# orthogonal projection onto a manifold

Let $$(M,g)$$ be a Riemannian manifold. By the Nash imbedding Theorem we can find an isometric embedding $$\varphi:M\rightarrow\mathbb R^s$$ for sufficiently large $$s$$.
In this sense we may view $$M$$ as a subset of $$\mathbb R^s$$.
Now here is my question:

For points $$v\in\mathbb R^s$$ which are close enough to $$M$$, how can we define the orthogonal projection of $$v$$ onto $$M$$?

If $$M$$ was a linear subspace of $$\mathbb R^s$$, then this would be clear.

In general this is not well defined, since there is no obvious way to project the origin onto the sphere.
But for points $$v\in\mathbb R^s$$ such that the distance $$\mathrm{dist}(v,M)$$ is very small, there should be a way to uniquely define this projection.

The sufficiently close condition should be understood as a condition such that the projection is well defined so that things like above cannot happen.
I think in general it is necessary to assume that $$M$$ is a closed subset of $$\mathbb R^s$$, but is there a way to define this projection without this assumption?

The adequate concepts to make you idea precise are those of the normal bundle and of tubular neigborhoods of a smooth submanifold $$M \subset \mathbb R^s$$. See for example

John M. Lee, Introduction to Smooth Manifolds (p. 139 ff)

and

I think you do need $$M$$ to be closed for this to work. Consider $$M$$ the open unit disk (in $$\mathbb{R}^2$$) embedded into $$\mathbb{R}^3$$ like an almost closed taco, ie the points $$(1,0)$$ and $$(-1,0)$$ are mapped to the same point in $$\mathbb{R}^3$$ (note that they are not part of $$M$$). You don't get a well defined orthogonal projection around that point.

• but that embedding will not be isometric, right? The pullback of the standard metric on $\mathbb R^3$ will not agree with the metric on the unit disc. Aug 29, 2019 at 12:42
• You can do this with a flat piece of paper, so the embedded disk is isometric to the disk in $\mathbb{R}^2$. Aug 29, 2019 at 12:44
• But even in this example, there’s a neighborhood of $M$ on which the projection is well defined. The neighborhood just won’t contain the problematic point. Aug 29, 2019 at 14:51
• @JackLee You can't find a universal $\varepsilon$ so that all points with distance less than $\varepsilon$ are in the tubular neighborhood. The distance bound has to shrink to zero when you approach the problematic point. Aug 30, 2019 at 6:35
• @quarague — That’s true, but the OP didn’t ask for a neighborhood of uniform size. Aug 30, 2019 at 13:15