# Weyl tube formula for manifolds with boundary

In 1939 H. Weyl proved the following non-trivial theorem. Let $$(M^n, g)$$ be a closed smooth Riemannian manifold. Fix an isometric imbedding $$\iota\colon M\to \mathbb{R}^N$$ into a Euclidean space (now such an imbedding is known to exist e.g. by Nash Theorem; Weyl actually used weaker results on local imbeddings). Consider the volume in $$\mathbb{R}^N$$ of the $$\varepsilon$$-neighborhood of $$\iota(M)$$. The first (elementary) observation of Weyl was that this volume is a polynomial in $$\varepsilon$$ for small values of $$\varepsilon$$: $$vol_N(\iota(M)_\varepsilon)=\varepsilon^{N-n}\sum_{i=0}^n \kappa_{i,n,N} V_i(M)\varepsilon^{n-i},$$ where $$\kappa_{i,n,N}$$ are appropriately chosen normalizing coefficients depending only on $$i,n,N$$ but not on $$(M,g)$$. The non-trivial result of Weyl says that $$V_i(M)$$ are independent of the imbedding $$\iota$$. More precisely he has shown that $$V_i(M)$$ equals to the integral over $$M$$ of some explicit polynomial in the Riemann curvature tensor.

Now let $$(M,g)$$ be a compact smooth Riemannian manifold with boundary. One can chose as previously an isometric imbedding $$\iota\colon M\to \mathbb{R}^N$$. It is well known (and not hard to show) that, as previously, volume of the $$\varepsilon$$-neighborhood of $$\iota(M)$$ is a polynomial for small $$\varepsilon$$. Is it true that its coefficients (appropriately normalized) are independent of the imbedding $$\iota$$? If yes, what are explicit expressions for them? A reference would be most welcome.

• Yes it is true, but do not ask me to write a formula. Look at the proof of Weyl tube formula and try to modify it (it is straightforward). – Anton Petrunin Aug 6 '19 at 19:53