Question

Let \(\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) be the map $$ (x, y) \rightarrow(u, v, w) \text { where } u=\sin (r v), r=x+y, w=2 $$ For the 1 -form \(\omega=w_{1} \mathrm{~d} u+w_{2} \mathrm{~d} v+w_{3} \mathrm{dw}\) on \(\Omega^{3}\) evaluate \(\varphi^{\circ} \omega\). For any function \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}\) verify Theorem 16.2, that \(\mathrm{d}\left(\varphi^{*} f\right)=\varphi^{*} \mathrm{~d} f\).

Short answer

The pullback of \(\omega\) under \(\varphi\) is evaluated by computing the differentials of the transformation \(u, v, w\) in terms of original variables \(x, y\), and substituting them into the expression for \(\omega\). Theorem 16.2 is confirmed by showing that the pullback of the differential of \(f\), and the differential of the pullback of \(f\) are the same.

Step-by-step solution

Evaluate Pullback of the 1-Form

Evaluate \(\varphi^{*} \omega\). Given that \(\omega = w_1 du + w_2 dv + w_3 dw\), we need to compute \(w_i\) and \(du, dv, dw\). The \(w_i\) are the coefficients of \(du, dv, dw\) respectively which aren't explicitly given, so for a general solution we keep them as is. The differentials \(du, dv, dw\) under the transformation are computed using the method of derivatives with respect to \(x, y\). Recall that \(du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy\), \(dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy\), and \(dw = \frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy\). Using the given map, the derivatives for you to compute are: \(\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}\), \(v\) is left as is because it is not defined in terms of \(x, y\), and \(dw = 0\) (since \(w\) is constant). Substituting these values back into the expression for \(\omega\), gives the result for \(\varphi^{*} \omega\).

Apply and Confirm Theorem 16.2

To apply and confirm Theorem 16.2, which states that \(\mathrm{d}\left(\varphi^{*} f\right) = \varphi^{*} \mathrm{~d} f\), we need to calculate both sides of the equality. Start by calculating \(\mathrm{d}\left(\varphi^{*} f\right)\). \(\varphi^{*} f\) represents the composition of the function \(f\) with \(\varphi\), visually represented as \(f(\varphi(x, y))\). The differential of this composition \(d\left(\varphi^{*} f\right)\) is given by the chain rule. Next, calculate the other side of the equation, \(\varphi^{*} \mathrm{~d} f\). For a function \(f: \mathbb{R}^3 \rightarrow \mathbb{R}\), the differential \(df\) would be \(\frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv + \frac{\partial f}{\partial w} dw\). To calculate its pullback under \(\varphi\), substitute the differentials of \(u, v, w\) under \(\varphi\) into \(\mathrm{~d} f\). If both sides of the theorem calculate to the same differential form, then Theorem 16.2 is confirmed.

Key concepts

Pullback

The pullback is a fundamental operation in differential geometry and calculus on manifolds. It allows us to map differential forms from one space to another through a smooth function. In the context of the given exercise, the pullback \(\varphi^* \omega\) involves transforming the 1-form \(\omega\) defined on \(\R^3\) to the space \(\R^2\) using the function \(\varphi\). This is done by substituting the expressions for \(u, v,\) and \(w\) given by \(\varphi\) into the differentials \(du, dv,\) and \(dw\).
For the map \(\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\), we have transformations defined by \(u=\sin(rv)\) where \(r=x+y\) and \(w=2\). During the pullback process, each differential such as \(du\) is rewritten in terms of the original variables \(x\) and \(y\) using partial derivatives, like \(du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy\). This results in a new differential form that expresses how \(\omega\) behaves back in the domain of \(\R^2\).
This operation is vital because it connects the way forms transform with the concept of derivative through a multi-dimensional context.

Chain Rule

The chain rule is crucial when dealing with compositions of functions, especially for derivatives of such compositions. In our context, this rule is applied when differentiating a composite function such as \(f(\varphi(x, y))\). The chain rule states that if you have a composition of functions, the derivative of this composition can be found by multiplying the derivatives of the nested functions.
For the composition \(f(\varphi(x, y))\), we express this differential in the form \(d(\varphi^* f)\). By using the chain rule, it ensures each component defined through \(\varphi\) is adequately accounted for. The differentiation becomes a multiplication of derivatives, allowing us to easily handle complex mappings.
This theorem is vital to confirm that the operation inside the geography of calculus is coherent when applied to transformations across different spaces.

Differential

In mathematics, a differential is a tool used to describe how functions change. A differential associated with a function gives us the infinitesimal change in the function's output as you make infinitesimal changes in input. Given a function \(f \colon \mathbb{R}^{3} \rightarrow \mathbb{R}\), the differential, denoted \(df\), is expressed as \(df = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv + \frac{\partial f}{\partial w} dw\).
When working with the map \(\varphi\), the challenge is translating these component differentials \(du, dv, dw\) into those relevant to the domain \(x, y\). This translation is a crucial part of assessing and transforming calculus operations between spaces.
Differentials are key in the pullback process because they allow a representation of a form that lives in a new transformed space defined by the map.

Tensors

Tensors are generalizations of scalars, vectors, and matrices and serve as a fundamental concept in fields like continuum mechanics and differential geometry. Think of them as multi-dimensional arrays of numbers following certain transformation rules. Tensors provide a framework to discuss forms and their transformations with respect to different coordinate changes.
In the given problem, the differential forms themselves can be seen as tensorial objects, specifically 1-forms, which are a type of tensor that can be pulled back. When we express an object such as the 1-form \(\omega\) consisting of components like \(w_{1} du + w_{2} dv + w_{3} dw\), each differential term represents tensorial operations reflecting how the transformation should be handled across dimensions.
Understanding the tensor nature of differentials and forms allows for deeper insight into how these mathematical objects behave under transformations like those prompted by pullbacks.