Ngô Quốc Anh

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.

2 Comments »

  1. 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

    • 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


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