0
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ One way to convince yourself is to write it using local coordinates on both $M$ and $N$. $\endgroup$
    – Deane Yang
    Commented Mar 31, 2018 at 4:45

2 Answers 2

Reset to default
2
$\begingroup$

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 $u^* TN$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ The sloppy notation that is necessary in differential geometry sometimes makes it difficult to communicate with certain analysts of a very classical training, because their objects are always functions with a clearly specified domain and range. $\endgroup$
    – Ben McKay
    Commented Mar 31, 2018 at 8:55
1
$\begingroup$

I do not really follow the question. Since everything is local, the map $u$ can be effectively replaced by its derivative $D u$, which is a matrix of dimension $\dim(N)\times \dim(M)$. The map may be bad and the matrix may send different elements to the same element - but the image of the map itself is well defined. Once defined, the second fundamental form on $N$ gives a symmetric bilinear form on $T^{*}(N)$ at $u_{x}$. This is well-defined and you may write it out in matrix form if you like. The evaluation then is just matrix multiplication.

Improve this answer
$\endgroup$
2
  • $\begingroup$ But doesn't the second fundamental form act on a pair of vector fields on $N$? Here we only have 2 tangent vectors. $\endgroup$
    – BigbearZzz
    Commented Mar 31, 2018 at 0:35
  • $\begingroup$ @BigbearZzz: I do not understand your confusion. The second fundamental form can be written as a $\dim(N)\times \dim(N)$ matrix. So its evaluation on $(v,u)$ is just $vMu^{T}$. $\endgroup$
    – Bombyx mori
    Commented Mar 31, 2018 at 0:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .