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In the appendix of the textbook of Group Theory in Physics by Wu-Ki Tung, the transpose of a matrix is defined as the following, Eq.(I.3-1)

$${{A^T}_i}^j~=~{A^j}_i.$$

This is extremely confusing for me, since in the case of Lorentz transformation ${\Lambda_\nu}^\mu$ is considered in the text (eg. Ch.10) as a matrix, and one can show that (eg. see Eq.(2.3.10) of The Quantum Theory of Fields Vol.1 by Steven Weinberg)

$${{(\Lambda^{-1})}^\nu}_\mu ~=~g_{\mu\sigma}{\Lambda^\sigma}_\alpha g^{\alpha\nu}.$$

In particular, it is defined on the very same line (of the above equation in Weinberg)

$${\Lambda_\mu}^\nu ~=~ {{(\Lambda^{-1})}^\nu}_\mu.$$

The above definition is quite natural, since it can be viewed as that the metric tensors $g_{\mu\nu}$ were used to raise and lower and corresponding subscript and superscript of the original ${\Lambda^\sigma}_\alpha$.

Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\Lambda_\mu}^\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix. However, it seems confusing, since in the book of Tung, it is mentioned explicitly that $g_{\mu\nu}$ can be used to obtain covariant tensor from contravariant tensor (see Appendix I) and ${\Lambda^\sigma}_\alpha$ is treated as a tensor (see for instance Ch.10). So it seems there is some confliction or how should one correctly understand the meaning of transpose defined in (I.3-1), which can be rewritten as $${{A}_i}^j~=~{(A^T)^j}_i.$$

Here is a summary of my confusion: it comes from the two ways that ${\Lambda_\mu}^\nu$ is related to the original matrix ${\Lambda^\nu}_\mu$. (1) Tung implies that ${\Lambda_\mu}^\nu = {(\Lambda^T)^\nu}_\mu$, and (2) ${\Lambda_\mu}^\nu \equiv g_{\mu\sigma}{\Lambda^\sigma}_\alpha g^{\alpha\nu} = {{(\Lambda^{-1})}^\nu}_\mu $, provided one treats ${\Lambda_\mu}^\nu$ as a mixed tensor. The question is: are they consistent?

According to Oscar Cunningham's explanation, I understand that the definition introduced in Tung's textbook leads to some contradiction.

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  • $\begingroup$ Comment to the question (v4): It seems Weinberg is not discussing the definition of a transposed matrix, but merely uses properties of a Lorentz transformation. $\endgroup$
    – Qmechanic ♦
    Commented Nov 2, 2014 at 13:39
  • $\begingroup$ Exactly, however Weinberg defines the symbol ${\Lambda_\mu}^\nu$ as inverse of a matrix, which is used in Tung to define the transpose of the (same) matrix. $\endgroup$
    – gamebm
    Commented Nov 2, 2014 at 17:43
  • $\begingroup$ No, Weinberg is not giving a definition (beyond lowering and raising indices with the metric). He is using a property of Lorentz matrices. $\endgroup$
    – Qmechanic ♦
    Commented Nov 2, 2014 at 18:11
  • $\begingroup$ Please correct me if I am wrong. ${{(\Lambda^{-1})}^\nu}_\mu ~=~g_{\mu\sigma}{\Lambda^\sigma}_\alpha g^{\alpha\nu}$ is a property, then ${\Lambda_\mu}^\nu ~=~ {{(\Lambda^{-1})}^\nu}_\mu$ says it is consistent with "obtaining ${\Lambda_\mu}^\nu$ by raising and lowering indices of ${\Lambda^\sigma}_\alpha$". The latter can be (sort of) viewed as a definition of ${\Lambda_\mu}^\nu$, especially looked side by side with the conventions introduced by Tung (I.3-1). $\endgroup$
    – gamebm
    Commented Nov 2, 2014 at 18:20
  • $\begingroup$ No, Tung is defining $(\Lambda^T)_{\mu}{}^{\nu}$, which in principle is different from $\Lambda_{\mu}{}^{\nu}$. $\endgroup$
    – Qmechanic ♦
    Commented Nov 2, 2014 at 18:25

2 Answers 2

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Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\Lambda_\mu}^\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix.

The symbol $\Lambda_\mu{}^\nu$ is not defined to be the transpose of the original matrix. The transpose of the original matrix is ${\Lambda^T}_\nu{}^\mu$ (assuming that the original matrix is $\Lambda^\mu{}_\nu$). You have to keep the "$^T$". So long as you use "$^T$" to tell the difference between the matrix and its transpose, everything should work out with no inconsistencies.

