# Index notation for inverse Lorentz transform

Hi all, just had a question about tensor/matrix notation with the inverse Lorentz transform. The topic was covered well here, but I’m still having trouble relating this with an equation in Schutz Intro to GR...

So I can use the following to get an eqution for the inverse:
$$x^{\overline{\mu}}x_{\overline{\mu}}=\Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda^{\beta}_{\;\overline{\mu}}x_{\beta}$$
And therefore
$$\Lambda^{\beta}_{\;\overline{\mu}}\Lambda^{\overline{\mu}}_{\;\alpha}=\delta^{\beta}_{\;\alpha}$$
This equation is just the one in ch2 from Schutz. But I can just as well reason as follows:
$$x^{\overline{\mu}}x_{\overline{\mu}}=\eta_{\overline{\mu}\overline{\nu}}x^{\overline{\mu}}x^{\overline{\nu}}=\eta_{\overline{\mu}\overline{\nu}}\Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda^{\overline{\nu}}_{\;\beta}x^{\beta}=\Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda_{\overline{\mu}\beta}x^{\beta}$$
And therefore
$$\Lambda_{\overline{\mu}\beta}\Lambda^{\overline{\mu}}_{\;\alpha}=\eta_{\beta\alpha}$$
Or
$$\Lambda_{\overline{\mu}}^{\;\ \beta}\Lambda^{\overline{\mu}}_{\;\alpha}=\delta^{\beta}_{\;\alpha}$$
Taken together, we seem to have
$$\Lambda_{\overline{\mu}}^{\;\ \beta}=\Lambda^{\beta}_{\;\overline{\mu}}$$
Is this correct? It seems wrong to me, and it seems that I might’ve confused my tensor and matrix indices, I’m just not sure how...

TeethWhitener
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Hi all, just had a question about tensor/matrix notation with the inverse Lorentz transform. The topic was covered well here, but I’m still having trouble relating this with an equation in Schutz Intro to GR...

So I can use the following to get an eqution for the inverse:
$$x^{\overline{\mu}}x_{\overline{\mu}}=\Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda^{\beta}_{\;\overline{\mu}}x_{\beta}$$
And therefore
$$\Lambda^{\beta}_{\;\overline{\mu}}\Lambda^{\overline{\mu}}_{\;\alpha}=\delta^{\beta}_{\;\alpha}$$
This equation is just the one in ch2 from Schutz.
Everything after this is fine. It's this part that's incorrect. In particular:
$$x^{\overline{\mu}}x_{\overline{\mu}}\neq \Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda^{\beta}_{\;\overline{\mu}}x_{\beta}$$
It might help to write out the sums on each side explicitly to see where you've gone wrong.

Everything after this is fine. It's this part that's incorrect. In particular:
$$x^{\overline{\mu}}x_{\overline{\mu}}\neq \Lambda^{\overline{\mu}}_{\;\alpha}x^{\alpha}\Lambda^{\beta}_{\;\overline{\mu}}x_{\beta}$$
It might help to write out the sums on each side explicitly to see where you've gone wrong.

Thanks very much for your reply, this issue has been bothering me for days now! Is the problem that I should’ve written this as

$$x^{\overline{\mu}}x_{\overline{\mu}}= \Lambda^{\overline{\mu}}_{\;\alpha}(\vec{v})x^{\alpha}\Lambda^{\beta}_{\;\overline{\mu}}(-\vec{v})x_{\beta}$$

I wasn’t sure whether putting the bars over indicies already implied this (and indeed, without the -v things turn out bad). Propagating that forward, we would then have

$$\Lambda_{\overline{\mu}}^{\;\ \beta}(\vec{v})=\Lambda^{\beta}_{\;\overline{\mu}}(-\vec{v})$$

Which still looks off to me (do the sides not differ by a transpose?)

vanhees71
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It's a bad habit to indicate the components wrt. to different bases not with different symbols but different indices. Strictly speaking it's wrong. That said, let's do a more careful analysis. By definition the Lorentz transformation between contravariant vector components of the same vector ##x## is given by
$$\bar{x}^{\mu}={\Lambda^{\mu}}_{\rho} x^{\rho},$$
and the matrix ##\Lambda## is by definition a Lorentz transformation iff
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}.$$
The transformation rule for covariant components most easily follow from the rule to lower and raise indices:
$$\bar{x}_{\mu} = \eta_{\mu \rho} \bar{x}^{\rho} = \eta_{\mu \rho} {\Lambda^{\rho}}_{\sigma} x^{\sigma} = \eta_{\mu \rho} {\Lambda^{\rho}}_{\sigma} \eta^{\sigma \nu} x_{\nu}={\Lambda_{\mu}}^{\nu} x_{\nu}.$$
Then from the Lorentz property of the matrix ##\Lambda## you easily derive
$${\Lambda^{\mu}}_{\rho} {\Lambda_{\mu}}^{\sigma}=\delta_{\rho}^{\sigma}, \quad(*)$$
i.e., the covariant components transform cogrediently to the contravariant components as it must be after all since covariant components are components of a linear form (dual vector) and contravariant components are components of a vector.

