Determing Hodges Maps by their Essential Algebraic Properties Let $M$ be a complex manifold of real dimension $2N$, equipped with a Hermitian metric. The corresponding Hodge map $\ast$ has the following properties:
(i)   It is a ${\bf C}$-linear map $\ast:\Omega^k(M) \to \Omega^{2N-k}(M)$;
(ii)  $\ast(\Omega^{(p,q)}(M)) = \Omega^{(N-p,N-q)}$(M);
(iii) $\ast^2 = (-1)^{k}$ on $\Omega^k(M)$.
Now I would guess that there exist other maps on $\Omega(M)$ with these properties which do not arise as Hodge maps from some Hermitian metric. So my question is, do there exist extra  (algebraic) properties of $\ast$, which when put together with $(i),(ii)$, and $(iii)$, determine all the Hodge maps, but without ever explicitly mentioning metrics.
 A: Not so much an answer as two comments, one picky and the other possibly helpful (but together exceeding the character limit for a comment).


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*In (i), you need $*$ to be linear over functions otherwise it is not algebraic (tensorial) at all.

*There are clearly necessary conditions for $*$ to be a Hodge map for some Hermitian metric.  We fix a volume form as $*1$ and then the metric on $1$-forms is defined by $(\omega_1,\omega_2)*1=\omega_1\wedge *\omega_2$.  At this point, we get two conditions: the metric so defined had better be non-degenerate and positive definite.  This is an open condition.  Moreover, $\star 1$ must now be the wedge of an orthonormal basis of $1$-forms and that is an algebraic condition relating $\star$ on $0$-forms with $\star$ on $1$-forms.  For $N=1$, this condition is vacuous: $\star$ on $1$-forms defines a conformal structure and then $\star$ on $0$-forms simply fixes a scale to get a metric.  However, when $N>1$, $\star$ on $k$-forms for $2\leq k\leq N$ is already determined by the metric we have found and so there must be compatibility conditions.
