What are Good Books on Tensors for Understanding Einstein's Field Equation?

[Alaindevos]

I'm looking for good books on Tensors.
I have "Introduction to Tensor Analysis and the Calculus of Moving Surfaces" from Pavel Grinfeld.
But i look for others.

[Mentor Note: Thread moved from the Relativity forum]

 

[Orodruin]

I’d say something, but I am biased.

 

[berkeman]

Then I'll say it...

https://www.amazon.com/Mathematical...attias-ebook/dp/B086H3LMZF/?tag=pfamazon01-20

 

[Orodruin]

Then I'll say it...
View attachment 342904
https://www.amazon.com/Mathematical...attias-ebook/dp/B086H3LMZF/?tag=pfamazon01-20

Say, that cover looks a lot like half of my profile image … weird!

 

[WWGD]

I'm looking for good books on Tensors.
I have "Introduction to Tensor Analysis and the Calculus of Moving Surfaces" from Pavel Grinfeld.
But i look for others.

Maybe you can tell us the perspective you're interested in: Math, Engineering, Physics /Relativity?

 

[Orodruin]

Maybe you can tell us the perspective you're interested in: Math, Engineering, Physics /Relativity?

If I was not hallucinating, this was originally posted in the relativity forum. That would indicate a physics perspective.

 

[WWGD]

If I was not hallucinating, this was originally posted in the relativity forum. That would indicate a physics perspective.

It only seems to be under "Science and Math textbooks".

 

[Orodruin]

It only seems to be under "Science and Math textbooks".

Yes, it was moved.

 

[WWGD]

Yes, it was moved.

And you liked to move it, move it.

 

[Orodruin]

And you liked to move it, move it.

Oh, I don’t have those kinds of powers any more …

But back to topic.

 

[berkeman]

If I was not hallucinating, this was originally posted in the relativity forum. That would indicate a physics perspective.

Good point; I did the move, and probably should have added a note to provide more context to the question. I'll add that into the OP now.

 

[berkeman]

And you liked to move it, move it.

I've tried several ways to parse this, but no joy so far. Not reverse-Polish, not Yoda-speak, not a single character typo that I can find... I'm getting dizzy.

 

[WWGD]

I've tried several ways to parse this, but no joy so far. Not reverse-Polish, not Yoda-speak, not a single character typo that I can find... I'm getting dizzy.

You and my therapist, who's jumped out of his window a few times, coincidentally in the middle of our sessions.

 

[berkeman]

You and my therapist, who's jumped out of his window a few times, coincidentally in the middle of our sessions.

But you know it's a first floor office, so you've never been impressed by this therapy tactic...

 

[WWGD]

Still, for anything related to Tensors from an Algebraic perspective, Keith Conrad from University of Connecticut has some great popularization
His Web Page:

https://kconrad.math.uconn.edu/

His expository papers:

https://kconrad.math.uconn.edu/blurbs/

 

[Alaindevos]

Some context. I just want to understand "Einsteins Field equation".
But as it is formulated in the form of tensors, i need to understand tensors.
Or abstract index notation ?
[ PS : I had vector calculus at University, but that's a different beast as the space was Euclidean]

 

[WWGD]

Some context. I just want to understand "Einsteins Field equation".
But as it is formulated in the form of tensors, i need to understand tensors.
Or abstract index notation ?
[ PS : I had vector calculus at University, but that's a different beast as the space was Euclidean]

John Lee's book on Riemannian Geometry may also be helpful .https://link.springer.com/book/10.1007/978-3-319-91755-9

 

[Ibix]

Some context. I just want to understand "Einsteins Field equation".
But as it is formulated in the form of tensors, i need to understand tensors.
Or abstract index notation ?
[ PS : I had vector calculus at University, but that's a different beast as the space was Euclidean]

In that case, almost any introductory GR book will provide enough of a grounding for you. I learned from Sean Carroll's online lecture notes originally, but you probably want to look at multiple sources. That one has the advantage of being free.

 

[Orodruin]

Some context. I just want to understand "Einsteins Field equation".
But as it is formulated in the form of tensors, i need to understand tensors.
Or abstract index notation ?

Abstract index notation is just a notational convention within tensor calculus/differential geometry. It is not a separate subject in itself.

[ PS : I had vector calculus at University, but that's a different beast as the space was Euclidean]

To quote Yoda:
"No! No different! Only different in your mind."

