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Hurkyl
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#16
Jun15-08, 09:32 PM
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Recall the basic computational formula for the coordinate representation [T] of a linear transformation T:
[T(v)] = [v] [T]
(Incidentally, this is the main reason I prefer the convention where vectors are column vectors; so that this identity isn't 'backwards')

Or, with indices, if w = T(v), then [itex]w^i = T^i_j v^j[/itex]

If you want to try working it out again, then don't read below this point.






Okay, the calculations work out to:
[tex][v]_{B'} [G]_{B'} = [G(v)]_B' = [G(v)]_B \Lambda = [v]_B [G]_B \Lambda = [v]_{B'} \Lambda^{-1} [G]_B \Lambda[/tex]
and so
[tex][v]_{B'} \left( [G]_{B'} - \Lambda^{-1} [G]_B \Lambda \right) = 0[/tex]
which implies, because this is true for all coordinate tuples [itex][v]_{B'}[/itex],
[tex][G]_{B'} = \Lambda^{-1} [G]_B \Lambda[/tex]


Your observation was correct -- this is different than the constraint we needed on G in order for our naive attempt at defining an inner product to work!


The notion of an inner product clearly make sense -- but it is not so clear that inner products bear a nice relation to linear transformations of tangent vectors. Now that we've uncovered a contradiction, how would you proceed in developing differential geometry?