This is a very simple question that I got confused. Is Riemannian metric a symmetric 2-tensor or symmetric 2-tensor field? Wikipedia says that it is a (0,2) tensor but my book says it is a tensor field. Are these things the same? What am I missing? (By the way, if it is a tensor field, I do not know what a symmetric tensor field is.)
Riemannian metric
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differential-geometry
1 Answers
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It's a symmetric 2 tensor field(*) on the manifold (let's say, $M$), and a symmetric 2 tensor when restricted to to the tangent space $T_pM$ for some $p \in M$. But people are often sloppy with this distinction.
Symmetry just means that (if the metric is denoted by $g$) $g(p)(v,w) = g(p)(w,v) \quad \forall p \in M, v,w \in T_pM$ which in local coordinates is equivalent to the statement that $g_{ij} dx^i\otimes dx^j = g_{ji} dx^i\otimes dx^j $
(*) more precisely: "(0,2) tensor field" or "covariant tensor field"
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0covariant, sorry, this was wrong first. – 2011-12-25