I Transpose and Inverse of Lorentz Transform Matrix

[vanhees71]

There is no isomorphism between the Lie algebra of the proper orthochronous Lorentz transformation and the Lie group.

 

[cianfa72]

There is no isomorphism between the Lie algebra of the proper orthochronous Lorentz transformation and the Lie group.

Surely, in fact we were talking about the canonical isomorphism between Linear transformation (such as Lorentz transformations in $\mathbb{R}^4$) and (1,1) tensors.

 

[PeterDonis]

The point to be highlighted is that actually there exist a natural (or canonical) isomorphism between the two that requires no choice about the basis.

See for example the first answer here https://math.stackexchange.com/questions/1108842/why-is-a-linear-transformation-a-1-1-tensor

the canonical isomorphism between Linear transformation (such as Lorentz transformations in $\mathbb{R}^4$) and (1,1) tensors.

The isomorphism referred to requires you to view the reals as a vector space. But from the point of view of tensor calculus, the reals--the numbers you get when you contract tensors with vectors and covectors--are not a vector space. They are just a field.

 

[ergospherical]

Every field is a vector space over itself.

 

[PeterDonis]

Every field is a vector space over itself.

As a matter of math, yes, this is true. But as a matter of physics, sometimes the vector space structure of a field is part of the physical model, and sometimes it's not.

 

[ergospherical]

As a matter of math, yes, this is true. But as a matter of physics, sometimes the vector space structure of a field is part of the physical model, and sometimes it's not.

I don't understand the distinction, could you explain? Any field $(F, \times, \cdot)$ is automatically a vector space over itself because all the vector space axioms are contained within the field axioms. It doesn't have anything to do with Physics?

 

[cianfa72]

The isomorphism referred to requires you to view the reals as a vector space. But from the point of view of tensor calculus, the reals--the numbers you get when you contract tensors with vectors and covectors--are not a vector space. They are just a field.

Do you mean $\mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} =\mathbb{R}^4$ as vector space ?

 

[PeterDonis]

Do you mean $\mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} =\mathbb{R}^4$ as vector space ?

For the case of Lorentz transformations in 4D spacetime, yes, you would have to consider $\mathbb{R}^4$ as a vector space. But note that it is not the same vector space as Minkowski spacetime; i.e., it is not the vector space that the Lorentz transformations are normally viewed as acting on.

Note also that, for the case of tensors, as defined by how they contract with other tensors to form scalars, the scalars are just reals; they are elements of $\mathbb{R}$, not $\mathbb{R}^N$ (where $N$ is the number of dimensions of the space or spacetime being considered).

 

[PeterDonis]

Any field $(F, \times, \cdot)$ is automatically a vector space over itself because all the vector space axioms are contained within the field axioms.

But $\mathbb{R}$ is not the only possible vector space over the field $\mathbb{R}$. And in the case of the physics we are discussing, the vector space in question is Minkowski spacetime, which is not $\mathbb{R}$, and not even $\mathbb{R}^4$ (which is not the same vector space as $\mathbb{R}$, but considered as a vector space over a field, is a vector space over the field $\mathbb{R}$), but is a different vector space over the field $\mathbb{R}$ (since the scalars you get when you contract Minkowski vectors and covectors are reals--this is the same point I made to @cianfa72 in my previous post just now). So the vector space structure being used to model the physics in this case is not the vector space structure of $\mathbb{R}$ (even though it automatically has one, as you point out), it's the vector space structure of Minkowski spacetime.

 

[ergospherical]

Minkowski spacetime, which is ... not even $\mathbb{R}^4$

This is a bit of a digression... but Minkowski spacetime is $M = (\mathbb{R}^4, \eta)$ with $\eta$ a symmetric, non-degenerate bilinear form of signature (-+++), and where the Lorentz group acts on the tangent space $T_{p}M$ at every point.

 

[PeterDonis]

This is a bit of a digression... but Minkowski spacetime is $M = (\mathbb{R}^4, \eta)$ with $\eta$ a symmetric, non-degenerate bilinear form of signature (-+++), and where the Lorentz group acts on the tangent space $T_{p}M$ at every point.

As a manifold with metric, yes. But we're talking about Minkowski spacetime as a vector space.

