I am reading the proof of this theorem from Andreas Arvanitoyeorgos and I cannot get some points in it, highlighted below.

Theorem. The map $\phi \to d\phi_0(1)$ defines a one-to-one correspondence between one-parameter subgroups of $G$ and $T_eG$.

Proof. Let $v \in T_eG$ and $X^v_g=(dL_g)_e(v)$ be (the value of) the corresponding left-invariant vector field.

We need to find a smooth homomorphism $\phi_v: \mathbb R \to G$.

Let $\phi: I \to G$ be the unique integral curve of $X^v$ such that $\phi(0)=e$ and $d\phi_t=X^v_{\phi(t)}$.

This curve is a homomorphism because if we fix an $s \in I$ such that $s+t \in I$ for all $t \in I$ then the curves

$$t \to \phi(s+t)$$


$$t \to \phi(s)\phi(t)$$

satisfy the previous equation (the second curve by the left-invariance of $X^v$), and take the common value $\phi(s)$ when $t=0$. By uniqueness of the integral curve then


where $s,t \in I$.

Extend to $\mathbb R$ and define $\phi_v(t)=\phi(t/n)^n$.

The map $v \to \phi_v$ is the inverse of $\phi \to d\phi_0(1)$ and this completes the proof.

So my questions are these:

  1. We need to find a smooth homomorphism $\phi_v: \mathbb R \to G$. (Why?)

    In particular this part is not clear to me: The map $v \to \phi_v$ is the inverse of $\phi \to d\phi_0(1)$ and this completes the proof.

  2. The map $t \to \phi(s)\phi(t)$ satisfies the integral equation of the vector field by the left-invariance of $X^v$ (Why?)

    I see that given the left-invariant vector field we can have $dL_{\phi(s)}X^v_{\phi(t)}=X^v_{\phi(s)\phi(t)}$ and $d(\phi(s)\phi(t))=Y_{\phi(s)\phi(t)}$

    But I cannot get it that $Y_{\phi(s)\phi(t)}=X^v_{\phi(s)\phi(t)}$ so that we have $d(\phi(s)\phi(t))=X^v_{\phi(s)\phi(t)}$ to say that $\phi(s)\phi(t)$ is the same integral curve.

  3. (What 1 in $d\phi_0(1)$ here refer to? Is it $n=1$?)


1) One way to show that a map constitutes a 1-1 correspondence is by showing that it has an inverse. This is what is done here: elements of $T_eG$ are tangent vectors to $G$ at the unit element, and one-parameter subgroups of $G$ are smooth homomorphisms $\mathbb R\to G$. Note that this is a notion more general than a closed 1-dimensional subgroup, for which the theorem wouldn't hold. The inverse should map elements of $T_eG$ to 1-parameter subgroups.

2) An integral curve is a curve that is everywhere tangent to a given vector field. From existence and unicity theorems of ordinary differential equations we get that through every point there is a unique integral curve. $\phi$ is the unique integral curve that passes through $e$ at $t = 0$, so $\phi(0) = e$ and $\dot\phi(0) \equiv d\phi_0 = X^v_e$. If instead we denote this curve $\Phi^e$, and more generally $\Phi^g$ for the unique integral curve that passes through $g$ at $t = 0$, then clearly we have $\Phi^{\phi(s)}(t) = \phi(s + t)$. The translated curve $g\phi$ passing through $g$ at $t = 0$ is an integral curve of the translated vector field, which is the same vector field by translation invariance, so it must be the unique integral curve passing through $g$. Now take $g = \phi(s)$.

3) In physics this would often just be written $\dot\phi(0)$, but strictly speaking, since $\phi$ is a smooth map $\mathbb R\to G$, its derivative in 0 is a map from the tangent space of $\mathbb R$ at 0, which is $\mathbb R$ again, to the tangent bundle of $G$ at $\phi(0) = e$, i.e. $T_eG$. Evaluating this in $1\in\mathbb R$ (or if you prefer $1\in T_0\mathbb R$) gives an element of $T_eG$. You could take any other fixed non-zero element, if you adapt your construction of $\phi_v$ accordingly.

  • $\begingroup$ Thanks a lot @doetoe. All clear now. Can you explain more your last line. How to adapt the construction for other non-zero elements? $\endgroup$ – user56963 Oct 11 '14 at 6:25
  • $\begingroup$ @VictorVahidiMotti I corrected a small typo: where I wrote $\dot\phi(1)$ I should have said $\dot\phi(0)$. More generally, $d\phi_0(\alpha)$ corresponds to $\dot\phi(0)\alpha$. If you take such an $\alpha\ne1$ you can construct $\phi_v$ as before, only now the inverse of the map is given by the map $v\mapsto \phi_{\alpha^{-1}v}$ $\endgroup$ – doetoe Oct 11 '14 at 21:09
  • $\begingroup$ Thanks again. Two related questions. So 1 here refers to the tangent vector at point 0 in R. But so is any other real number because 1 is the basis of the tangent space of R at 0, right? And we define the differential as a smooth function between tangent spaces of manifolds, I mean the definition of the differential of a one-parameter subgroup in its general Lie group sense. I see that for the one-parameter subgroup one can use the notion of differential as defined in calculus but again we take differential as defined in Lie groups, a function between two tangent spaces, right? $\endgroup$ – user56963 Oct 12 '14 at 13:23
  • $\begingroup$ @VictorVahidiMotti The first question: correct. $\endgroup$ – doetoe Oct 12 '14 at 13:57
  • $\begingroup$ The second: the total derivative of a map $\Phi:X\to Y$ between manifolds/Lie groups at a point $x\in X$ is a linear map $D_x\Phi$ of $d\Phi_x$ from the tangent space at $x$ to the tangent space at the image point $\Phi(x)$. The total derivative $D\Phi$ itself is a smooth function between what are called the tangent bundles, which can equivalently be seen as a smooth map on $X$ mapping $x$ to this linear map, which gives the best linear approximation to $\Phi$ at $x$. The latter is the common definition in calculus, where $X$ and $Y$ are $\Bbb R^n$ and $\Bbb R^m$, as are their tangent spaces. $\endgroup$ – doetoe Oct 12 '14 at 13:58

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