5
$\begingroup$

I am following lecture notes on SR. The author writes that the following is equivalent:

$$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda^\mu}_\rho{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}. \tag{1}$$ This surprises me, because

$$ {(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu.\tag{2} $$

And so I expected it to be $$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda_\rho}^\mu{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}.\tag{3}$$ Why is this wrong?

Improve this question
$\endgroup$
3
  • $\begingroup$ I think you're put an extra flip in there somewhere. Just think about it like matrices: $ABC = D$ means $A_{ij} B_{jk} C_{k\ell} = D_{i\ell}$. Therefore $A^T BC = D$ means $B_{jk} A_{ji} C_{k\ell} = D_{i\ell}$ where I used the commutativity of multiplication, and replaced $A_{ij}$ with $A_{ji}$. This is exactly the same as equation (1). $\endgroup$
    – knzhou
    Commented Apr 29, 2017 at 23:08
  • 5
    $\begingroup$ The upstairs/downstairs index placement is a red herring that can only add confusion; it's not a 'real' index because $\Lambda$ is not a tensor. $\endgroup$
    – knzhou
    Commented Apr 29, 2017 at 23:09
  • $\begingroup$ @knzhou By your first comment, I believe you confirm that you're puzzled too, correct? Flipping the indecies is exactly what I propose, but that's what the lecture notes do not. $\endgroup$
    – Mikkel Rev
    Commented Apr 29, 2017 at 23:21

1 Answer 1

Reset to default
12
$\begingroup$

  1. OP's three equations should read $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad (\Lambda^T)_{\rho}{}^{\mu}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} ,\tag{1'}$$ $$ (\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu} ,\tag{2'} $$ $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad \Lambda^{\mu}{}_{\rho}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} .\tag{3'}$$

  2. In more detail: Let $V$ be $n$-dimensional $\mathbb{R}$-vector space with a basis $(e_{\mu})_{\mu=1, \ldots, n}$. Let $V^{\ast}$ be the dual vector space with the dual basis $(e^{\ast \nu})_{\nu=1, \ldots, n}$. Let $$\Lambda~=~e_{\mu}~ \Lambda^{\mu}{}_{\nu}\otimes e^{\ast \nu}~ \in~V\otimes V^{\ast}~\cong~{\cal L}(V;V)$$ be a linear map from $V$ to $V$. Let us call the positions of the indices on $\Lambda^{\mu}{}_{\nu}$ for the NW-SE convention, cf. a compass rose. Let $$\Lambda^T~=~e^{\ast \nu}~ (\Lambda^T)_{\nu}{}^{\mu}\otimes e_{\mu}~\in~V^{\ast}\otimes V~\cong~{\cal L}(V^{\ast};V^{\ast})$$ be the transposed linear map from $V^{\ast}$ to $V^{\ast}$. Note that $(\Lambda^T)_{\nu}{}^{\mu}$ is written in the SW-NE convention. Let $$\eta~=~e^{\ast \mu}~\eta_{\mu\nu}\odot e^{\ast \nu}~\in~{\rm Sym}^2V^{\ast}~=~V^{\ast}\odot V^{\ast}$$ be an (indefinite) metric, i.e. an invertible element in the symmetrized tensor product. A (pseudo)orthogonal map $\Lambda$ satisfies by definition $$\Lambda^T\eta\Lambda~=~\eta.$$ See also this related Phys.SE post.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Wouldn't you have then an ambiguous notation? Wouldn't you use ${\Lambda_\mu}^\nu$ for refer to both $\Lambda^T$ and $\Lambda^{-1}$? $\endgroup$
    – falgenint
    Commented May 8, 2017 at 9:41
  • $\begingroup$ @falgenint: Thanks for the feed-back. $\Lambda_{\mu}{}^{\nu}$ does by definition not refer to $(\Lambda^T)_{\mu}{}^{\nu}$ to avoid ambiguity. Rather we use the metric to raise and lower indices, so by definition $\Lambda_{\mu}{}^{\nu}~:=~\eta_{\mu\rho}~\Lambda^{\rho}{}_{\sigma}~(\eta^{-1})^{\sigma\nu}$, which happens to be equal to $((\Lambda^{-1})^T)_{\mu}{}^{\nu}$ if $\Lambda$ is (pseudo)orthogonal. $\endgroup$
    – Qmechanic ♦
    Commented May 8, 2017 at 10:05
  • $\begingroup$ Sorry, I made a mistake. I wanted to say: wouldn't you use ${\Lambda^\mu}_\nu$ for refer to both $\Lambda^T$ and $\Lambda^{-1}$? I mean, if (using (2')) this is correct: ${((\Lambda^{-1})^T)_{\mu}}^{\nu}={(\Lambda^{-1})^\nu}_\mu={\Lambda_\mu}^\nu$ it follows ${((\Lambda^{-1})^T)_\mu}^{\nu}={((\Lambda^{T})^{-1})_\mu}^{\nu}$ and from this you can demonstrate that $\Lambda \Lambda^T=I$ with $I$ the identity matrix. Is this correct? $\endgroup$
    – falgenint
    Commented May 8, 2017 at 11:06
  • $\begingroup$ The last equation is not correct. When removing indices from a tensor equation, one should re-introduce/put back the metric tensor $\eta$ in pertinent places. $\endgroup$
    – Qmechanic ♦
    Commented May 8, 2017 at 11:59
  • $\begingroup$ So I guess I can't say that $(\Lambda^{-1})^T=(\Lambda^T)^{-1}$ although ${((\Lambda^{-1})^T)_\mu}^{\nu}={((\Lambda^{T})^{-1})_\mu}^{\nu}$? $\endgroup$
    – falgenint
    Commented May 8, 2017 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.