- #106
Careful
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Are you pretending now that you never heard of a correspondence between a certain class of frames and a Riemannian metric ?? BTW selfadjoint, the map \beta is NOT smooth at all - if you would care to look at it you will see that there is an infinity replaced by a zero.selfAdjoint said:Once you have the inner product on V, say <,> you know there is a unique map [tex]\alpha^*[/tex] satisfying [tex]\alpha^*(v) = \langle v,\alpha(v) \rangle[/tex]. This is independent of any basis. Of course you can exhibit it in any basis but that is not part of its definition. It is as smooth as [tex]\alpha[/tex] but clearly no smoother; if [tex]\alpha[/tex] is not injective at some point [tex]\sigma \in V[/tex] then [tex]\langle \sigma,\alpha(\sigma)\rangle[/tex] is clearly undefined. Smoothness of [tex]\beta [/tex] is obtained by the [tex]\chi[/tex] approximation.
**if [tex]\alpha[/tex] is not injective at some point [tex]\sigma \in V[/tex] then [tex]\langle \sigma,\alpha(\sigma)\rangle[/tex] is clearly undefined. **
Utterly false: if we assume V=W then this expression is perfectly well defined for all \alpha.
I promise you that when I make a stupid mistake I shall notify you so that you may scold upon me. But please, stop this ridiculous game about first grade algebra.
Cheers,
Careful