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I'm sure I'm going to miss some mathematical detail in this, but I want to make sure that I have the gist of the $\mathfrak{so}(1,3) \to \mathfrak{su}(2)\oplus\mathfrak{su}(2)$ relationship correct.

$\mathfrak{so}(1,3)$ is formed of the six generators of proper orthochronous Lorentz group, $\mathrm{SO}(1,3)$, three $J_{i}$ and three $K_{i}$, with the Lie brackets:

$$ [J_i,J_j] = i\varepsilon_{ijk}J_k \\ [K_i,K_j] = -i\varepsilon_{ijk}J_k \\ [J_i,K_j] = i\varepsilon_{ijk}K_k $$

If you redefine these as $A_p = \frac{1}{2}(J_p + iK_p)$ and $B_p = \frac{1}{2}(J_p - iK_p)$, these new generators span a subset of complexification of $\mathfrak{so}(1,3)$, $\mathfrak{so}(1,3)_{\mathbb{C}}$, that is in fact isomorphic to $\mathfrak{so}(1,3)$.

The new Lie brackets are then given by:

$$ [A_p,A_q] = i\varepsilon_{pqr}A_r \\ [B_p,B_q] = i\varepsilon_{pqr}B_r \\ [A_p,B_q] = 0 $$

so in fact the $\{A_p\}$ are isomorphic to $\mathfrak{su}(2)$, as are the $\{B_p\}$ independently. Hence $\mathfrak{so}(1,3)$ is isomorphic to a restriction of the full $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ Lie algebra to just generators of the form:

$$ A_p = \sigma_p \otimes I \\ B_p = I \otimes \sigma_p $$

Then, by rearranging $A_p = \frac{1}{2}(J_p + iK_p)$ and $B_p = \frac{1}{2}(J_p - iK_p)$ we get:

$$ J_p = A_p + B_p \\ K_p = -i(A_p - B_p) \\ $$

from which it follows that:

$$ J_p = \sigma_p \otimes I + I \otimes \sigma_p \\ K_p = -i(\sigma_p \otimes I - I \otimes \sigma_p) $$

which is the standard result that's on the Wikipedia page (barring the extra minus sign I seem to have in $K_p$...).

Hence, you can find the representations of $J_p$ and $K_p$ in terms of the representations of $\mathfrak{su}(2)$, which are well-known. What I don't understand is why you have the freedom to choose different dimensions for the representations of the two copies of $\mathfrak{su}(2)$. Is it just because the tensor product preserves the dimension of the product of square matrices under commutation?

gautampk
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1 Answers1

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I am not sure about your conventions, but it is not true that $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ are isomorphic or that $\mathfrak{so}(1,3)$ is a subalgebra of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. The both have dimension $6$ and they both have the same complexification, namely $\mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)$. However, $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is a compact real form, while $\mathfrak{so}(1,3)$ is non-compact (which can be seen from that fact that the Killing form is definite respectively indefinite).

Nonetheless, the real Lie algebras $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ have the same complex representations. This is because any complex representation of a real Lie algebra uniquely extends to the complexification, so in both cases complex representations are equivalent to complex representations of $\mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)$. In the language of $\mathfrak{so}(1,3)$ the fact that there are "two components" of a representation is best seen when building up all representations from spinors.

Andreas Cap
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  • But I thought the direct sum $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is 9 dimensional, since it consists of all ordered pairs of elements? – gautampk Apr 12 '17 at 09:57
  • Actually, I think I've got it. The complexification of $\mathfrak{su}(2)$ (viewed as a real space) is a six dimensional space with basis ${\sigma_{p},i\sigma_{p}}$ (since the basis of $\mathbb{C}$ as a vector space is ${1,i}$). These items satisfy the same commutation relations as $J_p$ and $K_p$ so then $\mathfrak{sl}(2,\mathbb{C}) \equiv \mathfrak{so}(1,3)$. However, I'm unsure as to how to get to the standard form from $J_p = \sigma_p$ and $K_p = i\sigma_p$. – gautampk Apr 12 '17 at 10:05
  • Since $\mathfrak{su}(2)$ has real dimension three, its complexification has complex dimension $3$. The direct sum of Lie algebras equals the direct product, so $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ has real dimension $6$, which is the same as the real dimension of $\mathfrak{so}(1,3)$. The natural point of view would be that $\mathfrak{so}(1,3)$ is a real form of $\mathfrak{so}(4,\mathbb C)$, but this happens to be isomorphic to $\mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)$. – Andreas Cap Apr 12 '17 at 10:12
  • Okay, if $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is a 6 dimensional vector space and the basis has the same commutation relations as $\mathfrak{so}(1,3)$ then it follows that $\mathfrak{so}(1,3)\equiv\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ surely? What more is there to show for an isomorphism between Lie algebras other than the Lie brackets being the same and them both being equal dimensional vector spaces over the same field (the reals)? – gautampk Apr 12 '17 at 10:48
  • Just to be clear, $\mathrm{SO}(1,3)$ is the proper orthochronous (identity) component of the Lorentz group that is connected but I think not compact. $\mathfrak{so}(1,3)$ is its Lie algebra. – gautampk Apr 12 '17 at 10:53
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    If there were bases for which the commutation relations are the same, then the Lie algebras would indeed be isomorphic. But such bases do not exist (and if you think you have found them, then there is an error in your computations). (As I said, I don't understand your conventions, in particular, I don't know why there are $i$'s in the commutation relations for $\mathfrak{so}(3,1)$.) – Andreas Cap Apr 12 '17 at 11:34