Does the Hodge star operator determine the metric?

Question

Let $M$ be a closed oriented smooth $4$-manifold with two metrics $g$ and $\tilde g$. Consider the Hodge star operators on $2$-forms $$ \star:\Omega^2(M) \to \Omega^2(M) \quad \text{and} \quad \tilde \star:\Omega^2(M) \to \Omega^2(M) $$ induced by $g$ and $\tilde g$ respectively. Suppose $\star=\tilde \star$, can we prove that $g$ is conformal to $\tilde g$?

Answer 1

As Malkoun observed, your problem is pointwise. So let $V$ be a 4-dimensional vector space, and $g$ and $\hat{g}$ be inner products. Since we are working with two inner products, by the spectral theorem they can be simultaneously diagonalized. So there exists a basis $e_1, e_2, e_3, e_4$ in which $$ g(e_i, e_j) = \delta_{ij}, \quad \hat{g}(e_i, e_j) = a_i^2 \delta_{ij} $$ where $a_i > 0$.

The Hodge operators satisfy

$$ *(e_1\wedge e_2) = e_3 \wedge e_4, \quad \hat{*}(a_1a_2 e_1\wedge e_2) = a_3 a_4 e_3 \wedge e_4 $$ etc. Our hypothesis is that the two Hodge operators are equal, hence you require the three equalities

$$ a_1 a_2 = a_3 a_4, \quad a_1 a_3 = a_2 a_4, \quad a_1 a_4 = a_2 a_3 $$

As you observed there is a conformal degree of freedom, hence we can WLOG set $a_1 = 1$. We find therefore $$ \implies a_2 = a_3 a_4 \implies a_3 = a_3 a_4^2 \implies a_4 = 1 \implies a_2 = a_3 = 1 $$ hence as you surmised, equality of the Hodge operators acting on two forms implies that the two metrics are equal up to conformal factor.

Answer 2

First remark is that the statement you are trying to prove reduces to the pointwise statement and thus to a linear algebra statement.

Let us say we are given 2 inner products $g$ and $\tilde{g}$ on $\mathbb{R}^4$. Then there is a $g$-orthonormal basis of $\mathbb{R}^4$, say $v_1, \ldots, v_4$, with respect to which $\tilde{g}$ is diagonal, say $$ \tilde{g} = \operatorname{diag}(\lambda_1, \ldots, \lambda_4).$$ Of course, we have that $\lambda_i > 0$, for $i = 1, \ldots, 4$.

We know that $$ * v^1 \wedge v^2 = v^3 \wedge v^4,$$ where $v^1, \dots, v^4$ (with upper indices) form the basis of $(\mathbb{R}^4)^*$ that is dual to $v_1, \dots, v_4$.

To calculate $\tilde{*} v^1 \wedge v^2$, recall the formula $$ \alpha \wedge \tilde{*} \alpha = \tilde{g}(\alpha, \alpha) \operatorname{vol}_{\tilde{g}},$$ where $\alpha$ is any element of $\Lambda^2(\mathbb{R}^4)^*$ and $$ \operatorname{vol}_{\tilde{g}} = \sqrt{\lambda_1 \dots \lambda_4} v^1 \wedge \dots \wedge v^4$$ is the volume form in $\Lambda^4(\mathbb{R}^4)^*$ with respect to $\tilde{g}$.

But $$ \lVert v^1 \wedge v^2 \rVert_{\tilde{g}}^2 = (\lambda_1 \lambda_2)^{-1},$$ so that $$ \tilde{*} v^1 \wedge v^2 = \frac{\sqrt{\lambda_3 \lambda_4}}{\sqrt{\lambda_1 \lambda_2}} v^3 \wedge v^4.$$ We now use the assumption that $* = \tilde{*}$ and deduce that $$ \lambda_1 \lambda_2 = \lambda_3 \lambda_4,$$ so that $$ \frac{\lambda_1}{\lambda_3} = \frac{\lambda_4}{\lambda_2}.$$ Similarly, if we calculate the two Hodge stars applied to $v^1 \wedge v^4$, instead of $v^1 \wedge v^2$, we then get $$ \lambda_1 \lambda_4 = \lambda_3 \lambda_2,$$ so that $$ \frac{\lambda_1}{\lambda_3} = \frac{\lambda_2}{\lambda_4}.$$ We deduce that $$ \frac{\lambda_4}{\lambda_2} = \frac{\lambda_2}{\lambda_4},$$ so $\lambda_2^2 = \lambda_4^2$. But all the $\lambda_i$ are positive, so that $\lambda_2 = \lambda_4$. Repeating this argument for various elements of the standard basis of $\Lambda^2(\mathbb{R}^4)^*$, we get that $$ \lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = \lambda, \text{ say}.$$ In other words, $\tilde{g} = \lambda{g}$.

We thus conclude that in the original problem, the two metrics $g$ and $\tilde{g}$ are conformal to each other.