# Hilbert manifold embedding in a Hilbert space

I've been working with Hilbert manifolds lately, and stumbled on the theorem which says that every smooth Hilbert manifold can be embedded in its model space. My question is if there is a way to explicitly give this embedding. The case I'm interested in is that of the loopspace of a manifold M.

• What do you mean by "explicitly"? This would depend on "how explicitly" your manifold is given to you. – Moishe Kohan Nov 12 '17 at 1:13
• Well i mean explicity in the case of the loopspace of a manifold. does a formula for the embedding exist? – Netivolu Nov 12 '17 at 9:22
• It's still unclear to me. However, if you are given an embedding $M\to R^n$, then the loop space $L(M)$ is a submanifold in the pre-Hilbert space $C(S^1, R^n)$ (with the $L_2$-norm); then take the completion of the latter. – Moishe Kohan Nov 12 '17 at 9:49
• well lets make it a little more precise. If $M = S^1$ then the theorem says that their exists an embedding $i:H^{r}(S^1,S^1) \rightarrow H^{r}(S^1,\mathbb{R} )$. So my question was more how does this embedding look like and how does it look for a general manifold – Netivolu Nov 12 '17 at 10:50

As you are interested in the Sobolev completion $L(S^1)=C^\infty(S^1, S^1)\subset H^r(S^1, S^1)$, the solution is to observe that $$H^r(S^1, R^2)\cong H^r(S^1, R)\oplus H^r(S^1, R)\cong H^r(S^1, R).$$ Hence, we obtain an embedding $$L(S^1)\to H^r(S^1, R).$$ One noncanonical thing about this embedding is an isomorphism $$\phi: H^r(S^1, R)\oplus H^r(S^1, R)\cong H^r(S^1, R)$$ for which there are infinitely many choices. You may choose your favorite orthonormal basis $\{e_i\}$ in $L^2(S^1)$ and then obtain an orthogonal isomorphism $\phi$ induced by your favorite bijection $$\psi: {\mathbb Z}^2\to {\mathbb Z},$$ $$(e_i, e_j)\mapsto e_{\psi(i,j)}.$$ If this is not explicit enough, you have to state your notion of "explicit" in much greater detail than "a formula".