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I keep on asking about this book, and it is just that when I think I understand something, I look back and something else goes back to sounding obscure and unmotivated.

I have learned a lot from the many presentations online by Pavel Grinfeld, ranging from his talks on linear algebra to differential equations. So I am convinced there is a good explanation to his approach to teaching this topic.

This is the definition that I have trouble with:

5.7 The Contravariant Basis

The contravariant basis $\bf Z^i$ is defined as

$${\bf Z}^i = Z^{ij}{\bf Z}_j$$

From the book:

enter image description here

The ${\bf Z}_j$ expression denotes the covariant basis of the tangent space at a point, aka $\partial_j,$ while $Z^{ij}=g^{ij}$ is the contravariant metric tensor.

From the book:

enter image description here

He considers the contravariant basis ${\bf Z}^{i}$ to be in $T_pM$ (my interpretation) and mutually orthogonal to ${\bf Z}_{j},$ i.e. $${\bf Z}_{j}{\bf Z}^{i}=\delta_j^i$$

From the book:

enter image description here

The metric tensor matrix $Z_{ij}$ is the result of the dot products ${\bf Z}_{i}\cdot {\bf Z}_j$

From the book:

enter image description here

and $Z^{ij}$ simply its inverse.

First question is whether it makes any sense to then decompose these basis vectors ${\bf Z}_{j}$ into any components, since it seems like they are the most basic, or primitive elements in the construct, and they are naturally linearly independent of each other. On a lecture on the metric and curvature of the sphere he derives the metric tensor very nicely from a purely spatial or geometric reasoning on the changes of the basis vectors tangential to azimuthal and polar coordinates.

If the answer, is 'no, it doesn't make sense to try and decompose these ${\bf Z}_{j}$ basis vectors into components, what is really going on in the definition of contravariant basis vectors above?

The metric tensor (or its inverse) are just scalars, so is the operation tantamount to a linear transformation of the vector ${\bf Z}_{j}$ in proportional to the dot products in the metric tensor on a given row of the metric tensor matrix? But how is this carried out on a concrete example?

Antoni Parellada
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    What in the world is he doing writing $Z^{ij}$ for $g^{ij}$?! This is wrong and beyond misleading. Saying mutually orthogonal is also ridiculous and wrong. These are dual basis vectors. He has a superb pedigree, so I find this alarming. The covariant basis vectors live in the dual space, so cannot be a linear combination of the contravariant basis vectors. Either you are misinterpreting or he is.making a mess of it all. – Ted Shifrin Jun 17 '22 at 03:01
  • @TedShifrin I included now screen shots of pertinent definitions in the OP to hopefully help understand his teaching technique or notation. – Antoni Parellada Jun 17 '22 at 03:21
  • @TedShifrin This minute of his talk: here. – Antoni Parellada Jun 17 '22 at 03:25
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    @TedShifrin yup, he calls $\frac{\partial}{\partial x^i}$ ($=\mathbf{Z}_i$) the covariant basis, and $g^{ij}\frac{\partial}{\partial x^j}$ ($=\mathbf{Z}^i$)the contravariant basis vectors, all the while remaining within the tangent space (or $\Bbb{R}^n$ in his case). Also to OP: I think I've written 2 (or maybe more) answers about similar issues recently from this text by now, so perhaps take a look at them? And regarding the question about decomposing the basis vectors: you can always write one set of basis vectors as linear combinations of others, but here I'm not sure why you would. – peek-a-boo Jun 17 '22 at 03:59
  • @peek-a-boo Your answers are great. In fact I think I sent you a +100 bounty, but I don't know whether you saw it. You will understand that it is not easy to just reconcile disparate concepts based on a few excellent paragraphs, and that I still believe PG has to make sense at some level. All his material is great. – Antoni Parellada Jun 17 '22 at 04:04
  • @peek-a-boo This is your answer I bountied: https://math.stackexchange.com/a/4462558/994433 – Antoni Parellada Jun 17 '22 at 04:06
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    oh, I didn't realize it was you who asked the question, nor that you awarded a bounty (thank you btw). I can definitely understand things not always being crystal clear, but this book is literally the only one I've seen that uses the metric to associate a new set of basis vectors within the tangent space, and I absolutely do not see the point of doing this. The standard math books use the metric to go back and forth between the tangent and cotangent space. Old-timey physics books talk all about raising and lowering the components of a vector/tensor, but I've never seen this before. – peek-a-boo Jun 17 '22 at 04:13
  • @peek-a-boo I keep on thinking about the analogy of column vectors (as contravariant vectors) and row vectors (as covariant or the dual) to understand what he is trying to teach. You can dot a row vector with a column vector. But again, this would apply to components. Hard to see with basis vectors. – Antoni Parellada Jun 17 '22 at 04:45
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    The components of the tensor $\mathbf{Z}_j$ are tensors, not scalars / numbers /functions. Further, you need to stop referring to a "dual space" or $T_p^*$ as this is misleading and confusing. Grinfeld is using a pretty common biorthogonal system for $\mathbb{R}^n$, see https://en.wikipedia.org/wiki/Dual_basis. For example, variants of this setup is used in fluid- and material mechanics. So no forms, no manifolds, no tangent spaces.This setup is easily extended to a more "advanced" setup with minior changes to the definitions. – ContraKinta Jun 17 '22 at 12:47
  • @ContraKinta Can you elaborate on what you mean by the first sentence. They are basis vectors. Invariants, yes, and hence tensors, but... – Antoni Parellada Jun 17 '22 at 12:51
  • The object $\mathbf{Z}_j$ transforms like a tensor, hence it is. Cogito ergo Tensor. Grinfeld probably shows this by attaching a base. – ContraKinta Jun 17 '22 at 12:54
  • @peek-a-boo Re: decomposing the basis vectors, my issue is that I imagine the ${\bf Z}^i=Z^{ij}{\bf Z}_j.$ – Antoni Parellada Jun 17 '22 at 12:55
  • Grinfeld says in the first lecture itself that his goal is cut through the formalism , this also means "hacking" in the sense that what he would be doing doesn't make sense but gives right answer for case he consider – Clemens Bartholdy Jun 17 '22 at 15:51
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    Ivo Terek has the most wonderful notes on tensors (and also on a lot of other good stuff) which explain things much better and are waaaay, way more clear than this. I'm gonna sound a little rude here but I'd suggest you stop trying to make sense of these absolute messes physicists make and try to learn from the perspective of actual mathematicians. John Lee and O'Neill also explain these things very weill in their Riemmanian (and semi-Riemannian) geometry books. Here's the link for Ivo's notes: https://www.asc.ohio-state.edu/terekcouto.1/texts/tensors.pdf – Matheus Andrade Jun 17 '22 at 23:51

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