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Following from this question and the links within, I have a couple of questions about the use of $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ for the classification of finite real restricted Lorentz algebra irreps, and I suspect they are closely related.

  1. Since there is a one-to-one correspondence between the finite-dimensional irreps of $\mathfrak{so}^+(1,3)$ and the irreps of $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$, why does this not imply an isomorphism between the two algebras?

  2. Since there is an isomorphism between $\mathfrak{so}^+(1,3)_\mathbb{C}$ and $[\mathfrak{su}(2) \oplus \mathfrak{su}(2)]_\mathbb{C}$, why does this not imply an isomorphism between $\mathfrak{so}^+(1,3)$ and $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$?

Qmechanic
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    Do you think all $\mathfrak{so}(n)$ should be isomorphic to $\mathfrak{so}(k, n - k)$ for any $k$? – Connor Behan Dec 11 '21 at 22:22
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    I guess I know that they're not, but don't understand why. Perhaps where I'm going wrong is conflating bijections and isomorphisms, where the latter has to preserve structure, e.g. the commutation relations, etc.? – tomdodd4598 Dec 12 '21 at 09:15

1 Answers1

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Much of a qualitative structure of the representations of a these algebra is determined only by the complexifications of the algebras—that is, their tensor products as real Lie algebras with the complex field $\mathbb{C}$. Many different real Lie algebras can collapse to the same thing when complexified; for example, $\mathfrak{su}(n)\otimes\mathbb{C}\cong\mathfrak{sl}(n)\otimes\mathbb{C}$ for all $n$.

This is a very powerful fact, and Weyl used it to categorize the finite-dimensional irreducible representations of semisimple Lie algebras. Given a (finite-dimensional) semisimple $\mathfrak{g}$, the complexification $\mathfrak{g}\otimes\mathbb{C}$ is isomorpic to the complexification $\mathfrak{h}\otimes\mathbb{C}$ of another Lie alebra $\mathfrak{h}$ that is the semisimple Lie algebra of a compact Lie group. Complete reducibility of the representations of compact topological groups is relatively easy to prove; using the Haar measure, one can integrate over the entire group, so the proofs of complete reducibility for finite groups that use an averaging over the entire group can be carried over with minimal modification. And Weyl's unitary trick makes it possible to carry this result over to any semisimple Lie algebra, even if the Lie group it corresponds to is not compact.

For the algebras specifically mentioned in the question, there is the general isomorphism of complexified Lie algebras: $\left[\mathfrak{su}(2)\oplus\mathfrak{su}(2)\right]\otimes{\mathbb{C}}\cong\mathfrak{so}(n,4-n)\otimes\mathbb{C},$ for any $0\leq n\leq4$. Note that this immediately also implies $\mathfrak{so}(4)\otimes\mathbb{C}\equiv\mathfrak{so}(4,0)\otimes\mathbb{C}\cong\mathfrak{so}(3,1)\otimes\mathbb{C}$. However, the real Lie algebras $\mathfrak{so}(4)$ and $\mathfrak{so}(3,1)$ are quite different; they describe infinitesimal rotations in four space dimensions and infinitesimal Lorentz transformations in three space and one time dimension, respectively. $\mathfrak{so}(4)$ acts as an infintesimal linear transformation on $[x_{1},x_{2},x_{3},x_{4}]\rightarrow[x_{1}',x_{2}',x_{3}',x_{4}']$ that leaves $x_{1}'^{2}+x_{2}'^{2}+x_{3}'^{2}+x_{4}'^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ invariant. The Lorentz group is an infinitesimal linear transformation that instead leaves the spacetime interval $x_{0}'^{2}-x_{1}'^{2}-x_{2}'^{2}-x_{3}'^{2}=x_{0}^{2}-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}$ invariant. However, replacing the real parameters of these linear transformations with complex ones results in the same complexified algebra. These means that the same "quantum numbers" can be used to characterize the irreducible representations of the all the $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, $\mathfrak{so}(4)$, and $\mathfrak{so}(3,1)$ algebras (among others). The representation theories are very similar, in that tensor products of representations break down into irreducible representations in the same way for each algebra. [Working in $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, this means that the representations can be broken down using two sets of Clebsch-Gordon coefficients.] However, the detailed structures of the individual representations are different.

Buzz
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  • You don’t seem to distinguish between $\mathfrak{so}$ and $\mathfrak{so}^+$, is that on purpose? Thanks. – Bcpicao Nov 18 '23 at 21:23
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    @Bcpicao $\mathfrak{so}$ and $\mathfrak{so}^+$ are the same (meaning that the latter notation is essentially never used). A Lie group $G$ may have multiple components, and the quotients $G_{0}$ or $G^{+}$ are formed as $G/Z$ with a finite subgroup $Z$. However, this global structure [just like the global structural difference between $SO(3)$ and $SU(2)$] cannot be discerned from the structure of the Lie algebra $\mathfrak{g}$. – Buzz Nov 19 '23 at 18:57