The metric tensor in SR obeys the transformation law (I am using Schutz's bar notation for different frame indices):


If I'm not wrong, the second Lorentz matrix can be contracted with the metric like so:


This seems wrong since the Lorentz matrix elements are usually written with one upper and one lower index, and I don't understand how $\Lambda_{\mu\bar{\beta}}$ could be written in a matrix representation.

From various sources, I am also told that the above equations are equivalent to matrix multiplication with the transpose of $\Lambda$:


  • Does contracting the Lorentz matrix transpose it?

  • What are the differences between $\Lambda^\mu_{~\bar{\nu}}$, $\Lambda^\bar{\mu}_{~\nu}$, $\Lambda_{\mu\bar{\nu}}$ and $\Lambda^{\mu\bar{\nu}}$? Which of them are inverses and which are transposes?

  • $\begingroup$ Would you prefer a mathematical insight as to what's going on or a more pragmatic approach as to how the calculations in the index notations work out? $\endgroup$
    – KilianM
    Dec 30, 2020 at 10:35
  • $\begingroup$ @KilianM I understand how to apply Einstein's Summation Convention so I get the answers through calculation. I'm more confused about the intuition and yes, insight, behind what's going on - like the effects of index lowering/raising on the matrix representation $\endgroup$ Dec 30, 2020 at 15:13

3 Answers 3


Note that $\Lambda$ is not itself a tensor. It is a matrix of transformation coefficients.

A tensor is well-defined in the absence of any specified frame, and its components can be defined in one frame or another. The Lorentz transformation, by contrast, is used to calculate how tensor components change from one frame to another.

Contracting one up- and one down-index on a second-rank tensor gives a zero-rank tensor, which is to say an invariant scalar. Contracting one up- and one down-index on a non-tensor such as $\Lambda^a_{\; b}$ will give some sort of mathematical result, but it is not guaranteed to be a quantity of any particular interest. Similarly, the combination of metric with $\Lambda^a_{\; b}$ can be used to get a quantity with all indices down, but it is not a tensor and it is best not to play these index gymnastics with non-tensors. Just let $\Lambda^a_{\; b}$ be what it is.

For a second-rank tensor, lowering the first index corresponds to pre-multiplying by the metric in matrix language; lowering the second index corresponds to post-multiplying by the metric. The transpose operation corresponds to reversing the order of the indices (N.B. without moving them up or down). For example $$ \Lambda^a_{\; \mu} \Lambda^b_{\; \nu} F^{\mu\nu} = \Lambda^a_{\; \mu} F^{\mu\nu} \Lambda^b_{\; \nu} = \Lambda^a_{\; \mu} F^{\mu\nu} (\Lambda^{\! \rm T})^{\;\; b}_{\nu} $$ and in matrix notation this combination can be written $$ \Lambda F \Lambda^{\! \rm T} $$

  • $\begingroup$ I think I get that you shouldn't play around too much with lowering/raising indices on the Lorentz matrix, but I was wondering more about how the act of lowering/raising is interpreted as transformations of the matrix representation i.e. Transposing/Inversing the matrix. How does one get the transposed result from the lowered index $\Lambda_{\mu\bar{\nu}}$ above? $\endgroup$ Dec 31, 2020 at 3:49
  • $\begingroup$ @HexiangChang I added some more information to address these points. $\endgroup$ Dec 31, 2020 at 9:32

First a brief introduction to tensors. An $(r,s)$ tensor $t$ on a $K$-vector space $V$ is just a multilinear map from $s$ copies of $V$ and $r$ copies of the dual vector space $V^*$ to the underlying field $K$: $$t:\underbrace{V^*\times...\times V^*}_{r-times}\times\underbrace{V\times...\times V}_{s-times}\to K$$ The dual vector space is just the set of all linear maps from $V$ to $K$. The set of all $(r,s)$ tensors on $V$ is commonly denoted $T^r_s(V)$.