The way I like to introduce things like the covariant derivative, metric tensor, Christoffel symbols, etc., is to do it in Euclidean space using curvilinear coordinates. This makes the geometrical interpretation clearer and more grounded in what people are familiar with. It makes things a bit easier once you move on to general curved spaces to have that kind of intuition built. Then you can focus on the particulars like more general connections, curvature, etc.

 

[Alaindevos]

I currently own the book "Introduction to Riemannian Manifolds" (John Lee).
But it's above my head. I cannot read it. Understand all the symbols.
Nabla_underscore_X - nabla_underscore_Y - [X,Y]
The last i think is a commutator.
I need an introduction to this book ...

 

[Orodruin]

Understand all the symbols.

What do you mean by this? If you understand what all the symbols mean, getting the meaning should not be very difficult.

Note that you can also use the LaTeX features of the forum to post more readable equations, eg,

$$
T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]
$$

 

[robphy]

You might find some helpful notes among:
Notes by https://en.wikipedia.org/wiki/Robert_Geroch
https://home.uchicago.edu/~geroch/Course Notes
https://home.uchicago.edu/~geroch/

 

[Muu9]

 

[Sagittarius A-Star]

I'm looking for good books on Tensors.

Dirac's book "General Theory of Relativity" might be helpful. It has less that 70 pages.
https://archive.org/details/DiracGeneralTheoryRelativity/mode/2up

 

[mathwonk]

I am a mathematician, and to me, if you already know what the tangent bundle is, and its dual the cotangent bundle, and hence know what tangent and cotangent vector fields are, then the 20 page explanation in Spivak's chapter 4, of his Comprehensive introduction to differential geometry, is as clear as it gets. Still, it is demanding and difficult, because the subject is complicated.

Basically, for each vector space V, one can construct new spaces of functions from it. One has linear functions V-->k, where k is the real numbers, and one has "multi linear" functions Vx...xV-->k, which are linear in each variable separately. These are called (point wise) "k-tensors".

Then one takes families of these spaces. Namely, at each point p of space, one considers the vector space of tangent vectors based at p. This family of tangent spaces, one for each point p, gives the "tangent bundle" of the space. Correspondingly one has the k-fold tensor bundle, a family assigning the space of k tensors at p, to each point p.

A k-tensor "field" is a choice of a k-tensor at each point p of space, This defines a function on k-vector fields. I.e. if at each point p we have chosen k vectors, then at each point p, we let our chosen k-tensor operate on the chosen k-tuple of tangent vectors, obtaining a number.

This gives us a function from k-tuples of tangent vector fields to numbers, which is characterized by the fact that it is linear in each variable separately, over the smooth functions.

I.e. a function times a function is a function, and a function times a vector field is a vector field; and if you let your tensor field act on a k-tuple of vector fields, getting a function, then multiply by a given function, you get the same result as if you first multiplied any one of the vector fields by your function, and then let the tensor field act on the resulting k-tuple.

I.e. given a tensor field, a k-tuple of vector fields, and a function, you can combine them to get a function in two ways: first multiply the function by any one of the vector fields, and then act on the resulting k-tuple of vector fields by the tensor field; or let the tensor field act first on the original k-tuple of vector fields, and then multiply by the function. The result is the same function.

The main result about tensor fields is that any such operation on k-tuples of vector fields that is (multi) linear in that way over the functions, comes from a k-tensor field.

You see why this is complicated. But I recommend spivak.

The really complicated part is the notation, because you need to assign multiple indexes to bases of k-tuples of linear functions.

But the idea is just that, given k tangent vectors at each point, a tensor assigns a number at each point. I.e., a k-tensor field takes k-vector fields to functions, just as when k=1, a (dual, or co-) vector field takes a vector field to a function.

One can also make this sound more complicated by considering a k-tensor field and an m-vector field, together as a "mixed" (m,k) [or (k,m)?] tensor field. Then letting some components of the tensor field act on some components of the vector field, is called "contracting", since some pairs of tensor and vector components are replaced by numbers. Hence contracting one pair of components of an (m,k) field, gives a smaller, (m-1,k-1) field.

In these mixed fields, the tensor components are called covariant, and the vector components are called contravariant, because their local coordinate expressions transform differently under changes of coordinates, (presumably either by the Jacobian matrix of partials or its transpose?). (A quick look at Orodruin's insight article suggests these different types of components are distinguished by upper or lower indices, in local coordinates. Then I suspect "contraction" just means multiplying a coefficient with a given upper index times one with corresponding lower index.)

