Correspondence between one-parameter subgroup and left-invariant vector field.

Question

Given a one-parameter subgroup of a Lie group $G$, we can show that the one-parameter subgroup $\phi : \mathbb{R} \to G$ defines a unique left-invariant vector field $X: \frac{d\phi^\mu(t)}{dt} = X^\mu(\phi(t))$.

However, we know that even in an arbitrary manifold, $M$, given an arbitrary vector field $V$, we can get an one-parameter transformation $\psi(t,x_0): \mathbb{R} \to M$, an integral curve of $V$ going through the point $x_0 \in M$ at $t=0$.

But this one-parameter transformation through $x_0$ does not uniquely define a vector field everywhere on the manifold $M$ (at least from what I think), whereas it does in the case of Lie group $G$.

However, from the proof that I am looking at, I cannot see any special role of the Lie structure of $G$, in fact, the book does not even say $G$ is a lie group so I am not sure if the proof is in fact only applicable to Lie group (the proof does use the Lie structure of $\mathbb{R}$.) However, I do doubt that the proof is applicable to any arbitrary manifold, since I think that an integral curve through $x_0$ only uniquely defines the vectors along the curve, but not at all points on the manifold.

Can someone help me understand this whole confusion better?

Answer 1

Any vector field on a general manifold corresponds to a first order ordinary differential equation on the manifold. Its solution is well known to have group property, and defines a local one-parameter group action: $$ \phi:W\subset\mathbb R\times M\to M $$ that is, $$ \begin{split} \phi(0,x)&=x, \forall x\in M\\ \phi(s,\phi(t,x))&=\phi(s+t,x), (s+t,x)\in W \end{split} $$ and the domain $W$ is an open subset of $\mathbb R\times M$ that contains $\{0\}\times M$.

The same situation applies to a Lie group, but the result is more beautiful, since the domain $W$ is all of $\mathbb R\times G$ for a left-invariant vector field (i.e., it admits a global one-parameter group action). Your one-parameter subgroup is actually the integral curve that passes through the identity element, say $e$ of $G$: $$ \phi':\mathbb R\to G, t\mapsto\phi(t,e) $$ So I guess you need to be clearly aware of the difference and connection between one-parameter group action and one-parameter subgroup.

Answer 2

I don't quite understand your definition of the left-invariant vector field (notably you didn't explain what $\mu$ is and how it can be used as exponent both to $\phi$ and to $X$). However a basic description of the situation is not hard to grasp.

A one-parameter subgroup $\phi$ is a parametrised curve in $G$ that passes through the identity element $e\in G$ (at $t=0$). It has a tangent vector at $e$, which is an element of the tangent space at $e$. The requirements for a one-parameter subgroup make that $\phi$ is entirely determined by this tangent vector, and can be recovered by exponentiation; this means the set of one-parameter subgroups is naturally in bijection with the tangent space of $G$ at $e$.

Now describing a vector field means giving a tangent vector at every point of $G$, whereas $\phi$ only gives such vector on the points it passes through. But a Lie group, as opposed to a manifold, has an intrinsic symmetry: it acts on itself by left-multiplication, and this action is transitive (any chosen point $a$ can be sent to another chose point $b$ by an appropriate group element acting). In fact the action is simply transitive: there is exactly one group element that gets you from $a$ to $b$. In particular, for any point $g\in G$ there is exactly one group element whose action sends $e$ to $g$; not surprisingly, that group element in $g$ itself. Now if one gives a tangent vector $X_e$ at $e$, one can use the tangent map of the action (left-multiplication by $g$) to obtain a tangent vector $X_g$ at $g$. Doing so for all $g\in G$ gives a vector field $g\mapsto X_g$ on $G$. Using the fact that left-multiplication is an action, one easily checks that this field is left-invariant (i.e., the tangent map to the left-multiplication that sends $a$ to $b$ sends $X_a$ to $X_b$). Also all left-invariant vector fields are like this, since the way we obtained $X_g$ from $X_e$ was just an instance of the left-invariance requirement. In summary, the tangent space of $G$ at $e$ is naturally in bijection with the space of left-invariant vector fields on $G$. Composing with the earlier bijection, the latter is also in bijection with the set of one-parameter subgroups of$~G$.

(One checks easily that, for those points of $G$ through which the curve passes, the vector field associated to a one-parameter subgroups coincides with the tangent vectors of the curve.)

What differs in the case of a manifold is that there is no given simply transitive action to transport vectors by. This means the is no notion of (left-)invariance, and consequently it is not possible to go from a set of vectors defined only on the points of the curve to a vector field defined everywhere.