Integral Notation: Abuse or Necessity?

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Discussion Overview

The discussion revolves around the notation used in integrals involving differential forms, specifically whether the integral notation should include the function evaluated at a point or if it should remain in its general form. Participants explore the implications of these notations in the context of differential geometry and exterior calculus.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the integral notation should be written as ##\displaystyle \int\limits_U f(x^1,\dots, x^n)~dx^1\wedge\cdots\wedge dx^n## or as ##\displaystyle \int\limits_U f\wedge dx^1\wedge\cdots\wedge dx^n##.
  • Another participant argues that both notations are correct, suggesting that the first notation is a simplified version of the second, which can be further simplified by computing the exterior product.
  • A later reply emphasizes the distinction between the function ##f(x)## as an output and ##f## as the function itself, indicating a preference for writing the integral without inserting specific coordinates into the function.
  • Some participants discuss the implications of treating ##f## as a 0-form and the conditions under which simplifications can be made in the notation.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of the integral notations, with no consensus reached on which form is preferable or if both are equally valid.

Contextual Notes

The discussion highlights potential limitations in understanding the implications of the notations, particularly regarding the treatment of functions as 0-forms and the simplification of exterior products.

Mandelbroth
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Quick question:

When we write an integral ##\displaystyle \int\limits_U f(x^1,\dots, x^n)~dx^1\wedge\cdots\wedge dx^n##, we really mean ##\displaystyle \int\limits_U f\wedge dx^1\wedge\cdots\wedge dx^n##, right?
 
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Both notations are right. The first one is already simplified, whereas the second one can be simplified further by explicitely computing the exterior product and you will end up with the first version. So I actually think the first version is cleaner.

The differential form in your integral is an assignment of an element of [itex]\Lambda(T^*_x U)[/itex] to every point [itex]x[/itex] of [itex]U[/itex]. [itex]\Lambda(T^*_x U)[/itex] is a vector space and [itex]f(x)[/itex] is just a coefficient (which is different at every point).
 
rubi said:
Both notations are right. The first one is already simplified, whereas the second one can be simplified further by explicitely computing the exterior product and you will end up with the first version. So I actually think the first version is cleaner.

The differential form in your integral is an assignment of an element of [itex]\Lambda(T^*_x U)[/itex] to every point [itex]x[/itex] of [itex]U[/itex]. [itex]\Lambda(T^*_x U)[/itex] is a vector space and [itex]f(x)[/itex] is just a coefficient (which is different at every point).
I'm sorry, I should have specified what I mean.

I'm talking mainly about the fact that ##f(x)## is the output of ##f##, whereas ##f## is the function.
 
Mandelbroth said:
I'm sorry, I should have specified what I mean.

I'm talking mainly about the fact that ##f(x)## is the output of ##f##, whereas ##f## is the function.
In that case I would write [itex]f \,\mathrm d x_1\wedge\ldots\wedge \mathrm d x_n[/itex], however, omitting the first wedge. If you know that [itex]f[/itex] is a 0-form, then you can simplify the exterior product. But you are right: Inserting the coordinates into f gives you the form at a point already, whereas the form itself should be written without inserting [itex]x[/itex] into [itex]f[/itex].
 

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