2 slightly changed question to fit the context given by answers

On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.

I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?

I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).

So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.)

EDIT: Thanks, Petya, I guess I need to restrict to smooth singular chains. Also thanks Gonçalo for pointing out that I appear to really be talking about currents - I will take a look at the book! My remaining question: first of all, it seems to me like in the context of a compact Riemannian manifold, the space of k-currents is naturally identified with the space of k-forms via the inner product. So in this space, is the set of currents given by integration over a smooth k-submanifold a proper subspace? I understand that the point of currents is that in the general case they are broader, but in the compact case it seems like maybe that doesn't happen, and I'm having difficulty understanding the statements about the mass norm that seem to concern this question.

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# Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds

On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.

I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?

I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).

So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.)