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  • I want to precisely understand in what sense is (if it is!) $SL(2,\mathbb{C})$ the "complexified" version of $SU(2)$?

Can I think of it like choosing a natural matrix basis of the real three dimensional Lie algebra of $SU(2)$ (say the Pauli matrices) and looking at the vector space they would span over $\mathbb{C}$ (i.e look at the vector space of matrices spanned by linear sums of Pauli matrices with complex coefficients) and then exponentiate it down?

  • Is there a natural action of $SL(2,\mathbb{C})$ on the $5$ dimensional irreducible representation of $SU(2)$? If yes then how does one best understand the quotient space and precisely in what way does this action respect the representation.

    (I would be interested about the general picture if there is any behind such actions, I chose the above particular example since it is most relevant to my current pursuits.)

Anirbit
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    Anirbit- While this is a good question, I'm not sure its a good question for MathOverflow. Have you read some basic textbooks on this stuff, like Fulton and Harris? – Ben Webster Dec 13 '10 at 06:32
  • $SL(2,{\mathbb{C}})$ is the complexification of $SU(2)$. You can find some information and references here: http://eom.springer.de/c/c024210.htm – Theo Buehler Dec 13 '10 at 06:33
  • By the way, the answer is yes. – Ben Webster Dec 13 '10 at 06:33
  • The general picture is as follows: take the complexified Lie Algebra of SU(n). Of course, this is in a natural way sl(n), which integrates up to a connected Lie Group, namely SL(n). FOr reference, see comments above. – Sebastian Dec 13 '10 at 07:44
  • @Ben I have a passing acquaintance with the subject (enough to understand the answers that others are typing in) but of course I have never done formal courses in it since my official education is not in Mathematics. – Anirbit Dec 13 '10 at 14:38
  • Both act in a natural (and the same) way on the space of homogenous polynomials of a given degree in two (complex variables). The action $SU(2)$ is unitary while the action of $SL(2,\mathbb{C})$ is not. Physisicst sometimes call these rep. of degree $2s$ the spin $s$ representation. Remember that $SU(2) = Spin(3)$ and $SL(2,\mathbb{C})=Spin(3,1)$ are the double cover of the 3 dimensional rotation and 4 dimensional Lorentz group, respectively. – Marcel Bischoff Dec 13 '10 at 14:56
  • @Marcel Thanks! I guess one can think of the spherical harmonic functions to give the natural vector spaces on which both $SL(2,\mathbb{C})$ and $SU(2)$ act. – Anirbit Dec 13 '10 at 16:21
  • @Marcel Doesn't the usual angular momentum basis $\vert l,m \rangle$ span the same representation as above? – Anirbit Dec 13 '10 at 16:34
  • I am not sure if spherical harmonics makes sense for $SL(2)$. Yes they span the same representation if i am not wrong you can decompose the polynoms like this: for example for spin 1/2 you have $|1/2,-1/2\rangle = x_1$, $|1/2,1/2\rangle = x_2$ and for spin 1 you have $|1,-1\rangle = x_1^2$ $|1,0\rangle = x_1x_2$ and $|1,1\rangle = x_2^2$ etc modulo signs and normalization constants. – Marcel Bischoff Dec 13 '10 at 16:59

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I am not an expert in this field but I think that the general picture looks more or less like that:

1. $\mathfrak{sl}(N,\mathbb{C}) = \mathfrak{su}_{\mathbb{C}}(N)$ ($\mathfrak{su}(N)$ is a compact real form of $\mathfrak{sl}(N,\mathbb{C} )$). Therefore, there is one to one correspondence between finite dimensional complex representations of Lie algebras: $\mathfrak{sl}(N,\mathbb{C})$ and $\mathfrak{su}(N)$. Since $SU(N)$ is compact, all finite dimensional complex representations of $\mathfrak{su}(N)$ are unitary and therefore completely reducible (finite dimensional complex representations of $\mathfrak{sl}(N,\mathbb{C})$ share this property). Moreover - there is one to one correspondence between complex, irreducible, finite dimensional representations of $\mathfrak{su}(N)$ and $ \mathfrak{sl}(N,\mathbb{C})$.
2. Since $SL(N,\mathbb{C})$ is simply- connected (see http://en.wikipedia.org/wiki/General_linear_group ), each finite dimensional irreducible complex representation of $\mathfrak{sl}(N,\mathbb{C})$ gives irreducible representation of $SL(N,\mathbb{C})$ (by matrix exponentiation of a Lie algebra representation) and vice-versa. The same thing can be said about $SU(N)$ and $\mathfrak{su}(N)$. As a result you have a natural action of $SL(N,\mathbb{C})$ on each irreducible complex representation of $SU(N)$ (note that given complex Lie group representation is irreducible iff corresponding representation of a Lie algebra is irreducible).

Michał Oszmaniec
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  • @Michal Thanks for your answer. You have some insights about the specific action I was looking for? (the second part of my question) – Anirbit Dec 13 '10 at 14:36
  • I don't know exactly what "quotient space" you have in mind. As for action of $SL(2,\mathbb{C}) $: it is given by the formula: $\tilde{\Phi}(e^{X})=e^{\tilde{\phi}(X)}$, where $X\in\mathfrak{sl}(2,\mathbb{C})$ and $\tilde{\phi}$ is the extension of the irreducible complex representation $\phi$ of a Lie algebra $\mathfrak{su}(2)$ to a complex represenation of $\mathfrak{sl}(2,\mathbb{C})$: if $X=\alpha +\imath \beta,\ \alpha,\beta\in\mathfrak{su}(2)$, $\tilde{\phi}(\alpha +\imath \beta)= \phi(\alpha) + \imath \phi(\beta)$. – Michał Oszmaniec Dec 13 '10 at 16:30
  • Only kind of compatibility that I can think of at the moment is the following: $\tilde{\Phi}|_{SU(2)}=\Phi$. Where $Phi$ is an irreducible complex representation of $SU(2)$ induced by $\phi$. – Michał Oszmaniec Dec 13 '10 at 16:36