Skip to main content
3 of 3
added 2 characters in body
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

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.

Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741