The minimal function certainly does not always exist mathematically. I haven't done this type of analysis for a long time but couldn't you just take say C^1([0,1]) as your space of functions with an action functional given by S(f)=\int_{0}^1 f(x) dx ? Surely this can't have a local min/max...
The technical reason is simply that you don't get a group (or groupoid if you want to define it on all composable paths not just the paths with some fixed beginning=end point) because this multiplication is not associative, inverses don't exists and an identity does not exist unless you take...
I'm not sure if you are aware of it or not but in my opinion one of the best sources that deals with variational problems and proves, among many other things, Noether theorems in the very general setting of local functionals defined for the infinite jet bundle of any fibered manifold (so in the...
Edit: Better answer above. My answer wasn't really applicable to the question because the theorem I quoted requires too many assumptions on the category being localized.
By the way, the theorem mathwonk mentioned which says that all right exact functors (at least functors between categories of...
By definition F(u) is the smallest field containing F as well as u . Since F(u) is a field it is closed under multiplication and addition. So, if it contains u then it contains u^n for any n, hence it contains au^n for any a\in F and so must also contain all sums of these terms ie...
Yeah, all this categorified differential geometry stuff seems somewhat esoteric still even among mathematicians (of course algebraic geometers have been doing this since Grothendieck so it's nothing new in that field) but the group of people associated with the nLab view higher categories as the...
I’m not sure entirely what you mean by a structure of the same type as E but if you are asking whether given any morphism in a category, there exists a “preimage” of every “point” which is again an object in the category then the answer is no in general. I put preimage and point in quotations...
In case anyone reads this thread, there is an obvious minor error in my previous post.
I wrote
\begin{align*}
\left(K^k_ce_b(L^\alpha_k)-K^j_b e_c(L^\alpha_j) \right) e_\alpha
&=e_c(K^j_b)L^\alpha_je_\alpha-e_b(K^k_c)L^\alpha_ke_\alpha\\
&=e_c(K^j_b)\frac{\partial}{\partial x^j}...
I should warn you that I almost certainly made some indexing errors in following.
Write e_b=K^i_b\frac{\partial}{\partial x^i} and \frac{\partial}{\partial
x^i}=L^b_i e_b so that K^j_bL_j^\beta=\delta^\beta_b since these are inverse
matrices. Then
\begin{align*}...
Well yes, I chose a frame of the tangent bundle coming from a coordinate chart, which implies that the Lie derivative vanishes, since the conventions I am familiar with reserve the symbol \Gamma^a_{bc} only for these types of bases. You can simply replace the special frame...
You haven't really asked a question we can answer yet. What is it you want to see proven about the torsion tensor or the curvature tensor and where are we supposed to start from? If you want to see a derivation of the components of the tensor what is the definition of the torsion tensor you are...
I can take a look at your work if you would like but I'm probably the wrong person to ask about doing actual hands on computations...I usually completely mess it up the first few times I try.
It may be helpful to check your work to observe that as graded rings...
Yeah my first post was nonsense.
I'm not sure I understand this. Do you mean this is true only in the special case of a complex manifold (or maybe even more specifically a Calabi-Yau manifold)? It seems to me for an arbitrary real manifold the vanishing of the first betti class does not control...
The proper way to formulate the question for a general manifold, and hence in a way that makes sense for Calabi Yau manifolds in particular, is to ask if there exists a nowhere vanishing section of the tangent bundle (ie. a a nowhere vanishing vector field.) This is always true for a Calabi Yau...
Yeah computing all cup products by hand will certainly yield a proof but I don't think you need to do anything more complicated than compute the homology groups. The simplest idea that comes to mind is just to compute the Euler characteristics which we know are multiplicative so you only need to...