This question may be a bit naive, but I am relatively new to the study of differential topology. I am currently reading through John Lee's Smooth Manifolds. He devotes a section to the theory of integral curves of vector fields on manifolds, and the associated flow on the manifold. In the case of a complete vector field, the flow corresponds to a smooth $\mathbb{R}$-action on $M$. In later sections, vector fields on $M$ get fit into the more general context of smooth sections of bundles over $M$ (vector bundles, tensor bundles, etc.) In this language, certain smooth sections of the tangent bundle (ie vector fields) induce smooth $\mathbb{R}$ actions on $M$ (in the form of the flow). I am wondering if this phenomenon in anyway generalizes to smooth sections of other types of bundles. More explicitly, are there other contexts in which a smooth section of a bundle over $M$ induce some smooth Lie group action on $M$? Or is this phenomenon specific to vector fields? Thank you in advance!
2026-04-07 02:20:22.1775528422
Relationship between smooth sections and group actions on manifold
120 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
There are 1 best solutions below
Related Questions in SMOOTH-MANIFOLDS
- Smooth Principal Bundle from continuous transition functions?
- Possible condition on locally Euclidean subsets of Euclidean space to be embedded submanifold
- "Defining a smooth structure on a topological manifold with boundary"
- Hyperboloid is a manifold
- The graph of a smooth map is a manifold
- A finite group G acts freely on a simply connected manifold M
- An elementary proof that low rank maps cannot be open
- What does it mean by standard coordinates on $R^n$
- Partial Differential Equation using theory of manifolds
- Showing that a diffeomorphism preserves the boundary
Related Questions in GROUP-ACTIONS
- Orbit counting lemma hexagon
- Showing a group G acts on itself by right multiplication
- $N\trianglelefteq G$, $A$ a conjugacy class in $G$ such that $A\subseteq N$, prove $A$ is a union of conjugacy classes
- Show that the additive group $\mathbb{Z}$ acts on itself by $xy = x+y$ and find all $x\in\mathbb{Z}$ such that $xy = y$ for all $y\in\mathbb{Z}$.
- Number of different k-coloring of an $n\times m$ grid up to rows and columns permutations
- How to embed $F_q^\times $ in $S_n$?
- orbit representatives for the group of unipotent matrix acting on the set of skew-symmetric matrices
- $S_n$ right-action on $V^{\otimes n}$
- Interpretation of wreath products in general and on symmetric groups
- Regarding action of a group factoring through
Related Questions in VECTOR-FIELDS
- Does curl vector influence the final destination of a particle?
- Using the calculus of one forms prove this identity
- In a directional slope field, how can a straight line be a solution to a differential equation?
- Partial Differential Equation using theory of manifolds
- If $\nabla X=h \cdot \text{Id}_{TM}$ for a vector field $X$ and $h \in C^{\infty}(M)$, is $h$ constant?
- Equivalent definition of vector field over $S^2$
- Study of a " flow "
- Extension of a gradient field
- how to sketch the field lines of $F(x,y)=(\sin y,-\sin x)$?
- Is a vector field a mathematical field?
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Find $E[XY|Y+Z=1 ]$
- Refuting the Anti-Cantor Cranks
- What are imaginary numbers?
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- How do we know that the number $1$ is not equal to the number $-1$?
- What are the Implications of having VΩ as a model for a theory?
- Defining a Galois Field based on primitive element versus polynomial?
- Can't find the relationship between two columns of numbers. Please Help
- Is computer science a branch of mathematics?
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- Generator of inertia group in function field extension
Popular # Hahtags
second-order-logic
numerical-methods
puzzle
logic
probability
number-theory
winding-number
real-analysis
integration
calculus
complex-analysis
sequences-and-series
proof-writing
set-theory
functions
homotopy-theory
elementary-number-theory
ordinary-differential-equations
circles
derivatives
game-theory
definite-integrals
elementary-set-theory
limits
multivariable-calculus
geometry
algebraic-number-theory
proof-verification
partial-derivative
algebra-precalculus
Popular Questions
- What is the integral of 1/x?
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- Is a matrix multiplied with its transpose something special?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- How to tell if a set of vectors spans a space?
- Calculus question taking derivative to find horizontal tangent line
- How to determine if a function is one-to-one?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- Is this Batman equation for real?
- How to find perpendicular vector to another vector?
- How to find mean and median from histogram
- How many sides does a circle have?
Among bundles over $M$, $TM$ is somewhat special in that vector fields correspond to differential equations on $M$, while there is no such correspondence for sections of other bundles.
There is, however, a generalization of the notion of a generator of a Lie group action. The space of vector fields $\mathfrak{X}M$ is a Lie algebra under Lie brackets, and a Lie algebra homomorphism $\varphi:\mathfrak{g}\to\mathfrak{X}M$ can, under certain conditions, give rise to a Lie group action $\theta:G\times M\to M$ where $G$ is a Lie group whose lie algebra is $\mathfrak{g}$. In this way, Lie group actions correspond not with sections of $TM$, but with Lie subalgebras of $\mathfrak{X}M$. In the case of an $\mathbb{R}$-action generated by a vector field $V$, the relevant lie algebra morphism is $t\mapsto tV$.