Get $M$ a manifold and the normal component of ∇ yields the negative of the second fundamental form of M. In particular, one defines the second fundamental form by $II(X,Y)= -(\nabla_X Y )^n$ ,Taking the trace of the bilinear form $ II(X,Y)$ the tangent space of M yields the mean curvature vector is given by $Tr(II(X,Y))=H$.
Now if we write $T =T^t + T^n$, where $T^t$ is the tangential component of T on M and $T^n$ is its normal component, then since $ T=d\phi(\partial / \partial t) $ and $M_t=\phi(M,t)$ deformations of M so,
$ \sum_{i=1}^n $ $ei \lt T^n,ei \gt = \lt T^n, H \gt $
I used the divergence to get Trace but $T^n$ vanished, suggestions please!
Can you check in Geometric Analysis, Peter Li