**Motivation**: In the theory of harmonic maps between manifold, we often see the characterization 
$$
\Delta\_g u = -g^{ij}A(u)\left(\frac{\partial u}{\partial x^i},\frac{\partial u}{\partial x^j}\right)
$$
when we are working in the extrinsic viewpoint. 



> 
> **Question**: Let $M,N$ be two Riemannian manifolds, $u:M\to N$ be a smooth map. Let $A$ denotes the *second fundamental form* of $N$, viewing as an embedded submanifold of $\Bbb R^k$. How is the expression
>  $$
> A(u)\left(\frac{\partial u}{\partial x^i},\frac{\partial u}{\partial x^j}\right)
> $$
>  rigorously defined?
> 
> 
> 


First of all, $u$ need not be injective, so there can be a point $y\in N$ such that $u(x\_1)=u(x\_2)=y$. Doesn't this means that $\frac{\partial u}{\partial x^i}(y)$ is really not a vector field on $N$?


I think this non-uniqueness can be fixed for $x$ such that $du(x)$ is of full rank, so that $u$ is locally an embedding there. When $du(x)$ degenerates, however, I don't know how to make sense of it.



What does $A(u)\left(\frac{\partial u}{\partial x^i},\frac{\partial u}{\partial x^j}\right)$ mean, exactly?
You are correct that $\partial u/\partial x\_i$ is not a vector field on $N$. It is a section of $u^\*TN$, i.e. associating to each point $x \in M$ a vector $\partial u/\partial x\_i \in T\_{u(x)} N$. So we take the second fundamental form $A$ of $N$, which is a section of $S^2(T^\*N) \otimes T^{\perp} N$, and we pull it back to form $u^\*A$, which you denote $A(u)$, a section of $S^2(u^\* T^\*N) \otimes u^\* T^{\perp} N$. Then we plug in two sections $\partial u/\partial x\_i, \partial u/\partial x\_j$ of $u^\* TN$.


