Riemannian Geometry – Distance Function on a Product Manifold

differential-geometrygeneral-topologymanifoldsmetric-spacesriemannian-geometry

Let $(M_1,g_1)$ and $(M_2,g_2)$ be two Riemannian manifolds. Let $d_1$ and $d_2$ be the respective induced Riemannian distance functions, i.e. metrics in the sense of metric spaces.

Let $(M_1\times M_2,g_1\oplus g_2)$ be the product Riemannian manifold, and let $d$ be the induced distance function on $M_1\times M_2$.

Let $d_{\text{prod}}$ be the product metric on $M_1\times M_2$ in the metric sense, i.e.

$$d_{\text{prod}}((x_1,x_2),(y_1,y_2))=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$

Question 1: Is it always true that
$$d=d_{\text{prod}}?$$

Comment: This seems to be true if either $M_1$ or $M_2$ is a manifold where any two points are connected by a unique geodesic (e.g. $\mathbb{R}$). I suspect that in general we only have $d\leq d_{\text{prod}}$. But what is an illustrative example?

Question 2: If not, is there anything that we can say about the relationship between $d$ and $d_{\text{prod}}$? (E.g. does there exist $C>0$ such that $\frac{1}{C}d\leq d_{\text{prod}}\leq Cd$; do they induce the same topology on $M_1\times M_2$?)

Best Answer

Let:

  • $p_1,q_1$ be two points in $M_1$.
  • $p_2,q_2$ be two points in $M_2$.
  • $c_1: [0,l_1] \rightarrow M_1$ be any parametrized by arc-length piecewise regular smooth curve in $M_1$ connecting $p_1$ and $q_1$.
  • $c_2: [0,l_2] \rightarrow M_2$ be any parametrized by arc-length piecewise regular smooth curve in $M_2$ with unit-speed connecting $p_2$ and $q_2$.
  • $\hat{c}=(\hat{c}_1,\hat{c}_2): [0,1] \rightarrow M_1 \times M_2$ be any piecewise regular smooth curve in $M_1\times M_2$ connecting $(p_1,p_2)$ and $(q_1,q_2)$.

Clearly $c: [0, 1] \ni t \mapsto \left( c_1(l_1t), c_2( l_2t) \right) $ is a piece-wise smooth connecting $(p_1,p_2)$ with $(q_1,q_2)$ and $\text{length}(c)=\sqrt{l_1^2+l_2^2}$. Hence, $d_{g_1 \oplus g_2}\left( (p_1,p_2),(q_1,q_2) \right) \le \sqrt{l_1^2+l_2^2}$. In consequences,$$d_{g_1 \oplus g_2}\left( (p_1,p_2),(q_1,q_2) \right) \le \sqrt{ d_{g_1}\left( p_1,q_1 \right)^2+d_{g_2}\left( p_2,q_2 \right)^2}.$$ Now, consider $\hat{c}$. We see that $\hat{c}_1,\hat{c}_2$ must also be piecewise smooth curves. Let $\hat{l_1},\hat{l_2}$ be the length of $\hat{c}_1$ and $\hat{c}_2$, respectively. We have:

\begin{align} \mathrm{length}(\hat{c}) &= \int_0^1 \sqrt{ |\hat{c}'_1(t)|^2+|\hat{c}'_2(t)|^2 } \mathrm{d}t \\ &\ge \frac{1}{ \sqrt{ \hat{l}_1^2+\hat{l}_2^2}} \left( \hat{l}_1\int_0^1 |\hat{c}'_1(t)| \mathrm{d}t+\hat{l}_2\int_0^1 |\hat{c}'_2(t)| \mathrm{d}t \right) =\sqrt{ \hat{l}^2_1+\hat{l}_2^2} \end{align}

Thus, $\text{length}(\hat{c}) \ge \sqrt{ d_{g_1}\left( p_1,q_1 \right)^2+d_{g_2}\left( p_2,q_2 \right)^2}$. Therefore, $$d_{g_1 \oplus g_2}\left( (p_1,p_2),(q_1,q_2) \right) \ge \sqrt{ d_{g_1}\left( p_1,q_1 \right)^2+d_{g_2}\left( p_2,q_2 \right)^2}.$$ Hence, the desired conclusion.

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