# There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold


Let $X,Y$ be Riemannian manifolds, such that $\dim(X) > \dim(Y)$.

I am trying to prove the following statement (mentioned by Gromov in his book on metric geometry):

There is no arcwise isometry (i.e length preserving map) from $X$ to $Y$.

However, the naive attempt to prove this hits an obstacle which I do not see how to pass:

Suppose by contradiction $f:X \to Y$ is an arcwise isometry. Then $f$ is $1$-Lipschitz, hence differeniable almost everywhere (by Rademacher's theorem).

Question: Let $p \in X$ be a point where $f$ is differentiable. Is $df_p:T_pX \to T_{f(p)}Y$ an isometry?

(This will imply the claim of course).

Here is what happens when trying to show this naively:

Let $v \in T_pX$, and let $\al:[0,1] \to X$ be a smooth path s.t $\al(0)=p,\dot \al(0)=v$.

Then $|\dot \alpha(s)| = |\dot \alpha(0)|+\Delta(s)=|v|+ \Delta(s)$ where $\lim_{s \to 0} \Delta(s) =\Delta(0)= 0$, thus

$$(1) \, \, L(\alpha|_{[0,t]})=\int_0^t |\dot \alpha(s)| ds=t|v|+\int_0^t \Delta(s) ds.$$

$\al$ is Lipschitz and $f$ is $1$-Lipschitz, so $\ga:= f \circ \al$ is Lipschitz. By theorem 2.7.6 in the book A course in metric geometry'' ( Burago,Burago,Ivanov) it follows that:

$$(2) \, \, L(\ga|_{[0,t]})=\int_0^t \nu_{\ga}(s) ds,$$

where $\nu_{\ga}(s):=\lim_{\e \to 0} \frac{d\left( \ga(s),\ga(s+\e) \right)}{|\e|}$ is the speed of $\ga$.

Note that $\nu_{\ga}(0)= |\dot \ga(0)|=|df_p(v)|$.

Using the assumption $f$ preserves lengths, we would now like to compare $(1),(2)$ and take derivatives at $t=0$, to get $$|v|=\frac{d}{dt}\left. \right|_{t=0} L(\al|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]})=|df_p(v)|.$$

However, it seems that the last equality is false in general; Even when the speed of a Lipschitz curve and the derivative of its length exist at a point, they do not need to be equal.

It seems that the only thing we can say is that $\frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]}) \ge \nu_{\ga}(0) =|df_p(v)|$, so we are left with $|v| \ge |df_p(v)|$ which doesn't help.

Is there a way to "fix" the proof above?

• Any metric-defined notion of dimension would work. For example Lebesgue covering dimension is defined for metric spaces. It behaves as expected for manifolds and metric isometries of this sort give you a lower bound on the dimension of the target. So it's an immediate contradiction. All the results you need are in Munkres's point-set topology textbook, for example. – Ryan Budney Dec 13 '16 at 22:35
• Doesn't this follow from the fact that an isometry is a homeomorphism? – Igor Rivin Dec 14 '16 at 1:21
• @IgorRivin: Asaf's definition of isometry allows for maps to be not bijections -- he allows covering spaces, for example. It also allows for isometric embeddings. He's trying to make a calculus-ish argument that the domain's dimension needs to be less than or equal the target space's. – Ryan Budney Dec 14 '16 at 7:19
• @RyanBudney I might be dense, but since $d(f(x), f(y)) = d(x, y),$ how can an isometry fail to be bijective? – Igor Rivin Dec 14 '16 at 15:19
• @IgorRivin These length-preserving maps need not be metric isometries, take for instance $\alpha(t)=(\cos t,\sin t)$. Also, even when they are injective isometries, they are not always surjective ($(x \to (x,0)$). – Asaf Shachar Dec 14 '16 at 16:12

Your proof is correct, but you need to add words "amost everywhere" at ane more place.

We use Rademacher's theorem and lemma about length of curve, which says that if a curve parametrized by length then its velocity is 1 almost everywhere, see 2.7.4 in Metric Geometry by Burgo, Burago and Ivanov.

Fix a chart $U_{\subset\mathbb{R}^n}\to X$ and a vector $v\in \mathbb{R}^n$. Let $g$ be the induced metric on $U$.

Note that from the lemma we get that $|d_pv|=|v|_g$ almost everywhere. Repeat the same for $N=\tfrac12\cdot n\cdot(n+1)$ vectors $v_1,\dots,v_N$ in general position. We get $|d_pv_i|=|v_i|_g$ for almost all $p$ and all $i$. It follows that for all $w$,the identity $|d_pw|=|w|_g$ holds for almost all $p$. Hence $\dim X \le \dim Y$.