In matrix notation (*) implies
$$\Lambda^{-1} = \eta \Lambda^t \eta.$$

• Demystifier
Orodruin
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It's a bad habit to indicate the components wrt. to different bases not with different symbols but different indices. Strictly speaking it's wrong.
I know we disagree on this, so I do not think you can say that it is strictly wrong. It is a matter of notation and a question of what one considers to be coordinate dependent and what not. I don't think notation can be "wrong" - but it can be more or less convenient and lucid. My preference is to use the symbol to denote a particular tensor, which is coordinate independent and therefore should be represented by the same symbol regardless of the coordinates used. The coordinate dependent quantities are the components and therefore I prefer to denote this using primes or other decorations. Of course, when you do not have tensors (or mean different tensors depending on the coordinate system, such as the basis vectors) and your quantities do depend on the coordinate system chosen then this also needs to be underlined, examples being the coordinates themselves or the basis vectors. Anyway, I think we have had this discussion before and we are probably not going to agree this time either ... • Ibix
vanhees71
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My argument is that ##x^{\mu}## and ##x^{\mu'}## are the same four real numbers for both ##\mu## and ##\mu'## running from 0 to 3, while ##x^{\prime \mu}## are (in general) different numbers from ##x^{\mu}## when ##\mu## is running from 0 to 3. When I learnt the theory of relativity first, I remember also to have used a textbook from the library and I didn't understand it only because of this strange notation to mark the changed components by varying the labels rather than the symbol.

Of course a tensor is coordinate independent, and this shows again, how important it is to label the symbols not the indices. For a vector we have e.g.
$$\underline{x}=x^{\mu} \underline{e}_{\mu} = x^{\prime \mu} \underline{e}_{\mu}'.$$
On the other hand, if one gets used to the "labeling-indices convention", maybe it's not such a big issue. I've always avoided it not to confuse myself ;-).

• Ibix
Orodruin
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Instead of just having us repeat the same conversation, let me just link it:

On the other hand, if one gets used to the "labeling-indices convention", maybe it's not such a big issue. I've always avoided it not to confuse myself ;-).
That is interesting. I converted to it after giving it a significant amout of thought. I avoid priming the symbol precisely to not confuse myself or having to write the actual tensor in boldface or in a different font.

• vanhees71
Thanks vanhees71 and Orodruin! Your posts have been a great help, and I think I’m more on my way to understanding the differences between these two notations now. I am reading Schutz GR, which uses one notation, and Srednicki QFT, which uses the other...

Reading for example Carroll, as far as I can understand, said ##{\Lambda^{\mu}}_{ \nu}## the NW-SE index notation for the Lorentz matrix ##\Lambda## he puts by definition:

$${\Lambda_{\nu}}^{ \mu} := {(\Lambda^T)^{\mu}}_{\nu}$$
##\Lambda## is really a bunch of numbers stored in a square matrix and ##\Lambda^T## is its transpose. Then sticking to the NW-SE index notation we represent them as ##{\Lambda^{\mu}}_{ \nu}## and ##{(\Lambda^T)^{\mu}}_{\nu}## respectively. The next step is the 'stipulation' made in the definition above.

Is the above correct ? Thank you.

• Demystifier
It does have motivation though. The transposed operator satisfies ##\eta(\mathbf{u},\Lambda \mathbf{v}) = \eta(\Lambda^T \mathbf{u}, \mathbf{v})## for any ##\mathbf{u},\mathbf{v} \in \mathbf{R}^4##. In index notation, ##\eta_{\mu \nu} u^{\mu} {{\Lambda^{\nu}}_{\rho}} v^{\rho} = \eta_{\rho \nu} {(\Lambda^T)^{\nu}}_{\mu} u^{\mu} v^{\rho}## thereby implying ##\eta_{\mu \nu} {{\Lambda^{\nu}}_{\rho}} = \eta_{\rho \nu}{(\Lambda^T)^{\nu}}_{\mu}##. Applying the inverse metric to both sides gives\begin{align*}
{\Lambda^{\sigma}}_{\rho} = \eta_{\rho \nu} {(\Lambda^T)^{\nu}}_{\mu} \eta^{\mu \sigma}
\end{align*}and now one formally raises and lowers the indices on ##\Lambda^T## by putting ## \eta_{\rho \nu} {(\Lambda^T)^{\nu}}_{\mu} \eta^{\mu \sigma} = {(\Lambda^T)_{\rho}}^{\sigma}##.

