Determing Hodges Maps by their Essential Algebraic Properties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:54:07Z http://mathoverflow.net/feeds/question/120528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120528/determing-hodges-maps-by-their-essential-algebraic-properties Determing Hodges Maps by their Essential Algebraic Properties John McCarthy 2013-02-01T16:02:36Z 2013-03-18T10:22:00Z <p>Let $M$ be a complex manifold of real dimension $2N$, equipped with a Hermitian metric. The corresponding Hodge map $\ast$ has the following properties:</p> <p>(i) It is a ${\bf C}$-linear map $\ast:\Omega^k(M) \to \Omega^{2N-k}(M)$;</p> <p>(ii) $\ast(\Omega^{(p,q)}(M)) = \Omega^{(N-p,N-q)}$(M);</p> <p>(iii) $\ast^2 = (-1)^{k}$ on $\Omega^k(M)$.</p> <p>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.</p> http://mathoverflow.net/questions/120528/determing-hodges-maps-by-their-essential-algebraic-properties/120717#120717 Answer by Fran Burstall for Determing Hodges Maps by their Essential Algebraic Properties Fran Burstall 2013-02-03T23:41:35Z 2013-02-04T07:30:02Z <p>Not so much an answer as two comments, one picky and the other possibly helpful (but together exceeding the character limit for a comment).</p> <ol> <li><p>In (i), you need $*$ to be linear over functions otherwise it is not algebraic (tensorial) at all.</p></li> <li><p>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.</p></li> </ol>