Second derivative of the volume of the $\varepsilon$-neighbourhood of a submanifold

Question

Let $M$ be a $n$-dimensional compact Riemannian manifold, and $N$ a smooth submanifold of $M$ of dimension strictly less than $n$.

Denote by $N_{\varepsilon}$ the $\varepsilon$-neighbourhood of $N$ - that is, the set $\{p \in M \ | \ d(x, N) < \varepsilon\}$.

Here $d$ denotes the Riemannian distance.

Let $\text{Vol}$ denote the Riemannian volume measure on $M$.

Question:

Define the function $f: [0, \infty) \to \mathbb R_+$ by $f(\varepsilon) := \text{Vol}(N_{\varepsilon})$.

Is it true that

$$\lim_{\varepsilon \to 0+} f’’(\varepsilon)$$

exists? If so, can we find an expression for it in terms of the Riemannian metric on $M$ and the embedding of $N$ in $M$?

Answer 1

I guess we are interested in closed submanifolds; otherwise the question does not have much sense. Suppose $k=\mathrm{codim}\, N$.

Note that $f(\varepsilon)=O(\varepsilon^k)$, so $f''(0)=0$ if $k\ge 3$. It remains to consdier two cases $k=2$ and $k=1$.

Given a normal vector $v$ to $N$ denote by $\rho(v)$ the jacobian of the expontential map from the normal bundle $E$ to $N$ to $M$. Note that $$f(\varepsilon)=\int\limits_{x\in N}\ \int\limits_{v\in B_\varepsilon\subset E_x} \rho(v).$$ Note that $\rho$ is smooth and $\rho(0)=1$.

If $k=2$, then it follows that $f''(0)=2\cdot\pi\cdot \mathrm{vol}\,N$.

For $k=1$, we get $f''(0)=0$,