# How do a bunch of integrals make an n-simplex or an n-cube?

• I
• George Keeling
In summary, Carroll discusses the parallel propagator in matrix notation and shows that it is given by a series of integrals. He explains that the nth term in the first series is an integral over an n-dimensional right triangle or simplex. The expression for the second term, for example, is an integral over a right angle triangle with short sides equal to ##\lambda-\lambda_0##. The long side of the triangle has the equation ##\eta_1 = \eta_2##, so to calculate the multiple integral over the triangle, one must first hold ##\eta_2## constant and then integrate over ##\eta_1## from ##\lambda_0## to ##\eta_2##, and then integrate
George Keeling
Gold Member
TL;DR Summary
Problem with series of integrals in parallel propagator. Parallel transporting vectors.
This question arises from Carroll's Appendix I on the parallel propagator where he shows that, in matrix notation, it is given by$$P\left(\lambda,\lambda_0\right)=I+\sum_{n=1}^{n=\infty}T_n$$and$$T_n=\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\eta_n}\int_{\lambda_0}^{\eta_{n-1}}{\ldots\int_{\lambda_0}^{\eta_2}A\left(\eta_n\right)A\left(\eta_{n-1}\right)\ldots A\left(\eta_1\right)}d^n\eta}$$$$\frac{1}{n!}\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\lambda}\int_{\lambda_0}^{\lambda}{\ldots\int_{\lambda_0}^{\lambda}\mathcal{P}\left[A\left(\eta_n\right)A\left(\eta_{n-1}\right)\ldots A\left(\eta_1\right)\right]}d^n\eta}$$where ##\mathcal{P}## orders the matrices ##A\left(\eta_i\right)## so that ##\eta_n\geq\eta_{n-1}\geq\ldots\geq\eta_1##.

Carroll says that the ##n##th term in the first series of integrals is an integral over an ##n##-dimensional right triangle or ##n##-simplex:$$T_1=\int_{\lambda_0}^{\lambda}{A\left(\eta_1\right)d\eta_1}$$$$T_2=\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\eta_2}A\left(\eta_2\right)A\left(\eta_1\right)d\eta_1d\eta_2}$$$$T_2=\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\eta_3}\int_{\lambda_0}^{\eta_2}A\left(\eta_3\right)A\left(\eta_2\right)A\left(\eta_1\right)d^3\eta}$$and illustrates them thus:

I can see how a 2-simplex with short sides equal is half a square and how that carries on for more dimensions but I can't see how the expression, for example, for ##T_2## is an integral over a right angle triangle, let alone one with short sides equal ##\lambda-\lambda_0##.

It would seem that what exactly ##A## is is immaterial. I tested calculating ##T_2## both ways for three cases$$A\left(x\right)=3x^2,A_{\ \ \rho}^\mu\left(\lambda\right)=\left(\begin{matrix}0&\sin{\psi}\cos{\psi}\\-\cot{\psi}&0\\\end{matrix}\right),\ A_{\ \ \rho}^\mu\left(x\right)=\left(\begin{matrix}2x&3x^2\\4x^3&5x^4\\\end{matrix}\right)$$the first two gave the same answer by both methods and the third did not (I may easily have made a mistake on that.) The second is what ##A_{\ \ \rho}^\mu## would be for propagating a vector on the surface of a sphere at constant latitude ##\psi##. It was much easier to use the second method (over an ##n##-cube) and that did indeed parallel propagate vectors round lines of constant latitude very well.

If anybody can help me understand why the integrals are integrals over ##n##-simplices or ##n##-cubes and how ##\mathcal{P}## orders the matrices ##A\left(\eta_i\right)## so that ##\eta_n\geq\eta_{n-1}\geq\ldots\geq\eta_1## (the ##\eta_i## seem to be moving targets) I would greatly appreciate it.

George Keeling said:
I can't see how the expression, for example, for ##T_2## is an integral over a right angle triangle, let alone one with short sides equal ##\lambda-\lambda_0##.
The long side of the triangle has the equation ##\eta_1 = \eta_2##, so to calculate the multiple integral over the triangle you first hold ##\eta_2## constant and then integrate over ##\eta_1## from ##\lambda_0## to ##\eta_2## (from the left edge to the long side), and then you integrate the result over ##\eta_2## from ##\lambda_0## to ##\lambda##.

George Keeling
Thank you @ergospherical I think I understand now. The multiple integrals over the square are more readily comprehensible.

## 1. How are integrals used to construct an n-simplex or an n-cube?

Integrals are used in the process of finding the volume of an n-dimensional shape, such as an n-simplex or an n-cube. By integrating over the bounds of each dimension, the total volume of the shape can be calculated.

## 2. What is the relationship between integrals and n-simplices or n-cubes?

The integral of a function over a certain domain can be thought of as the sum of infinitely many infinitesimal rectangles. These rectangles can be used to approximate the volume of an n-dimensional shape, such as an n-simplex or an n-cube.

## 3. How do integrals account for the different dimensions of an n-simplex or an n-cube?

The bounds of the integral correspond to the dimensions of the shape. For example, in a 3-dimensional n-cube, the integral would have 3 bounds, one for each dimension. This allows the integral to account for the different dimensions of the shape.

## 4. Can integrals be used to construct other n-dimensional shapes?

Yes, integrals can be used to construct a variety of n-dimensional shapes, not just n-simplices or n-cubes. For example, integrals can also be used to find the volume of an n-sphere or an n-torus.

## 5. Is there a specific formula for using integrals to construct an n-simplex or an n-cube?

Yes, there are specific formulas for calculating the volume of an n-simplex or an n-cube using integrals. These formulas involve integrating over the bounds of each dimension, and can be found in most calculus textbooks or online resources.

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