 

[vanhees71]

Minkowski spacetime is an affine pseudo-Euclidean (Lorentzian) space, i.e., a set called "points" and a vector space with a fundamental bilinear form of signature (1,3) (or equivalently (3,1)). For two points $A$ and $B$ there is always a unique vector $\overrightarrow{AB}$. The vectors can be identified with directed straight lines connecting two points. For three points $A$, $B$, and $C$ vector addition is defined as $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$. Two vectors $\overrightarrow{AB}$ and $\overrightarrow{A'B'}$ are equal, if one can parallel shift the corresponding directed straight lines into each other.

The full symmetry group, connected smoothly to the identity is the proper orthochronous Poincare group, generated by translations (homogeneity of spacetime, no spacetime point is distinguished from any other), and the Lorentz group, which is again generated by the rotation-free boosts (building NOT a subgroup), describing the invariance of the special-relativistic laws of nature when switching from one inertial reference frame to another inertial reference frame, and rotations (building a subgroup), describing the isotropy of space as observed by any inertial observer.

 

[cianfa72]

The full symmetry group, connected smoothly to the identity is the proper orthochronous Poincare group, generated by translations (homogeneity of spacetime, no spacetime point is distinguished from any other), and the Lorentz group, which is again generated by the rotation-free boosts (building NOT a subgroup), describing the invariance of the special-relativistic laws of nature when switching from one inertial reference frame to another inertial reference frame, and rotations (building a subgroup), describing the isotropy of space as observed by any inertial observer.

ok, from the point of view of the representation of a Lorentz group element (i.e. a Lorentz transformation) by a (Lorentz) matrix we need to select a basis in the vector space part of the definition of $M$ as affine space (as said above in this thread actually an orthonormal basis), right ?

 

[vanhees71]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Sure, a Lorentzian (inertial) frame of reference can be defined by choosing an arbitrary point $O$ (origin) of Minkowski space and 4 Minkowski-orthonormal basis vectors, $\bvec{e}_{\mu}$ with $\bvec{e}_{\mu} \cdot \bvec{e}_{\nu}=\eta_{\mu \nu}$. Then under a Poincare transformation the spacetime coordinates transform as
$$x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu},$$
where ${\Lambda^{\mu}}_{\rho}$ is a Lorentz-transformation matrix with $\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ and $a^{\mu}$ arbitrary four-vector components.

[EDIT: corrected the typos $L \rightarrow \Lambda$ mentioned in #65.

 

[cianfa72]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Sure, a Lorentzian (inertial) frame of reference can be defined by choosing an arbitrary point $O$ (origin) of Minkowski space and 4 Minkowski-orthonormal basis vectors, $\bvec{e}_{\mu}$ with $\bvec{e}_{\mu} \cdot \bvec{e}_{\nu}=\eta_{\mu \nu}$. Then under a Poincare transformation the spacetime coordinates transform as
$$x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu},$$

A doubt on coordinates $x^{\mu}$ and $x^{\prime \mu}$. They are used for the components of vectors (in a given orthonormal basis $\{ \bvec{e}_{\mu} \}$) that live into the 'translation' vector space involved in the definition of $M$ as affine space.

where ${\Lambda^{\mu}}_{\rho}$ is a Lorentz-transformation matrix with $\eta_{\mu \nu} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ and $a^{\mu}$ arbitrary four-vector components.

I think $\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}$ you meant.

Here as four-vector components $a^{\mu}$ do you mean the components of an arbitrary vector $\mathbf a$ that lives in the 'translation' vector space in the same basis $\{ \bvec{e}_{\mu} \}$ ?

In other words $x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu}$ actually is not a linear transformation of the 'translation' vector space into itself (since the term $a^{\mu}$), right ?

 

[vanhees71]

The full group of Minkowski space as an affine space includes space-time translations, i.e., you can shift the origin arbitrarily around without changing the physics, i.e., the physical laws look the same at any time and any place.

 

[cianfa72]

The full group of Minkowski space as an affine space includes space-time translations, i.e., you can shift the origin arbitrarily around without changing the physics, i.e., the physical laws look the same at any time and any place.

