description/proof of that for Lie group and 'vectors spaces - linear morphisms' isomorphism from double-product \(C^\infty\) manifold tangent space onto direct sum of constituent tangent spaces, composition of differential of multiplication at identity after inverse of isomorphism is addition of arguments
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of Lie group.
- The reader knows a definition of differential of \(C^\infty\) map between \(C^\infty\) manifolds with boundary at point.
- The reader admits the proposition that for any finite-product \(C^\infty\) manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.
Target Context
- The reader will have a description and a proof of the proposition that for any Lie group and the 'vectors spaces - linear morphisms' isomorphism from the double-product \(C^\infty\) manifold tangent space at the identity onto the direct sum of the constituent tangent spaces at the identities, the composition of the differential of the multiplication at the identity after the inverse of the isomorphism is the addition of the arguments.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the Lie groups }\}\)
\(\phi\): \(: T_{(1, 1)}(G \times G) \to T_1G \oplus T_1G\), \(= \text{ the 'vectors spaces - linear morphisms' isomorphism }\)
\(m\): \(: G \times G \to G\), \(= \text{ the multiplication map }\)
\(d m_{(1, 1)}\): \(: T_{(1, 1)}(G \times G) \to T_1G\), \(= \text{ the differential }\)
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Statements:
\(d m_{(1, 1)} \circ \phi^{-1}: T_1G \oplus T_1G \to T_1G, (v_1, v_2) \mapsto v_1 + v_2\)
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2: Proof
Whole Strategy: Step 1: see that \(\phi (v) = (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) and \(d m_{(1, 1)} \circ \phi^{-1}\) is linear; Step 2: see that \(d m_{(1, 1)} \circ \phi^{-1} ((v^1, 0)) = v^1\) and \(d m_{(1, 1)} \circ \phi^{-1} ((0, v^2)) = v^2\); Step 3: conclude the proposition.
Step 1:
For each \(v \in T_{(1, 1)}(G \times G)\), \(\phi (v) = (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) where \(\pi_j: G \times G \to G\) is the projection into the \(j\)-th component, by the proposition that for any finite-product \(C^\infty\) manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.
As \(\phi\) is a 'vectors spaces - linear morphisms' isomorphism, \(d m_{(1, 1)} \circ \phi^{-1}\) is linear.
Step 2:
Let us think of any \((v^1, 0) \in T_1G \oplus T_1G\).
\(v := \phi^{-1} ((v^1, 0))\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G \times G, t \mapsto (\gamma^1 (t), 1)\), where \(v^1 = d \gamma^1 (t) / d t \vert_0\).
\(d m_{(1, 1)} (v) = d (m \circ \gamma (t)) / d t \vert_0 = d (\gamma^1 (t) 1) / d t \vert_0 = d \gamma_1 (t) / d t \vert_0 = v^1\).
That means that \(d m_{(1, 1)} \circ \phi^{-1} ((v^1, 0)) = v^1\).
Let us think of any \((0, v^2) \in T_1G \oplus T_1G\).
\(v := \phi^{-1} ((0, v_2))\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G \times G, t \mapsto (1, \gamma^2 (t))\), where \(v^2 = d \gamma^2 (t) / d t \vert_0\).
\(d m_{(1, 1)} (v) = d (m \circ \gamma (t)) / d t \vert_0 = d (1 \gamma^2 (t)) / d t \vert_0 = d \gamma^2 (t) / d t \vert_0 = v^2\).
That means that \(d m_{(1, 1)} \circ \phi^{-1} ((0, v_2)) = v_2\).
Step 3:
As \(d m_{(1, 1)} \circ \phi^{-1}\) is linear, \(d m_{(1, 1)} \circ \phi^{-1} ((v_1, v_2)) = d m_{(1, 1)} \circ \phi^{-1} ((v_1, 0) + (0, v_2)) = d m_{(1, 1)} \circ \phi^{-1} ((v_1, 0)) + d m_{(1, 1)} \circ \phi^{-1} ((0, v_2)) = v_1 + v_2\).
