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Suppose $M \subseteq \mathbb R^n$ is an $m$-dimensional compact embedded manifold, possibly with boundary, with typically $n$ much larger than $m$. (Realistic values I am interested in: $n$ somewhere between $1000$ and $100000$ and $m$ up to, say, 30.) The manifold $M$ has an induced measure from $\mathbb R^n$.

We call a subspace $V \subseteq \mathbb R^n$ with orthogonal projection $\pi_V : \mathbb R^n \to V$ representative if the set $$ \{m \in M \mid \#\pi_V^{-1}(\pi_V(m)) > 1\} $$ has measure zero. That is, the projection restricted to $M$ is "injective except for a set of measure zero". My question is:

What is the least dimension of such a $V$ in general?

Clearly it must be greater than $m$ (just consider the sphere in $\mathbb R^{m+1}$); does $m+1$ suffice? If not, are $\mathcal O(m)$ dimensions sufficient?

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  • $\begingroup$ It's a foundational result in knot theory that two-dimensional $V$ works for $m=1$,$n=3$ - see e.g. math.stackexchange.com/questions/446091/…. My hunch (based mostly on the proof of the Whitney embedding theorem) is that something around $2m$ will work in general. $\endgroup$ – Anthony Carapetis Apr 26 '17 at 13:17
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    $\begingroup$ In fact I think the proof of the easy Whitney embedding theorem given by Hirsch (Differential Topology, Theorem 3.5) can be very slightly modified (just making the projection orthogonal) to show that $\pi_V|_M$ is an embedding for almost every $(2m+1)$-dimensional subspace $V$. $\endgroup$ – Anthony Carapetis Apr 26 '17 at 13:42
  • $\begingroup$ @AnthonyCarapetis Thank you very much! If you want to put that into an answer, I'd be happy to accept. (Although I still wonder if $n+1$ suffices for reasons of curiosity.) $\endgroup$ – Mees de Vries Apr 26 '17 at 13:55
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The proof given of Whitney's easy embedding theorem by Hirsch in Differential Topology works by iterating the following result:

If $M^m$ is an embedded submanifold of $\mathbb R^q$ and $q>2m+1$ then there is a vector $v \in S^{q-1} \setminus \mathbb R^{q-1}$ such that the projection $p_v$ along $v$ on to $\mathbb R^{q-1}$ is an embedding $M^m \to \mathbb R^{q-1}$.

Since $v$ is transverse to $\mathbb R^{q-1}$, the projection along $v$ from $\mathbb R^{q-1}$ on to $v^\perp$ is an isomorphism; and composing this with $p_v$ we get the orthogonal projection $\pi_v : M^m \to v^\perp$, which is thus an embedding. Thus iterating this we get a $2m+1$-dimensional subspace $V$ such that the orthogonal projection on to $V$ is an embedding on $M$.

Now, since we don't need an embedding, we can see how much further similar reasoning can get us. Let's assume we have an embedding $M \to \mathbb R^n$ and try to find a representative projection on to a $n-1$-dimensional subspace.

Let $\Delta = \{(p,p) : p \in M\}$ be the diagonal of $M \times M$ and define the secant map $\sigma: M \times M \setminus \Delta \to S^{n-1}$ by $$\sigma(p,q)=\frac{p-q}{|p-q|}.$$

Then for any direction $v \in S^{n-1}$, $\pi_{v^\perp}(p)=\pi_{v^\perp}(q)$ if and only if $\sigma(p,q)= \pm v$; so the singular set you want to be measure-zero can be described as $\pi_1 X \cup \pi_2 X$, where $X = \sigma^{-1}(v)$ and $\pi_i$ are the projections on to the factors of $M \times M$. Thus it suffices to find a $v$ such that $\dim X < m$.

By Sard's theorem we can choose $v$ to be a regular value of $\sigma$, so that $X$ is a submanifold of $M\times M \setminus \Delta$. Either $X$ is empty and we're done, or the dimension of $X$ is just the dimension of the kernel of $D \sigma$ at a point in $X$. Since $v$ is a regular value we know $D\sigma$ has rank $n-1$, and thus (by rank-nullity) $\dim X = 2m - n + 1$. This is less than $m$ exactly when $n > m + 1$ as you conjectured.

Unfortunately, we can't just iterate this argument like we do with the Whitney embedding theorem, since we need the $M$ we start with to be a submanifold, but the projection might not be a submanifold. Thus if $n >> m$ the best I can do for you is $2m$, by doing this once after applying the embedding theorem.

However, the fact that the single-dimension reduction works all the way down to $m+1$ makes me think it's quite likely that there's some way to get around this, and that $m+1$ dimensions always work. Perhaps some more geometrically intricate argument with a Grassmannian instead of $S^{n-1}$, or perhaps making the iteration work by relaxing the regularity hypotheses. Anyone have any ideas?

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