Surely, my point was to better understand the notation used for Lorentz transformations. To me it seems we are basically 'turning' the affine space into a vector space by selecting a given point in it. Hence the coordinates $x^{\mu}$ and $x^{\prime \mu}$ entering in $x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu}$ are really the coordinates for the 'translation' vector space part of the affine space definition (i.e. the operator $+$ inside it is actually a sum of two vectors in the 'translation' vector space and is not the sum of a vector and a point in the affine space we started with).

Thanks for your patience

 

[vanhees71]

$$\newcommand{\bvec}[1]{\boldsymbol{#1}}$$
Well, $x^{\prime \mu}$ are the spactime coordinates in an inertial reference frame $\Sigma'$, which is defined by the corresponding origin $O'$ and the tetrad $\bvec{e}_{\mu}'$ and $x^{\rho}$ are the spacetime coordinates in an inertial frame $\sigma$ with origin $O'$ and the tetrad $\bvec{e}_{\rho}$.

Let $P$ be an arbitrary point in Minkowski space. The relation of the coordinates is given as follows:
$$\overrightarrow{OP}=x^{\rho} \bvec{e}_{\rho}, \quad \overrightarrow{O'P}=x^{\prime \mu} \bvec{e}_{\mu}'.$$
Now
$$x^{\prime \mu} \bvec{e}_{\mu}'=\overrightarrow{O'P}=\overrightarrow{O'O}+\overrightarrow{OP}=\bvec{a}+x^{\rho} \bvec{e}_{\rho}=a^{\prime \mu} \bvec{e}_{\mu}' + \bvec{e}_{\mu}' {\Lambda^{\mu}}_{\rho} x^{\rho}.$$
From this you get
$$x^{\prime \mu}=a^{\prime \mu}+ {\Lambda^{\mu}}_{\rho} x^{\rho}.$$

 

[cianfa72]

ok thanks.

Sorry to come back to the original topic: Lorentz transformations are actually isometries of $M$ that leave fixed the (chosen) origin. Hence we can consider them as (linear) isometries on the 'translation' vector space (say it $V$) part of the definition of $M$ as affine space, I believe.

Now, if the group of Lorentz transformations was itself a vector space on $\mathbb {R}$, then we could 'apply' the canonical isomorphism as described in the aforementioned link in order to 'identify' a Lorentz transformation (i.e. its matrix representation into a given basis of $V$) with a (1,1) tensor element of $V^{*} \otimes V$.

I was thinking that the group of Lorentz transformations actually is not itself a vector space on $\mathbb {R}$ since the multiplication of a Lorentz transformation for a real number is not a Lorentz transformation anymore.

Maybe that is the reason why we cannot 'build' the canonical isomorphism as above.

 

[vanhees71]

The Lorentz group is a subgroup of the symmetry group of Minkwoski space as an affine space. The Lorentz transformations themselves don't form a vector space. As a Lie group their Lie algebra is a vector space though.

 

[cianfa72]

The Lorentz group is a subgroup of the symmetry group of Minkwoski space as an affine space. The Lorentz transformations themselves don't form a vector space.

So is that the reason why we cannot establish a canonical isomorphism between Lorentz transformations (hence Lorentz matrices in an orthonormal basis) and the vector space of (1,1) tensor built on $V$ (i.e. the translation vector space) ?

 

[vanhees71]

I don't know, how you come to the idea in the first place. What would this be good for? Only because you can represent 2nd-rank tensors as well as Lorentz transformations by $4\times 4$ matrices doesn't say that there is a "canonical" (i.e., basis independent) isomorphism between them.

 

[cianfa72]

Sorry, I've really difficulty to catch your point.
If the set of Lorentz transformations was itself a vector space why we could not apply the 'argument' in the aforementioned link (1st answer) to build such canonical isomorphism?

I do not understand if it really no makes sense or if, as agreed with you, there is actually no business to do that.

 

[PeterDonis]

To me it seems we are basically 'turning' the affine space into a vector space by selecting a given point in it.

That's one way of viewing what you are doing when you pick a particular coordinate chart on Minkowski spacetime, yes. But note that the choice of coordinates does not just fix an origin, it fixes the directions of the axes (four of them). So it does more than just turn the affine space into a vector space; it specifies a particular basis of the vector space.

