# Intrinsic vs. Extrinsic notions of Harmonic maps.

I have just began studying the theory of harmonic maps between manifolds recently. So far, I've come across 2 ways to define harmonic maps between (Riemannian) manifolds.

Let $u:(M,\gamma)\to(N,g)$ be a map, says $C^1$ for simplicity, between 2 Riemannian manifolds of dimension $m$ and $n$ respectively. The metrics on them are given by $\gamma=(\gamma_{\alpha\beta})$ and $g=(g_{ij})$.

Intrinsic definition: $u$ is said to be harmonic if, in coordinate chart, $$\Delta u^k=\gamma^{\alpha\beta}\Gamma^k_{ij} \frac{\partial u^{i}}{\partial x^{\alpha}}\frac{\partial u^{j}}{\partial x^{\beta}}$$ where $(\gamma^{\alpha\beta})=(\gamma_{\alpha\beta})^{-1}$.

On the other hand, by Nash embedding theorem, we can view $N$ as an isometrically embedded submanifold of $\Bbb R^K$ some large $K$. In this case, we may define harmonicity as follow.

Extrinsic definition: $u$ is said to be harmonic if \begin{align} \Delta u =& A(u)(\nabla u,\nabla u) \\ :=& \sum_{\alpha=1}^m A(u)(\frac{\partial u}{\partial x^{\alpha}},\frac{\partial u}{\partial x^{\alpha}}) \end{align} where $A$ is the second fundamental form of $N$ in $\Bbb R^k$.

Here are some of my questions:

-In both cases, $\Delta$ is the Laplace-Beltrami operator right?

-For the extrinsic definition, is $\frac{\partial u}{\partial x^{\alpha}}$ just the usual $(\frac{\partial u^1}{\partial x^{\alpha}},\dots,\frac{\partial u^K}{\partial x^{\alpha}})\in \Bbb R^K$?

-How do we show that both expressions are equivalent?

PS. My background is mostly in analysis (in Euclidean spaces) with some basic knowledge of Riemannian geometry and differential geometry. Note that I'm not quite good with differential geometry yet, as you can see, so I wouldn't be able to understand if the answer is too advanced, e.g. something involving Hodge isomorphism.

• In your second question: $\nabla u$ cannot be the usual gradient simply because $u$ is not defined on any open subset of $\mathbb R^K$. Jan 15, 2018 at 20:17
• @MarianoSuárez-Álvarez I looked into it a bit deeper and modified the question accordingly. Does my question make sense now? Jan 15, 2018 at 21:16

Yes, $\Delta$ is just the Laplace-Beltrami operator associated to $\gamma,$ and these two definitions of harmonicity are equivalent. (Note: Depending on your sign conventions for $\Delta$ and $A,$ either of both of these equations might need a minus sign.)
In the extrinsic case where we write $\Delta u$ for $u: M \to \mathbb R^K,$ the Laplacian is acting on the component functions $u^i : M \to \mathbb R.$ The derivative $\partial u /\partial x^\alpha$ is to be understood literally as $Du(\partial/\partial x^\alpha) \in TN;$ though strictly speaking the given extrinsic formula is correct only at points where the basis $\partial/\partial x^\alpha$ is orthonormal: the general extrinsic coordinate formula is more correctly written $$\Delta u = \gamma^{\alpha \beta} A(u)(\frac{\partial u}{\partial x^{\alpha}},\frac{\partial u}{\partial x^{\beta}}).$$ To be clear, yes, $\partial u/\partial x^\alpha$ is just the ordinary vector of partial derivatives $(\partial u^1/\partial x^\alpha, \ldots)$ when thought of as a tangent vector to $\mathbb R^K,$ but it's important to remember that it's also well-defined as a tangent vector to $N$, since this makes it a natural input to the second fundamental form $$A(u(x)) : T_{u(x)} N \times T_{u(x)} N \to (T_{u(x)} N)^\perp.$$
The usual way to see the equivalence of these two characterizations is via the variational definition: harmonic maps are exactly the critical points of the Dirichlet energy $E(u)=\int |\nabla u|^2.$ You can find a proof using this method in Chapter 1 of Analysis of harmonic maps and their heat flows by Lin and Wang.
For a more direct computational verification, we need to relate the extrinsic and intrinsic components of $u$; so we let $\phi : N \to \mathbb R^K$ denote the isometric embedding, so that the extrinsic equation is written explicitly as $$\Delta (\phi^a \circ u) = \gamma^{\alpha \beta}A^a_{ij}(u)\frac{\partial u^i}{\partial x^{\alpha}}\frac{\partial u^j}{\partial x^{\beta}}.$$ I'm going to use the conventions $\Delta = \nabla^\alpha \nabla_\alpha$ and $A = \nabla^{\mathbb R^K} - \nabla^N,$ so my sign in the intrinsic equation won't match yours.
Expanding the LHS using the usual coordinate formula for the Laplace-Beltrami operator and the chain rule we find $$\Delta (\phi^a \circ u) = \frac{\partial \phi^a}{\partial u^i}\Delta(u^i)+\frac{\partial^2 \phi^a}{\partial u^i \partial u^j}\frac{\partial u^i}{\partial x^{\alpha}}\frac{\partial u^j}{\partial x^{\beta}}\gamma^{\alpha \beta}.$$
From the definition of the second fundamental form we have $$A_{ij}^a(u) = \frac{\partial^2 \phi^a}{\partial u^i \partial u^j}-\frac{\partial \phi^a}{\partial u^k}\Gamma^k_{ij}(u),$$ so we find $$\Delta (\phi^a \circ u) -\gamma^{\alpha \beta}A^a_{ij}(u)\frac{\partial u^i}{\partial x^{\alpha}}\frac{\partial u^j}{\partial x^{\beta}} = \frac{\partial \phi^a}{\partial u^k}\left(\Delta (u^k)+\gamma^{\alpha \beta}\Gamma^k_{ij}(u)\frac{\partial u^i}{\partial x^{\alpha}}\frac{\partial u^j}{\partial x^{\beta}}\right).$$ Since $D\phi$ is injective, the LHS vanishes if and only if the parenthesized term on the RHS does, so we're done.