Norm of the gradient of a function between Riemannian manifolds.

Question

Let $(M, g), (N, h)$ be two Riemannian manifolds and $u: M \to N$ a smooth function. I would like to know how to show that, for $x \in M$, $$|\nabla u|^2(x) = g^{\alpha \beta}(x)h_{ij}(u(x)) \frac{\partial u^i}{\partial x^\alpha} \frac{\partial u^j}{\partial x^\beta}.$$

I know that in the case $f: M \to \mathbb R$, we define (locally) $$\nabla f = (g^{\alpha \beta} \partial_\alpha f )\partial_\beta,$$ which comes from the fact that we want $$g_p(\nabla f|_p, v) = df_p(v), \quad \forall v \in T_pM.$$ I guess we can define the gradient more or less the same way for $u: M \to N$, i.e. $$g(\nabla u, v) = du(v), \quad \forall v \in TM,$$ but now $du(v)$ is in $TN$ which means that the $g$ in the left hand side isn't really our Riemannian metric anymore (as it is not valued in $\mathbb R$). Admitting that this $g$ has locally the same components as the Riemannian metric, we would have locally $$g \nabla u = du \quad \Leftrightarrow \quad \nabla u = g^{-1}du$$ which implies $$\nabla u = \left(g^{ik} \frac{\partial u^k}{ \partial x^j}\right) \frac{\partial }{\partial y^i} \otimes dx^j.$$ It seems to more or less make sense with the form of $|\nabla u|^2$ above but it lacks details (in my opinion). Could any one of you explain clearly what's happening here ?

Answer 1

My preferred way of viewing $du$ is as a section of the vector bundle $T^*M \otimes f^*TN \to M$. Hence, if you end up contracting a metric $\langle \cdot, \cdot \rangle$ on this vector bundles coming from $T^*M$ and $f^*TN$ respectively, then you end up precisely with the expression $$ \langle du(x),du(x) \rangle=|d u(x)|^2 = g^{\alpha \beta}(x)h_{ij}(u(x)) \frac{\partial u^i}{\partial x^\alpha} \frac{\partial u^j}{\partial x^\beta} $$ in local coordinates. Now, we have $$ |\nabla u|^2=|du|^2 $$ since the Riesz isomorphism (or musical isomorphism, the thing identifying vectors and their duals) is isometric.

Answer 2

It follows more or less from the definitions, once you realize what are the definitions:

(1) Given a smooth mapping $u :M\to N$ and $x\in M$, the differential $\nabla u(x)$ (or $du(x), (u_*)_x)$ is a linear map $ du_x : T_p M\to T_{u(p)} N$. If you use the basis $$\left\{\frac{\partial}{\partial x^1}\bigg|_x, \cdots, \frac{\partial}{\partial x^n}\bigg|_x \right\}$$ of $T_pM$ and $$\left\{\frac{\partial}{\partial y^1}\bigg|_{u(x)}, \cdots, \frac{\partial}{\partial y^m}\bigg|_{u(x)} \right\}$$ of $T_{u(x)}N$ respectively, the local representation of this linear map is just $$ du_x \left(\frac{\partial}{\partial x^\alpha}\bigg|_x\right) = \sum_{i=1}^m \frac{\partial u^i}{\partial x^\alpha} (x) \frac{\partial}{\partial y^i}\bigg|_{u(x)}.$$

(2) Both $T_pM$, $T_{u(x)}N$ have a metric given by $g(x)$ and $h(u(x))$ respectively.

(3) In general for any linear map $L : (V, g)\to (W, h)$, one can define the norm of $L$ by $$ |L|^2_{g, h} := \sum_{\alpha=1}^n h(Le_\alpha, Le_\alpha),$$ where $\{e_1, \cdots, e_n\}$ is an orthonormal basis of $(V, g)$. One can check that this definition is independent of the orthonormal basis $\{e_1, \cdots, e_n\}$ chosen, and if $\{ v_1, \cdots, v_n\}$, $\{ w_1, \cdots, w_m\}$ are any basis of $V, W$ repsectively and $$ L v_\alpha = \sum_{i=1}^m L^\alpha_i w_i,$$ then

$$ |L|^2_{g, h} = \sum_{i, j=1}^n \sum_{\alpha, \beta = 1}^m g^{\alpha\beta} h_{ij} L^i_\alpha L^j_\beta,$$

where $h_{ij} = h(w_i, w_j)$ and $g^{\alpha\beta}$ is the inverse of $g_{\alpha\beta} = g(v_\alpha, v_\beta)$.