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Is the space of $k$-forms on a compact Riemannian manifold $M$ with the inner product given by $$(\alpha,\beta)=\int \alpha \wedge *\beta=\int g(\alpha,\beta)dv$$ which is called «$L^2$ product in $\Omega^k$» equivalent to the space $L^2$ on $M$?

Sorry I didn't understand properly what was the space $L^2(\Omega^p(M))$ so my question was misleading. Now I think I do, and it is the space of $p$-forms that have coefficients in $L^2$ in a any chart. My question would be then that if we extend what I called the «L^2 product» to the direct sum $\Omega(M):=\sum \Omega^p$ by saying that $\Omega^p$ is orthogonal to $\Omega^q$ for $p \neq q$ and then to $L^2(\Omega(M))$ (I think you can do this because on a compact manifold the smooth forms are dense in the $L^2$ ones right?) then what it is the relation of this space with the space $L^2(M)$. Is it a direct sum of copies of it?? thanks

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What exactly do you mean by equivalent? Both are Hilbert spaces, to there will probably an isomorphism (not a natural one, I'd assume), but I guess that's about it. Why should it, anyway? $k-$ forms and functions on manifolds are quite different kinds of objects. – user20266 Aug 19 '12 at 11:42
thanks I edited the question – inquisitor Aug 19 '12 at 17:37

They are the same when $k=0$, and can be identified if $k=\mathrm{dim}\,M$, but in general different.

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