In this short note, we will prove the following generalization of Hsiung-Minkowski formulas for hypersurfaces in the Euclidean space.
Theorem 1 Let be a closed hypersurface, then for any and , we have
Here is the normalized -th mean curvature, , is the tangential component of the position vector onto . (If , we use the convention that and the second term on the RHS of this equation is understood to be zero. )
Let us fix some notations. Let be a hypersurface and . Let be a local orthonormal frame on and let be the unit outward normal of . Let and be the connections on and respectively. We define the the shape operator by and is defined by . By abusing of notation, we will also denote the second fundamental form by .
We define the -th mean curvature and the normalized -th mean curvature of by
respectively. We use the convention that .
Following [Reilly], we define the -th Newton transformation of as
Alternatively, can be defined recursively by (see e.g. [Reilly])
We use the convention that , for example. Unless otherwise stated, repeated indices will be summed over .
We will need the following two lemmas.
Proof: The equations (2) and (3) require the Codazzi equation which holds for any hypersurface in a space form. (2) is well-known and can be found e.g. in [Reilly]. To prove (1), let us assume that are the principal directions with principal curvatures .
The third line follows from the fact that is symmetric in the indices , .
Proof: See e.g. [LWX] Lemma 2.
0.2. Main result
We can now state and prove our main result.
Theorem 4 Let be a closed hypersurface, then for any and , we have
Here is the tangential component of the position vector onto .
Proof: We first assume . In the following computation, we will omit in the integral symbol. Using Lemma 2 and integration by parts, we compute
where we have used Lemma 3. Using , we have
This is equivalent to
We now consider the case where . Define the function on . Then it is easy to check that where is the Euclidean metric. Let be a local orthonormal frame on , and be the restriction of on . Then we have and . From this we see that
So we have
This completes the proof.
Corollary 5 For , suppose is convex (i.e. ), then we have
Here is the region bounded by .