Hence the coordinates $x^{\mu}$ and $x^{\prime \mu}$ entering in $x^{\prime \mu}={\Lambda^{\mu}}_{\rho} x^{\rho} + a^{\mu}$ are really the coordinates for the 'translation' vector space part of the affine space definition

No, they aren't. The specification of coordinates, as above, specifies a set of basis directions as well as an origin.

 

[PeterDonis]

So it does more than just turn the affine space into a vector space; it specifies a particular basis of the vector space.

And to follow on with this, Lorentz transformations (the $\Lambda$ part of the transformation equation you wrote) are changes of basis; they switch from one basis of the vector space to another, without changing the vector space itself (i.e., they rotate the coordinate axes without changing the origin).

 

[cianfa72]

And to follow on with this, Lorentz transformations (the $\Lambda$ part of the transformation equation you wrote) are changes of basis; they switch from one basis of the vector space to another, without changing the vector space itself (i.e., they rotate the coordinate axes without changing the origin).

Sorry to bother you: so the operator $+$ inside
$$x^{\prime \mu}=a^{\prime \mu}+{\Lambda^{\mu}}_{\rho} x^{\rho}$$ is actually a sum of two vectors or it is a sum of a point in the affine space and the vector ${\Lambda^{\mu}}_{\rho} x^{\rho}$ ? I believe the former.

 

[PeterDonis]

Sorry to bother you: so the operator $+$ inside
$$x^{\prime \mu}=a^{\prime \mu}+{\Lambda^{\mu}}_{\rho} x^{\rho}$$ is actually a sum of two vectors or it is a sum of a point in the affine space and the vector ${\Lambda^{\mu}}_{\rho} x^{\rho}$ ? I believe the former.

Strictly speaking, it's a sum of real numbers. The equation you write is actually four equations, one for each value of $\mu$. Each equation is an equation in real numbers. The $\Lambda$ term on the RHS expands to a summation of four terms in each equation, one for each value of $\rho$.

If you want to interpret the four equations together as one equation, then it would appear to be a vector equation, but the vector space is not well defined, since the equation includes a change of origin. One could interpret it as a combination of a change of basis in the vector space of vectors from the original (unprimed) origin, plus a function from the first vector space (unprimed) to a second vector space (of vectors from the primed origin) that preserves the basis, but that still doesn't match either of the things you described, and the operator $+$ in this case would have to be interpreted as a sloppy shorthand for combining two different and incommensurable operations.

 

[cianfa72]

One could interpret it as a combination of a change of basis in the vector space of vectors from the original (unprimed) origin, plus a function from the first vector space (unprimed) to a second vector space (of vectors from the primed origin) that preserves the basis, but that still doesn't match either of the things you described, and the operator $+$ in this case would have to be interpreted as a sloppy shorthand for combining two different and incommensurable operations.

That's was actually my point to 'reduce' it to a sum of vectors that belong to the translation vector space part of the definition of affine space.

 

[PeterDonis]

That's was actually my point to 'reduce' it to a sum of vectors that belong to the translation vector space part of the definition of affine space.

There is no such thing as "the translation vector space part of the affine space" that I'm aware of. An affine translation can be viewed as a map between vector spaces; that's what I was describing. But a map between vector spaces is not the same thing as a vector space.

 

[cianfa72]

There is no such thing as "the translation vector space part of the affine space" that I'm aware of.

See for instance here https://en.m.wikipedia.org/wiki/Affine_space in the section Definition.

 

[PeterDonis]

See for instance here https://en.m.wikipedia.org/wiki/Affine_space in the section Definition.

Ok, so if we consider the translations as a vector space acting on a set, then we have:

The set is a set of points, which we are viewing as points in spacetime.

The vector space is the space of translations, which are maps from the set of points into itself. Note that this vector space is not Minkowski spacetime! It's just $\mathbb{R}^4$, or more precisely its additive component, considered as an additive vector space.

