Question regarding isomorphisms in low rank Lie algebras

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.

Solutions Collecting From Web of "Question regarding isomorphisms in low rank Lie algebras"

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$^{-1}$.