# Riemannian metric on submanifold of $\mathbb{R}^n$

Let $$U$$ be an open bounded domain of $$\mathbb{R}^n$$, with smooth boundary $$\partial U$$.

I'm confusing about definition of standard Riemannian metric on $$\partial U$$ as $$(n-1)$$ submanifold. Is it defined by the scalar product of $$\mathbb{R}^n$$ or any other product? Is the Riemannian inner product coincides with the Euclidean one in this case (maybe an identification of $$(x_1,\dots, x_{n-1}) \simeq (x_1,\dots, x_{n-1},0)$$ is used)? Finally, if $$\nabla_{\partial U}$$ is the tangential gradient on $$\partial U$$ why we have this identity on $$\partial U$$ for any smooth function $$f\in C^\infty(\overline{U})$$, $$|\nabla f|^2=|\nabla_{\partial U} f|^2 + |\partial_n f|^2,$$ where $$n$$ is the normal vector field on $$\partial U$$.

Abstractly, the metric on $$\Sigma:=\partial U$$ is defined by pullback along the inclusion $$i:\Sigma\hookrightarrow \Bbb R^n$$. This just means that if $$u$$ and $$v$$ are geometrically tangent to $$\Sigma$$ in $$\Bbb R^n$$, then $$g(u,v):=u\cdot v$$, the Euclidean dot product.

For $$p\in\Sigma$$ we can take an orthonormal basis of $$T_p\Bbb R^n$$ that is $$e_1,\dotsc,e_{n-1},n$$, where $$e_1,\dotsc,e_{n-1}$$ are tangent to $$\Sigma$$ and $$n$$ is a choice of normal. Then $$\nabla f=\sum_{i=1}^{n-1}(e_if)e_i+(\partial_nf)n$$, so dotting this with itself and using the fact that this is an orthonormal basis gives your claim.

In response to the comment, I'll elaborate. I now realize there's a conflict of notation so we'll rename the ambient space to $$\Bbb R^d$$. The notation $$e_if$$ is just the action of $$e_i$$ on the (smooth) function $$f$$ as a tangent vector. The notation $$\partial_nf$$ is just code for $$nf$$, as well.

Now $$\nabla f$$ is the total gradient of $$f$$ on $$\Bbb R^d$$. In the PDE world, this means it's $$(\partial_1f,\dotsc,\partial_df)$$, where $$\partial_i:=\partial/\partial x^i$$. In differential geometric language, $$\nabla f=\sum_{i=1}^d (\partial_if)\partial_i.$$ (For other ambient manifolds, you need to worry about the metric here.) But in fact, if $$v_1,\dotsc,v_d$$ is any basis of $$T_p\Bbb R^d$$, then $$\nabla f=\sum_{i=1}^d(v_if)v_i$$. What I wrote above is this in the particular case that $$v_1,\dotsc, v_{d-1}$$ are an orthonormal basis of $$T_p\Sigma$$ and $$n=v_d$$ is a normal vector.

Now $$\nabla_\Sigma f$$ is by definition the part of $$\nabla f$$ tangent to $$\Sigma$$. By above expansion, we indeed have $$\nabla_\Sigma f=\sum_{i=1}^{d-1}(e_if)e_i$$ if $$e_1,\dotsc, e_{d-1}$$ is an orthonormal basis of $$T_p\Sigma$$. In a coordinate basis of $$\Sigma$$, we have the formula $$\nabla_\Sigma f=g^{ij}\partial_jf\partial_iF,$$ where $$F$$ is the inclusion $$\Sigma\hookrightarrow \Bbb R^d$$.

• @Migalobe $g^{ij}$ is the matrix inverse of $g_{ij}$ Jun 11 '20 at 19:22
• @Migalobe You need to interpret correctly what the coordinates $x^i$ are. You have an abstract manifold $M$ and it's embedded into Euclidean space by the map $F$. The $x^i$ refer to coordinates on $M$. So we have $$g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})=\frac{\partial F}{\partial x^i}\cdot \frac{\partial F}{\partial x^j}.$$ This is just writing in coordinates the definition $g=F^*\delta$, where $\delta$ is the ambient Euclidean metric. Jun 14 '20 at 16:23
• @Migalobe Do you know the difference between $g$ and $g_{ij}$? The embedding $F$ is implicitly contained in the equation $g(u,v)=u\cdot v$ because this is equivalent to $g=F^*\delta$. In my post I said $u$ and "geometrically tangent" to $\Sigma=F(M)$, which means it's the pushforward of a vector on $M$. You can identify $M$ and $\Sigma$ using $F$, but that doesn't mean $F$ won't appear in some formulas. Jun 14 '20 at 19:09