Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Correspondence between one-parameter subgroups of $G$ and $T_eG$

+ 2 like - 0 dislike
1968 views

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

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

and

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

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

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

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 $n=1$?)


This post imported from StackExchange Physics at 2014-10-11 09:48 (UTC), posted by SE-user Victor Vahidi Motti

asked Oct 10, 2014 in Mathematics by Victor Vahidi Motti (20 points) [ revision history ]
recategorized Oct 11, 2014 by dimension10

1 Answer

+ 4 like - 0 dislike

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.


This post imported from StackExchange Physics at 2014-10-11 09:48 (UTC), posted by SE-user doetoe

answered Oct 10, 2014 by doetoe (125 points) [ revision history ]
edited Oct 11, 2014 by doetoe
Thanks a lot @doetoe. All clear now. Can you explain more your last line. How to adapt the construction for other non-zero elements?

This post imported from StackExchange Physics at 2014-10-11 09:48 (UTC), posted by SE-user Victor Vahidi Motti

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...