Classical tensor analysis and Tensors on Manifolds

Question

I learned tensors the bad way (Cartesian first, then curvilinear coordinate systems assuming a Euclidean background) and realize that I am in very bad shape trying to (finally) learn tensors on manifolds.

In particular, I am struggling to understand which properties of "classical" tensor algebra / analysis carry over into tensors on manifolds. By "classical" I mean the the one that assumes a background 3D Euclidean space and allows general curvilinear coordinate systems in it.

1) For instance, I have seen equations of the following sort in physics-type books $$ \mathbf{v} = v^i \mathbf{e_i} = v_i \mathbf{e^{*i}} $$ The argument is that if a non-degenerate metric tensor exists, it also has an inverse, i.e. $$ g^{ij} \equiv g_{ij}^{-1} \text{ and } g_{ij} g^{jk} \equiv \delta_i^k\\ v^i \equiv g^{ij} v_j $$ and thus $v_j$ are the "covariant" components of the vector $\mathbf{v}$.

My question is: Can the equation $$ \mathbf{v} = v^i \mathbf{e_i} = v_i \mathbf{e^{*i}} $$ be made meaningful for tensors (at each point) on a manifold by proceeding in a similar way? I suspect yes but I am not sure.

One could introduce a (0,2) metric tensor on the manifold. But is its "inverse" automatically a (2,0) tensor as in the "classical" case? Also, which metric? I understand they are not unique?

3) Is there some good resource (a book perhaps) which clearly helps one to transition to tensors on manifolds?

Thank you.

Answer 1

1), 2): yes. On Manifolds you can introduce a metric, which makes the manifold a Riemmannian manifold (assuming the metric is positive definite). A metric, by definition, is a (0,2) tensor field which defines a scalar product on the tangent bundle, i.e. the metric in a point $p$ is a scalar product on the tangent space at $p$. This, in local coordinates, has a representation which is usually denoted $(g_{ij})$ and corresponds to what you called the metric tensor.

If the metric is positiv definite, this matrix representation is invertible and as you wrote, the inverse is usually denoted $(g^{ij})$. The raising and lowering of indices (making contravariant tensors covariant and vice versa) works the same way you wrote it down. This is nothing but the fact that on a Euclidean vector space $E$ there is a natural isomorphism between the vector space and it's dual, induced by the metric (i.e. if $v$ is a vector, $w\mapsto \langle v, w \rangle$ is a linear map on $E$ and each linear map arises that way).

As for 3), most books on Riemannian geometry should do the job. Which one suits you best depends on you. A very comprehensive description of these things is to be found in Spivaks treatise 'A comprehensive introduction to Differential Geometry' ;-)

Answer 2

Given your background, it is useful to remember that a manifold is a space that is almost Euclidean in the neigbourhood of each point. This means that locally tensor calculus on manifolds is not that different to working with curvilinear coordinates on euclidean spaces and most of your intuitions from working with curvilinear coordinates should carry over. In particular dual bases, index manipulation and metric tensors work the same.

However you generally may not have a globally consistent coordinate system for an entire manifold, but this is similar to working with polar or spherical coordinates that have critical points at the origin. Also much of the algebraic work can be done from using abstract index notation, where you do not actually commit to a coordinate system until you need numerical answers.

The other main difference is curvature, ie $$\nabla_i\nabla_j X^l \neq \nabla_j\nabla_i X^l$$ unlike in Euclidean spaces. The main new element then is the Riemann curvature tensor, $R$, defined by $$ R_{ij}{}^k{}_l X^l \triangleq 2\, \nabla_{[i} \nabla_{j]}X^k = \nabla_{i}\nabla_{j}X^k -\nabla_{j} \nabla_{i}X^k$$ and the derived Ricci and scalar curvatures, which are all zero in Euclidean spaces regardless of the coordinate system.

To get a mental picture of how tensors on manifolds work I really recommend ch 14 of Penrose's Road to Reality.