Why is $\operatorname{SO}(4)$ not a simple Lie group? $\DeclareMathOperator\SO{SO}$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 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 $A_4$, but I'm more centered on the particularity of 4 dimensions.
 A: 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 $\star : \Lambda^k(\mathbb{R}^n) \to \Lambda^{n-k}(\mathbb{R}^n)$ 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.
A: The question is somewhat open-ended — more about beauty than truth. So I will offer two rather different narratives.

One way a group $G$ can fail to be simple is that it has a representation which is "too small", so the action of $G$ on the representation is forced to have a kernel. If you are very open minded about what "representation" means, then of course this is the same as being non-simple; but I really mean a classical representation (as if there were not exceptional Lie groups). Of course, by "a representation" I mean a nontrivial one, and I'm perfectly happy with projective representations.
In the case of the groups $\mathrm{SO}(n)$, most of the fundamental representations are essentially wedge powers of the defining $n$-dimensional representation. But there are two fundamental representations, namely the "half-spin" representations, of dimension $2^{(n-2)/2}$. (This is for $n$ even. When $n$ is odd, the full-spin representation is simple of dimension $2^{(n-1)/2}$.)
So this representation could in principle provide non-simplicity when $2^{(n-2)/2} \lesssim n$. This happens when $n \lesssim 8$.
Now you can ask: does it in fact provide non-simplicity? This requires thinking more carefully about what "too small" means. If $G$ were $\mathrm{SU}(n)$, then certainly a representation of dimension $<n$ would be too small. But we care about $G = \mathrm{SO}(n)$, which is a bit smaller than $\mathrm{SU}(n)$.
This also reminds that different types of representations — real, complex, and quaternionic — have different qualitative sizes. Let's decide that a complex $m$-dimensional representation has "size" $m$. Unpacked, this is a map $G \to \mathrm{U}(m)$, and $\dim \mathrm{U}(m) = m^2$. So by extension a real representation $G \to \mathrm{O}(m)$ would have "size" $\sqrt{\dim \mathrm{O}(m)} \approx m/\sqrt{2}$. Finally, a quaternionic representation $G \to \mathrm{Sp}(2m)$ has "size" $\sqrt{\dim \mathrm{Sp}(2m)} \approx 2m/\sqrt{2} = \sqrt{2}m$.
And remember that we care about $G = \mathrm{SO}(n)$, so our comparison is the vector representation of size $n/\sqrt{2}$.
Which of the spin representations are real, complex, or quaternionic? The answer for even $n$ is: for $n=2,6 \mod 8$ the half-spin representations are complex; for $n=0 \mod 8$ the half-spin representations are real; for $n=4 \mod 8$ the half-spin representations are quaternionic. For odd $n$: for $n=1,7 \mod 8$, the full-spin representation is real; for $n=3,5 \mod 8$, the full-spin representation is quaternionic.
Taking this into account, you get the following approximate sizes:

*

*$n=2$: vector $\approx 2/\sqrt{2} \approx 1.4$; spinor $\approx 1$.

*$n=3$: vector $\approx 3/\sqrt{2} \approx 2.1$; spinor $\approx \sqrt{2} \approx 1.4$.

*$n=4$: vector $\approx 4/\sqrt{2} \approx 2.8$; spinor $\approx \sqrt{2} \approx 1.4$.

*$n=5$: vector $\approx 5/\sqrt{2} \approx 3.5$; spinor $\approx 2\sqrt{2} \approx 2.8$.

*$n=6$: vector $\approx 6/\sqrt{2} \approx 4.2$; spinor $\approx 4$.

*$n=7$: vector $\approx 7/\sqrt{2} \approx 4.9$; spinor $\approx 4\sqrt{2} \approx 5.6$.

*$n=8$: vector $\approx 8/\sqrt{2} \approx 5.6$; spinor $\approx 4\sqrt{2} \approx 5.6$.

*$n=9$: vector $\approx 9/\sqrt{2} \approx 6.4$; spinor $\approx 8\sqrt{2} \approx 11.3$.

*$n \geq 10$ large: vector grows linearly; spinor grows exponentially.