In index notation, ##\eta_{\mu \nu} u^{\mu} {{\Lambda^{\nu}}_{\rho}} v^{\rho} = \eta_{\rho \nu} {(\Lambda^T)^{\nu}}_{\mu} u^{\mu} v^{\rho}##.
As far as I can understand we get the above result since ##\eta_{\mu \nu}## is symmetric, i.e. ##\eta_{\mu \nu} = \eta_{\nu \mu}##.

The main point is that even if ## {{\Lambda^{\nu}}_{\rho}}## and ##{(\Lambda^T)^{\nu}}_{\mu}## are not actually (1,1) tensors anyway, as you pointed out, we can formally raise and lower their indexes employing the metric (tensor) ##\eta_{\mu \nu}## or ##\eta^{\mu \nu}##.

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vanhees71
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I think you are overcomplicating things. By definition a Lorentz-trafo matrix obeys
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}_{\sigma}}=\eta_{\rho \sigma}.$$
From this you immediately have
$$\eta_{\rho \sigma} \eta^{\rho \alpha}=\delta_{\sigma}^{\alpha} = (\eta_{\mu \nu} \eta^{\rho \alpha} {\Lambda^{\mu}}_{\rho}) {\Lambda^{\nu}_{\sigma}} = {\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}.$$
This means that
$${(\Lambda^{-1})^{\alpha}}_{\nu} = {\Lambda_{\nu}}^{\alpha}.$$
In matrix notation the last two equations
$$(\hat{\eta} \hat{L} \hat{\eta})^{\text{T}} \hat{L}=\hat{\eta} \hat{L}^{\text{T}} \hat{\eta} = \mathbb{1} \; \Rightarrow \; \hat{L}^{-1} = \hat{\eta} \hat{L}^{\text{T}} \hat{\eta}.$$
Of course I've used ##\eta_{\mu \nu}=\eta_{\nu \mu}##, i.e., ##\hat{\eta}=\hat{\eta}^{\text{T}}## and ##\hat{\eta}^{-1} = (\eta^{\mu \nu})=\hat{\eta}##.

pervect
Staff Emeritus
Hi all, just had a question about tensor/matrix notation with the inverse Lorentz transform. The topic was covered well here, but I’m still having trouble relating this with an equation in Schutz Intro to GR...

Taken together, we seem to have
$$\Lambda_{\overline{\mu}}^{\;\ \beta}=\Lambda^{\beta}_{\;\overline{\mu}}$$
Is this correct? It seems wrong to me, and it seems that I might’ve confused my tensor and matrix indices, I’m just not sure how...

I'm not sure this will help, but I use the convention in MTW, where by definition, one always writes the transformation matrix with indices running from "northwest" to "southeast", i.e. ##\Lambda^a{}_b##. I can dig up the page reference in MTW if I really have to.

Then components of a vector transform as ##\bar{u}^a = \Lambda^a{}_b u^b## and covectors transform as ##\bar{u}_a = \Lambda^b{}_a u_b##. The vector u itself doesn't change upon a change of basis, of course, but the components do. In this sense the transformation matrix ##\Lambda## is not a tensor, it's an object that deals with change of basis, it's not itself a geometric object independent of basis like a vector is. You can't change basis with an object that is blind to them.

The only difficulty with this approach is when one has to communicate with someone follows a convention from a different textbook. I basically attempt to rework foreign notation into a notation I'm comfortable with.

• vanhees71
vanhees71
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Of course you can make a Lorentz transformation a basis-independent object when defining it as a linear map of vectors in the Minkowski-vector space: ##x \mapsto \Lambda x##, which leaves the Minkowski product of any two vectors unchanged, ##(\Lambda x) \cdot (\Lambda y)=x \cdot y##.

The components wrt. an arbitrary basis ##b_{\mu}## and the corresponding dual basis ##b^{\nu}## is then given by
$$y^{\mu} = b^{\mu} y = b^{\mu} \Lambda x = b^{\mu} \Lambda (b_{\nu} x^{\nu}) = {\Lambda^{\mu}_{\nu}} x^{\nu}.$$
Note that ##b_{\mu}## doesn't need to be necessarily a pseud-Cartesian (aka Lorentzian) basis here.

