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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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In summary, the conversation discusses the differences between vector-valued forms and rank (n,p) tensors, as well as the use of wedge products and exterior derivatives with vector-valued forms. It is also mentioned that index suppression can sometimes cause confusion, but the simpler notation can make it easier to understand the geometric meaning of the equations.

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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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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.

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quasar987

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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?

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Ben Niehoff

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Matterwave said: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.

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Ben Niehoff

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Matterwave said: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.

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A rank (n,p) tensor is a mathematical object that represents a multilinear mapping between vector spaces of dimension n and p. It can also be thought of as a higher-dimensional generalization of vectors and matrices.

A vector is a tensor of rank (1,0), meaning it can be represented as a list of numbers in a single dimension. Tensors of higher rank (n,p) have multiple dimensions and can represent more complex relationships between vector spaces.

In physics and engineering, tensors are used to describe physical quantities and their relationships in a coordinate-independent manner. They are particularly useful in describing the properties of materials, such as stress and strain in solid mechanics, and the electromagnetic field in electromagnetism.

A contravariant tensor is one that transforms in the opposite way as the coordinate system is changed, while a covariant tensor transforms in the same way. This is related to the concept of basis vectors and dual vectors in linear algebra.

While it is difficult to visualize tensors of higher rank (n,p), tensors of rank (2,0) and (0,2) can be represented as matrices, and tensors of rank (3,0) and (0,3) can be represented as 3D arrays. However, for tensors of higher rank, it is more useful to think of them in terms of their mathematical properties rather than trying to visualize them.

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