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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 24, 2011 at 17:03 vote accept John McCarthy
Feb 24, 2011 at 2:17 history closed Deane Yang
Tim Perutz
Johannes Ebert
S. Carnahan
no longer relevant
Feb 23, 2011 at 23:56 history edited Qfwfq CC BY-SA 2.5
edited body
Feb 23, 2011 at 20:57 answer added Misha Verbitsky timeline score: 3
Feb 22, 2011 at 20:51 comment added John McCarthy Ok, I see now what's going on. Thanks a lot Johannes. Sorry for asking a question before I understood waht I was asking about. I think the best thing to do with this question would be to close it.
Feb 22, 2011 at 18:45 comment added Johannes Nordström The whole question seems to stem from a misunderstanding. $J$ is an endomorphism of $TM$. It acts on the dual space $T^*M$ by the dual map, and on $k$-forms (possibly complex-valued) by multilinear extension. $\Lambda^{1,0}T^*M$ is by definition the $+i$ eigenspace of $J$ in $T^*M \otimes \mathbb{C}$, and $\Lambda^{0,1}T^*M$ the $-i$ eigenspace. $J$ acts on $\Lambda^{p,q}T^*M$ as multiplication by $i^{p-q}$. The $(1,0)$-part of a 1-form $\alpha$ is $\frac{1}{2}(\alpha - iJ\alpha)$, so $\partial f = \frac{1}{2}(df + iJ(df))$ for a function. The expressions in the question seem incorrect.
Feb 22, 2011 at 17:06 comment added John McCarthy I'm looking at the $J$ as an operator on $\Omega^1(M)$. Then if $f$ is a smooth function, $\partial(f)$ is a (1,0)-form. Operating on $\partial(f)$ by $J$ will, if I understand it correctly, will send it to the $\Omega^{(0,1)}(M)$ forms, which are spanned by elements of the form $\overline{\partial}(g)$, for $g$ also a smooth function. Thus, if $J$ is just multiplication by $i$, we get my statement above.
Feb 22, 2011 at 17:00 comment added Andrea Ferretti I think you should rewrite your question and explain your notation. I do not understand what $\omega$, $g_j$ and $h_j$ are. Furthermore I do not understand what your last comment about deriving a function has to do with the issue at hand.
Feb 22, 2011 at 16:50 comment added John McCarthy It just doesn't seem right to me that for any smooth function $f$, we have $i\partial(f) = \sum_{j=1}^k g_j \overline{\partial}(h_j)$, for some other smooth functions $g_i,h_i$.
Feb 22, 2011 at 16:46 comment added John McCarthy Really, it's that simple, $J(\omega) = i.\omega$?
Feb 22, 2011 at 16:40 comment added Andrea Ferretti Indeed it is often the case that there more than one complex structure on a complex manifold. In any case, once you fix a complex structure, the almost complex structure associated to it is multiplication by i. How is this definition not global?
Feb 22, 2011 at 16:24 comment added John McCarthy .... I mean of course canonical not cancoical.
Feb 22, 2011 at 16:23 comment added John McCarthy @Gunnar For your first question: The definition of an almost complex structure $J$ is certainly global, but the construction of the canonical $J$ for a complex manifold is what I'm interested in. Surely, there is more than one almost complex structure on a complex maniold, ie $J^2 = $id$_{T_M}$ does not define it uniqely. So I suppose my question is how does one identify the cancoical one globally? For your second: As I said just above, I am assuming that my manifold is complex, and so, I certainly have $\text{d}=\partial + \overline{\partial}$.
Feb 22, 2011 at 16:14 comment added Gunnar Þór Magnússon Isn't defining an almost complex structure $J$ as a section of $End(T_M)$ which satisfies $J^2 = -id_{T_M}$ pretty global? Also, to derive your last equation you need to know that $d = \partial + \bar \partial$, which is equivalent to $J$ being integrable. And what is your $\omega$?
Feb 22, 2011 at 15:58 history asked John McCarthy CC BY-SA 2.5