Apologies if this helps no one. The physicists here are certainly better sources, but I hoped to understand it better myself by writing this paragraph.

As to the physics, one presumably has some physical phenomena which see k-tuples of tangent vectors, gobble them up, and spit out numbers.

 

[mathwonk]

Well a quick look at Wikipedia shows that the output of a tensor can be more general than a function, e.g. the Riemann curvature tensor ( a slight elaboration on the formula in post #21) takes a triple of vector fields and outputs a vector field.

Namely the commutator of two vector fields is a vector field, and the covariant derivative of one vector field by another is a vector field. So given three vector fields, we can first differentiate the third by each of the first two, and then commute the results, or we can commute the first two and then differentiate the third wrt this commutator field. The results are the same iff the space is flat, i.e. not curved. I.e. space is locally euclidean iff the Riemann curvature tensor, the difference of these two procedures, is zero.

Presumably (?) one interesting aspect of this construction is that some of these individual operations are not linear over the functions, i.e. some of these derivative operations are not themselves "tensors", but the resulting combination of them is, i.e. is a tensor.

well I seem to have stated that wrong. The riemannian curvature seems to assign to two vector fields, the tensor that maps a third vector field to a vector field. I better let the experts, e.g. Orodruin, weigh in.

 

[mathwonk]

well I was enjoying the excerpt of Biennow on amazon https://www.amazon.com/Mathematical...attias-ebook/dp/B086H3LMZF/?tag=pfamazon01-20

until it ran out just starting tensors, but discovered that physicists like tensors so much they use them to represent objects that mathematicians think of more intrinsically. E.g. the first example given there of a tensor, p.70, is a mixed (1,1) tensor that is actually just a linear transformation. But since a linear transformation is represented in coordinates by a matrix, with two indices, it can also be thought of as a tensor of type (1,1). This uses the isomorphism of the tensor space VtensorV* with the space L(V,V) of linear transformations from V to V.

I.e. one can construct a linear transformation from V to V out of a linear function f from V to k (reals), plus a given vector w in V. I.e. one can send v to f(v).w. This would send all vectors to multiples of w, but by adding up several of these, one can get any linear transformation. I.e. every linear map from V to V has form f1w1 + f2w2 +....+fnwn. This is the linear map associated to the tensor f1(tens)w1+...+fn(tens)wn, under the map V(tens)V*-->L(V,V).

One reason this is somewhat unnatural to a mathematician is the fact that it works only in finite dimensions, i.e. not all linear transformations of an infinite dimensional V to itself have this special form. I.e. this means that to me, although using tensors to represent linear maps is notationally useful, it may not help as much to understand them. Of course physicists come equipped with a lot more understanding of what is happening than do mathematicians.

The second "example" is just a tensor product v(tens)w of two vectors, living in a tensor product space which is not described at all, except the formula 2.7 at the bottom of page 71, tells you that in such a product, scalars pull out of each entry separately, i.e. tensor product of vectors is "bilinear".

This property is actually the defining characteristic of a tensor product. I.e. it is easy to come away with the impression that a tensor product is something that looks like v(tens)w, or a sum of such, but the real meaning is that it is something satisfying equation (2.7). (Oh yes, it is also 'biadditive", i.e. (u+v)(tens)w = u(tens)w + v(tens)w. This is probably on p. 72.)

This book looks indeed interesting and helpful, but I have exhausted my free page limit.

 

[Orodruin]

This book looks indeed interesting and helpful, but I have exhausted my free page limit.

I think it should be pointed out that I wrote it with a physics student audience in mind. A mathematician might find it horrendously non-rigorous and the focus is more on trying to convey an understanding to how to use the concepts to model physics rather than being rigorous in the mathematical sense.

The riemannian curvature seems to assign to two vector fields, the tensor that maps a third vector field to a vector field. I better let the experts, e.g. Orodruin, weigh in.

Generally, yes, that would be the definition, which typically takes the form $R(X,Y) Z = \ldots$. The exact usage depends on the application. In some situations you might think of it as a map from 2 vectors to a (1,1) tensor. In others better as a map from 3 vectors to a vector, etc,

 

[mathwonk]

You are reminding me of how stunned I was when my professor asked why I was so sure, when given a function f(x), that x was the variable and f the function. Why not define x(f) to be f(x), and then f was the variable?

