How should I show that the Lie algebra $\mathfrak{sl}(4,\mathbb{C})$ is isomorphic to the Lie algebra $\mathfrak{so}(6,\mathbb{C})$? It is the Problem.3 of the "S.-T. Yau College Student Mathematics Contests 2022 Algebra and Number Theory"
The problem is

Give a direct proof that the Lie algebra $\mathfrak{sl}(4,\mathbb{C})$ is isomorphic to the Lie algebra $\mathfrak{so}(6,\mathbb{C})$.(You should construct a Lie algebra homomorphism and prove that it is an ismorphism; you should not use Dynkin diagrams or the classification theory of simple Lie algebras).

 A: As you surely suspect, there is not a "systematic" approach to these anomalous facts. :)
One example-documentation of many of these anomalous/sporadic isomorphisms/isogenies, etc. is at
https://www-users.cse.umn.edu/~garrett/m/v/sporadic_isogenies.pdf
That list/discussion is not exhaustive, because we could also do "quaternion" forms... not to mention maybe p-adic, etc. :)
A: I thought I'd give a pared down explanation with less Lie group focus and fewer (in my opinion) unneeded choices. The action of $\mathfrak{g} = \mathfrak{sl}_4$ on $V = \mathbb{C}^4$ naturally extends to an action on $\bigwedge^2V$ via:
$$ X \cdot (v\wedge w) = Xv \wedge w + v\wedge Xw.$$
To see where this comes from we only need to differentiate the diagonal group action $g\cdot(v\wedge w) = gv \wedge gw$. Now $\dim\bigwedge^2V = 6$ and it comes with a natural bilinear form $(a\wedge b, c\wedge d) := a\wedge b \wedge c\wedge d \in \bigwedge^4V \cong \mathbb{C}$. This last identification of $\bigwedge^4V \cong \mathbb{C}$ is a choice but it doesn't actually matter which choice we make as $\mathfrak{so}_n$ doesn't change if we rescale the bilinear form.
So now we have to show that $\mathfrak{sl}_4$ indeed acts as $\mathfrak{so}_6$ on this space. In other words, that it is skew for the bilinear form:
$$ (X\cdot\alpha,\beta) + (\alpha,X\cdot\beta) = 0$$
Take $\alpha = a \wedge b, \beta = c \wedge d$. Then the above reads:
$$ (Xa\wedge b \wedge c\wedge d) + (a\wedge Xb \wedge c\wedge d) + (a\wedge b \wedge Xc\wedge d) + (a\wedge b \wedge c\wedge Xd) = 0$$
Note this is just the natural action of $\mathfrak{sl}_4$ on $\bigwedge^4V$ the left hand side and that this equals zero for any $a,b,c,d$ is exactly the statement that the action on $\bigwedge^4V$ is trivial or equivalently that $\mathrm{tr} X =0$ which is exactly our definition of $X\in\mathfrak{sl}_4$.
For this last idea involving the trace we can find this by differentiating the  true definition of determinant:
$$ga\wedge gb \wedge gc\wedge gd = \det(g)a\wedge b \wedge c\wedge d$$
So to be precise we have found a map $\mathfrak{sl}_4 \to \mathfrak{so}_6$ and to see that this is an isomorphism we note that it is injective (not quite trivial but not hard) and both our Lie algebras have the same dimension. In fact as Torsten points out we only need to show that the map is not the zero map since both of them are simple Lie algebras.
Finally, note that this isomorphism descends to the real forms as well. For example:
$$ \mathfrak{su}(4) \cong \mathfrak{so}(6)$$
$$ \mathfrak{sl}(2,\mathbb{H}) \cong \mathfrak{so}(5,1)$$
$$ \mathfrak{su}(2,2) \cong \mathfrak{so}(4,2)$$
$$ \mathfrak{sl}(4,\mathbb{R}) \cong \mathfrak{so}(3,3)$$
$$ \mathfrak{su}(3,1) \cong \mathfrak{so}^*(6)$$
A: This is essentially an example of an accidental isomorphism, an isomorphism between a classical matrix Lie group and a spin group. (Equivalently, said isomorphisms at the lie algebra level, or complexified versions of these isomorphisms.) These are a special case of exceptional isomorphisms, which are isomorphisms between small objects in families of objects which do not generalize to the rest of the families (for example, only so many special orthogonal and special linear groups are isomorphic). These include finite simple groups especially. (Some use "accidental" and "exceptional" interchangeably or a little differently, but this is my preference to distinguish them.) While there is some general theory for determining these isomorphisms, such as Dynkin diagrams for complex lie algebras, there are typically more interesting (and mysterious) explanations lurking around that only work for small sets or low dimensions (an illustration of the Law of Small Numbers in action).