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  • $\begingroup$ I understood the following: first rewrite (I.3-1) as ${A_i}^j={(A^T)^j}_i$, then use it as a definition of (element $(i,j)$ of) a new form of matrix on the left (at this moment I do not assume it can be obtained by raising and lowering indices) by the (element $(i,j)$ of original matrix ${B^j}_i$ on the) right hand side of the expression. The above "definition" is compared to that in Weinberg. Since intuitively (which can be the origin of my misunderstanding), one may define the following ${(A^T)^j}_i = {A^i}_j$ and ${(A^T)_j}^i = {A_i}^j$, which is obviously different from that in Tung. $\endgroup$
    – gamebm
    Commented Nov 3, 2014 at 11:54
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    $\begingroup$ You can't have ${(A^T)^j}_i={A^i}_j$ because you can only set upper indices equal to upper indices and vice-versa. (The reason for this rule is that the indices transform differently, so if you set them equal in one basis they wouldn't be equal in another.) $\endgroup$
    – Oscar Cunningham
    Commented Nov 3, 2014 at 12:06
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    $\begingroup$ For me there are two things: matrix and tensor. For a tensor, one may only lower a superscript using a metric. But I was thinking in terms of matrix (and forget about tensor for a sec). For a matrix $M_{i,j}$, its transpose is $M^T_{i,j}=M_{j,i}$, where $M_{i,j}$ is the $(i,j)$ component of a matrix $M$. Now let us adopt the convention that the $(i,j)$ component of a matrix is denoted by ${\Lambda^i}_j$ (and pretend that we know nothing about tensor, and in fact, the definition of matrix and that of its transpose do not depend on that of tensor). $\endgroup$
    – gamebm
    Commented Nov 3, 2014 at 12:20
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    $\begingroup$ So what is the $(i,j)$ component of the tranpose of this matrix? To me, it will be ${\Lambda^j}_i$. Therefore ${(\Lambda^T)^i}_j={\Lambda^j}_i$ $\endgroup$
    – gamebm
    Commented Nov 3, 2014 at 12:20
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    $\begingroup$ I am sorry I did not understand your explanation. (You meant that the original matrix is ${\Lambda^\nu}_\mu$, right?) If ${\Lambda_\mu}^\nu$ is not defined as the transpose of ${\Lambda^\nu}_\mu$. What shall I understand ${\Lambda_\mu}^\nu = {(\Lambda^T)^\nu}_\mu$ (or ${(\Lambda^T)_\mu}^\nu = {\Lambda^\nu}_\mu$)? Thanks! $\endgroup$
    – gamebm
    Commented Nov 3, 2014 at 17:23
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In terms of $4\times 4$ matrices, elements of the representation of SO(3,1) group must obey the following relation:

$$ g\cdot A^{T} \cdot g = A^{-1}, $$

where $g = \text{diag}(1, -1, -1, -1) = g^{-1}$.

Does it answer your question?

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  • $\begingroup$ Thanks for the answer. As I understood, when expressed in matrix components, the equation you wrote is exactly ${{(\Lambda^{-1})}^\nu}_\mu ~=~g_{\mu\sigma}{\Lambda^\sigma}_\alpha g^{\alpha\nu}$. Then no. I edited the question, trying to make its statement more clearly. $\endgroup$
    – gamebm
    Commented Nov 2, 2014 at 17:48
  • $\begingroup$ @gamebm you speak of defining $A^T$ and $A^{-1}$ as is you have some freedom in those definitions. The truth is they are already defined for any $A$. Can you reformulate the essence of your question, again? $\endgroup$
    – Prof. Legolasov
    Commented Nov 2, 2014 at 19:12
  • $\begingroup$ My question is concerning the convention introduced in the textbook by Tung. So it concerns some specific notation used in that textbook. It is defined (or at least, equivalently) there ${{A}_i}^j~=~{(A^T)^j}_i$, which is (to me) a convention to rewrite the component of the transpose of a matrix in terms of component of another matrix, however, those two matrices are (also) related to each other by the notations for (transform between) covariant and contravariant vectors, so it seems to me that something is not quite consistent (due to the lack of freedom). $\endgroup$
    – gamebm
    Commented Nov 2, 2014 at 20:45
  • $\begingroup$ Everything is consistent. You just have to remember that ${A_i}^j \neq {A^j}_i$. The order of indices matters. $\endgroup$
    – Prof. Legolasov
    Commented Nov 2, 2014 at 21:43
  • $\begingroup$ Yes, I am aware of this. The confusion comes from the two ways that ${\Lambda_\mu}^\nu$ is related to the original matrix ${\Lambda^\nu}_\mu$. (1) According to Tung, which literally says ${\Lambda_\mu}^\nu = {(\Lambda^T)^\nu}_\mu$, and (2) ${\Lambda_\mu}^\nu \equiv g_{\mu\sigma}{\Lambda^\sigma}_\alpha g^{\alpha\nu} = {{(\Lambda^{-1})}^\nu}_\mu $ if one treats ${\Lambda_\mu}^\nu$ as a mixed tensor. Are they consistent? $\endgroup$
    – gamebm
    Commented Nov 3, 2014 at 12:02

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