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I am trying to construct a particular manifold locally using a metric, Can I simply take the inner product of my basis vectors to first achieve some metric.
Let me shorten it a little further, can I take the inner product of basis vectors and form a suitable metric?What do you have and where do you want to go to? Where are your basis vectors from? What is the space?
Your question is far too vague to be answered.
Inner product is scalarYes, if the inner product is one, i.e. positive definite, then it defines angles, lengths and distance. However, manifolds are usually not flat, which is why I wondered what your inner product is.
Sorry, but it is completely unclear what you mean by this. You have still not answered the question in #2: What do you have and where do you want to go?Let me shorten it a little further, can I take the inner product of basis vectors and form a suitable metric?
Assume I am in some surface of a differentiable manifold embedded in a 3 dimensional space. Given that I have some covariant basis vectors along the surface, is it as simple as taking the inner product of basis along the surface to construct a metric? How do you describe the surface in terms of the ambient 3d space using normal surface vectors?Sorry, but it is completely unclear what you mean by this. You have still not answered the question in #2: What do you have and where do you want to go?
Well forget about the ambient space. What would be the best way of finding a metric if you could only work with the basis vectors of the manifold?You don't have covariant basis vectors of the surface, only of its tangent space which is a different one at each point. If you want to transport the metric onto the surface, you can either use the outer metric, since it is an embedded surface, or line integrals on the surface. See also connections. The outer metric is more or less worthless, as it does not measure an actual distance, which would be a geodesic.