Laplacian of a submanifold in an Euclidean space

Question

let $M^n$ be a smooth an $n$-dimensional sub-manifold of a $\mathbb{R}^{m}$ ($n<m$). Denote by $\nabla^{M}$ and $\nabla^{\mathbb{R}^{m}}$ be the gradient of $M$ and $\mathbb{R}^{m}$ respectively. Similarly for $\Delta^{M}$ and $\Delta^{\mathbb{R}^{m}}$.

Please can you explain for me the relation between $\nabla^{M}$ and $\nabla^{\mathbb{R}^{m}}$ and that between $\Delta^{M}$ and $\Delta^{\mathbb{R}^{m}}$?

For example if I have $f$ a smooth function defined on $\mathbb{R}^{m}$. then $$\nabla^{M}(f\mid_{M})=??(\nabla^{\mathbb{R}^{m}}f)^{T}$$ $(\,)^T$ represents the tangential part.

Is that true? What about Laplacian?

Answer 1

I don't think that for Laplacian we might have a relationship like the gradients. Consider $\mathbb{R}^2$ and the unit circle $S^1$. Suppose $f(x,y) = x $. Then $ \nabla_{R^2}f= 0 $ however on the circle with induced metric $\nabla_{S^1}f = -\cos(\theta) \neq 0$. In general we know that, Laplacian of $f$ on $M$ is the sum of second derivatives of $f\restriction _M$ along the geodesics. On $\mathbb{R}^m$ geodesics are just lines. On the other hand, on a Riemannian submanifold of $\mathbb{R}^m$ geodesics might be very different then lines. In that sense, I think we can always create a function which has zero Laplacian on $\mathbb{R}^m$ but non-zero Laplacian on $M$.

Answer 2

The gradient $\nabla^M f$ is the unique vector tangent to $M$ such that $$\langle \nabla^M f, V \rangle = df(V) = V(f)$$ for all vector fields $V$ tangent to the manifold $M$ in question. Since $\langle (\nabla f)^T, V \rangle = \langle\nabla f,V\rangle$ when $V$ is tangent to $M$, we conclude that your formula for the submanifold gradient is correct.

For the Laplacian, there's no direct relationship, even in the flat case $\mathbb R^n \subset \mathbb R^m$: the issue is that $\Delta^M$ cannot capture second derivatives in the normal directions. However, we can get a useful formula in terms of the ambient derivatives. We have $$\Delta^M f = \mathrm{tr}_{TM} \nabla^M df$$ (where $\mathrm{tr}_{TM}$ means tracing over an orthonormal frame of the subspace $T_xM \subset \mathbb R^m$); and we know the extrinsic and intrinsic covariant derivatives are related by $\nabla^{\mathbb R^m}_X Y = \nabla^M_XY+A(X,Y)$ where $A$ is the second fundamental form. Thus the intrinsic second derivative is $$\nabla^M d f(X,Y) =X(Yf) - df(\nabla^M_X Y) = X(Yf)-df(\nabla^{\mathbb R^m}_XY) + df(A(X,Y)),$$ which we recognise as $\nabla^{\mathbb R^m}df(X,Y) + df(A(X,Y)),$ and so taking the trace we find $$\Delta^M f = \mathrm{tr}_{TM}\nabla^{\mathbb R^m}df+df(H),$$ where $H = \mathrm{tr} A$ is the mean curvature vector of $M$. If we evaluate at a point $x$, the first term can be written as $\Delta (f|_{T_x M}),$ where we are thinking of the tangent space $T_x M$ in the sense of analytic geometry, i.e. as a genuine subset of $\mathbb R^m$ touching $M$ at $x$.