I.e. if F is the space of all k valued functions on the set X, then we have a pairing FxX-->k, taking the pair <f,x> to f(x). So we could consider this pairing as a mapping from X to functions from F to k, taking x to the function x:F-->k whose value at f is f(x).

So anytime we have a pairing XxYxZ-->k, we can consider it as a function from XxYxZ to k, or a function from X to functions from YxZ to k, or a function from XxY to functions from Z to k, or........

In particular, if V is a vector space and V* is its dual space of linear functions from V to k, we have a pairing VxV*-->k, so we can consider V as a space of linear functions from V* to k, i.e. there is a natural map V-->(V*)*. Here I contradict my earlier remark about naturality, since this natural map, which I usually think of as a natural isomorphism, is only isomorphic in finite dimensions.

So I guess I should think of the map V(tens)V*-->L(V,V) as equally natural, even though not always surjective.

Thank you!

 

[mathwonk]

Ah yes, proceeding from Orodriuin's guidance, we can see what kind of tensor the Riemann curvature should be:

We need the basic insight that the map VxV-->V(tens)V sending <v,w> to v(tens)w, is bilinear, and that thus composing with any linear map out of V(tens)V yields another bilinear map out of VxV. Indeed all bilinear maps out of VxV occur this way.
Hence Bil(VxV,W) ≈ L(V, L(V,W)) ≈ L(V(tens)V,W).
Moreover, we have seen that L(X,Y) ≈ X*(tens)Y, for all X,Y.

In particular, when W = k, we have [V(tens)V]* ≈ L(V(tens)V,k) ≈ Bil(VxV,k) ≈ L(V,L(V,k)) ≈ L(V,V*) ≈ V*(tens)V*.

Hence a tensor that sends a pair of vectors bilinearly to a linear map from vectors to vectors, belongs to L(V(tens)V, L(V,V)) ≈ [V(tens)V]* (tens)L(V,V)
≈ V* (tens)V* (tens) V* (tens) V.

Thus in local coordinates it is a linear combination of
dx^j (tens) dx^k (tens) dx^l (tens) ∂/∂x^i. i.e. is a tensor of type (3,1).

Since also V*(tens)V* (tens) V* (tens)V ≈ (V(tens)V(tens)V)* (tens)V ≈ L(V(tens)V(tens)V, V) ≈ Trilin(VxVxV, V), we can regard, as Orodruin said, our Riemann tensor, i.e. our element of V*(tens)V*(tens)V*(tens)V, as taking three vectors (trilinearly) to a vector.

so for computations, basic results seem to be that L(X,Y) ≈ X*(tens)Y, and (X(tens)Y)* ≈ X*(tens)Y*. then use that the dx^j are a basis for V* and ∂/∂xi are a basis for V, when V = R^n has coordinates x^j. and apparently tensors of type (m,n) have m elements from V* and n from V.

A simpler example, the dot product, i.,e,. a Riemannian metric, is a bilinear product from two vectors to a number, hence belongs to Bil(VxV,k) ≈ (V(tens)V)* ≈ V*(tens)V*, so is a tensor of type (2,0) and a linear combination of dx^j(tens)dx^I, in local coordinates.

[oops, I used "k" both for the real numbers and as an index.]

does this scan?

 

[mathwonk]

Ok, so I think of it this way now: in physics there are various operations on (fields of) vectors taking them linearly or multi-linearly to other (fields of) vectors, and it is useful for calculations to represent these operations as tensor fields, hence entirely in terms of tensor products of dx^j and ∂/∂x^I in local coordinates.

Moreover, as long as the value at a point of the output of the operation, depends only on the values of the inputs at this point, (which is not usually true for derivative operations), this can always be done, using these rules: (finite dimensions only!)

1) L(X,Y) ≈ X*(tens)Y, for all X,Y,
2) and (XtensY)* ≈ X*(tens)Y*.
3) Also, Bil(XxY, Z) ≈ L(X(tens)Y, Z).

Hence, Bil(XxY,Z) ≈ L(X(tens)Y ,Z) ≈
(XtensY)*(tens)Z ≈ X*(tens)Y*(tens)Z.

alternately,
Bil(XxY, Z) ≈ L(X, L(Y,Z)) ≈ X*(tens)L(Y,Z) ≈ X*(tens)Y*(tens)Z.

Similarly, Trilin(XxYxZ,W) ≈ X*(tens)Y*(tens)Z*(tens)W.
(the Riemann curvature tensor lives here.)

