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Chapter 8: Problem 26

Suppose \(F: G \rightarrow H\) is a Lie group homomorphism. Show that the kernel of \(F_{*}: \operatorname{Lie}(G) \rightarrow \operatorname{Lie}(H)\) is the Lie algebra of Ker \(F\) (under the identification of the Lie algebra of a subgroup with a Lie subalgebra as in Theorem 8.46).

Short Answer

Expert verified
The kernel of \( F_* \) is the Lie algebra of \( \ker(F) \).

Step by step solution

01

Define the Kernel of F

The kernel of a Lie group homomorphism, Ker \( F \), consists of all elements \( g \in G \) such that \( F(g) = e_H \), where \( e_H \) is the identity element of the Lie group \( H \). This kernel is a normal subgroup of \( G \).
02

Use Lie Algebra Homomorphism

Consider the differential (or pushforward) of the homomorphism \( F \) at the identity, noted as \( F_{*} : \operatorname{Lie}(G) \to \operatorname{Lie}(H) \). This maps from the Lie algebra of \( G \) to the Lie algebra of \( H \).
03

Identifying Kernel of F_*

The kernel of \( F_{*} \), noted as \( \ker(F_{*}) \), consists of all vectors \( X \in \operatorname{Lie}(G) \) such that \( F_{*}(X) = 0 \) in \( \operatorname{Lie}(H) \).
04

Relate Kernel to Lie Algebras

By Theorem 8.46, the Lie algebra of a Lie group (or subgroup) contains the tangent space at the identity element. The kernel of \( F_{*} \) corresponds to the Lie algebra of the identity component of \( \ker(F) \). Thus, \( \ker(F_{*}) = \operatorname{Lie}(\ker(F)) \).
05

Conclusion

We conclude that the kernel of the Lie algebra homomorphism \( F_{*} \) is the Lie algebra of the Lie group \( \ker(F) \), as \( \ker(F_{*}) = \operatorname{Lie}(\ker(F)) \). This follows naturally from the above steps and the definition of the differentials of maps between Lie groups.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kernel of Homomorphism
In the simplest terms, the kernel of a homomorphism is a concept from algebra that helps in understanding the structure of groups. When we're dealing with Lie group homomorphisms, we refer to the kernel as
  • the set of all elements in Lie group \( G \) that are mapped to the identity element in Lie group \( H \).
  • This is represented as \( \ker(F) \).
The kernel
  • plays a crucial role because it is always a normal subgroup of the Lie group. A normal subgroup means that any group element can be conjugated with an element from this subgroup, and the result will still be in that subgroup.
Understanding the kernel is essential in examining the properties of a Lie group homomorphism and grasping the behavior of the group structure beyond the homomorphic mapping.
Lie Algebra
Lie algebras provide a vector space framework with a particular multiplication rule called a Lie bracket. This bracketing operation
  • must satisfy properties such as bilinearity, alternativity, and the Jacobi identity.
  • Lie algebras are tied closely to Lie groups, acting as the infinitesimal version of Lie groups.
A Lie algebra
  • gives a linear approximation to the local structure of a Lie group near the identity element.
These algebras encapsulate
  • the tangent space at the identity component, simplifying complex studies of group symmetries to manageable linear algebra problems.
Lie Group Differential
A Lie group differential, or the pushforward, is an important tool when studying homomorphisms. When you consider a homomorphism \( F: G \to H \) between Lie groups, the differential maps the corresponding Lie algebras:
  • It is denoted by \( F_{*}: \operatorname{Lie}(G) \to \operatorname{Lie}(H) \).
This mapping translates the continuous group structure into terms that can be interpreted through linear transformations.
  • The differential captures how tiny movements around the identity in one group relate to movements in another group.
Understanding this concept helps in simplifying the study of group homomorphisms as it reduces continuous transformations to linear interactions.
Tangent Space at Identity
The tangent space at the identity of a Lie group is a pivotal component in understanding Lie algebras. For a Lie group \( G \), its tangent space at the identity element \( e_G \) contains:
  • The derivatives of all curves within the group that pass through \( e_G \).
This tangent space
  • provides a linear 鈥榮lice鈥� of the group around its identity,
  • serving as the building block of the Lie algebra of \( G \).
Through this space, we can visualize the essence of how the group behaves locally and how algebraic structures are mirrored in the group鈥檚 geometry.
Normal Subgroup of Lie Group
In the context of Lie groups, a normal subgroup is an important component as it maintains coherence in the group structure relative to conjugations. A normal subgroup \( N \) of a Lie group \( G \) has the property that:
  • For any element \( g \) in \( G \) and any element \( n \) in \( N \), the element \( gng^{-1} \) also lies in \( N \). This property makes normal subgroups crucial in the analysis of quotient groups.
Normal subgroups allow for decomposing the group into simpler, more digestible parts while retaining symmetry. They frequently appear when discussing kernels of homomorphisms, indicating that such kernels divide the group structure in a uniform manner. This division is helpful in exploring different group characteristics and functions.

