Htykuut

Suppose $ F $ is a homogeneous polynomial function of degree $ m $ on Euclidean space $ mathbb{R}^{n+1} $. Restrict $ F $ to the unit sphere $ S^n $ and we get a function on $ S^n $ denoted by $ f $, prove:

$(1)$ $ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 $;

$(2)$ $ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-m(m-1)f-mnf .$

Where $ nabla_{S^n}, nabla_{mathbb{R}^{n+1}} $ are gradients on $ S^n $ and $ mathbb{R}^{n+1} $ respectively, $ Delta_{S^n}, Delta_{mathbb{R}^{n+1}} $ are Laplace operators on $ S^n $ and $ mathbb{R}^{n+1} $ respectively.

Hint: Use Euler's homogeneous function theorem $ langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle=mF(z) $.

The question above comes from my Riemannian geometry textbook. Can someone give me a hint about how to deal with $ nabla_{S^n}f $ ? I can't find any direct relation between $ nabla_{S^n}f $ and $ nabla_{mathbb{R}^{n+1}}F $. Though we can use local coordinates to expand $ nabla_{S^n}f $ and compute directly, I am still looking forward to a more elegant way to do this.

Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?

Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$

for all $v in T_pN$.

In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition

$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$

where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,

$$ P(v) = v - left<v, p right>p $$

and so

$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$

This handles the first part. I'll leave the second part to you.

Thanks levap for the nice answer, here is what I did based on the hint given by levap:

$(1)$ Seeing that $ langle nabla_{S^n}f, X rangle=X(f) $ where $ Xin mathfrak{X}(S^n) $ by the definition of gradient. Now we play the trick that $ X(f)=operatorname{d}f(X)=operatorname{d}(Fcirc i)(X) $ where $ i $ is the inclusion map $ i: S^nhookrightarrow mathbb{R}^{n+1} $. And
$$ operatorname{d}(Fcirc i)(X)=operatorname{d}(F)circoperatorname{d}(i)(X)=operatorname{d}F(tilde{X})=tilde{X}(F)=langle nabla_{mathbb{R}^{n+1}}F, tilde{X} rangle=langle (nabla_{mathbb{R}^{n+1}}F)^{T}, Xrangle $$
where $ tilde{X}=X $ on $ S^n $.

Hence we have $ nabla_{S^n}f=(nabla_{mathbb{R}^{n+1}}F)^{T} $. Therefore, $$|nabla_{mathbb{R}^{n+1}}F|^2=|nabla_{S^n}f|^2+|(nabla_{mathbb{R}^{n+1}}F)^{perp}|^2=|nabla_{S^n}f|^2+langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle^2=|nabla_{S^n}f|^2+m^2F^2(z) $$ where $ zin S^n $
which is the same as $$ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 .$$

$(2)$ We have
begin{align*} Delta_{mathbb{R}^{n+1}}F&=operatorname{div}(nabla_{mathbb{R}^{n+1}}F)\
&=operatorname{div}left((nabla_{mathbb{R}^{n+1}}F)^{T}+(nabla_{mathbb{R}^{n+1}}F)^{perp}right)\
&=operatorname{div}(nabla_{mathbb{R}^{n+1}}F)^{T}+operatorname{div}(nabla_{mathbb{R}^{n+1}}F)^{perp}\
&=Delta_{S^n}f+operatorname{div}(nabla_{mathbb{R}^{n+1}}F)^{perp}\
&=Delta_{S^n}f+operatorname{div}(langle nabla_{mathbb{R}^{n+1}}F,z rangle z)\
&=Delta_{S^n}f+langle nabla_{mathbb{R}^{n+1}}F,z rangleoperatorname{div}(z)+z(langle nabla_{mathbb{R}^{n+1}}F, z rangle )\
&=Delta_{S^n}f+mfoperatorname{div}(z)+z(mF(z))\
&=Delta_{S^n}f+mfn+mz(F(z))\
&=Delta_{S^n}f+mnf+mlangle nabla_{mathbb{R}^{n+1}}F, z rangle\
&=Delta_{S^n}f+mnf+mmf .
end{align*}
Hence $$ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-mmf-mnf .$$

Edit:

To be perfectly clear, we need to prove the fact that $ operatorname{div}(nabla_{mathbb{R}^{n+1}}F)^{T}=Delta_{S^n}f=operatorname{div}(nabla_{S^n}f) $. If we pick local orthogonal coordinates $ { E_i }_{i=1}^{n+1} $ at $ zinmathbb{R}^{n+1} $ such that $ E_{n+1}=z $ and $ { E_i }_{i=1}^n $ is naturally local orthogonal coordinates at $ z $ restricted to the unit sphere $ S^n $. Since $(nabla_{mathbb{R}^{n+1}}F)^{T} $ is perpendicular to $ z $, then by the definition of divergence,
begin{align*} operatorname{div}(nabla_{mathbb{R}^{n+1}}F)^{T}-operatorname{div}(nabla_{S^n}f)&=sum_{i=1}^{n+1}langle nabla_{E_i}(nabla_{mathbb{R}^{n+1}}F)^{T}, E_i rangle-sum_{i=1}^nlangle nabla_{E_i}(nabla_{mathbb{R}^{n+1}}F)^{T}, E_irangle \
&=langle nabla_{E_{n+1}}(nabla_{mathbb{R}^{n+1}}F)^{T}, E_{n+1}rangle\
&=0 .
end{align*}

Another note for $ operatorname{div}(z)=0 $. Since in Euclidean space $ mathbb{R}^{n+1} $, $ z=(x_1, x_2,..., x_n, sqrt{1-sum_{i=1}^nx_i^2}) $ and $ operatorname{div}(z)=sum_{i=1}^{n}frac{partial x_i}{partial x_i}+frac{partial sqrt{1-sum_{i=1}^nx_i^2}}{partial x_{n+1}}=sum_{i=1}^n1+0=n. $