The condition for being a Lorentz transformation follows from
$$(\Lambda x) \cdot (\Lambda y) = (b_{\mu} {\Lambda^{\mu}}_{\rho} x^{\rho}) \cdot (b_{\nu} {\Lambda^{\nu}}_{\sigma} y^{\sigma}) = b_{\mu} \cdot b_{\nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} x^{\rho} y^{\sigma}=g_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} x^{\rho} y^{\sigma} \stackrel{!}{=} g_{\rho \sigma} x^{\rho} y^{\sigma}.$$
Since this must hold for all vectors ##x## and ##y## one must have
$$g_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=g_{\rho \sigma}.$$

Then components of a vector transform as ##\bar{u}^a = \Lambda^a{}_b u^b## and covectors transform as ##\bar{u}_a = \Lambda^b{}_a u_b##.
Sorry I'm confused. In the expression ##\Lambda^b{}_a u_b##, ##\Lambda^b{}_a## is actually the same as ##\Lambda^a{}_b## employed as 'transponse' in the 'dummy index sum' , or is it actually the inverse of ##\Lambda^a{}_b## ?

Don’t forget how to form matrix equations, e.g.

##A_{ij} v_j = (Av)_i##
and
##A_{ij} v_i = A^T_{ji} v_i = (A^T v)_j##

• vanhees71
##A_{ij} v_j = (Av)_i##
and
##A_{ij} v_i = A^T_{ji} v_i = (A^T v)_j##
So ##\Lambda^b{}_a u_b## is really the matrix product of the transponse of the coefficients stored in the matrix ##\Lambda^a{}_b## multiplied by the components ##u_b## (i.e. the covector components in the associated basis in dual space) ?

vanhees71
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Usually the ##u_{\mu}## are put into a "row vector", i.e., the ##1 \times 4##-matrix ##\tilde{u}=(u_0,u_1,u_2,u_3)=u^{\text{T}} \hat{\eta}##. Then the transformation law ##u_{\mu}'=u_{\rho} {\Lambda^{\rho}}_{\nu}## in matrix-vector notation simply reads ##\tilde{u}'=\tilde{u} \hat{\Lambda}##. I don't think that the matrix-vector notation is good in relativity, because it's (a) limited to at most 2nd-rank tensor anyway and (b) the simple notation of co- and contravariant components of tensors through the vertical placement of the indices in the Ricci calculus is lost.

pervect
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Sorry I'm confused. In the expression ##\Lambda^b{}_a u_b##, ##\Lambda^b{}_a## is actually the same as ##\Lambda^a{}_b## employed as 'transponse' in the 'dummy index sum' , or is it actually the inverse of ##\Lambda^a{}_b## ?
MTW's treatment is on pg 66. It relates to the Lorentz transformation specifically, but it should work in general. Their notation was slightly different than mine, so it's probably clearest if I stick to the textbook notation.

MTW said:
The key entities in the Lorentz transformation are the matrices ##\Lambda^\alpha{'}{}_\beta## and ##\Lambda^\beta{}_{\alpha'}##; the first transforms coordinates from an unprimed frame to a primed frame, while the second goes from primed to unprimed".

This is different from what I originally posted and also than vanhees71 notation.

MTW then proceeds to derive a number of results from the coordinate transformations, including how to transform both vectors and one-forms from primed to unprimed frames, and vica-versa.

They note that a transformation from an unprimed frame to a primed frame, followed by / composed with a second transformation from the primed frame back to the unprimed frame, must be an identity operation and write the appropriate math.

While I could type in more specifics of the transformation laws, and will if there is enough interest, I'd rather just summarize their rules.

MTW said:
One need never memorize the index positions in these transformation laws. One need only line up the indices so that (1) free indices on each side of the equation are in the same position; and (2) summed indices appear once up and once down. Then all will be correct! (Note: the indices on ##\Lambda## always run "northwest to southeast.")

If one follows MTW's conventions, all transformation matrices have indexes that run from "northwest to southeast". And, as they note, one gets the correct results using their notation, and there is IMO less room for confusion.

• vanhees71
They note that a transformation from an unprimed frame to a primed frame, followed by / composed with a second transformation from the primed frame back to the unprimed frame, must be an identity operation
So basically in MTW notation the matrix ##{\Lambda ^{\beta}}_{\alpha'}## is actually the inverse of ##{\Lambda ^{\alpha'}}_{\beta}## ?

