I Transpose and Inverse of Lorentz Transform Matrix

[cianfa72]

A tensor is not a linear transformation but a multi-linear map $T:V^m \times V^{*n} \rightarrow \mathbb{R}$. The corresponding tensor components are given by putting the corresponding basis vectors and their dual-basis vectors into the "slots":
$${T_{\mu_1\ldots \mu_m}}^{\nu_1\ldots \nu_n}=T(e_{\mu_1},\ldots,e_{\mu_m},e^{\nu_1},\ldots,e^{\nu_n}).$$

Consider the multi-linear map $T:V^{*} \times V \rightarrow \mathbb{R}$, $T \in V \otimes V^{*}$

$$T^{\mu}{}_{\nu} \text{ } e_{\mu} \otimes e^{\nu}$$
If we contract it with a vector $v = v^{\alpha} e_{\alpha}$ (basically filling in its correspondent slot) we get the vector $T^{\mu}{}_{\nu}v^{\nu} e_{\mu}$ that is any linear transformation of the vector space $V$ in itself (an endomorphism) is actually a (1,1) tensor, dont'you ?

 

[vanhees71]

The Lorentz-transformation matrix is just the transformation from one pseudo-Cartesian (in the following called "Lorentzian") basis (and its corresponding dual basis) to another. It has thus no basis-independent meaning and hence its matrix elements are no tensor components.

This reminds me that I should also write the general mathematical part in the Appendix about tensor algebra and calculus in Minkowski space...

 

[cianfa72]

The Lorentz-transformation matrix is just the transformation from one pseudo-Cartesian (in the following called "Lorentzian") basis (and its corresponding dual basis) to another. It has thus no basis-independent meaning and hence its matrix elements are no tensor components.

Sorry (I'm not an expert )...even if we think of it from an active point of view (as a mapping of a vector space in itself) ?

Just to be clear: I'm not saying Lorentz-transformation matrix has a physical basis-indipendent meaning, though..

 

[ergospherical]

Just to be clear: I'm not saying Lorentz-transformation matrix has a physical basis-indipendent meaning, though..

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

The active point of view: A Lorentz transformation $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ is an isometry (read: preserves the scalar square of vectors) of spacetime and the Lorentz matrix $\Lambda = [{\Lambda^{\mu}}_{\nu}]$ of this transformation in some orthonormal basis $\{ \hat{\mathbf{e}}_{\mu} \}$ is the matrix that satisfies $L(\hat{\mathbf{e}}_{\nu}) = {\Lambda^{\mu}}_{\nu} \hat{\mathbf{e}}_{\mu}$.

[The set of all these transformations $L$ (along with a binary operation $\circ$) forms the Lorentz group $O(3,1)$]

 

[hartmutneff]

In my opinion, the clean way of doing this is to take the transpose of metric type matrices only. So of a matrix with either to upper or two lower indices.

For the coefficients of tensors, there is no rule of where the upper or the lower indices have to go, with one exception. If you want to use the convention of how to multiply matrices and vectors, so the row times column rule. There is no need to follow this convention, as you can always write down the full tensor equations with all their indices. But if you want to follow the convention, then you will have to stick to some rules to be consistent.
The basic rule for this convention is, that upper indices run along columns and lower indices along rows. That immediately poses a problem with a metric, with two lower indices. A metric doesn't have an upper index that runs along columns. Hence a metric can't really be written as a matrix.

When it comes to the orthogonality equation of the Lorentz transformation, people often fix this by swapping the indices of the Lorentz matrix horizontally and denote this by the transpose. I think the consistent way of doing this is to raise the first index of the metric instead. That is twisting the notation as well, but less so than swapping indices horizontally. With one upper index, the metric and the whole equation can be written in matrix form. Then one can swap the index of the Lorentz transformation between upper left and lower right corner to get the transpose.

So again, I think the confusion with this equations is due to the fact that for this kind of tensors, namely vectors, dual vectors and (1,1) tensors representing linear transformations, one would like to use the matrix vector multiplication convention.

