# Equivalence between vector field and generator of a group of translations

I've been reading Olver's Applications of Lie groups to differential equations and did not understand the excerp where the autor explains that "[...] every nonvanishing vector field is locally equivalent to the infinitesimal generator of a group of translations."

In fact, I did not understand what is the relation between this affirmation and the Proposition 1.29.

If anyone there is reading this book or knows how to answer to this question, please I would appreciate very much your help.

## 1 Answer

There are two points ( and I think this is already in Olver on that page, so I'm not sure how much I'm going to help you here )

1. if a vector field $X$ has $X(p) \neq 0$ then there exists some system of coordinates for which $X = \frac{\partial}{\partial x^1}$ on some neighborhood of $p$. This is the very unexciting cannonical form for a non-vanishing vector field. I think Olver is nice to read on all this in his other text Equivalence, Invariants, and Symmetry. Chapter 1 is a great overview of manifold theory.
2. The first coordinate derivative is a vector which generates a translation. As described, the exponential of $X$ which when acting on the coordinate function $x^j$ either does nothing ($j \neq 1$) or translates $x^1 \mapsto x^1+a$. Formally, $$exp(aX)x^1 = (I+aX+\frac{1}{2}a^2X^2+ \cdots )x^1 = x^1+a$$ as $X(x^1)=1$ so all the higher derivatives vanish.

It's a local equivalence because the result that $X$ can be written as the first coordinate derivative is only possible in some neighborhood. There is much more to say about all this, I hope you continue asking questions...

• Thanks, Cook. Now I'm seeing the other Olver's book ( where it is better ). Well, I was wondering if is there any ( or could exist ) condition where we could find that a vector field could generate a rotational flow, for example... I mean, I'm not sure about it; I don't' even know if this question makes any sense. – Poli Tolstov Jun 4 '14 at 12:54
• @Poli yes, it is possible, that is hidden in the change of coordinates. I think the example to think about would be $x\partial_y-y\partial_x$ which is up to a scalar multiple $\partial_{\theta}$. In the polar coordinates that vector field generates a translation in the angular coordinate whereas in $x,y$ it is what we would call a rotation. That's the trick, the symmetry becomes a translation whilst the non-linear aspect is tucked away in the coordinate map. – James S. Cook Jun 4 '14 at 15:08
• About the Theorem (on the book you mentioned): there he says that you consider a point $x$ and in this point the vector field $\bf{v}$ is nonvanishing. After, he says that in a neighborhood of this point there is another point $y$ that is represented by a local coordinate system where we can write $y=(y^{1},...,y^{m})$, right? This vector field $\bf{v}$ can be written in terms of this new local coordinates $\{{y^{i}}\}$ finding that its "face" is $\bf{v}={\partial}/{\partial}y^{1}$... I've other question: what is the necessity of knowing that the $\bf{v}$ is nonvanishing in a point $x$? – Poli Tolstov Jun 4 '14 at 15:11
• Well, I suppose the point $y$ and the coordinate map $y$ ought to be distinguished, there might be a slight abuse of notation. Notational quibbles aside, the point it that if we follow his flow-construction then the $y$ coordinate system reproduced $v$ as $\partial/\partial y^1$ at the point where $v$ is non-vanishing. If $v$ were zero then that wouldn't allow the flow I think. Oh, the example I mentioned in the last comment seems to be on page 29 of the Yellow Olver book. – James S. Cook Jun 4 '14 at 15:19
• So the form of this new vector field (in terms of this new coord. system) is $v={\partial}/{\partial}y^1$ in a neighborhood of $y$ that includes the point $x$? I am asking this because in my former question I was not worried about the 'nonvanishing' fact, but that, in first place, we consider the vector field evaluated in a initial point $x$. If we consider that the necessity of the initial evaluation of the $v$ in one point is due to the fact we need to guarantee the new $v$ is the same even in a neighborhood where it is defined, I think the answer for the 1st question (this comment) is yes. – Poli Tolstov Jun 4 '14 at 15:43