description/proof of that for 2 Lie group homomorphisms between Lie groups, map as multiplication of 1st homomorphism and multiplicative inverse of 2nd homomorphism has constant rank
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 %structure kind name% homomorphism.
- The reader knows a definition of rank of \(C^\infty\) map between \(C^\infty\) manifolds with boundary at point.
- The reader admits the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank.
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:
\(G_1\): \(\in \{\text{ the Lie groups }\}\)
\(G_2\): \(\in \{\text{ the Lie groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(f'\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(h\): \(: G_1 \to G_2, g \mapsto f (g) (f' (g))^{-1}\)
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Statements:
\(\forall g \in G_1 (Rank ({d h}_g) = Rank ({d h}_1))\)
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2: Note
\(h\) is not any Lie group homomorphism in general, because \(h (g g') = f (g g') (f' (g g'))^{-1} = f (g) f (g') (f' (g) f' (g'))^{-1} = f (g) f (g') (f' (g'))^{-1} (f' (g))^{-1} = f (g) h (g') (f' (g))^{-1}\), which is not guaranteed to equal \(h (g) h (g') = f (g) (f' (g))^{-1} f (g') (f' (g'))^{-1}\), unless \(G_2\) is Abelian.
As an immediate corollary, \(f\) has a constant rank, because \(f'\) can be taken to be \(f' = 1\), which is a Lie group homomorphism, and \(h (g) = f (g) (f' (g))^{-1} = f (g) 1^{-1} = f (g) 1 = f (g)\), so, \(h = f\).
3: Proof
Whole Strategy: Step 1: see that \(h (g_0 g) = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h (g)\) and take \(h': G_1 \to G_2, g \mapsto h (g_0 g), = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h = h \circ L_{g_0}\); Step 2: see that \({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0)} \circ {d L_{f (g_0)}}_1 \circ {d h}_1 = {d h}_{g_0} \circ {d L_{g_0}}_1\); Step 3: conclude the proposition.
Step 1:
\(h\) is \(C^\infty\), because \(f\) and \(f'\) are \(C^\infty\), and the multiplicative inverse map and the multiplication map are \(C^\infty\): it is \(: G_1 \to G_1 \times G_1 \to G_2 \times G_2 \to G_2 \times G_2 \to G_2, g \mapsto (g, g) \mapsto (f (g), f' (g)) \mapsto (f (g), (f' (g))^{-1}) \mapsto f (g) (f' (g))^{-1}\).
For each \(g_0 \in G_j\), let \(L_{g_0}: G_j \to G_j, g \mapsto g_0 g\) and \(R_{g_0}: G_j \to G_j, g \mapsto g g_0\) be the left translation and the right translation by \(g_0\), which are some diffeomorphisms.
For each \(g_0, g \in G_1\), let us see that \(h (g_0 g) = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h (g)\).
\(h (g_0 g) = f (g_0 g) (f' (g_0 g))^{-1} = f (g_0) f (g) (f' (g_0) f' (g))^{-1} = f (g_0) f (g) (f' (g))^{-1} (f' (g_0))^{-1} = f (g_0) h (g) (f' (g_0))^{-1} = L_{f (g_0)} (h (g)) (f' (g_0))^{-1} = R_{(f' (g_0))^{-1}} (L_{f (g_0)} (h (g)))\).
So, let us take \(h': G_1 \to G_2, g \mapsto h (g_0 g), = R_{(f' (g_0))^{-1}} \circ L_{f (g_0)} \circ h = h \circ L_{g_0}\), which is \(C^\infty\) as a composition of \(C^\infty\) maps.
Step 2:
\({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0) h (1)} \circ {d L_{f (g_0)}}_{h (1)} \circ {d h}_1 = {d h}_{g_0 1} \circ {d L_{g_0}}_1\).
As \(h (1) = f (1) (f' (1))^{-1} = 1 1^{-1} = 1 1 = 1\), \({d h'}_1 = {d R_{(f' (g_0))^{-1}}}_{f (g_0)} \circ {d L_{f (g_0)}}_1 \circ {d h}_1 = {d h}_{g_0} \circ {d L_{g_0}}_1\).
Step 3:
\({d R_{(f' (g_0))^{-1}}}_{f (g_0)}: T_{f (g_0)}G_2 \to T_{f (g_0) (f' (g_0))^{-1}}G_2\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(R_{(f' (g_0))^{-1}}\) is a diffeomorphism.
\({d L_{f (g_0)}}_1: T_1G_2 \to T_{f (g_0)}G_2\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(L_{f (g_0)}\) is a diffeomorphism.
\({d h}_1: T_1G_1 \to T_1G_2\).
\({d h}_{g_0}: T_{g_0}G_1 \to T_{h (g_0)}G_2\).
\({d L_{g_0}}_1: T_1G_1 \to T_{g_0}G_1\), which is a 'vectors spaces - linear morphisms' isomorphism, because \(L_{g_0}\) is a diffeomorphism.
By the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map, \(Rank ({d h'}_1) = Rank ({d h}_1) = Rank ({d h}_{g_0})\).
