Can every complete $d$-dimensional Riemannian manifold be represented as a subset $U \subset \mathbb{R}^d$ with some smooth metric $G$

Question

Let $M$ be an arbitrary complete $d$-dimensional Riemannian manifold. Let $\langle \cdot, \cdot\rangle$ denote the standard inner product on $\mathbb{R}^d$. Does there exist an open set $U \subset \mathbb{R}^d$ and a symmetric positive definite function $G : U \rightarrow \mathbb{R}^{d \times d}$ such that the Riemannian manifold $(U, g_x)$ is isomorphic to $M$, where $g_x$ is the inner product $g_x(u,v) = \langle u, G(x) v\rangle$? For example, the Poincare half-plane model does this for hyperbolic space of dimension $2$. My question is can every complete manifold be represented in this way?

Answer 1

No; there will typically not even exist any immersion $M\to\mathbb{R}^d$. For instance, if $M=S^1$, then there is no immersion $M\to \mathbb{R}$. More generally, if $M$ is nonempty and compact and $d>0$ then $M$ cannot immerse in $\mathbb{R}^d$ because the image of an immersion $M\to\mathbb{R}^d$ would have to be both compact and open.

Note that your question is equivalent to asking whether $M$ embeds smoothly into $\mathbb{R}^d$, because if $f:M\to\mathbb{R}^d$ is an embedding with image $U$ then you can simply transport the metric of $M$ along $f$ to get a metric on $U$ which then differs from the standard metric by some function $G$.