0

Consider the Lorentz group $G$ consisting of Lorentz transformations $\Lambda :\mathbb{R}^4\to \mathbb{R}^4$ which satisfy $\Lambda^T\eta\Lambda=\eta$ where $\eta$ is the minkowski metric whose matrix form is given by $$\eta_{\mu\nu}:=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$$And consider the $Lie(G)$ with generators $J^{\mu\nu}=-J^{\nu\mu}$ defined by ($\mu,\nu$ take value from 0 to 3)$$(J^{\mu\nu})^\alpha_\beta=i(\eta^{\mu\alpha}\delta^{\nu}_\beta-\eta^{\nu\alpha}\delta^\mu_\beta)$$ We further define $J^i=i\epsilon^{i}_{jk}J^{jk}$, $K^i=J^{0i}$, it is easy to check (here i, j take values from 1 to 3) they also form a basis for the vector space $Lie(G)$. Now we define $J^{-}$ and $J^{+}$ as follows: $J^{-i}=J^{i}-iK^i$, $J^{+i}=J^i+iK^i$ and it is easy to verify the following commutation relations: $$[J^{-i},J^{-j}]=i\epsilon^{ijk}J^{-k}$$$$[J^{+i},J^{+j}]=i\epsilon^{ijk}J^{+k}$$$$[J^{-i},J^{+j}]=0$$ so we see that the generators of $Lie(G)$ can be separated into $J^{-}$ and $J^{+}$ such that either of them is a basis for a representation of the vector space $Lie(SO(3))$.

Question 1: Can we say, in this case, that $Lie(G)\cong Lie(SO(3))\oplus Lie(SO(3))$ ?

Now if we want to investigate further the complete set of finite dimensional irreducible representations of $Lie(G)$, we label $J^{-},J^{+}$ by $j_-$ and $j_+$ such that when $j_-=0$, it corresponds to the trivial representation. And when $j_-=\frac{1}{2}$ we have the spinor representation $J^{-i}=\frac{\sigma_i}{2}$ where $\sigma$ are pauli matrices. I am told that for those numbers $(j_-,j_+)$ we have a dimension $d$ with $d=(2j_{-}+1)(2j_{+}+1)$.

Question 2 : what does this dimension $d$ represent?

Rescy_
  • 2,002
  • 1
    Re Question 2, do you know what the dimension of a representation is? Because I think that is it. Or I do not understand this question. – Torsten Schoeneberg Oct 24 '22 at 18:09
  • I looked up the definition of the dimension of a representation. Indeed you are right, I agree $d$ is the dimension of the Lie algebra representation. Thanks! – Rescy_ Oct 24 '22 at 18:22
  • 2
    As to question 1, the answer is no. Those Lie algebras have the same dimension and even better have the same complexification: $\mathfrak{so}(4,\mathbb{C}) \cong\mathfrak{so}(3,\mathbb{C})\oplus \mathfrak{so}(3,\mathbb{C})$ but they are not isomorphic themselves. Indeed the right hand side is compact but the left hand side is not. – Callum Oct 24 '22 at 23:39
  • 2
    Expanding on @Callum's comment, the distinction between the non-compact $\mathfrak{so}(3,1)$ and the compact $\mathfrak{so}(4) \simeq \mathfrak{so}(3) \oplus \mathfrak{so}(3)$ as real Life algebras, while (as Callum notes) their complexifications are isomorphic, was the example which started my answer to https://math.stackexchange.com/q/3255785/96384. Note that the $J^{\pm}$ you cleverly construct are not in your original Lie algebra, but only in its complexification. They are probably the $A$s and $B$s I quote from Wikipedia in my answer there. – Torsten Schoeneberg Oct 25 '22 at 14:55
  • 1
    Besides the many related links in that answer, I point out https://math.stackexchange.com/a/4184237/96384 which shows why and how the representations of the non-isomorphic $\mathfrak{so}(3,1)$ and $\mathfrak{so}(4)$ look "the same", in particular can both be given by such pairs of numbers $(j_-,j_+)$. It's because both have complexification $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$. Yet they are different. On the representation level that only shows in how conjugation works on those representations. – Torsten Schoeneberg Oct 25 '22 at 15:32

0 Answers0