description/proof of that for Lie group homomorphism, if differential at identity Is \(0\), map is constant over connected component
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 differential of \(C^\infty\) map between \(C^\infty\) manifolds with boundary at point.
- The reader knows a definition of connected topological component.
- The reader admits 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.
- The reader admits the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds, if the map has the constant rank \(0\), the map is constant over each connected component.
Target Context
- The reader will have a description and a proof of the proposition that for any Lie group homomorphism, if the differential at the identity is \(0\), the map is constant over each connected component.
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:
\(G_1\): \(\in \{\text{ the Lie groups }\}\)
\(G_2\): \(\in \{\text{ the Lie groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the Lie group homomorphisms }\}\)
\(d f_1\): \(: T_1{G_1} \to T_1{G_2}\), \(= \text{ the differential at } 1\)
//
Statements:
\(d f_1 = 0\)
\(\implies\)
\(\forall T \in \{\text{ the connected components of } M_1\} (f \vert_T \in \{\text{ the constant maps }\})\)
//
2: Note
Especially, for the connected component that contains \(1\), \(T_1\), \(f \vert_T = 1\), because \(f \vert_T (1) = 1\).
3: Proof
Whole Strategy: Step 1: see that \(f\) is a constant rank map; Step 2: conclude the proposition.
Step 1:
Any Lie group homomorphism has a constant rank, by 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: the immediate corollary mentioned in the Note.
So, \(f\) has a constant rank, but as \(f\) has the rank, \(0\), at \(1\), \(f\) has the constant \(0\) rank.
Step 2:
Let \(T\) be any connected component of \(M_1\).
By the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds, if the map has the constant rank \(0\), the map is constant over each connected component, \(f \vert_T\) is constant.