$\mathrm{SL}_4\mathbb{R}$ acts on $\Bbb R^4$, and consequently acts on $\Lambda^2\Bbb R^4$ (which is $6$-dimensional). This inherits an inner product from $\mathbb{R}^4$ by polarizing and linearizing Grammian determinants, i.e. extending
$$ \langle\alpha\wedge\beta,\gamma\wedge\delta\rangle=\det\left(\begin{bmatrix} | & | \\ \alpha & \beta \\ | & |\end{bmatrix}^T\begin{bmatrix}| & | \\ \gamma & \delta \\ | & |\end{bmatrix}\right)=\det\begin{bmatrix}\langle\alpha,\gamma\rangle & \langle\alpha,\delta\rangle \\ \langle\beta,\gamma\rangle & \langle\beta,\delta\rangle \end{bmatrix}, $$
but this is actually not what we're interested in! (It is, however, how you calculate angles between 2D planes in 4D, so I thought I'd mention it.) Instead, we can be interested in the symmetric bilinear form
$$ (\alpha\wedge\beta,\gamma\wedge\delta):=\alpha\wedge\beta\wedge\gamma\wedge\delta, $$
where we identify $\Lambda^4\mathbb{R}^4$ with $\mathbb{R}$ via $\lambda(e_1\wedge e_2\wedge e_3\wedge e_4)\leftrightarrow\lambda$. This is a pseudo-Euclidean inner product with signature $(3,3)$. Two canonical choices of 3D subspaces, in the bilinear form restricts to a positive- or negative-definite signature, are the eigenspaces $\Lambda^2_+$ and $\Lambda^2_-$ of the Hodge star operator $\star\substack{\curvearrowright\\~}\Lambda^2\mathbb{R}^4$.
To understand this, consider the oriented Grassmanian $\widetilde{\mathrm{Gr}}_2\mathrm{R}^4$ which paramatrizes the oriented 2D subspaces of $\mathbb{R}^4$. This may be identified with a subset of $\Lambda^2\mathbb{R}^4$: if an oriented 2D subspace has an ordered orthonormal basis $\{e_1,e_2\}$, it corresponds to $e_1\wedge e_2\in\Lambda^2\mathbb{R}^4$ (note this is well-defined: if it has another ordered orthonormal basis $\{f_1,f_2\}$, then $e_1\wedge e_2=f_1\wedge f_2$ within $\Lambda^2\mathbb{R}^4$). (Also note $\widetilde{\mathrm{Gr}}_2\mathbb{R}^4$ is a double cover of the unoriented Grassmanian $\mathrm{Gr}_2\mathbb{R}^4$ which may be identified with a subspace of projectivization $\mathbb{P}(\Lambda^2\mathbb{R}^4)$.) Taking orthogonal complements (of oriented subspaces) is an involution of $\widetilde{\mathrm{Gr}}_2\mathbb{R}^4$ which extends to a linear operator $\star$ on $\Lambda^2\mathbb{R}^4$ with eigenvalues $\pm1$ both of multiplicity $3$.
The $\pm1$-eigenspaces $\Lambda^2_{\pm}$ are spanned by elements of the form $e_i\wedge e_j+ e_k\wedge e_\ell$ with $ijk\ell$ an even or odd permutation of $1234$ respectively. Under the identification $\Lambda^2\mathbb{R}^4\cong\mathfrak{so}(4)$, where $\alpha\wedge\beta$ (with $\{\alpha,\beta\}$ orthonormal) is a right-angle rotation in the $\alpha\beta$-plane and vanishes on the complement, the $\pm1$-eigenspaces correspond to left/right isoclinic rotations, which in turn correspond to left- or right-multiplying $\mathbb{H}$ by unit quaternions (in $SO(4)$).
The aforementioned symmetric bilinear form restricts to a positive- or negative-definite form on $\Lambda^2_{\pm}$ respectively. Note $\Lambda_{\pm}^2$ are orthogonal with respect to the form (as well as the irrelevant Euclidean inner product mentioned earlier!), so the form has signature $(3,3)$. Now, $\mathrm{SL}_4\mathbb{R}$ respects this form, since
$$ (A\alpha\wedge B\beta,A\gamma\wedge A\delta)=A\alpha\wedge A\beta\wedge A\gamma\wedge A\delta=\det(A\alpha\,A\beta\,A\gamma\,A\delta)$$
$$ =(\det A)\det(\alpha\,\beta\,\gamma\,\delta)=(1)(\alpha\wedge\beta\wedge\gamma\wedge\delta)=(\alpha\wedge\beta,\gamma\wedge\delta). $$
Thus, we have a homomorphism $\mathrm{SL}_4\mathbb{R}\to\mathrm{SO}(3,3)$. We can check this is $2$-to-$1$ map with kernel $\{\pm I_4\}$, and the corresponding lie algebra homomorphism $\mathfrak{sl}_4\mathbb{R}\to\mathfrak{so}(3,3)$ is an isomorphism. Since
$$ \mathfrak{sl}_4\mathbb{C}=(\mathfrak{sl}_4\mathbb{R})\oplus(\mathfrak{sl}_4\mathbb{R})i, \qquad \mathfrak{so}_6\mathbb{C}=\mathfrak{so}(3,3)\oplus \mathfrak{so}(3,3)i, $$
we can extend the isomorphism to the complexifications, i.e. $\mathfrak{sl}_4\mathbb{C}\to\mathfrak{so}_6\mathbb{C}$.