Now let's unpack the full transformation equation in that light:

$$
\left( x^\prime \right)^\mu = \Lambda^\mu{}_\rho x^\rho + a^\mu
$$

We can view $a^\mu$ as an element of the above vector space (a translation). However, that vector space, as above, is a map between points. So the $+$ operator in the above is shorthand for "take the point referred to by $\Lambda^\mu{}_\rho x^\rho$ and apply the translation $a^\mu$ to it". So it's still not an addition of vectors. It's a shorthand for the translation operation.

 

[cianfa72]

The set is a set of points, which we are viewing as points in spacetime.

The vector space is the space of translations, which are maps from the set of points into itself. Note that this vector space is not Minkowski spacetime! It's just $\mathbb{R}^4$, or more precisely its additive component, considered as an additive vector space.

Yes, surely.

So the $+$ operator in the above is shorthand for "take the point referred to by $\Lambda^\mu{}_\rho x^\rho$ and apply the translation $a^\mu$ to it". So it's still not an addition of vectors. It's a shorthand for the translation operation.

Actually I prefer to think of such $+$ operator as the 'add' operation of the translation vector space (see post#68 by @vanhees71).

To me that sum actually 'implements'
$$x^{\prime \mu} \vec{e}_{\mu}'=\overrightarrow{O'P}=\overrightarrow{O'O}+\overrightarrow{OP}=\vec{a}+x^{\rho} \vec{e}_{\rho}=a^{\prime \mu} \vec{e}_{\mu}' + \vec{e}_{\mu}' {\Lambda^{\mu}}_{\rho} x^{\rho}.$$ This is really a sum of elements (vectors) belonging to the translation vector space.

For that reason I said $x^{\mu}$ and $x^{\prime \mu}$ are in practice the coefficients (i.e. the components) of vectors from translation vector space into the two given basis, respectively.

 

[vanhees71]

Sorry to bother you: so the operator $+$ inside
$$x^{\prime \mu}=a^{\prime \mu}+{\Lambda^{\mu}}_{\rho} x^{\rho}$$ is actually a sum of two vectors or it is a sum of a point in the affine space and the vector ${\Lambda^{\mu}}_{\rho} x^{\rho}$ ? I believe the former.

It's a sum of vector components. It's good to distinguish between vectors (tensors) and their components. The vectors and tensors don't depend on the choice of any basis, while the components always do since to introduce the components you need a basis first to define them.

 

[cianfa72]

It's a sum of vector components. It's good to distinguish between vectors (tensors) and their components. The vectors and tensors don't depend on the choice of any basis, while the components always do since to introduce the components you need a basis first to define them.

Yes, definitely. In fact in that equation basis vectors do not appear at all.

 

[PeterDonis]

This is really a sum of elements (vectors) belonging to the translation vector space.

No, it's not. As I said in post #81, it's a shorthand for the action of a single vector in the translation vector space, namely $a^\mu$, on the underlying set of points. The other terms in your equation are not vectors in the translation vector space; they are coordinate vectors, i.e., points in the underlying set (the spacetime) viewed as vectors (by choosing a particular origin). The translation $a^\mu$ shifts the origin from the unprimed one to the primed one.

A sum of elements in the translation vector space would be adding two different translations, say $a^\mu$ and $b^\mu$, to get a new translation. That's not what the equation you wrote is saying.

 

[cianfa72]

The other terms in your equation are not vectors in the translation vector space; they are coordinate vectors, i.e., points in the underlying set (the spacetime) viewed as vectors (by choosing a particular origin).

Maybe I could be wrong but...choosing a particular point (i.e. the origin) in the affine space does not get you a vector space namely the same 'translation' vector space ?

 

[vanhees71]

I'd rather say by choosing an arbitrary point $O$ there's a one-to-one mapping between the points of the affine manifold and the vectors of this affine manifold. This is part of the axioms defining what an affine manifold is: For any point $P$ there's a unique vector $\vec{x}=\overrightarrow{OP}$ and also the other way around: For any vector $\vec{x}$ there's a uniqe point $P$ such that $\overrightarrow{OP}=\vec{x}$. So you have a one-to-one mapping between points of the affine manifold and its vectors by choosing an arbitrary point $O$ as "the origin".

You can of course choose another origin $O'$ and you get another mapping between the points of the affine manifold and the vectors via $\vec{x}'=\overrightarrow{O'P}$.

