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Ali Taghavi's user avatar
Ali Taghavi's user avatar
Ali Taghavi's user avatar
Ali Taghavi
  • Member for 11 years, 5 months
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Generalization of Darboux's Theorem
@smyrlis your beautiful question is (indirectly) a motivation to give the following two questions: Question 1) Let $X$ be a topological space, by $CC(X,\mathbb{R})$ $(CC(X,\mathbb{C})$, we mean all functions from $X$ to $\mathbb{R}$ ($X$ to $\mathbb{C}$) which sendsopen con connected sets to connected sets, assume that $f,g \in CC(X,\mathbb{R}$, does this imply that $f+ig \in CC(X,\mathbb{C}$.Are $CC(X,\mathbb{R}$ and $CC(X, \mathbb{C}$, vector spaces?
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$Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$
considering these two graded structures, how can we compare the k theory of corresponding trivial homo genus algebras?(zeo degree algebras)?
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$Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$
In particular, does the graded structure which you mentioned is graded isomorphic to this structure which I mentioned?
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$Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$
@AlainValette Thank you very much for your comments. I consider the grading structure for $C^{*}_{red}( F_{2})= A_{0} \oplus A_{1}$ where $A_{0}$ is the banach space generated by even words and $A_{1}, with odd words.
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Generalization of Darboux's Theorem
So it is natural to ask:"Let $A$ be a subset of $\mathbb{R}^{n}$ which can be separated by no hyperplane or sphere, does it implies that $A$ is connected"?
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Generalization of Darboux's Theorem
@liviuNicolaescu thanks for the example. However it can be shown that this set can not be equal to $\nabla f[U]$, when $U$ is open connected set. In fact no sphere can separates $\nabla f[U]$. Without lose of generality assume the sphere which separates the image, is the unit sphere around 0. Let $a,b \in U$ and $\parallel \nabla f(a)\parallel <1$ and $\parallel \nabla f(b)\parallel >1$. Choose a unit speed curve $\gamma: [0,1] \rightarrow U$ which connect a to b and its velocity at end points is parallel to $\nabla f(a)$, $\nabla f(b)$ now apply Darboux to $f\circ \gamma$
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Number of disjoint simple closed geodesics
What about the remaining part of my question: for a fixed analytic metric g, is the invariant $m$ defined in my question, finite? For a natural number $n$, is there an analytic metric $g$ for which $m=n$?
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$Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$
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$Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$
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Number of disjoint simple closed geodesics
@AntonPetrunin Can you explain why $\gamma_{\infty}$ can not have vertex?
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Volume-preserving mappings in the torus $T^n$
I think $SL(n,\mathbb{Z})$ is discrete and the matrix degree is a homotopy invariant. so the component of the identity is completly determined.
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Number of disjoint simple closed geodesics
@AntonPetronin Thank you for the answer. I honestly admit that I need to some reference to underestand the detail of your proof. for example analytic extension, stability... could you please give a reference( with minimal necessary background). Ihave also some question: Is not possible that $\gamma_{\infty}$ would be a triangle (not necessarily closed geodesic)? What is the domain of $\ell$? Why $\ell$ can be well defined? Iapologize if my questions are elementary, but I need to a reference to underestand the details of your proof.Thanks again for your help
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Generalization of Darboux's Theorem
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Generalization of Darboux's Theorem
I think the third part is dimension of homology
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Generalization of Darboux's Theorem
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Generalization of Darboux's Theorem
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Number of disjoint simple closed geodesics
I emphasis on "real analytic" because my question is influenced by Ilyashenk Eckel theorem which says every analytic vector filed on $S^{2}$ has only a finit number of isolated closed orbit. In my question, closed geodesics play the role of closed orbit in the later theorem
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