The pullback metric under scaling

February 24, 2012

The pullback metric under scaling

Filed under: Uncategorized — Ngô Quốc Anh @ 20:11

Today, let us discuss a very simple question. Assume that $(M,g)$ is a Riemannian manifold and $\lambda>0$ is constant. We study the following so-called pullback metric $\lambda^\star g$ of $g$ under the scaling $x \mapsto \lambda x$ . Precisely, we aim to compare $\lambda^\star g$ and $g$ .

For simplicity, we follow definition of the pullback metric. Besides, by $\lambda$ we mean the map $x \mapsto \lambda x$ .

The pushforward $\lambda_\star$ by $\lambda$ . This is the first step. Suppose $X \in T_xM$ for some $x \in M$ . By definition, we have $\lambda_\star X \in T_{\lambda (x)}M$ which is just

$\displaystyle (\lambda_\star X) f = X ( f \circ \lambda), \quad \forall f \in C^\infty(M,\mathbb R).$

By the definition of derivation, in local coordinates, we clearly have

$\displaystyle X(f \circ \lambda ) = X(f \circ (\lambda {\rm id})) = \lambda X(f \circ {\rm id}) = \lambda X(f).$

The pullback metric $\lambda^\star g$ . By definition, we have

$\displaystyle \lambda^\star g (X, Y) = g(\lambda_\star X, \lambda_\star Y), \quad \forall X, Y \in T_xM.$

Obviously,

$\displaystyle g({\lambda _ \star }X,{\lambda _ \star }Y) = {\lambda ^2}g({\rm id}_*X,{\rm id}_ \star Y)= {\lambda ^2}g(X,Y).$

Thus, we can conclude that

$\displaystyle \lambda^\star g=\lambda^2 g.$

Comments (2)

2 Comments »

What do you mean by “the” map x \rightarrow \lambda x ? I don’t think this map is well defined (nor unique) on an arbitrary manifold. The best that I could think of is: fix a local coordinate system with image being the whole of \mathbb{R}, then define a multiplication map on the domain of this local coordinate system via the local coordinate system. But I don’t see the interest.

Alternatively, you have to assume that their exist an action of R on your manifold. Your map is then dependent on fixing this action.

Comment by me — March 5, 2012 @ 17:39

Reply
- Thanks for your interest in my post. I agree with you that the whole topic is somewhat ridiculous and probably doesn’t make any sense. Besides, thanks for your suggestion, I think that should be the best and that matches my thinking.
  
  At this moment, I am interested in some connection between $(M, f^\star g)$ and $(N, g)$ where $f : M \to N$ is a smooth map. In other words, how much we understand the pullback metric $f^\star g$ . But, I guess this is far way from what I really need and what I really can. I just need, as you pointed out, to deal with the case $f$ is the local coordinate map, i.e., $(N,g)\equiv (\mathbb R^n, ds^2)$ . To be precise, $f \equiv \exp_p$ .
  
  If I have time, I will write a note where I make use of this trivial stuff.
  
  Comment by Ngô Quốc Anh — March 5, 2012 @ 18:00
  
  Reply

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Ngô Quốc Anh

February 24, 2012