Reference request: $\operatorname{Sym}^2_0(T^*M) \simeq \Lambda_- \otimes \Lambda_+$

Question

I am looking for proof of the "well-known" result that for a $4$-dimensional Riemannian manifold $(M, g)$, we have an isomorphism $$ \operatorname{Sym}^2_0(T^*M) \simeq \Lambda_- \otimes \Lambda_+ $$ between the traceless, symmetric, covariant $2$-tensors on $M$ and the tensor product of the spaces of anti-self-dual and self-dual $2$-forms on $M$.

I saw in some papers on Einstein manifolds that the isomorphism is the map $$ \Lambda_- \otimes \Lambda_+ \ni \omega_- \otimes \omega_+ \mapsto \omega_- \omega_+ \in \operatorname{Sym}^2_0(T^*M), $$ where $\omega_- \omega_+$ is given, in coordinates, as $$ (\omega_- \omega_+)_{ij} = (\omega_-)_{ik} g^{kl} (\omega_+)_{lj} $$ (this is the notation from, for instance, the proof of Proposition 4.5.3 in "Wormholes in ACH Einstein manifolds" by Biquard and Rollin). However, none of these references provides a proof of this result. While I understand the definition of $\omega_- \omega_+$, I do not know how to prove that it is the desired isomorphism; I am not sure how to even check that $\omega_- \omega_+$ is a symmetric tensor.

Remark. I also posted this question on Stack Exchange here a month ago, but I have received no answers.

Answer 1

One reference is Donaldson and Sullivan's paper "Quasiconformal 4-manifolds". See Lemma 2.3. They prove a a little more. A traceless symmetric tensor is an infinitesimal change of metric which is changing the conformal class of the metric. They characterize which such products correspond to deformations which are positive definite. This is the four dimension analogue of studying conformal structures on two manifold by picking reference complex structure and parameterizing others by certain anti-linear maps from $\Lambda^{1,0}$ to $\Lambda^{0,1}$ or equivalently with a element of $(\Lambda^{1,0})^{\otimes 2}$

That said it is straightforward to check any number of ways that the isomorphism is correct. For example you can check that these are the same representation of $SO(4)$ say by computing characters. To check the given map does the job (though I think that you DO need to symmetrize), take an orthonormal basis $e^1, \ldots e^4$.Suppose for example take $\omega_+=e^1\wedge e^2+e^3\wedge e^4$ and $\omega_-=-e^1\wedge e^2+e^3\wedge e^4$ then $$ \omega_+\omega_-= (e^1\otimes e^1+e^2\otimes e^2-e^3\otimes e^3-e^4\otimes e^4). $$ Taking instead taking $\omega_-=e^1\wedge e^3+e^2\wedge e^4$ we get $$ \omega_+\omega_-= -(e^2\otimes e^3+e^3\otimes e^2)+(e^1\otimes e^4+e^4\otimes e^1). $$ Here the convention for symmetrization fixes things which are already symmetric.

Finally note that we can identify $\Lambda^+\otimes \Lambda^-$ with $Hom(\Lambda^+,\Lambda^-)$ to compare with the two dimensional case.

Answer 2

There is also a quick abstract proof via representation theory: $S^2_0(\mathbb{R}^4)$ is a 9-dimensional representation of $\mathrm{SO}(4)/\{\pm I_4\}\simeq \mathrm{SO}(3)\times\mathrm{SO}(3)$ and, hence, must be a sum of tensor products of representations of the two simple factors. Each of the two (commuting) $\mathrm{SU}(2)$s in $\mathrm{SO}(4)$ acts irreducibly on $\mathbb{R}^4$, so they each only preserve the one quadratic form, which is not in $S^2_0(\mathbb{R}^4)$. Hence each $\mathrm{SO}(3)\times\mathrm{SO}(3)$-irreducible piece of $S^2_0(\mathbb{R}^4)$ must be a sum of tensor products of nontrivial representations of the two factor $\mathrm{SO}(3)$s. Meanwhile, the lowest dimension nontrivial representation of $\mathrm{SO}(3)$ is $\mathbb{R}^3$. Hence $S^2_0(\mathbb{R}^4)$ must be isomorphic to the tensor product of the $3$-dimensional irreducible representations of the two factors.

A similar argument shows that $\Lambda^2(\mathbb{R}^4)\simeq\mathbb{R}^6$, which is also a representation of $\mathrm{SO}(4)/\{\pm I_4\}\simeq \mathrm{SO}(3)\times\mathrm{SO}(3)$, must be a sum of two irreducible pieces $\Lambda^2_\pm\simeq\mathbb{R}^3$, each of which is the tensor product of the trivial representation of one factor $\mathrm{SO}(3)$ with the $3$-dimensional representation of the other factor $\mathrm{SO}(3)$.

Hence $S^2_0(\mathbb{R}^4)\simeq \Lambda^2_+\otimes\Lambda^2_-$ as $\mathrm{SO}(4)/\{\pm I_4\}$-modules.