Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations:
\begin{equation}
f\_1(\vec x)=0, \\
\vdots \\
f\_m(\vec x)=0,
\end{equation}
where $\vec x$ are coordinates in $\mathbf R^n$.


Given a point $\vec x\_0$ in $M$, how can the Ricci scalar (calculated from the induced metric on $M$) at $\vec x\_0$ be expressed in terms of the functions $f\_1,...,f\_m$?



Ricci scalar of submanifold of $\mathbf R^n$
*This is an extended comment answering [Ricci scalar of sub-manifold of $\mathbf R^n$](https://mathoverflow.net/questions/430953/ricci-scalar-of-sub-manifold-of-mathbf-rn?noredirect=1#comment1109048_430953)*


Assuming the $f\_i$ are independent, at every point $x$ their gradients span the orthogonal complement to the tangent space of the manifold.


The second fundamental form is given by the normal projection of the ambient derivative. So if $V,W$ are tangent vector fields to $M$, then at the point $p$ you have that $II\_p(V,W)$ is the projection to the span of $\{\nabla f\_i\}$ of $\partial\_V W$. Using that the $f\_i$ are independent, this means $II\_p(V,W)$ is uniquely determined by the $m$ values
$$ \langle \partial\_V W, \nabla f\_i(p) \rangle $$
(The reconstitution formula, when $\nabla f\_i$ are orthogonal, is just to multiply this against $|\nabla f\_i(p)|^{-2} \nabla f\_i(p)$ and sum. When they are not, there's a matrix inversion involved which makes the formula messy.)


The above inner product, using that $\langle W,\nabla f\_i(p)\rangle = 0$ as $W$ is a tangent and $\nabla f\_i(p)$ is a normal, can be rewritten as
$ \langle W, \partial\_v \nabla f\_i(p)\rangle$ which is nothing more than the Hessian of $f\_i$ evaluated in the direction $V\otimes W$.




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More about the matrix inversion: the idea is that given any vector in the normal space you can write it as
$$ \sum \alpha\_i \nabla f\_i $$
You wish to reconstruct $\alpha\_i$ from the measuraments
$$ \beta\_j = \langle\sum\_i \alpha\_i \nabla f\_i, \nabla f\_j\rangle = \sum\_i \underbrace{\langle \nabla f\_j , \nabla f\_i \rangle}\_{= A\_{ji}} \alpha\_i $$
Linear independence guarantees that the matrix $A\_{ji}$ is invertible; in fact you have explicit formulas of this in terms of cofactors. These are highly multilinear expressions in terms of $\nabla f$, but do not involve Hessians.


For now, denote by $B\_{ji}$ the inverse matrix to $A\_{ji}$ above. So you can write
$$ II\_p(V,W) = \sum\_{i,j} B\_{ji} \nabla f\_j \nabla^2\_{V,W} f\_i $$


Note finally for the expression that shows up in the Gauss equations
$$ \langle II\_p(V,W), II\_p(X,Y)\rangle = \sum\_{i,j,k,l} \langle B\_{ji} \nabla f\_j \nabla^2\_{V,W} f\_i , B\_{kl} \nabla f\_k \nabla^2\_{X,Y} f\_l\rangle $$
the RHS can be reorganized as
$$ \sum\_{i,j,k,l} B\_{ji} B\_{kl} A\_{jk} \nabla^2\_{V,W} f\_i \nabla^2\_{X,Y} f\_l = \sum\_{i,l} B\_{li} \nabla^2\_{V,W} f\_i \nabla^2\_{X,Y} f\_l$$


So the formula for the Riemann curvature (with all indices lowered) is actually not too bad.


For the Ricci scalar you also need to take the trace, which requires computing the tangential projection using $\nabla f\_i$: you can do this just by subtracting the normal projection from the identity. This can complicate the expression quite a bit.


