Riemannian distance function on a product manifold 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 $M_1\times M_2$?)
 A: 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.
A: The answer to your second question is yes! To see this, first consider an admissible (piecewise smooth) curve $\gamma: \left[ a, b \right] \rightarrow M_1 \times M_2$. Then, it is of the form $\gamma \left( t \right) = \left( \gamma_1 \left( t \right), \gamma_2 \left( t \right) \right)$, where $\gamma_1: \left[ a, b \right] \rightarrow M_1$ and $\gamma_2: \left[ a, b \right] \rightarrow M_2$ are admissible curves. Now, consider the lengths of these curves. By definition, we have
$$L \left( \gamma \right) = \int\limits_{a}^{b} \| \gamma' \left( t \right) \| \ \mathrm{d}t,$$
$$L \left( \gamma_1 \right) = \int\limits_{a}^{b} \| \gamma_1' \left( t \right) \|_1 \ \mathrm{d}t,$$
$$L \left( \gamma_2 \right) = \int\limits_{a}^{b} \| \gamma_2' \left( t \right) \|_2 \ \mathrm{d}t.$$
Here, $\| \cdot \|, \| \cdot \|_1,$ and $\| \cdot \|_2$ are the norms induced by $g_1 \oplus g_2, g_1,$ and $g_2$ on the tangent spaces of $M_1 \times M_2, M_1,$ and $M_2$ respectively.
We know that the product Riemannian metric $g_1 \oplus g_2$ is given by
$$\left( g_1 \oplus g_2 \right)_{\left( p, q \right)} \left( \left( X_1, Y_1 \right), \left( X_2, Y_2 \right) \right) = \left( g_1 \right)_p \left( X_1, X_2 \right) + \left( g_2 \right)_q \left( Y_1, Y_2 \right).$$
This clearly gives
$$\| \gamma' \left( t \right) \| = \left( \| \gamma_1' \left( t \right) \|_1^2 + \| \gamma_2' \left( t \right) \|_2^2 \right)^{\frac{1}{2}}.$$
That is,
\begin{align}
L \left( \gamma \right) = \int\limits_{a}^{b} \| \gamma' \left( t \right) \| \ \mathrm{d}t &= \int\limits_{a}^{b} \left( \| \gamma_1' \left( t \right) \|_1^2 + \| \gamma_2' \left( t \right) \|_2^2 \right)^{\frac{1}{2}} \mathrm{d}t \\
&\geq \max \left\lbrace L \left( \gamma_1 \right), L \left( \gamma_2 \right) \right\rbrace. \\
&\geq \max \left\lbrace d_1 \left( \gamma_1 \left( a \right), \gamma_1 \left( b \right) \right), d_2 \left( \gamma_2 \left( a \right), \gamma_2 \left( b \right) \right) \right\rbrace.
\end{align}
Let us now denote $\gamma_1 \left( a \right) = p_1, \gamma_1 \left( b \right) = q_1, \gamma_2 \left( a \right) = q_1,$ and $\gamma_2 \left( b \right) = q_2$. That is, $\gamma_1$ is an admissible curve in $M_1$ joining $p_1$ and $q_1$, $\gamma_2$ is an admissible curve in $M_2$ joining $p_2$ and $q_2$, and $\gamma$ is an admissible curve in $M_1 \times M_2$ joining $\left( p_1, p_2 \right)$ and $\left( q_1, q_2 \right)$.
This gives that $d \left( \left( p_1, p_2 \right), \left( q_1, q_2 \right) \right) \geq d_{\infty} \left( \left( p_1, p_2 \right), \left( q_1, q_2 \right) \right)$, where I use the notation:
$$d_{\infty} \left( \left( p_1, p_2 \right), \left( q_1, q_2 \right) \right) = \max \left\lbrace d_1 \left( p_1, q_1 \right), d_2 \left( p_2, q_2 \right) \right\rbrace.$$
On the other hand, we have
\begin{align}
L \left( \gamma \right) &= \int\limits_{a}^{b} \left( \| \gamma_1' \left( t \right) \|_1^2 + \| \gamma_2' \left( t \right) \|_2^2 \right)^{\frac{1}{2}} \ \mathrm{d}t \\ &\leq C \left( \int\limits_{a}^{b} \left( \| \gamma_1' \left( t \right) \|_1^2 + \| \gamma_2' \left( t \right) \|_2^2 \right) \ \mathrm{d}t \right)^{\frac{1}{2}} \ \ \ \ (\text{by the Holder's inequality}) \\
&\leq C \left( L \left( \gamma_1 \right)^2 + L \left( \gamma_2 \right)^2 \right)^{\frac{1}{2}}.
\end{align}
In the last inequality, we have used the "reverse Holder inequality".
This gives us
$$d \left( \left( p_1, p_2 \right), \left( q_1, q_2 \right) \right)^2 \leq C^2 \left( L \left( \gamma_1 \right)^2 + L \left( \gamma_2 \right)^2 \right).$$
Notice that this is true for every admissible curve $\gamma_1$ joining $p_1$ and $q_1$, and $\gamma_2$ joining $p_2$ and $q_2$. So we can get
$$d \left( \left( p_1, q_1 \right), \left( p_2, q_2 \right) \right) \leq C \left( d_1 \left( p_1, q_1 \right)^2 + d_2 \left( p_2, q_2 \right)^2 \right)^{\frac{1}{2}} = C d_{prod} \left( \left( p_1, q_1 \right), \left( p_2, q_2 \right) \right),$$
in the notation of your question. That is, we have
$$d_{\infty} \leq d \leq C d_{prod}.$$
But one can easily prove that if $\left( X_1, d_1 \right)$ and $\left( X_2, d_2 \right)$ are metric spaces, then $d_{\prod}$ and $d_{\infty}$ are equivalent metrics (for instance, see here). Thus, there is some $A > 0$ such that
$$A d_{prod} \leq d \leq C d_{prod}.$$
That is $d$ and $d_{prod}$ are equivalent metrics and hence they give the same topology.
Also, this proof gives an idea (which need not be correct at the moment) that in general the two metrics need not be equal unless $A = C = 1$. However, at this moment I do not have a counterexample.
