Let $(M,g)$ be a Riemannian manifold and $d_g:M\times M\rightarrow \mathbb{R}$ the induced metric.
I am told that such induced metric is not $smooth$ on the diagonal $D\subset M\times M$, $D:=\{(m,m)\in M\times M\}$, but I can not figure out why is that. Could anyone explain it to me or point me out a reference where this is explained?
The intuition here is that the metric $d_g$ is zero on the diagonal but positive either side of it, which generates a "$|x|$" style singularity at which the metric is not differentiable. Consider, for instance, $d:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $d(x,y) = |x-y|$, fix $x$ to be some constant $x_0$ and examine what happens to the metric as $y$ passes through $x_0$.
More formally, let $p$ be a point in the manifold and set up a normal coordinate system in a neighbourhood of $p$. If $d_g: M \times M \to \mathbb{R}$ is smooth at $(p,p)$ then $f: M \to \mathbb{R}$ given by $f(q) = d_g(p,q)$ is smooth at $p$. Then in normal coordinates the map $f$ acts thus*: $$f(X^\mu) = \sqrt{\delta_{\mu \nu} X^\mu X^\nu }\,. $$ This is not smooth at $X^\mu = 0$.
*To be completely explicit about this: a point $q$ with normal coordinates $X^\mu$ lies unit parameter distance along the geodesic through $p$ with tangent vector $V_p$ (where $X^\mu$ are the components of $V_p$ in some chosen basis). If $V$ is the tangent vector at an arbitrary point on this geodesic, then $$ d_g(p,q) = \int_0^1 \mathrm{d}t\,|V| = |V_p| \,,$$ where we have used the fact that $|V|$ is constant along the geodesic. To get the form of $f$ above, we simply use that the metric at $p$ in normal coordinates is the Euclidean one.