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Most popular questions from this chapter

Prove that up to isomorphism, there are exactly one l-dimensional Lie algebra and two 2 -dimensional Lie algebras. Show that all three algebras are isomorphic to Lie subalgebras of \(\mathrm{g} \mathrm{I}(2, \mathbb{R})\).

Suppose \(F: M \rightarrow N\) is a smooth submersion, where \(M\) and \(N\) are positivedimensional smooth manifolds. Given \(X \in \mathfrak{X}(M)\) and \(Y \in \mathfrak{X}(N)\), we say that \(X\) is a lift of \(Y\) if \(X\) and \(Y\) are \(F\)-related. A vector field \(V \in \mathfrak{X}(M)\) is said to be vertical if \(V\) is everywhere tangent to the fibers of \(F\) (or, equivalently, if \(V\) is \(F\)-related to the zero vector field on \(N\) ). (a) Show that if \(\operatorname{dim} M=\operatorname{dim} N\), then every smooth vector field on \(N\) has a unique lift. (b) Show that if \(\operatorname{dim} M \neq \operatorname{dim} N\), then every smooth vector field on \(N\) has a lift, but that it is not unique. (c) Assume in addition that \(F\) is surjective. Given \(X \in \mathfrak{X}(M)\), show that \(X\) is a lift of a smooth vector field on \(N\) if and only if \(d F_{p}\left(X_{p}\right)=d F_{q}\left(X_{q}\right)\) whenever \(F(p)=F(q)\). Show that if this is the case, then \(X\) is a lift of a unique smooth vector field. (d) Assume in addition that \(F\) is surjective with connected fibers. Show that a vector field \(X \in \mathfrak{X}(M)\) is a lift of a smooth vector field on \(N\) if and only if \([V, X]\) is vertical whenever \(V \in \mathscr{X}(M)\) is vertical.

Let \(M\) be a nonempty positive-dimensional smooth manifold with or without boundary. Show that \(\mathfrak{X}(M)\) is infinite-dimensional.

Let \(g\) be a Lie algebra. A linear subspace \(\mathfrak{h} \subseteq g\) is called an ideal in \(\mathrm{g}\) if \([X, Y] \in \mathfrak{h}\) whenever \(X \in \mathfrak{h}\) and \(Y \in \mathrm{g}\). (a) Show that if \(\mathfrak{h}\) is an ideal in \(g\), then the quotient space \(g / \mathfrak{h}\) has a unique Lie algebra structure such that the projection \(\pi: \mathrm{g} \rightarrow \mathrm{g} / \mathrm{h}\) is a Lie algebra homomorphism. (b) Show that a subspace \(\mathrm{h} \subseteq \mathrm{g}\) is an ideal if and only if it is the kernel of a Lie algebra homomorphism. (Used on p. 533.)

Let \(M\) be a smooth manifold with or without boundary, let \(N\) be a smooth manifold, and let \(f: M \rightarrow N\) be a smooth map. Define \(F: M \rightarrow M \times N\) by \(F(x)=(x, f(x))\). Show that for every \(X \in \mathfrak{X}(M)\), there is a smooth vector field on \(M \times N\) that is \(F\)-related to \(X\).
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