Since both mappings between points and vectors are bijective there's a bivective map between the corresponding vectors, and these build a group, the group of translations:
$$\vec{x}'=\overrightarrow{O'P}=\overrightarrow{O'O}+\overrightarrow{OP}=\vec{x}+\vec{a}.$$
The translations build an Abelian group.

 

[PeterDonis]

choosing a particular point (i.e. the origin) in the affine space does not get you a vector space namely the same 'translation' vector space ?

No. Go read the last paragraph of my post #85 again, carefully.

 

[PeterDonis]

choosing a particular point (i.e. the origin) in the affine space does not get you a vector space namely the same 'translation' vector space ?

To complement my post #85, here's another way to look at it.

Suppose we leave out the translation $a^\mu$ and just consider the Lorentz transformation $\left( x^\prime \right)^\mu = \Lambda^\mu{}_\rho x^\rho$. As I have already said, this is a change of basis in a vector space. What vector space?

If the answer to your question quoted above were "yes", the change of basis would be in the translation vector space. But that won't work. As GR makes clear, the vector space in which this change of basis is being made is the vector space of tangent vectors at a point. A tangent vector is not a translation; roughly speaking, its magnitude is a rate of change in the direction in which the vector points. The Lorentz transformation $\Lambda$ preserves magnitudes of vectors, so it preserves the rate of change while rotating the directions (which is what a change of basis does).

Note also that, strictly speaking, what I said above for Lorentz transformations only works for timelike and spacelike vectors. The action of a Lorentz transformation on null vectors is fundamentally different: it doesn't rotate them, it dilates them. (Physically, this corresponds to the relativistic Doppler shift.) But such an action would make no sense if it were interpreted as acting on the translation vector space.

In other words, translation vectors, such as $a^\mu$ in your equation, are not Minkowski spacetime vectors; there is no such thing as timelike, spacelike, or null with translation vectors. A translation vector is just a "bare" element of $\mathbb{R}^4$, and its magnitude, i.e., the "distance" by which it shifts the origin of coordinates, is its "Euclidean" magnitude, not its "Minkowski" magnitude. (Trying to assign a Minkowski magnitude to a translation vector like $a^\mu$ would mean that, for example, the translation $a^\mu = (1, 1, 0, 0)$ would have a magnitude of zero, i.e., it would shift the origin by zero distance, which is obviously false.)

However, the vectors $x^\mu$ and $\left( x^\prime \right)^\mu$ are Minkowski spacetime vectors, since the Lorentz transformation acts on them--whereas it can't act on the translation vector $a^\mu$. You can't Lorentz transform a translation of the origin.

 

[pervect]

Conventions are different. MTW, for instance, uses the convention that all transformation matrices (including but not limited to the Lorentz transformation) obey the "northwest, southeast" convention.

Other textbooks do not necessarily follow MTW's convention, while I haven't read Tung, I assume you're correct that he doesn't follow it. I wouldn't say MTW's way is the only way, but I like the way they do it, it's simple and convenient. MTW's convention is to always write transformation matrices as $\Lambda^a{}_b$, which MTW describes as the "northwest-southeast" convention. I suppose though, that if you have a paper or book that doesn't follow the conventions, you have to figure out what they mean. As I always use MTW's approach, I don't know how to help with other approaches.

What I think is important is that one know how to transform vectors, how to transform co-vectors (one-forms), and that one know how to find the inverse of a transform. I can't think of anything else one needs to know. So if you know how to do these three things with whatever convetion it is that your source uses, you're set.

WIth MTW's approach, vectors transform as $u^a = \Lambda^a{}_b u^b$, one forms/ covectors transform as $\omega_b = \Lambda^a{}_b \omega_a$ and that for $\Lambda^{-1}$ to be the inverse of $\Lambda$, we must have $\Lambda^a{}_b (\Lambda^{-1})^b{}_c = \delta^a{}_c$, i.e. the composition of a transform and it's inverse must be the identity transform. I believe it also follows that $(\Lambda^{-1})^a{}_b \Lambda^b{}_c= \delta^a{}_c$, i.e. the order of the composition of a transform and it's inverse doesn't matter. But I could be mistaken, this is from memory.