 

[cianfa72]

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

Surely, my point was that even if do not 'attach' any physical quantity/property to it from the active point of view it is actually a (1,1) tensor, though.

In fact using the identification of a vector space $V$ with its double dual $V^{**}$ we get an isomorphism between Linear transformations of $V$ and (1,1) type tensors.

 

[ergospherical]

Surely, my point was that even if do not 'attach' any physical quantity/property to it from the active point of view it is actually a (1,1) tensor, though.

That is a mistake. The elements ${\Lambda^{\mu}}_{\nu}$ are not the elements of a tensor. Nonetheless you may still raise and lower its indices with the metric $\eta_{\mu \nu}$.

 

[cianfa72]

That is a mistake. The elements ${\Lambda^{\mu}}_{\nu}$ are not the elements of a tensor.

I'm really struggling with it

From the point of view of active linear transformation of $V$ into itself, ${\Lambda^{\mu}}_{\nu}$ is actually its representation in a chosen basis of $V$. Furthermore as linear transformation it is isomorphic to a specific (1,1) tensor.

Then upon a change of basis in $V$ (and implicitly in the dual $V^{*}$) the entries in ${\Lambda^{\mu}}_{\nu}$ change in the same way the components of the (1,1) tensor isomorphic to it, dont' you ?

 

[ergospherical]

Not every object with two indices is a rank-2 tensor! [Recall that a tensor is specifically a multilinear map of vectors and covectors to some field e.g. $\mathbb{R}$, and whose components in a given basis transform according to the tensor transformation law.]

Rather a Lorentz transformation is an element of the Lorentz group [its elements are endomorphisms of $\mathbb{R}^4$ which map one orthonomal basis to another orthonormal basis], and each has a matrix representation with respect to a given basis. That a Lorentz transformation is an isometry implies that, for arbitrary vectors $\mathbf{u},\mathbf{v} \in \mathbb{R}^4$,\begin{align*}
\eta(L(\mathbf{u}), L(\mathbf{v})) = \eta_{ab} [L(\mathbf{u})]^a [L(\mathbf{v})]^b &\overset{!}{=} \eta_{cd} u^c v^d = \eta(\mathbf{u},\mathbf{v})\\

\eta_{ab} {\Lambda^a}_c u^c {\Lambda^b}_d v^d &= \eta_{cd} u^c v^d \\

\eta_{ab} {\Lambda^a}_c {\Lambda^b}_d &= \eta_{cd} \\

{(\Lambda^T)_c}^a \eta_{ab}{\Lambda^b}_d &= \eta_{cd} \\

\Lambda^T \eta \Lambda &= \eta
\end{align*}which is the defining property of a Lorentz matrix $\Lambda$. Taking the determinant of both sides yields $|\Lambda|^2 = 1$; the sub-group of $O(3,1)$ with determinant $+1$ is called the proper Lorentz group and is denoted $SO(3,1)$. Meanwhile an orthochronous Lorentz transformation is one for which, given any timelike vector $\mathbf{u}$ [$\eta(\mathbf{u},\mathbf{u}) < 0$], we have $\eta(L(\mathbf{u}), \mathbf{u}) < 0$. You can show that a Lorentz matrix is orthochronous if ${\Lambda^0}_0 \geq 1$ [exercise!]. The group of Lorentz transformations which are both proper and orthochronous [called restricted Lorentz transformations] is denoted $SO^{+}(3,1)$; any Lorentz transformation can in fact be reduced to a restricted one.

 

[cianfa72]

Rather a Lorentz transformation is an element of the Lorentz group [its elements are endomorphisms of $\mathbb{R}^4$ which map one orthonomal basis to another orthonormal basis], and each has a matrix representation with respect to a given basis.

Take an element of the Lorentz group (i.e. a Lorentz transformation) and consider its matrix representation in a given not orthonormal basis. Is actually this matrix a 'Lorentz matrix' ?