Ok, how do we get from these vectors and linear maps to these numbers $g_{\mu\nu}$ that we're working with?

Well, we do what we always intuitively do namely, choose a basis. Say we chose some basis $\{e_i\}\subset V$ on our vector space, we can now express every vector $v\in V$ as a linear combination of these basis vectors: $v=v^ie_i$. The $v^i\in K$ are basically numbers and the position of the index so far is just convention. We can now do the same for the dual vector space, with a particularly clever choice of basis $\{e^i\}\subset V^*$, such that $e^{i}(e_j)=\delta^i_j$. Again, the position of the index is just convention in order not to mix up vectors and dual vectors, as these are fundamentally different objects!

With this index convention, we would now write the components of a vector as $v^i$ and of a dual vector as $v_i$.

For the tensors, it is very similar: $$ t(w_k^{(1)}e^k,...,w_l^{(r)}e^l,v^i_{(1)}e_i,...,v^j_{(s)}e_j)=w_k^{(1)}...w_l^{(r)}v^i_{(1)}...v^j_{(s)}t(e^k,...,e^l,e_i,...,e_j) $$ The $t(e^k,...,e^l,e_i,...,e_j)\equiv t^{k...l}_{i...j}$ are again just numbers, independent of what vectors and covectors the tensor is acting on (just the basis we chose). Similarly to the vectors and covectors, we can also express the tensor in terms of its components: $$ t = t^{k...l}_{i...j}e_k\otimes...\otimes e_l\otimes e^i\otimes...\otimes e^j $$

Ok now, what happens when we choose a different basis $\{b_i\}\subset V$ and a corresponding dual basis $\{b^i\}\subset V^*$? Since our new basis vectors (and new basis covectors) are still elements of the same underlying (dual-)vector space, we can express them as some linear combination: $$b_i = A_i^je_j \qquad \text{and} \qquad b^i=B^i_je^j$$ Similar to above, we can view the numbers $A_i^j$ and $B^i_j$ as components of a tensor: $$ A=A_i^je_j\otimes b^i \qquad \text{and} \qquad B=B^i_jb_i\otimes e^j $$ note: In physics, we typically don't view these transformations as tensors, however in I think this is quite useful for our discussion here. For more information see, e.g. (Is Lorentz transform a tensor?)

Now lastly what does contracting two tensors actually mean? Tensor contractions are really just defined for an individual tensor and are linear maps $C^k_l:T^r_s(V)\to T^{r-1}_{s-1}(V)$ defined by: $$ T^{\nu_1...\nu_r}_{\mu_1...\mu_s}e_{\nu_1}\otimes...\otimes e_{\nu_r}\otimes e^{\mu_1}\otimes...\otimes e^{\mu_s}\\ \mapsto T^{\nu_1...\nu_r}_{\mu_1...\mu_s} e^{\mu_l}(e_{\nu_k}) e_{\nu_1}\otimes...\otimes e_{\nu_{l-1}}\otimes e_{\nu_{l+1}}\otimes...\otimes e_{\nu_r}\otimes e^{\mu_1}\otimes...\otimes e^{\mu_{k-1}}\otimes e^{\mu_{k+1}}\otimes...\otimes e^{\mu_s}) $$ But thats no problem, since we can make one tensor out of two using the tensor product. In your particular case we have the "tensor" $\Lambda=\Lambda^{\nu}_{\mu}e_{\nu}\otimes b^{\mu}$ and the tensor $\eta=\eta_{\mu\nu}e^{\mu}\otimes e^{\nu}$: $$ \eta\otimes\Lambda = \eta_{\mu\nu}\Lambda^{\alpha}_{\beta}e^{\mu}\otimes e^{\nu}\otimes e_{\alpha}\otimes b^{\beta} $$ and $$ C^{\alpha}_{\nu}(\eta\otimes\Lambda)=\eta_{\mu\nu}\Lambda^{\alpha}_{\beta}e^{\nu}(e_{\alpha})e^{\mu}\otimes b^{\beta} = \eta_{\mu\nu}\Lambda^{\nu}_{\beta}e^{\mu}\otimes b^{\beta}\equiv \Lambda_{\mu\beta} e^{\mu}\otimes b^{\beta} $$ In GR (and SR) space-time is a Lorentzian manifold $L$, the vector space of interest is the tangent space $T_pL$ at some point $p\in L$ and the tensors you're considering have special meaning, however (as always) the maths doesn't care too much about the physics.