Note that both ##{\Lambda ^{\beta}}_{\alpha'}## and ##{\Lambda ^{\alpha'}}_{\beta}## are not tensors themselves.

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BTW I noticed that in the book "Spacetime and geometry - an introduction to general relativity" Sean Carroll does not introduce the symbol ##{\Lambda_{\mu'}}^{\nu}## anymore (as he did in his Lecture notes on GR).

See for instance eq 1.50 in the book vs eq 1.37 in the Lecture notes, namely:

##\omega_{\mu'} = {\Lambda^{\nu}}_{\mu'} \omega_{\nu}## vs ##\omega_{\mu'} = {\Lambda_{\mu'}}^{\nu} \omega_{\nu}##

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vanhees71
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MTW's treatment is on pg 66. It relates to the Lorentz transformation specifically, but it should work in general. Their notation was slightly different than mine, so it's probably clearest if I stick to the textbook notation.

This is different from what I originally posted and also than vanhees71 notation.

MTW then proceeds to derive a number of results from the coordinate transformations, including how to transform both vectors and one-forms from primed to unprimed frames, and vica-versa.

They note that a transformation from an unprimed frame to a primed frame, followed by / composed with a second transformation from the primed frame back to the unprimed frame, must be an identity operation and write the appropriate math.

While I could type in more specifics of the transformation laws, and will if there is enough interest, I'd rather just summarize their rules.

If one follows MTW's conventions, all transformation matrices have indexes that run from "northwest to southeast". And, as they note, one gets the correct results using their notation, and there is IMO less room for confusion.
I find the notation, where you put the primes on the index rather than the symbol denoting the tensor components the most confusing idea ever. Of course, if one knows that an author uses this confusing notion, it's easy to put it on the symbol in my mind, and then all makes sense again ;-)).

I find the notation, where you put the primes on the index rather than the symbol denoting the tensor components the most confusing idea ever.
As others said, the 'geometrical object' itself does not change. What changes are its components in a different basis. So the name of the object itself should not change, I believe.

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vanhees71
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I disagree, because an index is just taking the values 0, 1, 2, 3, no matter how it's named. For me the logic is to have a set of basis vectors ##\vec{b}_k## and another set of basis vectors ##\vec{b}_l'##. Correspondingingly I also have to name the comonents of a vector ##\vec{V}## with ##V^k## and ##V^{\prime l}## such that
$$\vec{V}=V^k \vec{b}_k = V^{\prime l} \vec{b}_l'.$$

pervect
Staff Emeritus
So basically in MTW notation the matrix ##{\Lambda ^{\beta}}_{\alpha'}## is actually the inverse of ##{\Lambda ^{\alpha'}}_{\beta}## ?

Note that both ##{\Lambda ^{\beta}}_{\alpha'}## and ##{\Lambda ^{\alpha'}}_{\beta}## are not tensors themselves.
That's my interpretation. What they actually write is:
[/quote=MTW]
$$\Lambda^{\alpha'}{}_{\beta} \Lambda^{\beta}{}_{\gamma'} = \delta^{\alpha'}{}_{\gamma'}$$
[/quote]

That's my interpretation. What they actually write is:
$$\Lambda^{\alpha'}{}_{\beta} \Lambda^{\beta}{}_{\gamma'} = \delta^{\alpha'}{}_{\gamma'}$$
At the end of the day it is the same approach taken from Sean Carroll in the book "Spacetime and geometry - an introduction to general relativity" (not in his Lecture notes on GR).

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vanhees71
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But that's not right. What this equation says in matrix notation is that ##\hat{\Lambda}^2=\mathbb{1}##, which would mean ##\hat{\Lambda}^{-1}=\hat{\Lambda}##, but that's not correct for almost all Lorentz matrices. Correct of course is ##\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta}## or ##\hat{\Lambda}^{-1}=\hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta}=\hat{\Lambda}##.

Ibix
But that's not right.
It is right in their notation, by definition. They are using ##\Lambda## for any Lorentz transform, and using decorators on the indices to identify which two frames it's transforming between. The swapping of the prime from one index to the other tells you that the two transforms are inverses.

I'm not sure I like the notation, but it doesn't seem to be ambiguous taken on its own terms.

vanhees71
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I always struggle with it when using MTW (which nevertheless is of course a great book).

It is right in their notation, by definition. They are using ##\Lambda## for any Lorentz transform, and using decorators on the indices to identify which two frames it's transforming between.
Yes, it took to me a long time to grasp it: same symbol ##\Lambda## for different Lorentz transformations !

vanhees71
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