From: ram@zedat.fu-berlin.de   
      
   ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:   
   >[[Mod. note -- I think that last subscript "mu" should be a "nu".   
   >That is, equations (0) and (1) should read (switching to LaTeX notation)   
   >$X := p_\mu p^\mu   
   >   =  p_\mu \eta^{\mu\nu} p_\nu$   
   >-- jt]]   
      
     Thanks for that observation!   
      
     In the meantime, I found the answer to my question reading a text   
     by Viktor T. Toth.   
      
     Many Textbooks say,   
      
                 ( -1  0  0  0 )   
   eta_{mu nu} = (  0  1  0  0 )   
                 (  0  0  1  0 )   
                 (  0  0  0  1 ),   
      
     but when you multiply this by a column (contravariant) vector,   
     you get another column (contravariant) vector instead of a row,   
     while the "v_mu" in   
      
   eta_{mu nu} v^nu = v_mu   
      
     seems to indicate that you will get a row (covariant) vector!   
      
     As Viktor T. Toth observed in 2005, a square matrix (i.e.,   
     a row of columns) only really makes sense for eta^mu_nu (which is   
     just the identity matrix). He then clear-sightedly explains that   
     a matrix with /two/ covariant indices needs to be written not   
     as a /row of columns/ but as a /row of rows/:   
      
   eta_{mu nu} = [( -1 0 0 0 )( 0 1 0 0 )( 0 0 1 0 )( 0 0 0 1 )]   
      
     . Now, if one multiplies /this/ with a column (contravariant)   
     vector, one gets a row (covariant) vector (tweaking the rules for   
     matrix multiplication a bit by using scalar multiplication for   
     the product of the row ( -1 0 0 0 ) with the first row of the   
     column vector [which first row is a single value] and so on)!   
      
     Exercise  Work out the representation of eta^{mu nu} in the same   
     spirit.   
      
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