Question

Show that if \(\rho: M^{\prime} \rightarrow M\) is a smooth map, then \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*} u^{\prime}, \rho_{*} v^{\prime}\right]\) for any vector fields \(u^{\prime}, v^{\prime}\) on \(M^{\prime}\).

Short answer

Question: Prove that if $\rho:M' \rightarrow M$ is a smooth map between two manifolds, and $u'$ and $v'$ are vector fields on $M'$, then \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right]\) where $[\cdot,\cdot]$ denotes the Lie bracket and $\rho_{*}$ denotes the differential of the smooth map $\rho$.

Step-by-step solution

Recall the definition of the Lie bracket

The Lie bracket of two vector fields \(u\) and \(v\) on a manifold \(M\) is defined as the commutator of the two vector fields, i.e. \([u,v] = uv - vu\), where \(uv\) and \(vu\) are the action of \(u\) on \(v\) and \(v\) on \(u\) respectively.

Recall the definition of the differential of a smooth map

Given a smooth map \(\rho:M' \rightarrow M\), the differential of \(\rho\) at a point \(p' \in M'\), denoted by \(\rho_{*}\), is a linear map from the tangent space at \(p'\), \(T_{p'}M'\), to the tangent space at \(\rho(p')\), \(T_{\rho(p')}M\). If \(u'\) is a vector field on \(M'\), then, by definition, \(\rho_{*}u'\) is a vector field on \(M\).

Calculate the action of two vector fields on a scalar function

Let \(u'^i \frac{\partial}{\partial x'^i}\) and \(v'^i \frac{\partial}{\partial x'^i}\) be local coordinate representations of two vector fields \(u'\) and \(v'\), and let \(f:M' \rightarrow \mathbb{R}\) be a smooth function. Expand the Lie bracket expression using the definition: \([u',v']f = u'v'f - v'u'f.\)

Calculate the components of the Lie bracket

To derive the expression for the Lie bracket of two vector fields \(u'\) and \(v'\) in local coordinates, use the chain rule and the commutator property: \([u',v']^i = u'^j \frac{\partial v'^i}{\partial x'^j} - v'^j \frac{\partial u'^i}{\partial x'^j}.\)

Calculate the pushforward of each side of the expression

We want to show that \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right].\) Expand both sides of this equation using the expressions we derived above: \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)^i =\rho_{*}(u'^j \frac{\partial v'^k}{\partial x'^j} - v'^j \frac{\partial u'^k}{\partial x'^j})\frac{\partial}{\partial x^i} \rho(x'^k)\) and \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right]^i = (\rho_{*}u')^j \frac{\partial}{\partial x^i} (\rho_{*}v')^k - (\rho_{*}v')^j \frac{\partial}{\partial x^i} (\rho_{*}u')^k.\)

Show the equality between these expressions

Using the chain rule, we can rewrite the differential of \(\rho\) in local coordinates: \((\rho_{*}u')^j = u'^k \frac{\partial \rho^j}{\partial x'^k}\) and \((\rho_{*}v')^j = v'^k \frac{\partial \rho^j}{\partial x'^k}\). Using these expressions, we deduce that \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right]^i = u'^k \frac{\partial \rho^j}{\partial x'^k} \frac{\partial}{\partial x^i} (v'^l \frac{\partial \rho^k}{\partial x'^l}) - v'^k \frac{\partial \rho^j}{\partial x'^k} \frac{\partial}{\partial x^i} (u'^l \frac{\partial \rho^k}{\partial x'^l}).\) Applying the chain rule, we get \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right]^i = u'^k v'^l \frac{\partial^2 \rho^j}{\partial x'^k \partial x'^l} \frac{\partial \rho^k}{\partial x'^l} - v'^k u'^l \frac{\partial^2 \rho^j}{\partial x'^l \partial x'^k} \frac{\partial \rho^k}{\partial x'^l}\) which simplifies to \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right]^i = \rho_{*}(u'^k \frac{\partial v'^l}{\partial x'^k} - v'^k \frac{\partial u'^l}{\partial x'^k}) \frac{\partial}{\partial x^i} \rho(x'^l) = \rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)^i.\) Since both expressions are equal, the original statement is proved: \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right].\)

Key concepts

Vector Fields

Vector fields are a fundamental concept in differential geometry and vector calculus. Think of a vector field as an assignment of a vector to every point in a space, typically a manifold. In this context, a vector is a quantity that has magnitude and direction. Imagine taking a smooth surface, like Earth's surface. A vector field would represent something like wind speed and direction at every point on this surface. Here's what to consider about vector fields:

• A vector field is often represented by expressions like \(u'\) or \(v'\), which in the context of our exercise, denote vector fields on a manifold \(M'\).
• Locally, vector fields can be represented in terms of their components with respect to a chosen coordinate system, like \(u'^i\) and \(v'^i\) in local coordinates.

Vector fields can interact with functions defined on manifolds using the concept of a directional derivative. This is basically how much a given function is increasing or decreasing in the direction of a vector field. Together, vector fields and the ways they interact with functions are a powerful tool in understanding the geometry of manifolds.

Smooth Map

A smooth map is an essential idea in differential geometry. It refers to a function that connects two manifolds in a way that is infinitely differentiable. This means there are no sharp corners or breaks in the map; everything changes smoothly. A smooth map doesn't just deal with values but also maintains the structure needed to handle derivatives and higher-order derivatives. Some key points about smooth maps include:

• They are used to create connections between different manifolds. In our exercise, \(\rho: M' \rightarrow M\) is an example of a smooth map from one manifold \(M'\) to another \(M\).
• The differential of a smooth map, noted as \(\rho_{*}\), is an important tool. It maps tangent vectors from one manifold to the other, creating a bridge between their respective geometries.

Understanding smooth maps is crucial for advanced topics in mathematics and physics because they allow transformations between different spaces that preserve the smoothness necessary for calculus.

Differential Manifolds

Differential manifolds are the playground in which differential geometry takes place. They generalize the ideas of curves and surfaces we are familiar with into more complex and higher-dimensional spaces. Essentially, a differential manifold is a space that looks locally like Euclidean space. This means that around every point, there is a region that resembles ordinary flat space, allowing the tools of calculus to be applied. Why are they important?

• A manifold allows us to do calculus in more abstract contexts. For instance, differential manifolds enable calculus on the surface of a globe, unlike on a plane.
• Manifolds can be of any dimension, not just the ones we can visualize, like 1D, 2D, or 3D. This idea is central in higher mathematics and theoretical physics.
• They are equipped with a structure that fits smoothly onto them, making the definition of smooth maps and vector fields possible within their setting.

In summary, by generalizing the idea of curves and surfaces, differential manifolds offer a flexible framework to handle problems in many fields that require sophisticated and abstract mathematical apparatus.