Remark: Although the derivative of a vector field at a point wrt another vector field, depends also on input values at nearby points, miraculously, certain commutator relations involving derivatives, e.g. the Riemann curvature, do depend only on the input values at the point, hence are actually tensors.

[to paraphrase another member, please take all this with a grain of salt.]

remark: rule 3) actually follows from rules 1),2) as follows:
Bil(XxY, Z) ≈ L(X,L(Y,Z))≈ X*(tens)L(Y,Z) ≈ X*(tens)Y*(tens)Z ≈ (XtensY)*(tens)Z ≈ L(X(tens)Y, Z).

 

[mathwonk]

To someone like me, i.e. me, who knows nothing of this, and never heard of Einstein's field equations before, wikipedia looks actually useful.

It is an equation between two tensors of type (0,2). (I have been calling them type (2,0)), i.e. 2-covariant tensors, linear combinations of dx^j(tens)dx^k.

Since the dot product is such a tensor, I think of such a tensor as a generalization of a dot product, i.e. a generalization of a Riemannian metric. Since it defines a bilinear pairing on pairs of tangent vectors, it tells us something about the shape of space.

The left side is a sum of two terms, each of which is derived from the given Riemannian metric on space, more precisely it is entirely determined (up to some constant multipliers) by that metric and its first and second derivatives. Thus the left side is a certain measure of the geometric curvature of space.
This is set equal to a tensor on the right side that is called the "stress-energy" tensor, something apparently determined by matter, radiation, and force fields, in space(time). Thus the equation says a certain geometric measure of the shape of space(time) is determined by the physical attributes or content of space(time).

This equation involves three tensors, so to understand it, one should address each one of them individually. One of the tensors is just (a scalar multiple of) the Riemannian metric tensor, so the first place to start is understanding that.

Then the second tensor on the same side, is the "Einstein" tensor, which is determined by the metric, and its first and second derivatives. It involves a certain "averaging" of the Riemann curvature tensor. I.e. it is a sum of the "Ricci" curvature, and the "scalar" curvature multiplied by the metric. Both of these curvatures are computed as a sort of "trace" starting from the Riemann curvature tensor.

I.e. thinking of that Riemann curvature tensor as a sort of matrix whose entries are bilinear forms, the Ricci curvature is the trace of that matrix, hence itself a bilinear form. Then thinking of that bilinear form as the composition of the metric with a linear map in the first variable, one further takes the trace of that map, a scalar.

In coordinates, one gets the (0,2) Ricci curvature as a contraction of the (1,3) Riemann curvature; and then one gets a scalar by further contracting this (0,2) tensor against the "inverse" of the metric (0,2) tensor, which apparently makes it a (2,0) tensor.(?) [as a mathematician, it looks to me as if matrix transposes are involved, rather than matrix inverses.]

Apparently, the thing to focus on, after the metric tensor, is the Riemann curvature tensor, and then its descendants, the Ricci and scalar curvatures. Finally one would study the stress -energy tensor.

I would suggest studying closely the covariant derivative, and how it leads to the Riemann curvature operator, and finally how that operator is expressed using tensor notation. I.e. in a sense, tensors are just a notation to express this concept, and understanding the concept seems to be the more crucial matter.

To a simple person like myself, second order mixed partial derivatives are equal in flat Euclidean space, but not in curved space. Thus the difference in the values obtained by taking second derivatives in different orders in two given directions, measures curvature of space in the direction of the plane they span.
A careful, intrinsic, elaboration of this idea, using "covariant derivatives", defines the Riemann curvature operator, which can be expressed as a tensor of "type (1,3)", hence contracts to a tensor of type (0,2), and further, using the metric, to a scalar.

As a quick check on the equations, what happens if the space is geometrically flat? Then it seems the Einstein tensor is zero, since it measures the curvature. Thus the left side seems to reduce to a constant multiple of the metric. What then is the right side, the stress energy tensor? Why is it also a multiple of the metric?

Apologies for posting on something I do not know anything at all about. This is obviously just speculation. Thanks to you all for introducing me to what tensors are used for in real life.

 

[Orodruin]

This equation involves three tensors, so to understand it, one should address each one of them individually. One of the tensors is just (a scalar multiple of) the Riemannian metric tensor, so the first place to start is understanding that.

That would be the cosmological constant term. You can either keep it on the LHS or move it to the RHS and consider it part of the stress-energy tensor (corresponding to a cosmological constant dark energy component).