  • $\begingroup$ Why are you allowed to treat the Lorentz matrix as a tensor? Does your result change when it isn't treated a tensor? ( $\endgroup$ Dec 31, 2020 at 3:45
  • $\begingroup$ You can construct a tensor from the components $\Lambda^{\mu}_{\nu}$ in the way i described it, however since it is then a tensor product of the old and new basis (co)vectors the components do not transform as we expect under change of basis. However when you want to talk precisely about these mathematical ideas like contraction, etc. it is quite hellpful to have a precise understanding of what we're doing. $\endgroup$
    – KilianM
    Dec 31, 2020 at 9:10

This took me quite a while to get at the beginning of relativism class but this may help: a general matrix multiplication like we all know it looks like this: $$[A][B] = \sum a_{ri}b_{ir}$$ Now the relativity version with einstein summation convenction would make this matrix multiplication look like this: $$[A][B] = a_{\nu\sigma}b^{\sigma\nu}$$ As you see this is the same as the previous statement with just the summation symbol dropped, and thus with the first index representing the row and the second representing the column. Now there are possibly people who have an explanation for how a (2,0) (i.e 'covariant') tensor is different from a (0,2) (i.e 'contravariant') tensor, however I just keep the rule in my head that you can only contract covariant with contravariant indices, and thus apply a metric of the form $g_{\mu \mu}$ ($g^{\mu \mu}$) if it is necessary to lower (rise) indices. To get to the form of matrices it is necessary to convert the tensor back to a matrix-form by contracting it with a metric. I will use the minkowski metric $\eta$ here for the demonstration. This brings us back to your question, your equation can be written as: $$\Lambda_{\sigma\alpha}\eta^{\mu \sigma}\Lambda^{\nu\sigma}\eta_{\beta\sigma}\eta_{\mu\nu} = \Lambda_{\sigma\alpha}\eta_{\beta\sigma}\Lambda^{\nu\sigma}\delta^{\sigma}_{\nu} = \Lambda_{\nu\alpha}\eta_{\beta\sigma}\Lambda^{\nu\sigma} \stackrel{\nu \leftrightarrow \beta}{=} [\Lambda]\cdot[\eta]\cdot[\Lambda]^T$$ Where I've used the fact that $$\eta^{\mu\sigma}\eta_{\beta\sigma} = \delta^{\mu}_{\sigma}$$ And just basic rising and lowering of indices via: $$\Lambda^{\mu}_{\;\alpha} = \Lambda_{\sigma\alpha}\eta^{\mu \sigma} \text{ and } \Lambda_{\mu}^{\;\alpha} = \Lambda^{\sigma\alpha}\eta_{\mu \sigma}$$

I could've made an index mistake somewhere but I hope this gives a kind of understanding and I hope it helps.

  • $\begingroup$ It's not really just 'naming' because the objects you're working with aren't matrices, but tensors. For calculations, it is most convenient to just work with the components of these tensors in a given basis and the lower and upper indices then have a precise meaning. $\endgroup$
    – KilianM
    Dec 30, 2020 at 11:05
  • $\begingroup$ I edited it, is it more correct now? $\endgroup$
    – nemo
    Dec 30, 2020 at 11:08
  • 1
    $\begingroup$ Please don't use two pairs of identical indices in one expression $\endgroup$
    – kaylimekay
    Dec 30, 2020 at 11:11

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