Timeline for A generalization of holomorphic functions
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 12, 2016 at 15:29 | comment | added | Willie Wong | In the 3x3 case: note that $A$ must have a real eigenvalue $\lambda$ with eigenvector $v$. Then for any function $\phi:\mathbb{R}\to\mathbb{R}$, the function $f(x) = v \phi(v^Tx)$ satisfies $$A(Df) = (Df)A = \lambda vv^T \phi'(v^Tx)$$ Now just pick a sequence $\phi_n$ of $C^1$ functions that converges uniformly to function that is not differentiable. | |
| Jul 12, 2016 at 15:10 | comment | added | Willie Wong | @AliTaghavi: I have no idea what your comment is trying to say. (1) $J$ is, as you defined, the matrix for the complex structure on $\mathbb{R}^2$. (2) Why must $f$ be a self map? You certainly didn't specify that in your question. In the generality of your question it makes more sense to think about $f$ as a map between complex manifolds $(U,J)$ and $(V,J')$ each with its own complex structure, and think about the relation $(Df)J' = J(Df)$. You can of course restrict to the case where it is a self map, but you don't have to. | |
| Jul 12, 2016 at 15:08 | comment | added | Ali Taghavi | @WillieWong another question: What about odd dimensional case; Is there a $3 \times 3$ matrix $A$ for which $\mathcal{S}_{A}$, the space of A_holomorphic functions from $R^3\to R^3$ is a closed subset of $C(R^3, R^3)$ with the topology of unfiorm convergence on compact subsets. | |
| Jul 12, 2016 at 15:01 | comment | added | Ali Taghavi | @WillieWong Thank you very much for your answer. But I do not understand the last part of your answer. Is not $J$ the usual complex structure of Euclidean reagons in $\mathbb{R}^{2n}$? And is not true to say that:" the preserving of this almost complex structure by a map f is equivalent to $DfJ=JDf$?I do not understand that why you think my formulation is not a good way for thinking on this problem?Thanks again for your very interesting answer. | |
| Jul 12, 2016 at 10:51 | vote | accept | Ali Taghavi | ||
| Jul 11, 2016 at 17:06 | comment | added | Willie Wong | @July: yes. I included the constant $c$ since in the original post, the matrices don't have to have determinant $\pm 1$. | |
| Jul 11, 2016 at 16:56 | comment | added | M.G. | It seems I've misunderstood you the first time. The constant $c$ is a matter of taste, i.e. it is independent of the choice of $P$, right? (I thought at first that it depended on $P$, which baffled me a little :-) ) | |
| Jul 11, 2016 at 14:26 | history | edited | Willie Wong | CC BY-SA 3.0 |
added 920 characters in body
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| Jul 11, 2016 at 14:06 | history | answered | Willie Wong | CC BY-SA 3.0 |