A description/proof of that curves on manifold as the \(C^\infty\) right actions of curves that represent same vector on Lie group represent same vector
Topics
About: manifold
About: Lie group
About: right action of Lie group on manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold.
- The reader knows a definition of Lie group.
- The reader knows a definition of \(C^\infty\) right action of Lie group on \(C^\infty\) manifold.
Target Context
- The reader will have a description and a proof of the proposition that any curves on any \(C^\infty\) manifold as the \(C^\infty\) right actions of curves that represent the same tangent vector on any Lie group represent the same tangent vector.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any \(C^\infty\) manifold, \(M\), with any \(C^\infty\) right action, \(\mu: M \times G \rightarrow M\), of any Lie group, \(G\), and any \(C^\infty\) curves, \(c_1 (t)\) and \(c_2 (t)\), on \(G\), that represent the same tangent vector on \(G\), the \(C^\infty\) curves, \(m c_1 (t)\) and \(m c_2 (t)\) where \(m \in M\), on \(M\), represent the same tangent vector on \(M\).
2: Proof
\(m c_i (t)\) is in fact a \(C^\infty\) curve, because it is a compound of \(C^\infty\) maps as \(c_i (t)\) is \(C^\infty\) and \(\mu\) is \(C^\infty\), where m is fixed.
For any chart around \(c_1 (0) = c_2 (0)\) on \(G\), \(\left.\frac{d c^j_1 (t)}{d t}\right|_{t = 0} = \left.\frac{d c^j_2 (t)}{d t}\right|_{t = 0}\) where \(c^j_i\) is the \(j\) component of \(c_i\) by the \(G\) chart.
For any chart around \(m c_1 (0) = m c_2 (0)\) on \(M\), \(\left.\frac{d (m c_i)^j}{d t}\right|_{t = 0} = \left.\frac{d \mu^j (m, c_i)}{d t}\right|_{t = 0} = \left.\left(\frac{\partial \mu^j}{\partial c^k_i}\frac{d c^k_i}{d t}\right)\right|_{t = 0}\) where \((m c_i)^j = \mu^j (m, c_i)\) is the \(j\) component of \(m c_i\) by the \(M\) chart. As \(\frac{\partial \mu^j}{\partial c^k_i}\) is a matter of the coordinates function and \(c_1 (0) = c_2 (0)\), \(\left.\frac{\partial \mu^j}{\partial c^k_1}\right|_{t = 0} = \left.\frac{\partial \mu^j}{\partial c^k_2}\right|_{t = 0}\).
As also \(\left.\frac{d c^j_1 (t)}{d t}\right|_{t = 0} = \left.\frac{d c^j_2 (t)}{d t}\right|_{t = 0}\), \(\left.\frac{d (m c_1)^j}{d t}\right|_{t = 0} = \left.\frac{d (m c_2)^j}{d t}\right|_{t = 0}\).