Then the second tensor on the same side, is the "Einstein" tensor, which is determined by the metric, and its first and second derivatives. It involves a certain "averaging" of the Riemann curvature tensor. I.e. it is a sum of the "Ricci" curvature, and the "scalar" curvature multiplied by the metric. Both of these curvatures are computed as a sort of "trace" starting from the Riemann curvature tensor.

I.e. thinking of that Riemann curvature tensor as a sort of matrix whose entries are bilinear forms, the Ricci curvature is the trace of that matrix, hence itself a bilinear form. Then thinking of that bilinear form as the composition of the metric with a linear map in the first variable, one further takes the trace of that map, a scalar.

In coordinates, one gets the (0,2) Ricci curvature as a contraction of the (1,3) Riemann curvature; and then one gets a scalar by further contracting this (0,2) tensor against the "inverse" of the metric (0,2) tensor, which apparently makes it a (2,0) tensor.(?) [as a mathematician, it looks to me as if matrix transposes are involved, rather than matrix inverses.]

When we talk about the metric inverse, we are generally referring to the natural map between tangent vectors and dual vectors implied by the metric. For example, given a tangent vector $V \in T_p M$, we have a corresponding dual vector $\omega_V \in T^*_p M$ given by
$$
\omega_V(X) = g(X,V)
$$
for all $X$. In other words, we are considering a map $g: T_p M \to T^*_p M$ implied by the metric. When there is a metric we will usually forego the distinction between tangent and dual vectors and talk about $V$ and $\omega_V$ interchangeably, just with contravariant/covariant components. We would refer to the components of $\omega_V$ as $V_a = g_{ab} V^b$.

In this sense, a (0,2) tensor can also be considered a map $g: T_pM \to T^*_pM$ and the inverse $g^{-1}$ would then be a (2,0) tensor, being a map $g^{-1}:T^*_pM \to T_pM$ such that
$$
g^{-1}(\xi, \omega_V) = \xi(V)
$$
for all dual vectors $\xi$.

Apparently, the thing to focus on, after the metric tensor, is the Riemann curvature tensor, and then its descendants, the Ricci and scalar curvatures. Finally one would study the stress -energy tensor.

The metric tensor is the focus all of the time. The Riemann curvature tensor, and therefore also the Ricci tensor and Ricci scalar, are generally written in terms of the metric components and their derivatives. This makes the Einstein field equation a non-linear coupled system of partial differential equations for the metric components. The stress-energy tensor on the right hand side depends on the matter content that you put into the physical system.

The Einstein field equations generally follow from extremising the Einstein-Hilbert action with additional terms describing the cosmological constant and the matter content. The EFEs drop out when you vary the action with respect to the metric components (or, more naturally, the inverse metric components). In addition to these equations, you will also need the equations of motion for the matter fields entering through the stress energy tensor. These are acquired by varying the action with respect to the matter fields.

I would suggest studying closely the covariant derivative, and how it leads to the Riemann curvature operator, and finally how that operator is expressed using tensor notation. I.e. in a sense, tensors are just a notation to express this concept, and understanding the concept seems to be the more crucial matter.

Yes, this is a crucial part of understanding general relativity. Not only because of the fact that the covariant derivative is necessary to understand the EFEs, but also because it is absolutely required to understand it in order to interpret the results once you have a solution.

In addition, you will ideally need the concepts of the Lie derivative, Killing fields, and derived symmetries to help you along in finding solutions to the EFEs as well as to study the dynamics of test particles in the resulting spacetime.

As a quick check on the equations, what happens if the space is geometrically flat, and there is no input from the stress-energy?

Then you get the Minkowski space of special relativity - at least locally (up to non-trivial topologies globally).

 

[mathwonk]

Orodruin to the rescue!
As Orodruin makes clear, a map from T to T*, (where T is the tangent space), corresponds to a tensor of type (0,2), say by rule 1), post #31, since L(T,T*) ≈ T*(tens)T*. Thus its inverse is a map from T* to T, and by the isomorphism L(T*,T) ≈ T**(tens)T ≈ T(tens)T, the inverse (not the transpose) corresponds to a tensor of type (2,0).

In fact, if as I said, the matrix of the (0,2) form is represented by the composition of (the matrix of) a map and (the matrix of) the metric, then to recover the matrix of that map, we would compose the (0,2) form with the inverse of the metric.

Finally, in this setting, all the matrices involved are symmetric, so transposes would not change anything. More to the point, taking the transpose of a map does not change the type of the corresponding tensor, (again by rule 1)).