Euler Lagrange equation for harmonic maps In the paper "The existence of minimal immersions of 2-spheres" by Sacks and Uhlenbeck the authors claim that the Euler Lagrange equation for the modified functional $E_\alpha(s) = \int_M (1 + |ds(x)|^2)^\alpha d\mu$, $\alpha>1$, is
$$
\Delta s + (\alpha - 1)\frac{(d^2s,ds)ds}{1+|ds|^2} + A(s)(ds,ds) = 0
$$
namely Equation (2).
As far as I understand, this is a compact form. But how should I interpret it?
Can anyone help me unravel the equation in order to write it neatly in components?
@John: what I mean is the following: if we were given the classical Euler-Lagrange equation for harmonic maps, i.e.
$$
\Delta s + A(s)(ds,ds) = 0,
$$
this could be re-written as, putting $\phi = i \circ s$,
$$
\Delta \phi^a(x) + g^{ij}(x) A^a_{s(x)} ( \frac{\partial \phi}{\partial x^i}; \frac{\partial \phi}{\partial x^j} ) =0,
$$
where $i$ is the immersion in $R^k$ of the target manifold, $a=1,2,...,k$ and $x^i$ are local coordinates on the first manifold.
Now I would like to do something similar to the equation above, but the middle term in the left hand side is troublesome.
 A: The notation $(d^2s,ds)ds$ is slightly confusing - there are in fact more contractions happening here than first meets the eye. To work out what's going on here, I looked at the previous equation in the paper, which suggests that the terms
$$\Delta s + (\alpha - 1)\frac{(d^2s,ds)ds}{1+|ds|^2}$$
should come from the expression $$(1+|ds|^2)^{1-\alpha} d^* ((1+ |ds|^2)^{\alpha - 1}ds).$$
When we apply the product rule to this expression, the Laplacian pops out immediately from one term, and the other term is (using notation $(M, \delta_{ij})$, $(N, g_{\mu \nu})$, and $s_{;ij} = \nabla_j s_i = \nabla_j (\partial s /\partial x^i)$)
$$(\alpha-1)\frac{\delta^{kl} s_{,k}^\lambda (2 s_{;il}^\mu s_{,j}^\nu g_{\mu \nu} \delta^{ij})}{1 +|ds|^2}. $$
It seems to me that the factor of $2$ (which arises from the product rule $D(u,u) = 2(Du, u)$) has been left out in the paper - I'm not sure what I'm missing here. Anyway, the key point here is that 
$$ (d^2 s, ds) = s^\mu_{;il} s^\nu_{,j}g_{\mu \nu}\delta^{ij} \in \Gamma(T^*M)$$
is just a 1-form on the sphere $M$, and $(d^2 s, ds)ds$ is the section of the pullback bundle $s^*TN$ obtained by contracting this once with $ds$:
$$ (d^2 s, ds)ds = \delta^{kl} s^{\lambda}_{,k} s^\mu_{;il} s^\nu_{,j}g_{\mu \nu}\delta^{ij} \in \Gamma(s^* TN).$$
You should be able to get this in terms of your $\phi : M \to \mathbb R^k$ pretty easily - just need to write the derivatives of $s$ in terms of the coordinate derivatives $\partial \phi^\mu / \partial x^i$ (the only non-trivial thing here should be an orthogonal projection onto $TN$ for the $s^\mu_{;il}$ term).
