Timeline for transformation properties of divergence (of a vector field)

Current License: CC BY-SA 2.5

12 events

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Dec 20, 2010 at 18:39	comment	added	Ben Sprott		$ \xymatrix@R+2em@C+2em{ Man \ar[r]^-G \ar[d]_-F & Top \ar[d]^-G \\ Vec \ar[r]_-k & Q } $ ..trying to get latex to work. Consider the commutative square which demonstrates how the functor $G: Man \rightarrow Vec $ takes a smooth map in Man to a linear map in Vec. I want to map the entire square, via some kind of forgetful transformation (I guess a natural transformation), to square S. Square S is the one demonstrating how a continuous map in Top goes to a linear map in the category of Topological Vector spaces. Is this sound? I may also want the adjoint.
Dec 20, 2010 at 18:31	comment	added	Ben Sprott		I was trained through John Lee's Topological Manifolds. Thus, to me, a manifold is a topological space with added structure, namely the differentiable structure. Thus, if we have a vector field, that is a functor $F: Man \rightarrow Vect$, and we can forget the differentiable structure on Man $G: Man \rightarrow Top$ we can consider the commutative square: $$ \xymatrix@R+2em@C+2em{ Man \ar[r]^-G \ar[d]_-F & Top \ar[d]^-G \\ Vec \ar[r]_-k & Q } $$ ie, what structure is left over in Q? Is Q a topological vector space?
Dec 14, 2010 at 1:03	vote	accept	Ben Sprott
Dec 8, 2010 at 4:45	answer	added	Bill Thurston		timeline score: 24
Dec 8, 2010 at 3:24	comment	added	Donu Arapura		As people have pointed out, the divergence of a vector field is not well defined without specifying some additional structure. If, however, you are willing to recast the problem in terms of closed $(\dim X-1)$-forms then it is diffeomorphism invariant notion. This is implicit in the identifications one makes in Euclidean space. But, it's hard to tell if this matches what you have in mind.
Dec 8, 2010 at 0:43	comment	added	Ryan Budney		@Ben Sprott: I removed the topological vs. smooth confusion in your post.
Dec 8, 2010 at 0:39	history	edited	Ryan Budney	CC BY-SA 2.5	take out the topological manifold confusion, update the title
Dec 7, 2010 at 23:58	comment	added	Willie Wong		Also, if $\phi: M\to N$ is a diffeomorphism between smooth manifolds, and let $\omega$ be a volume form on $N$ and $X$ a vector field on $M$, then by definition the following two are equivalent: (a) $\mathop{div}_{\phi^\omega}X = 0$ and (b) $\mathop{div}_\omega \phi_X$, where $\phi^\omega$ is the pull-back of $\omega$, and $\phi_X$ is the push-forward of $X$. So this essentially answers your first question. Please clarify the rest of your post, or the question may risk closure.
Dec 7, 2010 at 23:54	comment	added	Willie Wong		There are a lot of problems with your question. Metric? Is your manifold a Riemannian manifold? To talk about the divergence implies that your manifold is at least orientable (you need a fixed volume form). And since when is a dimension-reducing contraction a diffeomorphism?
Dec 7, 2010 at 23:53	comment	added	drbobmeister		How are you defining divergence? As I recall, to do so you need a volume form on the manifold in question.
Dec 7, 2010 at 23:24	comment	added	Mariano Suárez-Álvarez		Your manifold is probably a smooth manifold?
Dec 7, 2010 at 23:18	history	asked	Ben Sprott	CC BY-SA 2.5