Finally, we need to remember that $\sqrt{\dim \mathrm{SO}(m)}$ and $\sqrt{\dim \mathrm{Sp}(m)}$ are not quite exactly $m/\sqrt{2}$, but have some subleading corrections: our estimate for $\sqrt{\dim \mathrm{SO}(m)}$ is too large by a factor of $\sqrt{1-1/m}$, whereas our estimate for $\sqrt{\dim \mathrm{Sp}(m)}$ is too small by about the same factor.
Subleading corrections don't matter when $m$ is large, but do matter when $m$ is small. Taking these into consideration adjusts most of the cases in the above list, in particular, all the odd ones, where already there's an extra factor of $\sqrt{2}$ around: in all those cases, the spinor is not small enough to cause non-simplicity. The one remaining case is $n=4$, where $2.8$ is sufficiently lot bigger than $1.4$ that the subleading corrections just don't make up the difference.

Here is a completely different story. Pick up a compact Lie group $G$. Chevalley and Eilenberg calculated the de Rham cohomology $\mathrm{H}^\bullet(G;\mathbb{R})$ of (the underlying manifold of) $G$, showing that it is computed by a finite-dimensional complex built from $\mathfrak{g} = \mathrm{Lie}(G)$. (Method: write down the de Rham complex, and write down the subcomplex of $G$-invariant de Rham forms; by averaging, show that there is a deformation retract of the large complex onto its subcomplex; on the other hand, by translating, compare the subcomplex with something computed locally near the identity element.)
What you find is that $\mathrm{H}^1(G;\mathbb{R})$ counts the number of abelian factors of $G$. In particular, it vanishes if $G$ is semisimple. Assuming $G$ is semisimple, $\mathrm{H}^2(G;\mathbb{R})$ also vanishes. And $\mathrm{H}^3(G;\mathbb{R})$ counts the number of invariant symmetric bilinear forms on the adjoint representation. But, since the adjoint comes with a favourite symmetric bilinear form (the Killing form), this is the same as counting the number of endomorphisms of the adjoint representation, which is in turn the same as counting the number of simple factors of $\mathfrak{g}$.
So, in other words: a semisimple compact Lie group $G$ is simple exactly when $\mathrm{H}^3(G;\mathbb{R})$ is one-dimensional.
On the other hand, there is a fibre sequence $\mathrm{SO}(n) \to \mathrm{SO}(n+1) \to S^n$. So if you know the cohomology of $\mathrm{SO}(n)$, you can work out the cohomology of $\mathrm{SO}(n+1)$. Specifically, to build (a cell model for) $\mathrm{SO}(n+1)$, you take $\mathrm{SO}(n)$, and attach a cell in degree $n$, and then some more cells of higher degree. (Whence higher cells? If this fibre sequence were a product, then the cells of $\mathrm{SO}(n+1)$ would be the products of cells of $\mathrm{SO}(n)$ and of $S^n$. So far we've put in the cells of the form (cell)$\times$(zero-cell) and (zero-cell)$\times$(cell).)
This newly attached degree-$n$ cell either adds cohomology in dimension $n$, or it subtracts cohomology in dimension $n-1$; the latter is "more likely" if you were to just randomly attach a cell. Well, if you start at $\mathrm{SO}(2) = S^1$, then to get to $\mathrm{SO}(3) = \mathbb{RP}^3$ you add a cell in dimension 2, killing the degree-1 cohomology, and then some higher cell adds cohomology in dimension $3$. Now note that $\mathrm{SO}(3)$, like any semisimple Lie group, has no cohomology in degree 2. So when you build $\mathrm{SO}(4)$, you add a cell in degree 3, but there's no cohomology to subtract so this one must add.
So $\mathrm{H}^3(\mathrm{SO}(4);\mathbb{R})$ must be 2-dimensional!
Then you get to $\mathrm{SO}(5)$, and here is the one time you have to do a calculation: we're going to add a 4-dimensional cell, and it more likely than not will kill cohomology in dimension 3, but it might happen not to. Well, calculate: it does kill that cohomology! So $\mathrm{H}^3(\mathrm{SO}(5);\mathbb{R})$ is one-dimensional.
From here on out, you never have to calculate. You add cells of degree $>4$, and they can never touch $\mathrm{H}^3$. So $\mathrm{H}^3(\mathrm{SO}(>5);\mathbb{R})$ stays 1-dimensional.
All together you get this little blip, where exactly when $n=4$ is $\mathrm{H}^3(\mathrm{SO}(n);\mathbb{R})$ not one-dimensional.
