Vector valued forms


by Matterwave
Tags: forms, valued, vector
Matterwave
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#1
Mar6-12, 07:15 PM
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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 anti-symmetric bottom indices right? Is there a difference?

A (n,m) valued p form is a (n,m+p) tensor which is anti-symmetric in the lower p indices?
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Ben Niehoff
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#2
Mar6-12, 08:30 PM
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Yes, basically.

However, vector-valued forms can also take values in vector spaces other than [itex]T_xM[/itex]. For example, a gauge connection is a 1-form that takes values in the Lie algebra of some Lie group.
quasar987
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#3
Mar6-12, 08:39 PM
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More generally, vector valued p-forms take value in a vector bundle. I.e. they are elements of [tex]\Omega^p(M)\otimes C^{\infty}(E)[/tex]

Matterwave
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#4
Mar6-12, 11:44 PM
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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 k-form? It's a (n,p+k) tensor that's fully anti-symmetrized in the lower p+k indices?

To take the exterior derivative of a vector-valued 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?
Ben Niehoff
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#5
Mar7-12, 12:54 AM
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Quote Quote by Matterwave View Post
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 k-form? It's a (n,p+k) tensor that's fully anti-symmetrized in the lower p+k indices?
Yes. I tend to think of wedge products more formally, so I would say it "means" that you stick the forms next to each other with a wedge symbol in between. E.g., say [itex]\xi^a, \zeta^a[/itex] are vector-valued 1-forms, then

[tex]\xi^a \wedge \zeta^b = - \zeta^b \wedge \xi^a[/tex]
is a (2,0)-tensor-valued 2-form.

To take the exterior derivative of a vector-valued 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?
No, the exterior derivative is still defined the same way. So if

[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 matrix-valued 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 G-bundle, then [itex]\omega^a{}_b[/itex] can be related to the gauge connection.
Matterwave
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#6
Mar7-12, 01:00 AM
P: 2,078
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.
Ben Niehoff
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#7
Mar7-12, 01:26 AM
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Quote Quote by Matterwave View Post
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?
You just use some words to say what it is, like I did in my post above. Even if you chose to use index notation, you would still end up using the same words, because the purpose of writing things down is to clearly explain them to someone else.

I think the index suppression is messing with me when I try to read some of the material in this subject.
I think the notation is much cleaner this way. For example, if I want to take the covariant exterior derivative of a matrix-valued 2-form, the formula

[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 vector-valued forms, labelled by an index I? The more "decoration" each quantity has to carry around, the more difficult it is to read formulas.
Matterwave
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#8
Mar7-12, 01:36 AM
P: 2,078
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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