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A vector space $\mathfrak{g}$ over some field $F$ and kitted with a bilinear, antisymmetric and Jacobi-identity-fulfilling product ("Lie Bracket" or "commutator"). In physics, most often arises as the Lie algebra (tangent space to the identity) of a Lie group; in gauge theories, basis vectors of the gauge group's Lie algebra correspond to Noether currents and conserved quantities.

8 votes
2 answers
2k views

Number of the Generators of Poincare Group

It is said that the Poincare group, $P(3,1)$ has $10$ generators. $6$ of them are the generators of the Lorentz group, $O(3,1)$ and the other $4$ generators are the generators of $4D$ translational gr …
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3 votes
3 answers
3k views

Number of Parameters of Lorentz Group

We embed the rotation group, $SO(3)$ into the Lorentz group, $O(1,3)$ : $SO(3) \hookrightarrow O(1,3)$ and then determine the six generators of Lorentz group: $J_x, J_y, J_z, K_x, K_y, K_z$ from the r …
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1 vote
1 answer
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Verification of the Poincare Algebra

The generators of the Poincare group $P(1;3)$ are supposed to obey the following commutation relation to be verified: $$\left[ M^{\mu\nu}, P^{\rho} \right] = i \left(g^{\nu\rho} P^{\mu} - g^{\mu\rho} …
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10 votes
1 answer
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Generators of Poincare Groups

How can I determine the generators of the Poincare Group, $P(1,3)$ explicitly? Here $P(1,3)$ means a matrix Lie group.
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6 votes
1 answer
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$(1/2, 0)$ representation of the Lorentz Group $SO(1,3)$

Let us consider the $(j, j') = \left(\frac{1}{2}, 0\right)$ representation of $SO(1, 3)\cong SU(2) \otimes SU(2)$. $j = \frac{1}{2}$ corresponds to $SU(2)$ generated by $$ \tag{1} N_i^+ = \frac{1} …
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7 votes
2 answers
1k views

Rotation Group and Lorentz Group

It is often stated that rotations in the 3 spatial dimensions are examples of Lorentz transformations. But Lorentz transformations form a group named the Lorentz Group, $O(1,3)$ which is a group a $ …
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5 votes
3 answers
756 views

Poincare Generators in terms of Position and Momentum

The $10$ generators of the Poincare group $P(1;3)$ are $M^{\mu\nu}$ and $P^\mu$. These generators can be determined explicitly in the matrix form. However, I have found that $M^{\mu\nu}$ and $P^\mu$ a …
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