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 Emeritus Sci Advisor PF Gold P: 16,101 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 $w^i = T^i_j v^j$ If you want to try working it out again, then don't read below this point. Okay, the calculations work out to: $$[v]_{B'} [G]_{B'} = [G(v)]_B' = [G(v)]_B \Lambda = [v]_B [G]_B \Lambda = [v]_{B'} \Lambda^{-1} [G]_B \Lambda$$ and so $$[v]_{B'} \left( [G]_{B'} - \Lambda^{-1} [G]_B \Lambda \right) = 0$$ which implies, because this is true for all coordinate tuples $[v]_{B'}$, $$[G]_{B'} = \Lambda^{-1} [G]_B \Lambda$$ 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?