Nevermind
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Registered User
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2d |
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Vector field pull back from embedding Ok. Good point. So this "pullback" depends on the choice of $r$. |
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2d |
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Vector field pull back from embedding added 341 characters in body; added 1 characters in body |
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2d |
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Decalage isomorphism and algebra structure Did you read the paper arxiv.org/abs/math/0601312 ? |
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May 19 |
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Vector field pull back from embedding Don't need a preferred base-point I think. To be functorial, it should be enough to show that any choice of the retract gives the same vector field on $M$, but as I said at any $f(m)$ the equation $r\circ f=id_M$ gives $r'_*(x)=r_*(x)$ for any two $r'_*,r_*:T_{f(m)}N \to T_mM$. |
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May 19 |
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Vector field pull back from embedding If $dim(M)=dim(N)$ then $r=f^{-1}$. |
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May 19 |
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Vector field pull back from embedding Since $r\circ f=id_M$ every choice of $r$ should give the same $r_*$ on $f(m)$. And in the equi-dimensional case this is just the ordinary pullback of a vector field along a diffeomorphism, since embeddings are diffeomorphisms then... |
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May 19 |
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Vector field pull back from embedding So the question is if different choices of $r$ gives different vector fields on $M$. |
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May 19 |
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Vector field pull back from embedding sorry dude but $r_*$ is a map $T_nN \to T_r(n)M$. |
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May 19 |
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Vector field pull back from embedding So what you say is,that different choices of $r$ gives different vector fields on $M$? Hmm... Can you explain why? |
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May 19 |
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Vector field pull back from embedding Not a good style to first downtalk like "by the way ... makes no sense" and then not even say, WHY it makes no sense. |
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May 19 |
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Vector field pull back from embedding Ok it requires both the data of $f$ and $r$ and $r$ is not uniquely defined by $f$. But it is a well defined vector field on $M$ at least. Right? |
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May 19 |
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Vector field pull back from embedding Why has $X_{f(x)}$ to be in the image of $df_x$? |
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May 19 |
asked | Vector field pull back from embedding |
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May 4 |
revised |
Explicit Lie May structure on cosimplicial DG Lie algebras added 111 characters in body |
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May 4 |
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Explicit Lie May structure on cosimplicial DG Lie algebras I mixed it up... Replaced 'May' by 'Schechtman' as author. Sorry for that. |
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May 4 |
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Explicit Lie May structure on cosimplicial DG Lie algebras Corrected an author from May to Schechtman |
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May 3 |
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Explicit Lie May structure on cosimplicial DG Lie algebras The proof uses the Lie Eilenberg Zilber operad and doesn't make any reference to the maps I'm after. If one can read the k-ary maps from the action of this operad on the cosimpl. DG Lie algebra, then this is not at all obvious... At least to me. |
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May 3 |
asked | Explicit Lie May structure on cosimplicial DG Lie algebras |
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Apr 17 |
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Bisections in Kan Complexes Seems like I have taggin problems on this computer. Can someone please tag it with groupoids, higher category theroy, too? |
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Apr 17 |
asked | Bisections in Kan Complexes |
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Mar 20 |
awarded | ● Yearling |
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Mar 19 |
asked | Smooth function algebra on cartesian product and beyond |
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Mar 17 |
awarded | ● Critic |
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Mar 17 |
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Induced Riemannian metric on Jet-Manifold Robert, this is not an answer, because you don't give a proof or a reference for a proof. Bad style, not to reply |
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Mar 15 |
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Induced Riemannian metric on Jet-Manifold You use 1.) the identification $J^1(M,N) \simeq TN \otimes T^*M$ 2.) If $g$ is a metri on $M$,then there are induced metrics on $TM$ and $T^*M$ 3.) There is a metric on $TN \otimes T^*M? Right? -- If yes, the missing link for me is a reference to the construction in 3. |
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Mar 15 |
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Induced Riemannian metric on Jet-Manifold Thanks. Can you give references for a more in-deep look at the constructions you mentioned? What do you mean by "[..] induce a metric on this bundle [...]". Which bundle? $J^1(M,N)$? Then how can this be done? --You said there are many ways, but given at least one example would be required to qualify as an answer. |
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Mar 15 |
asked | Induced Riemannian metric on Jet-Manifold |
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Mar 3 |
accepted | Pseudo-Differentialforms |
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Feb 12 |
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Pseudo-Differentialforms Just using the absolute value here sounds a bit arbitrary to me. What's the reason for that and how to generalize? |
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Feb 12 |
answered | Pseudo-Differentialforms |
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Feb 11 |
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Pseudo-Differentialforms @alvarezpaiva: I'll try to define 'Density' as an answer to my own question. Please comment,to see if we mean the same. |
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Feb 11 |
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Pseudo-Differentialforms @Liviu: Why not post this as an answer? |
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Feb 7 |
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Pseudo-Differentialforms thanks, I will look on these references tomorrow, since I don't have access right now. |
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Feb 7 |
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Pseudo-Differentialforms What do you mean by a complicated definition? I want a definition that is natural. Those appear to me usually as the most easiest... |
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Feb 7 |
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Pseudo-Differentialforms Is your definition of a density functorial? |
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Feb 7 |
asked | Pseudo-Differentialforms |
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Dec 18 |
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Smooth submanifolds defined by Subrings Ah! Thanks for the answer David... |
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Dec 17 |
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Smooth submanifolds defined by Subrings For example, I think it is valid to say that the subring of constant functions (i.e. multiples of $1_{C∞(M)}$) is the function ring of the point? It arise as a pullback by the terminal morphism. |
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Dec 17 |
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Smooth submanifolds defined by Subrings I can't mark it as an answer because, what I actually want to discuss is more about when subrings are smooth function rings by itself. But I must confess, that the title is misleading. It should be called "Smooth Manifolds defined by Subrings" rather. Or is this almost never the case, for reasons I don't see? |
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Dec 17 |
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Smooth submanifolds defined by Subrings yeah right. I should have just written manifolds defined by subrings... |
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Dec 15 |
awarded | ● Self-Learner |
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Dec 15 |
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Smooth submanifolds defined by Subrings Thanks David. I'll think about it |
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Dec 15 |
awarded | ● Teacher |
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Dec 15 |
revised |
Smooth submanifolds defined by Subrings deleted 3 characters in body |
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Dec 15 |
answered | Smooth submanifolds defined by Subrings |
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Dec 15 |
asked | Smooth submanifolds defined by Subrings |
