# Pointwise equality of codimension-zero immersions

Suppose $$f,g: U \subset \mathbb{R}^n \to \mathbb{R}^n$$ are smooth immersions from a closed and bounded domain $$U$$ such that $$f|_{\partial U} = g|_{\partial U}$$. If $$f$$ and $$g$$ induce the same Riemannian metric on $$U$$ (i.e. $$\langle df, df \rangle = \langle dg, dg \rangle$$), does it follow that $$f = g$$ pointwise?

I suspect the answer is yes, and I have a sketch of a proof for $$n=2$$. In particular, the hypotheses imply that (at least locally), we have $$(g \circ f^{-1})^*\delta = \delta$$ where $$\delta$$ is the Euclidean metric on $$\mathbb{R}^2$$. Then, $$g\circ f^{-1}$$ is identity on the boundary, and to show that it is also identity on the interior we can consider any Euclidean triangle in $$f(U)$$ with two vertices fixed on (some convex part of) the boundary and the third in the interior. This triangle has two vertices and all edge lengths preserved under $$g\circ f^{-1}$$, therefore its third vertex is preserved also. If it is not clear enough from this, I think starting close to the boundary and ''bootstrapping'' the same procedure toward the interior of the domain should allow it to be used anywhere, but I have not thought carefully about it yet.

I know that codimension-zero immersions are quite rigid as a general rule, but so far I cannot come up with a proof nor a counterexample for this statement. Do you know of any arguments or references which might answer this question?

• What do you meany by an immersion in this case? The definition is clear in the interior, but not at the boundary. Do you require that for each boundary point $x\in \partial U$, the differential $d_xf$ has full rank $n$? In this case the proof is not difficult. But if you only require an immersion in the interior of $U$ and continuity in the closure of $U$ then one would need to work a bit harder. Commented Jul 23, 2021 at 18:06
• Thanks for your comment. I suppose that I mean $df$ has full rank everywhere -- up to and including the boundary. On the other hand, I don't see why this distinction determines the degree of difficulty in the proof. Would you please elaborate a little bit on this? Commented Jul 23, 2021 at 18:25

With the clarification of the definition, here is a proof:

Theorem. Suppose that $$M$$ is a smooth $$n$$-dimensional compact connected manifold with boundary, $$f, h: M\to {\mathbb R}^n$$ are immersions, i.e. differentials $$df_x, dh_x$$ have rank $$n$$ everywhere in $$M$$, such that $$f|\partial M= h|\partial M$$ and $$g=f^*(g_0)=h^*(g_0)$$ where $$g_0$$ is the standard flat metric on $${\mathbb R}^n$$. Then $$f=h$$.

Proof. Consider a point $$x$$ in the interior of $$M$$. Pick a unit tangent vector $$u\in T_xM$$ and let $$c$$ denote the maximal unit speed geodesic segment in $$(M,g)$$ such that $$c(0)=x, c'(0)=u$$. Maximality here mean that the domain $$J$$ of $$c$$ is maximal possible subinterval in $${\mathbb R}$$. Geodesic is understood in the Riemannian sense, i.e. $$\nabla_{c'}c'=0$$, where $$\nabla$$ is the LC-connection of $$g$$.

Lemma. $$J$$ is a compact interval $$[a,b]$$ such that $$c(a), c(b)\in \partial M$$.

Proof. There are several cases one has to exclude, I will eliminate the most interesting one, when $$J$$ is unbounded. Observe that $$f\circ c: J\to {\mathbb R}^n$$ has image contained in $$f(M)$$, which is a compact subset of $${\mathbb R}^n$$. But $$f\circ c$$ is a parameterized with the unit speed infinite Euclidean line segment. This is a contradiction. I will leave it to you to complete the proof in the case when $$J$$ has the form $$(a,b), (a,b], [a,b)$$. (You again use compactness of $$M$$ and the fact that the metric $$g$$ is defined in the entire $$M$$, including its boundary.) qed

Thus, we have, $$f\circ c, h\circ c: [a,b]\to {\mathbb R}^n$$ are isometric maps, which, by the hypothesis of the proposition, satisfy $$fc(a)=hc(a), fc(b)=hc(b)$$. Now, it is a simple fact of Euclidean geometry that $$f\circ c= h\circ c$$. In particular, $$f(x)=h(x)$$. qed

• Your argument is clearer than mine. Thanks for the instructive explanation. Commented Jul 23, 2021 at 19:37