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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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    thanks I edited the question2012-08-19

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