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I'm beginning to study $4$-dimensional Riemannian manifolds, in particular the decomposition induced by the Hodge $\star$-operator on the space of the differential two-forms. I'm considering $\mathbb{R}^4$ as an example: I understood that $\Lambda^2 \mathbb{R}^4=\Lambda_{+}\oplus \Lambda_{-}$, where $\Lambda_{\pm}$ is the eigenspace of $\star$ relative to the eigenvalue $\pm 1$.

The group $SO(4)$ acts on $\Lambda^2 \mathbb{R}^4$ in the standard way induced by the action of $SO(4)$ on $\mathbb{R}^4$, i.e. if $X\in SO(4)$ and $a\wedge b$ is a two-form, then $X(a\wedge b)=Xa\wedge Xb$. I've got three questions:

$1)$ Since $\Lambda^2 \mathbb{R}^4\cong so(4)$ (the Lie algebra of $SO(4)$), are there some relations between the action of $SO(4)$ on $\Lambda^2 \mathbb{R}^4$ and the adjoint action of $SO(4)$ on $so(4)$?

$2)$ What is the action of $SO(4)$ on $\Lambda^2 \mathbb{R}^4$ when restricted to $\Lambda_{\pm}$? I know that $so(4)\cong so(3)\oplus so(3)$, so I thought it could be useful to compute the action, but I don't know how to do it.

$3)$ I read that the restriction of the action to $\Lambda_{\pm}$ induces a surjective homomorphism $\phi:SO(4)\rightarrow SO(3)\times SO(3)$ which is two to one. How can I prove this and compute the homomorphism? I understood that there is a surjective homomorphism from $SU(2)$ to $SO(3)$ and that $SU(2)\times SU(2)$ is the universal cover of $SO(4)$ (a double cover, in particular): could it be useful?

Lukath
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  • What gave you the idea that $\bigwedge^2 \mathbb{R}^4 \cong \mathfrak{so}(4)$? It seems that the answer to that question could get you already half-way to the answer to your question 1 – Vincent Feb 18 '20 at 11:34
  • (...which of course is 'yes'.) – Vincent Feb 18 '20 at 11:39
  • @Vincent Basically, if ${e_1,...,e_4}$ is the canonical basis for $\mathbb{R}^4$, I considered the map from $so(4)$ to $\Lambda^2 \mathbb{R}^4$ given by $A=(A){ij}\longmapsto \frac{1}{2}A{ij}e_i\wedge e_j$ (using Einstein's convention on repeated indices), which is an isomorphism. I think I found that there's a correspondance between the two actions given by this map, but I wasn't quite sure. What about the other questions? I still don't have any idea. – Lukath Feb 18 '20 at 12:06

1 Answers1

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The induced action of $SL(4,\mathbb{R})$ on $\bigwedge^2 \mathbb{R}^4$ preserves the symmetric bilinear form induced by $\bigwedge^2 \mathbb{R}^4 \times \bigwedge^2 \mathbb{R}^4 \to \bigwedge^4 \mathbb{R}^4 \cong \mathbb{R}$. This form happens to have signature $(3,3)$, so $SL(4)$ actually lands in $SO(3,3)$, and the same applies to $SO(4)$. This is enough to see that $\mathfrak{so}(4)$ lands in a copy of $\mathfrak{so}(3)\oplus \mathfrak{so}(3)$, and you can conclude (2) by dimension counting. With a little care this should yield (3) as well.

For (1), the adjoint action of $SO(n)$ on $\mathfrak{so}(n)$ is equivalent as a representation to the action of $SO(n)$ on $\bigwedge^2(\mathbb{R}^n)$. This requires writing the right map $\bigwedge^2 \mathbb{R}^n \to \mathfrak{so}(n)$: $$ v \wedge w(x)= \langle x,v \rangle w - \langle x,w \rangle v .$$

Max
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  • Ok, thanks, I get the correspondance between actions. I still have some doubts though: I managed to write the action of $SO(4)$ restricted to $\Lambda_{\pm}$, making use of the action of $SO(4)$ on the two copies of $so(3)$, but how should this induce a surjective homomorphism from $SO(4)$ to $SO(3)\times SO(3)$? – Lukath Feb 19 '20 at 10:16
  • You can use some general topology facts to promote the map $SO(4) \to SO(3) \times SO(3)$ to a covering map. For instance, it is a local diffeomorphism from a compact manifold to a connected manifold. See here: https://math.stackexchange.com/questions/45990/when-is-a-local-homeomorphism-a-covering-map?lq=1 – Max Feb 20 '20 at 17:52