$\DeclareMathOperator\SO{SO}$I'm not asking for a proof but for some hint that might be helpful to understand this "anomaly" in 4 dimensions. I'm aware of the parallelism with the $A\_4$ finite group case as shown in the [answer](https://mathoverflow.net/a/145066) by @Benoit Kloeckner which I would also like to see elaborated on as I'm not sure I understood how the non-simplicity arises from the 3 decompositions into pairs of orthogonal planes in [The non-simplicity of $SO(4)$ and $A\_4$](https://mathoverflow.net/questions/145026/the-non-simplicity-of-so4-and-a-4), but I'm more centered on the particularity of 4 dimensions.



Why is $\operatorname{SO}(4)$ not a simple Lie group?
Here's a different approach than the ones that have been suggested so far. For any $n, k$ we can consider the action of $SO(n)$ on the exterior power $\Lambda^k(\mathbb{R}^n)$. These representations are almost all irreducible, *except* when $n = 2k$ and $k \ge 2$ is even; in this case the [Hodge star](https://en.wikipedia.org/wiki/Hodge_star_operator) $\star : \Lambda^k(\mathbb{R}^n) \to \Lambda^{n-k}(\mathbb{R}^n)$ restricts to an involutive automorphism (as an $SO(n)$-representation) of $\Lambda^k(\mathbb{R}^{2k})$ and hence splits it up into a direct sum of its $+1$-eigenspace and its $-1$-eigenspace, each of which is irreducible (I think). (When $k$ is odd the Hodge star squares to $-1$ so it no longer has real eigenvalues.)


These representations have dimension $\frac{1}{2} {2k \choose k}$, and the special feature of $SO(4)$ is that this dimension is smaller than the dimension $n = 2k$ of the defining representation iff $k = 2$. In the $k = 2$ case the exterior square $\Lambda^2(\mathbb{R}^4)$ is $6$-dimensional and the eigenspaces of the Hodge star are $3$-dimensional. This exhibits the double cover $SO(4) \to SO(3) \times SO(3)$ which shows that $SO(4)$ is not simple, and there is no need to know anything about the quaternions, spin groups, or spin representations.


More explicitly, writing $e\_1, e\_2, e\_3, e\_4$ for the standard basis of $\mathbb{R}^4$, and choosing the orientation $\omega = e\_1 \wedge e\_2 \wedge e\_3 \wedge e\_4$, the Hodge star $\star : \Lambda^2(\mathbb{R}^4) \to \Lambda^2(\mathbb{R}^4)$ by definition satisfies


$$\alpha \wedge (\star \beta) = \langle \alpha, \beta \rangle \omega$$


where $\langle \alpha, \beta \rangle$ is the Gram determinant inner product as described on Wikipedia; it is completely specified by saying that the wedge products $e\_i \wedge e\_j, i < j$ form an orthonormal basis. This lets us compute the action of $\star$ explicitly:


$$\star (e\_1 \wedge e\_2) = e\_3 \wedge e\_4$$
$$\star (e\_1 \wedge e\_3) = - e\_2 \wedge e\_4$$
$$\star (e\_1 \wedge e\_4) = e\_2 \wedge e\_3$$
$$\star (e\_2 \wedge e\_3) = e\_1 \wedge e\_4$$
$$\star (e\_2 \wedge e\_4) = - e\_1 \wedge e\_3$$
$$\star (e\_3 \wedge e\_4) = e\_1 \wedge e\_2.$$


The $+1$-eigenspace is spanned by $e\_1 \wedge e\_2 + e\_3 \wedge e\_4$, $e\_1 \wedge e\_4 + e\_2 \wedge e\_3$, and $e\_1 \wedge e\_3 - e\_2 \wedge e\_4$, while the $-1$-eigenspace is spanned by $e\_1 \wedge e\_2 - e\_3 \wedge e\_4$, $e\_1 \wedge e\_4 - e\_2 \wedge e\_3$, and $e\_1 \wedge e\_3 + e\_2 \wedge e\_4$. This corresponds to $3$ decompositions of $\mathbb{R}^4$ into pairs of orthogonal planes which means it presumably has something to do with the answer you quoted from the other MO question but I'm not sure what exactly.


