# Nevermind

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 2d comment Vector field pull back from embeddingOk. Good point. So this "pullback" depends on the choice of $r$. 2d revised Vector field pull back from embeddingadded 341 characters in body; added 1 characters in body 2d comment Decalage isomorphism and algebra structureDid you read the paper arxiv.org/abs/math/0601312 ? May19 comment Vector field pull back from embeddingDon'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 $r'_*,r_*:T_{f(m)}N \to T_mM$. May19 comment Vector field pull back from embeddingIf $dim(M)=dim(N)$ then $r=f^{-1}$. May19 comment Vector field pull back from embeddingSince $r\circ f=id_M$ 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... May19 comment Vector field pull back from embeddingSo the question is if different choices of $r$ gives different vector fields on $M$. May19 comment Vector field pull back from embeddingsorry dude but $r_*$ is a map $T_nN \to T_r(n)M$. May19 comment Vector field pull back from embeddingSo what you say is,that different choices of $r$ gives different vector fields on $M$? Hmm... Can you explain why? May19 comment Vector field pull back from embeddingNot a good style to first downtalk like "by the way ... makes no sense" and then not even say, WHY it makes no sense. May19 comment Vector field pull back from embeddingOk 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? May19 comment Vector field pull back from embeddingWhy has $X_{f(x)}$ to be in the image of $df_x$? May19 asked Vector field pull back from embedding May4 revised Explicit Lie May structure on cosimplicial DG Lie algebrasadded 111 characters in body May4 comment Explicit Lie May structure on cosimplicial DG Lie algebrasI mixed it up... Replaced 'May' by 'Schechtman' as author. Sorry for that. May4 revised Explicit Lie May structure on cosimplicial DG Lie algebrasCorrected an author from May to Schechtman May3 comment Explicit Lie May structure on cosimplicial DG Lie algebrasThe 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. May3 asked Explicit Lie May structure on cosimplicial DG Lie algebras Apr17 comment Bisections in Kan ComplexesSeems like I have taggin problems on this computer. Can someone please tag it with groupoids, higher category theroy, too? Apr17 asked Bisections in Kan Complexes Mar20 awarded ● Yearling Mar19 asked Smooth function algebra on cartesian product and beyond Mar17 awarded ● Critic Mar17 comment Induced Riemannian metric on Jet-ManifoldRobert, this is not an answer, because you don't give a proof or a reference for a proof. Bad style, not to reply Mar15 comment Induced Riemannian metric on Jet-ManifoldYou 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 $TN \otimes T^*M? Right? -- If yes, the missing link for me is a reference to the construction in 3. Mar15 comment Induced Riemannian metric on Jet-ManifoldThanks. Can you give references for a more in-deep look at the constructions you mentioned? What do you mean by "[..] induce a metric on this bundle [...]". Which bundle?$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. Mar15 asked Induced Riemannian metric on Jet-Manifold Mar3 accepted Pseudo-Differentialforms Feb12 comment Pseudo-DifferentialformsJust using the absolute value here sounds a bit arbitrary to me. What's the reason for that and how to generalize? Feb12 answered Pseudo-Differentialforms Feb11 comment 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. Feb11 comment Pseudo-Differentialforms@Liviu: Why not post this as an answer? Feb7 comment Pseudo-Differentialformsthanks, I will look on these references tomorrow, since I don't have access right now. Feb7 comment Pseudo-DifferentialformsWhat do you mean by a complicated definition? I want a definition that is natural. Those appear to me usually as the most easiest... Feb7 comment Pseudo-DifferentialformsIs your definition of a density functorial? Feb7 asked Pseudo-Differentialforms Dec18 comment Smooth submanifolds defined by SubringsAh! Thanks for the answer David... Dec17 comment Smooth submanifolds defined by SubringsFor 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. Dec17 comment Smooth submanifolds defined by SubringsI 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? Dec17 comment Smooth submanifolds defined by Subringsyeah right. I should have just written manifolds defined by subrings... Dec15 awarded ● Self-Learner Dec15 comment Smooth submanifolds defined by SubringsThanks David. I'll think about it Dec15 awarded ● Teacher Dec15 revised Smooth submanifolds defined by Subringsdeleted 3 characters in body Dec15 answered Smooth submanifolds defined by Subrings Dec15 asked Smooth submanifolds defined by Subrings