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 a 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.

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.

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 $k \ge 2$; 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).

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 $k = 2$ (the case $k = 1$ is degenerate since in this case the Hodge star has only one eigenspace). 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 a 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.

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 a 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.