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I am reading Brian Hall's book 'Lie Groups, Lie Algebras, & Representations' and on p.271 I find that in low rank Lie algberas there are some isomorphisms. For example, $\mathfrak{sl}(2;\mathbb{C})$ is isomorphic to $\mathfrak{so}(3;\mathbb{C})$. There is a nice way to understand this, which is quite standard (just the complexified Lie algbera version of the connection between $SU(2)$ and $SO(3)$) so I am not writing it here and can be found say on p.18 of the same book.

Is there a similar way to understand the Lie algbera isomorphism between $\mathfrak{sl}(4;C)$ and $\mathfrak{so}(6;C)$ ? What about the isomorphism between $\mathfrak{so}(5;C)$ and $\mathfrak{sp}(2;C)$ ?

Of course one way is to draw the Dynkin diagram and see, but I wish to see alternate ways of looking at these isomorphisms.

Even if there is no nice intuitive view, any one isomorphism with explicit formula will be helpful.

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You will like this: mathoverflow.net/questions/81344/… –  John Nov 13 '13 at 9:06

1 Answer 1

up vote 1 down vote accepted

I will list the coincidences of complex classical Lie algebras in increasing order:

  • $\mathfrak{u}(1) \overset\sim\to \mathfrak{so}(2)$. Trivial.

  • $\mathfrak{sl}(2) \overset\sim\to \mathfrak{so}(3)$. Look at the adjoint action of $\mathfrak{sl}(2)$ on itself; this preserves the Cartan pairing $\langle x,y\rangle = \mathrm{tr}(xy)$.

  • $\mathfrak{so}(3) \times \mathfrak{so}(3) \overset\sim\leftarrow \mathfrak{so}(4)$. Decompose the canonical action of $\mathfrak{so}(4)$ on $\mathbb C^6 \cong (\mathbb C^4)^{\wedge 2}$ into two three-dimensional submodules, and recognize them as the modules $\mathbf 3 \otimes \mathbf 1$ and $\mathbf 1 \otimes \mathbf 3$ of $\mathfrak{so}(3) \times \mathfrak{so}(3)$, where $\mathbf 1$ is the trivial module and $\mathbf 3$ is the defining representation. You can also write the map explicitly by recalling that $\mathfrak{so}(n)$ consists precisely of the antisymmetric $n\times n$ matrices. Finally, you can write $\mathfrak{so}(3) = \mathfrak{sl}(2)$, and then take $\mathbf 4 = \mathbf 2 \otimes \mathbf 2$.

  • .$\mathfrak{sp}(2) \overset\sim\to \mathfrak{so}(5)$. I take the domain to mean the maps preserving four-dimensional symplectic space (I would prefer to call this $\mathfrak{sp}(4)$, but there is much disagreement in the literature). Note that $\mathfrak{sp}(2)$ acts on $\mathbb C^4$, and hence on $(\mathbb C^4)^{\wedge 2} = \mathbb C^6$. Now, by definition, $\mathfrak{sp}(2)$ preserves an antisymmetric pairing on $\mathbb C^4$, meaning that as an $\mathfrak{sp}(2)$, this $\mathbb C^6$ has a one-dimensional trivial subrepresentation (spanned by the inverse matrix to the canonical pairing). Moreover, this $\mathbb C^6$ has a symmetric inner product given by the determinant pairing $\langle v_1\wedge w_1, v_2 \wedge w_2\rangle = \det(v_1,w_1,v_2,w_2)$, and $\mathfrak{sp}(4)$ preserves this pairing. The pairing of course restricts to the two pieces $\mathbb C^6 = \mathbf 5 \oplus \mathbf 1$, whence we get $\mathfrak{sp}(2) \to \mathfrak{so}(5)$.

  • For $\mathfrak{sl}(4) \overset\sim\to \mathfrak{so}(6)$, my answer to your other recent question follows exactly the previous bullet point: $\mathbf 4^{\wedge 2} = \mathbf 6$.

  • The deepest of the coincidences of low-dimensional complex Lie algebras is the nontrivial "triality" automorphism $\mathfrak{so}(8) \overset\sim\to \mathfrak{so}(8)$. The best explicit description of this that I know is a blog post by Jacques Distler.

A remark. The defining representations of $\mathfrak{sp}(2)$ and $\mathfrak{sl}(2)$ are, respectively, the spin representations of $\mathfrak{so}(5)$ and $\mathfrak{so}(3)$; the two defining representations of $\mathfrak{sl}(2) \times \mathfrak{sl}(2)$ and the defining and dual-defining representations of $\mathfrak{sl}(4)$ are respectively the spin representations of $\mathfrak{so}(4)$ and $\mathfrak{so}(6)$. I've also put the example $\mathfrak{u}(1) = \mathfrak{so}(2)$, which are both abelian one-dimensional. But the correct way to think about this is as the complexification of the map in which $e^{i\theta}$ acts as rotation by $2\theta$, not by $\theta$. Then the defining representation of $\mathfrak{u}(1)$ and its dual are the spin representation of $\mathfrak{so}(2)$; the defining representation of $\mathfrak{so}(2)$ is the sum of the squares of these. The spin representations of $\mathfrak{so}(8)$ are the two representations formed from the defining representation by precomposing with triality or triality$^{-1}$.

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Thanks a lot ! Could you please add some references where I can learn more about such stuff ? –  user90041 Nov 19 '13 at 15:02
    
One reference, but by no means the best, is the draft textbook that I started editing (based on lectures given at UC Berkeley --- none due to me), and hope to finish one day math.northwestern.edu/~theojf/LieQuantumGroups.pdf . The MO question linked above also contains some references. –  Theo Johnson-Freyd Nov 19 '13 at 21:52

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