# Confusion about what is meant by a metric being not admissible on a Riemannian manifold

Definition A Riemannian metric $$g$$ on a smooth manifold $$M$$ is a smoothly chosen inner product $$g_x:T_xM\times T_xM\to\mathbb{R}$$ on each of the tangent spaces $$T_xM$$ of $$M$$.

(taken from About the definition of a Riemannian metric)

The definition above suggests that one is free to choose an arbitrary (symmetric, positive definite) tensor as metric tensor.

Edit: I was confused about what it means for a given metric not to be admissible after reading Conditions for a given manifold to admit a given metric? Would that mean that certain choices of the tensor, yet conformant to the definition above, would not work or be permitted as a Riemannian metric? The rest of this question in original formulation must be understood under this assumption. (By the comments it is apparent that this assumption is actually false.)

First bunch of questions:

If we just take a "wrong" metric tensor and go with it, I assume that some "terrible things" will happen. What exactly are these terrible things? What inconsistent phenomena would occur? Would it break some of the metric axioms? Would the chart maps fail to have sufficiently differentiable or even discontinuous transition maps? (Geodesic on a Riemannian manifold with a random metric tensor, The space of Riemannian metrics on a given manifold. and Existence of a Riemannian metric inducing a given distance. are somewhat related to this question but not the same, e.g. special cases)

Edit: In the original text I had various misconceptions in the example below.

As a prototype example I imagine a sphere where we force an "everywhere-identity-matrix" metric tensor onto it.

Edit: What I meant is: Viewed from a chart one gets a metric tensor. Can we modify the metric on the manifold such that it is the standard scalar product ("identity matrix") on the chart. Sure, I know we can't because the sphere is not flat. The question is, what goes wrong if we just do so, and push it to the manifold.

I think we just would not be able to move it smoothly through the transition maps if we did. So here the soft question remains whether this is correct. Or would it break in a different way?

Edit: The remaining part of the question is obsolete now. I deleted it because it really does not make sense any more.

Some context:

I am asking this because at work we are machine-learning the metric tensor of a state space from exploration data and I am trying to improve mathematical justification of the method. If I model the state space as a Riemannian manifold I am able to show that if the exploration is free of certain biases the learned metric tensor indeed converges over time to a metric tensor that induces the standard scalar product under an isometric embedding into Euclidean space. I suppose this is a justification. However, I would like to understand the issues one has to expect if the learning did not sufficiently converge or is in fact not free of the mentioned biases.

• Any chance to explain the downvote? I know the question may be ill-posed and it is actually part of the question whether it is. I thought this would suffice as a disclaimer. Feb 18 '20 at 18:14
• As the definition suggests that one is free to choose an arbitrary (symmetric, positive definite) tensor as metric tensor. The issue is what properties do you want the metric to have? I did not downvote, but there are many things unclear in your post, for instance, what is a "location-independent scalar product"? Feb 18 '20 at 19:30
• Yes, sorry, I am right now busy with editing. I am now aware of the faults I think. Thank you for your reply! Basically it seems I misunderstood what is meant by "admit a given metric". Feb 18 '20 at 22:18
• "admit a given metric" is a meaningless expression. One can ask if a manifold admits a metric with some properties, say, a metric of positive curvature, etc. My suggestion is to pick up a differential geometry textbook and read few chapters. Feb 18 '20 at 22:21
• Part of your confusion may come from your assumption that a bilinear form is "the same thing" as a matrix. If you pick a basis for your vector space, you can represent any bilinear form with a matrix, but there is not any canonical way to do this. In particular, given an abstract vector space, it is completely meaningless to talk about "the bilinear form corresponding to the identity matrix". Feb 18 '20 at 22:41