You may want to check

• Thanks, however, I am still not sure this finishes the problem. As far as I can see, the problem is that we need to know the differential at a single fixed point preserevs lengths of vectors, and I do not see a trivial way to ensure this based on your argument. (I have written another "answer" below where I expressed my concerns explicitly). Also, I think it is interesting to know whether it's true that the differential is an isometry at every point of differentiability. Do you think this is true? – Asaf Shachar Dec 14 '16 at 8:10
• @AsafShachar for your second question. It is easy to construct a length pereserving map $\mathbb{R}\to\mathbb{R}$ which differential vanish at some point. Passing to product maps, you can do the same for $\mathbb{R}^n\to\mathbb{R}^n$. – Anton Petrunin Dec 14 '16 at 15:16
• @AsafShachar I updated the answer, hope it is clear now. – Anton Petrunin Dec 14 '16 at 15:34
• @Thanks. I thought I understood, though now I see that something is still not clear to me. I agree you can restrict yourself to a chart, and I also see why it would suffice to prove this for a fixed vector $v \in \mathbb{R}^n$. I am still not convinced how exactly do you deduce that (after fixing $v$) , $|df_p(v)|=|v|$ for almost every $p$ in $U$. Can you please elaborate on this? – Asaf Shachar Dec 22 '16 at 11:11
• @AsafShachar At almost all points the differential is defined and LINEAR. Does it help? – Anton Petrunin Dec 22 '16 at 21:15

I am completenig some details based on Anton's answer:

We prove the following theorem:

Let $X,Y$ be Riemannian manifolds, and let $f:X \to Y$ be a length preserving map. Then $df$ is an isometry almost everywhere.

The proof is composed of 3 steps:

Step I:

It is enough to prove the claim where $X=(\mathbb{R}^n,g)$ where $g$ is an arbitrary metric.

Step II:

Assume step I, i.e we consider $f:(\mathbb{R}^n,g) \to Y$. Let $v \in \mathbb{R}^n$ fixed . Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.

Step III:

Conclude that $df$ is an isometry almost everywhere.

We begin with proofs of steps I,III, since these are easy.

Proof of step I:

$f$ is $1$-Lipschitz, hence by Rademacher's Thm it is differentiable a.e. Since any manifold admits a countable cover of charts, we finished.

Proof of step III:

$f:(\mathbb{R}^n,g) \to Y$.

Let $v_1,...,v_n$ be a basis for $\mathbb{R}^n$. By applying step II for every $v \in Q=\{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, we get that $|df_x(v)|_{f(x)}=|v|_x$ for all $v \in Q$ and for almost every $x \in \mathbb{R}^n$. Now it is a simple linear algebra fact that at all these points $x$, $df_x$ is an isometry. Indeed,

$$\langle df_x(v_i),df_x(v_j) \rangle = \frac{1}{2}(|df_x(v_i)+df_x(v_j)|^2 - |df_x(v_i)|^2 - |df_x(v_j)|^2) = \frac{1}{2}(|df_x(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)$$ $$= \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle.$$

Comment: Here we see why restricting to a chart was so helpful! We can use the "same" basis vectors $v_i$ for different points $x$ (on a general non-parallelizable manifold, this is meaningless).

Proof of step II:

Fix $v \in \mathbb{R}^n$. Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.

Proof:

Fix $x \in \mathbb{R}^n$, and define $\alpha_x(t)=x+tv$. Putting $\gamma_x=f \circ \alpha_x$ (this is a Lipschitz path), we get by theorem 2.7.6 that $$(1): \, \, \nu_{\alpha_x}(t) =\frac{d}{dt}\left. \right|_{t=0} L(\alpha_x|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga_x|_{[0,t]})=\nu_{\ga_x}(t) \, \, \text{ for almost every t}$$

For any path $\beta$ in a Riemannian manifold, at every point of differentiability, it holds that $\nu_{\beta}(t)=|\dot \beta(t)|_{\beta(t)}$.

Specializing this for $\beta = \alpha_x$ (and combining $(1)$) we obtain

$$(1)': \, \, |v|_{\alpha_x(t)}=\nu_{\ga_x}(t) \, \, \text{ for almost every t}$$

Specializing this for $\beta = \gamma_x$ leads to

observation I:

$\nu_{\gamma_x}(t)=|df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}$ for those $t$ where $f$ is differentiable at $\alpha_x(t)$.

We define the sets

$B:=\{(x,t) \in \mathbb{R}^n \times \mathbb{R} | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \} \subseteq \mathbb{R}^{n+1},$

$B_x=\{ t \in \mathbb{R} | \, \, (x,t) \in B \} \subseteq \mathbb{R}$ and

$B^t= \{x\in\Bbb R^n: (x,t)\in B\} \subseteq \mathbb{R}^n$.

We also define $h:\mathbb{R}^{n+1} \to \mathbb{R}^n$ by $h(x,t)=\alpha_x(t)$.

$h(B)=\{\alpha_x(t) | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \, \text{ is differentiable at } \, \alpha_x(t) \}=\{\alpha_x(t) | \, \, |df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}=|v|_{\alpha_x(t)} \}$

where the last equality follows from observation I.

Hence, it is enough to show $h(B)^c$ has measure zero in $\mathbb{R}^n$.

This essentially follows from Tonelli's theorem:

Since $h$ is surjective, $h(B)^c \subseteq h(B^c)$. Thus, it is enough to show $\mu(B^c)=0$ in $\mathbb{R}^{n+1}$.

$$B^c=\{ (x,t) | \, \, \neg ( \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} ) \} \cup \{ (x,t) | \, \, f \, \text{ is not differentiable at } \, \alpha_x(t) \}:=A \cup D$$

$(1)'$ implies that $\lambda_1(A_x)=0$ for every $x \in \mathbb{R}^n$, hence (by Tonelli) $\mu(A)=0$.

$D^t = \{ x \in \mathbb{R}^n | f \, \text{ is not differentiable at } \, x+tv \} = W -tv$ (where $W \subseteq \mathbb{R}^n$ is the set of points of non-differentiability of $f$), so $\lambda_n(D^t)=\lambda_n(W)=0$

Again, Tonelli implies $\mu(D)=0$.