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Im a physics student an I have been studying Lie Groups and Lie algebras for some time from a mathematical point of view mostly following Hall's book. Thing is that the Highest Weight Theorem is ennunciated for complex semisimple Lie algebras and the irr. reps. are labeled with the weights that are of course positive integers. But in physics we work with $su(2)$ that is not a complex Lie algebra, and we label its irr. reps. by a Half Integer. In Hall it just says that

In the physics literature, the representations of $su(2)\cong so(3)$ are labeled by the parameter $l=m/2$. In terms of this notation, a representation of $so(3)$ comes from a representation of $SO(3)$ if and only if $l$ is an integer. The representations with $l$ an integer are called “integer spin”; the others are called “half-integer spin.”

So it just lets it as a notational thing. Thing is that this cannot be, Half integer in something key in physics, it just arises naturally. I thought it may be because the isomorphism $sl(2,\mathbb{C})\cong su(2)\oplus su(2)$ Like, i dont know, half integer + integer =integer but i dont even know how to show that, if true. Anything is largely apreciated.

Thomas Andrews
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olsrcra
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    You might be better suited posting this in physics.stackexchange.com – Sidharth Ghoshal Sep 17 '22 at 17:54
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    Isn't this answered in the quoted text? The representations with integer values come from $SO(3)$. I think the only thing you may not know is that the fundamental group $\pi_1(SO(3)) \simeq \mathbb{Z}/2\mathbb{Z}$, and $SU(2)$ is its universal cover. This is the reason for the denominator of $2$. – A. Thomas Yerger Sep 17 '22 at 18:01
  • at first i thought about that, but my question goes more with the highest weight theorem, why it can be applied to su(2) and why in this case there are half integer weights @SidharthGhoshal – olsrcra Sep 17 '22 at 18:02
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    @A.ThomasYerger and that of course has to do with the fact that it is its double cover as paul said right? – olsrcra Sep 17 '22 at 18:15
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    @olsrcra it is not that we get half-integer weights. It is just that the physics numbering system differs by a factor of 2 from the highest weight system. – Callum Sep 17 '22 at 22:01

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To my mind, the explanation is at the level of Lie groups, not Lie algebras, since the Lie algebras preserve slightly less information than the Lie groups. Here, while $su(2)\to so(3)$ is an isomorphism, the corresponding Lie group map $SU(2)\to SO(3)$ is $2$-to-$1$. So only about half the irreducibles of $SU(2)$ descend to $SO(3)$. The ones that do descend are easily distinguished (as it turns out) by some parity issues (equivalently, integer/half-integer).

The other incidental issue in your question, about real versus complex Lie algebras, is not directly related to this... but perhaps it is helpful to note that a complex repn of a real Lie algebra has a canonical extension to a repn of the complexification of that Lie algebra on the same complex vector space. (And, it's not really the case that the complexification of $su(2)$ is two copies of it... more accurate is that it's $s\ell(2)$).

EDIT: yes, by $s\ell(2)$ I mean the complex Lie algebra. And, the way complexification of Lie algebras (or lots of other things over-the-reals) work does product two copies of the original but only as real vector spaces... not as Lie algebras, for example. As a simpler example, the complexification of the ring (or field) $\mathbb R$, is $\mathbb C$, as a ring/field. As a real vector space, sure, it's $\mathbb R\oplus \mathbb R$, but not as a ring/field.

paul garrett
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  • I got the first, i think. I thought it was more at the lie algebra level and didnt understand quite well why half integer weights were posible. About the second part, $sl(2)$ is $sl(2,\mathbb{C})$? why is not the complexifcation two copies of it self? i dont think i did understand the second paragraph, my question is about why we can apply highest weight to su(2) – olsrcra Sep 17 '22 at 18:13
  • (i just saw the edit) okay, thank you very much. May you please explain more about the complexification and why i can apply highest weight theorem to $su(2)$ ? – olsrcra Sep 17 '22 at 18:34
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    Explaining "complexification" is a whole story in itself... but/and, after some details are smoothed-out, the "highest-weight theory" applied to a real Lie algebra, acting on complex vector spaces, instantly applies to the complexified Lie algebras. – paul garrett Sep 17 '22 at 21:28
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    @olsrcra perhaps the easiest way to see that highest weight theory works for real Lie algebras is as follows. If you have a representation $\rho$ of $\mathfrak{g}^\mathbb{C}$ you can restrict $\rho$ to $\mathfrak{g}$ and you get a representation of $\mathfrak{g}$. Conversely, you can extend a representation of $\mathfrak{g}$ to its complexification as $\rho(X+iY):=\rho(X)+i\rho(Y)$. Note here the vector space is assumed to be complex. Sometimes we can find a real subspace preserved by $\rho$ and then we get a rep on a real vector space but this doesn't always happen. – Callum Sep 17 '22 at 21:57