Lipschitz manifold and semi-algebraic is Lipschitz graph?

Question

It is known that there are Lipschitz manifolds that are not Lipschitz graphs and $C^1$ manifold is locally $C^1$ graph.

If $M\subset \mathbb{R}^m$ is a Lipschitz manifold (with the outer distance) and algebraic, then M is locally Lipschitz graph?

Answer 1

With the extra explanation, there is a simple counter example: round circle in the plane. I suspect though, that you actually meant to ask something else.

Edit: Here is a local example. Take a polygonal unknot $K$ in $R^3$ such that the projection of $K$ to any affine plane in $R^2$ is not 1-1. Let $M\subset R^4$ denote the cone over $M$ from a point $p\in R^4\setminus R^3$. Then $M$ is semialgebraic, Lipschitz manifold (the latter is since $K$ is a polygonal unknot). However, at the point $p$, $M$ is not the graph of any function of two variables with values in $R^2$ (for any choice of an affine plane in $R^4$).

Edit 2. Here are some further details.

Let $K$ be a nontrivial polygonal knot in $R^3$. Then (affine) projection of $K$ to every plane in $R^3$ has at least 2 crossings (actually, at least 3, but we need just 2). By subdividing edges of $K$ further, we can assume that none of the edges of $K$ appears in both of these crossings.

Now, remove an edge from $K$. The result is a finite polygonal chain $C$ which has the property that projection $\pi(C)$ of $C$ to every plane in $R^3$ has at least one crossing, i.e. the restriction $\pi|C$ is not injective.

Lemma. If $C$ is a finite polygonal chain in $R^3$, then there exists a polygonal $T\subset R^3$ unknot containing $C$ as a subset.

Proof. Take a piecewise-linear homeomorphism $f: R^3\to R^3$ which maps $C$ to a straight line segment $c$. Next, let $t\subset R^3$ be a triangle containing $c$ as an edge. Lastly, set $T=f^{-1}(t)$. qed

In the case of the chain $C$ that we constructed above, the unknot $T$ has the property that its projection to every plane in $R^3$ is non-injective. Next, embed $R^3$ in $R^4$ as the hyperplane $x_4=0$. Set $p=(0,0,0,1)\in R^4$ and let $M\subset R^4$ be the cone with the tip $p$ and the base $T$. In other words, $M$ is the union of 2-dimensional triangles $S_i$ such that $p$ is a vertex of each $S_i$ and the opposite edge of $S_i$ is an edge $e_i$ of $T$. In particular, $M$ is a 2-dimensional piecewise-linear manifold with boundary (its boundary is the unknot $T$). If you really do not like this boundary, just remove it. By construction, $M$ is (real) semi-algebraic as a finite union of semialgebraic subsets $S_i$. Since $T$ is an unknot, there exists a piecewise-linear homeomorphism of $R^3$ sending $T$ to a triangle. Coning-off this homeomorphism from $p$, we obtain a PL homeomorphism of $R^4$ sending $M$ to a convex surface $M'$ (cone over a triangle). This PL homeomorphism is clearly bilipschitz. Composing with a further PL homeomorphism of $R^4$ we flaten $M'$ to a 2-dimensional triangle.

Lastly:

Lemma. The surface $M$, locally, at $p$, is not a graph of any function.

Proof. Suppose it is a graph of a function $R^2\to R^2$. Hence, exist an affine projection $\pi: R^4\to R^2$ which is injective on $U\cap M$, where $M$ is a small neighborhood of $p$ in $R^4$. Let $H\subset R^4$ be a hyperplane given by equation $x_4=r$, where $r<1$ is sufficiently close to $1$ so that $$ L=H\cap M\subset U $$ Then $\pi|H: H\to R^2$ is injective on $L$. However, with identification of $H$ with $R^3$, the unknot $L$ is just a scaled-down copy of the unknot $T$. This means that there exists an affine projection $R^3\to R^2$ whose restriction to $T$ is 1-1. Contradiction.