
#1
Mar612, 07:15 PM

P: 2,045

Hi, I'm just wondering, a vector valued (or (n,0) tensor valued) p form is the same as a rank (n,p) tensor which is totally antisymmetric bottom indices right? Is there a difference?
A (n,m) valued p form is a (n,m+p) tensor which is antisymmetric in the lower p indices? 



#2
Mar612, 08:30 PM

Sci Advisor
P: 1,563

Yes, basically.
However, vectorvalued forms can also take values in vector spaces other than [itex]T_xM[/itex]. For example, a gauge connection is a 1form that takes values in the Lie algebra of some Lie group. 



#3
Mar612, 08:39 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

More generally, vector valued pforms take value in a vector bundle. I.e. they are elements of [tex]\Omega^p(M)\otimes C^{\infty}(E)[/tex]




#4
Mar612, 11:44 PM

P: 2,045

Vector valued forms
Ok, sounds good to me.
Now, what does it mean to take the wedge product of two of them together? For simplicity, what does it mean to take the wedge product of a (n,0) valued p form and a kform? It's a (n,p+k) tensor that's fully antisymmetrized in the lower p+k indices? To take the exterior derivative of a vectorvalued form, do we have to use parallel transport? In other words, we must first define a connection before we can define such an exterior derivative? 



#5
Mar712, 12:54 AM

Sci Advisor
P: 1,563

[tex]\xi^a \wedge \zeta^b =  \zeta^b \wedge \xi^a[/tex] is a (2,0)tensorvalued 2form. [tex]\xi^a = \xi^a{}_i \, dx^i[/tex] then [tex]d \xi^a = \frac{\partial}{\partial x^j} \xi^a{}_i \, dx^j \wedge dx^i.[/tex] However, there is another object called a "covariant exterior derivative", which can be defined as [tex]D \xi^a = d \xi^a + \omega^a{}_b \wedge \xi^b[/tex] for some matrixvalued connection form [itex]\omega^a{}_b[/itex]. If the vector bundle is the tangent bundle, then [itex]\omega^a{}_b[/itex] can be related to the Christoffel symbols. If the vector bundle is some principal Gbundle, then [itex]\omega^a{}_b[/itex] can be related to the gauge connection. 



#6
Mar712, 01:00 AM

P: 2,045

I see...so there seems to be some index suppression going on here. If we suppress the "form" indices, how are we supposed to know if an object is a vector valued 1 form or 2 form or 3 form etc? I think the index suppression is messing with me when I try to read some of the material in this subject.




#7
Mar712, 01:26 AM

Sci Advisor
P: 1,563

[tex]D R^a{}_b = d R^a{}_b + \omega^a{}_c \wedge R^c{}_b  R^a{}_c \wedge \omega^c{}_b[/tex] is much easier to understand than [tex](D R^a{}_b)_{\mu\nu\rho} = 3 \partial_{[\mu} R^a{}_{b\nu\rho]} + 3 \omega^a{}_{c[\mu} R^c{}_{b\nu\rho]}  3 \omega^c{}_{b[\mu} R^a{}_{c\nu\rho]}[/tex] because I don't have to stare at the formula and process all the tiny symbols to figure out what is being contracted with what, and what it means geometrically. The nice thing about differential forms, wedge products, and exterior derivatives is that they have simple geometric meanings, and you can visualize what is happening by looking at the equations. They also have simple rules of manipulation that make it easy to solve equations. The index notation, on the other hand, is more like a description of an algorithm for plugging things into a computer. Also, you may find reasons to use indices that have nothing to do with components relative to some basis. For example, what if I have a collection of vectorvalued forms, labelled by an index I? The more "decoration" each quantity has to carry around, the more difficult it is to read formulas. 



#8
Mar712, 01:36 AM

P: 2,045

Yea, I think the simpler notation will be much more useful when I become acquainted with the subject. It's just from first glance, it's hard for me to "get" exactly what each quantity is.



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