# local isometry for riemannian manifolds is not transitive

Let $(M_1,g_1)$ and $(M_2,g_2)$ be Riemannian manifolds of the same dimension, and let $\phi: M_1 \to M_2$ be a smooth map. We say that $\phi$ is a local isometry if $g_2 (\phi_* X, \phi_* Y ) = g_1 (X, Y )$ for all $m \in M_1$ and $X, Y \in T_m M_1,$ where $\phi_* : T_m M_1 \to T_{\phi(m)} M_2$ is the derivative of the map $\phi$ at $m.$

The relation of being locally isometric for Riemannian manifolds is not symmetric, it is of course reflexive: is it transitive?

• I'll admit I don't understand all the words in your question, but what issue comes up if you try to compose isometries? Sorry if this comment is stupid. – Patrick Da Silva Jun 22 '13 at 9:54
• honestly, I couldn't think of what goes wrong, but I've been told this should be the case, so I'd like to see a counterexample, or a proof that it is indeed transitive. – jj_p Jun 22 '13 at 10:47
• are you sure that it is not transitive? – user55449 Jun 22 '13 at 11:19
• ok, let me put it as a question, since I'm not that sure – jj_p Jun 22 '13 at 11:43
• Did you try to see what happens if you just plug in the definition, i.e., take local isometries $\phi\colon M_1 \to M_2$ and $\psi \colon M_2 \to M_3$ and check whether $\psi\circ \phi$ is a local isometry? – Martin Jun 22 '13 at 11:52

Let $\phi \colon (M_1,g_1) \to (M_2,g_2)$ and $\psi \colon (M_2,g_2) \to (M_3,g_3)$ be local isometries. The chain rule $(\psi \circ \phi)_\ast = \psi_\ast \circ \phi_\ast$ yields for all $X,Y \in T_{m}M_1$ that \begin{align*} g_{3}((\psi \circ \phi)_\ast X, (\psi \circ \phi)_\ast Y) & = g_{3}(\psi_\ast\phi_\ast X, \psi_\ast\phi_\ast Y) &&\text{chain rule}\\ &= g_{2}(\phi_\ast X, \phi_\ast Y) && \psi \text{ is a local isometry}\\ &= g_{1}(X,Y) && \phi \text{ is a local isometry} \end{align*} so $\psi \circ \phi \colon (M_1,g_1) \to (M_3,g_3)$ is a local isometry.