations can be. Every transformation of finite order (i.e. $\Lambda^n = I$ for some $n$) generates a discrete group of transformations, so elements of finite order are sometimes called discrete transformations. As Qmechanic explains, $T$ and $P$ generate a subgroup of coset ...

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Because you are looking only at the so-called global part, i.e. the part of the gauge transformation which resembles a group action. Recall that the vector bosons transform as $$A_\mu \to g A_\mu g^{-1} - (\partial_\mu g) g^{-1}$$ where the first part is the global part of the gauge transformation, which tells you that $A_\mu$ transform in the Adjoint ...

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Usually the first step in deriving the reps of Poincaire is to go to the rest frame of the particle. This amounts to choosing a basis where $P^0$ acting on the state is nonzero, and where the eigenvalue a of $P^i$ are zero. We can do this if the momentum is timelike, that is if the eigenvalue of $P_\mu P^\mu$ is negative (in -+++ signature). Furthermore the ...

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