 

[ergospherical]

I'd say no, a Lorentz matrix $\Lambda$ is specifically a representation of a Lorentz transformation in an orthonormal basis of $\mathbb{R}^4$, i.e. a matrix for which there exists a map $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ and an orthonormal basis $\{\hat{\mathbf{e}}_a \}$ such that $L(\hat{\mathbf{e}}_a) = {\Lambda^b}_a \hat{\mathbf{e}}_b$. That's to say the columns of $\Lambda$ are the images $L(\hat{\mathbf{e}}_a)$.

[But of course, you can represent $L$ in terms of any basis you wish!]

 

[cianfa72]

[But of course, you can represent $L$ in terms of any basis you wish!]

Maybe this was the missing point to me.
Hence a Lorentz transformation is isomorphic to a (1,1) tensor instead a Lorentz matrix (since restricted to represent a Lorentz transformation in an orthonormal basis) is not.

 

[ergospherical]

a Lorentz transformation is isomorphic to a (1,1) tensor

You should drop this idea, it's misleading! Lorentz transformations don't really have anything to do with tensors.

 

[cianfa72]

You should drop this idea, it's misleading! Lorentz transformations don't really have anything to do with tensors.

Sorry, surely I am wrong but...is a Lorentz transformation actually a linear transformation of $\mathbb {R}^4$ in itself ?
Is any linear transformation isomorphic to a (1,1) tensor ?

Hence...

 

[ergospherical]

I think @vanhees71 already said this; Lorentz transformations are endomorphisms $L: \mathbb{R}^4 \rightarrow \mathbb{R}^4$, whilst tensors are maps $T: {V^*}^p \times {V}^q \rightarrow \mathbb{F}$, i.e. they're different types of objects.

However, it is a result of linear algebra that vector spaces of equal dimension are isomorphic. So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

 

[cianfa72]

So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

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

 

[ergospherical]

Ah cool, yeah that's one worked example! Though I'm not sure how helpful it is to put a Lorentz transformation into correspondence with a $(1,1)$ tensor. The isomorphism between linear transformations and matrices (e.g. taking $L$ to $\Lambda$) seems like the only important one.

Maybe someone else can think of an example, though?

 

[cianfa72]

The isomorphism between linear transformations and matrices (e.g. taking $L$ to $\Lambda$) seems like the only important one.

Btw that isomorphism is not natural because the selection of a basis is needed.

 

[vanhees71]

That's a shame, because you'd be correct.
You can of course view it either actively or passively.

The active point of view: A Lorentz transformation $L : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ is an isometry (read: preserves the scalar square of vectors) of spacetime and the Lorentz matrix $\Lambda = [{\Lambda^{\mu}}_{\nu}]$ of this transformation in some orthonormal basis $\{ \hat{\mathbf{e}}_{\mu} \}$ is the matrix that satisfies $L(\hat{\mathbf{e}}_{\nu}) = {\Lambda^{\mu}}_{\nu} \hat{\mathbf{e}}_{\mu}$.

[The set of all these transformations $L$ (along with a binary operation $\circ$) forms the Lorentz group $O(3,1)$]

Another way to see that a Lorentz matrix are not tensor components is

I think @vanhees71 already said this; Lorentz transformations are endomorphisms $L: \mathbb{R}^4 \rightarrow \mathbb{R}^4$, whilst tensors are maps $T: {V^*}^p \times {V}^q \rightarrow \mathbb{F}$, i.e. they're different types of objects.

However, it is a result of linear algebra that vector spaces of equal dimension are isomorphic. So yeah, it's certainly possible to write down isomorphisms between $(1,1)$ tensors and endomorphisms on $\mathbb{R}^4$, although I'm not sure if that's helpful when dealing with Lorentz transformations...

It's not an isomorphism, but you can look at the (proper orthochronous) Lorentz group as a Lie group. Then the infinitesimal generators indeed behave like tensor components. Note, though, that you cannot write all Lorentz transformations as a single exponential of a generator. So the map is not surjective.

 

[cianfa72]

It's not an isomorphism

Which is not an isomorphism ?

 

[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.