Recent content by astros

I heared "connection one-form" but i see 2 components "two indices" ! {{\omega^{\alpha}}_\beta}'s are not one-forms and are not expressed in the natural basis (dx) they are the components of a 1-1 tensor and a form has never been a 1-1 tensor ! I think :biggrin:

No! it's like matrices... the trace of a product is not the product of traces. Now if you have a product of tensors which is an 1-1 tensor the trace is the contraction of these two indices, if you have more than 2 indices you must specify over which indices the trace must be done, and then you...

Sometimes one has a product of a symmetric tensor with another tensor which is not symmetric nor antisymmetric, then one can show that the antisymmetric part of the second is killed by the first, the same thing occurs for the antisymmetric case, this is why we need to antisymmetrise and...

Hi, I think that one can study other mathematical concepts independently of topology, but I'm not an expert on the subject, For this reason, I will content myself to give you this link, It has served me in the past ... But warn me when you'll get it, So I remove it! I hope that it will serve...

I think that one can even study the topological aspect of functions i.e: forms of degree 0 ! Furthermore one can also find algebraic structures for the algebra of forms as a whole (the graded algebra)... >> Can I say in this case that two distinct mathematical concepts "work together" to form...

Sorry,what do you mean by "mixed with topological properties" ?

Hi, Yes I'm studying BRST This book is precisely what I read now but I did not understand this section! The problem is how the derivation d becomes a derivation on H*(d') by means of the first condition? an after that becomes nilpotent by means of the seconde condition?

Hello, One says that an odd derivation d is a differential modulo another differential d' if this two things hold: [d,d']=0 d^2=-[d',d"] for some d" - The first condition says that d defines a derivative on H*(d'), but how? To have a well-defined action of this derivation on H*(d') we...

nobody!

Hello, I know that the CS Lagrangian is a topological invariant, in the sense that it does not depend on the connection we choose. OK, but a TFT is a field theory whose Lagrangian and all other observables do not depend on the metric, a connection in general is not uniquely defined by a metric...

hello and thank you for your assistance Emmanuel114: quite sour that I have all the articles of Witten but they are rather innaccessibles for a beginner (in any case for me they are!) Sauron: the book of which you speak is surely that of nash, I have started to read it, but it is necessary...

hello, I have just started to study topological field theories but my problem is that I do not find of books really specialized in the subject, can somebody inform me what I must do?

Hello, thank you for your interest: I have introduced ghosts to eliminate the Fadeev-Popov determinant, by using an integral over fermionic numbers (the ghosts, c), it is the same method as for the gauge d.A=0 except that in this case one finds free ghosts.

Hello, I proceed as usual, I found L=c_bar(d.d+2.A.d)c (excuse for latex!) which is not independant and hence can not be absorbed in the coefficient of the generating fonctional! one have then diagrams 2photons-ghost, I think that it's essential to have this ghost coupling to renormalize these...

Hello; By using the gauge d.A+A.A=0 to find the quantum Lagrangian density, I found a term of Ghost-Photon coupling.does it have a physical significance? And why one does not find it in the gauge d.A=0? Thank you.