Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.