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: \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*}u^{\prime}, \rho_{*} v^{\prime}\right].\)