description/proof of that for Lie group, differential of inversion at identity is inversion of argument
Topics
About: group
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.
- The reader admits the proposition that for any maps between any arbitrary subsets of any \(C^\infty\) manifolds with boundary \(C^k\) at corresponding points, where \(k\) includes \(\infty\), the composition is \(C^k\) at the point.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any Lie group, the differential of the inversion at the identity is the inversion of the argument.
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:
\(G\): \(\in \{\text{ the Lie groups }\}\)
\(i\): \(: G \to G\), \(= \text{ the inversion }\)
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Statements:
\(d i_1: T_1M \to T_1M, v \mapsto - v\)
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2: Proof
Whole Strategy: Step 1: let \(m\) be the multiplication map and let \(\phi: T_{(1, 1)}(G \times G) \to T_1G \times T_1G\) be the 'vectors spaces - linear morphisms' isomorphism, and think of \(m \circ (id, i): G \to G \times G \to G, g \mapsto m (id (g), i (g))\) and see that \(d (m \circ (id, i))_1 = d m_{(1, 1)} \circ \phi^{-1} \circ \phi \circ d (id, i)_1 = 0\); Step 2: see that \(\phi \circ d (id, i)_1 (v) = (v, d i_1 (v))\); Step 3: apply 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.
Step 1:
Let \(m: G \times G \to G\) be the multiplication map.
There is the 'vectors spaces - linear morphisms' isomorphism, \(\phi: T_{(1, 1)}(G \times G) \to T_1G \oplus T_1G, v \mapsto (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) 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.
Let us think of \(m \circ (id, i): G \to G \times G \to G, g \mapsto m (id (g), i (g))\) where \(id\) is the identity map.
\(m \circ (id, i)\) is \(C^\infty\) as a composition of \(C^\infty\) maps, by the proposition that for any maps between any arbitrary subsets of any \(C^\infty\) manifolds with boundary \(C^k\) at corresponding points, where \(k\) includes \(\infty\), the composition is \(C^k\) at the point.
\(d (m \circ (id, i))_1 = d m_{(1, 1)} \circ d (id, i)_1 = d m_{(1, 1)} \circ \phi^{-1} \circ \phi \circ d (id, i)_1 = 0\), because \(m \circ (id, i)\) is constant to \(1\).
Step 2:
For any \(v \in T_1G\), \(v\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G, t \mapsto \gamma (t)\).
\(\phi \circ d (id, i)_1 (v) = (d \pi_{1, (1, 1)} (d (id, i)_1 (v)), d \pi_{2, (1, 1)} (d (id, i)_1 (v))) = (d (\pi_1 ((id (\gamma (t)), i (\gamma (t))))) / d t \vert_0, d (\pi_2 ((id (\gamma (t)), i (\gamma (t))))) / d t \vert_0) = (d (id (\gamma (t))) / d t \vert_0, d (i (\gamma (t))) / d t \vert_0) = (d \gamma (t) / d t \vert_0, d (i (\gamma (t))) / d t \vert_0) = (v, d i_1 (v))\).
Step 3:
But \(d m_{(1, 1)} \circ \phi^{-1} ((v, d i_1 (v))) = v + d i_1 (v)\), by 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.
So, \(v + d i_1 (v) = 0\), so, \(d i_1 (v) = - v\).
