Let us consider, a closed Riemannian surface $(\Sigma,h)$ and a compact Riemannian manifold $(N,g)$ with dimension greater than $3$. If we are given a sequence of harmonic maps $u\_n:(\Sigma,h) \rightarrow (N,g)$ with bounded energy, i.e.
$$E(u\_n)=\int\_\Sigma \vert du\_n\vert^2 \, dv < C,$$
it is well know that we have an energy identity, that is to say there exists an harmonic map $u^\infty : (\Sigma,h) \rightarrow (N,g)$ and some bubbles, i.e. harmonic maps $\omega\_i:\mathbb{C} \rightarrow (N,g)$ such that 


$$\lim\_n E(u\_n)= E(u^\infty)+\sum\_{i} E(\omega\_i).$$
My question is: 



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When $\Sigma=\hat{\mathbb{C}}$, the answer is clearly yes, considering 
$$u\_n(z)=(z,nz).$$
But here, the fact that the conformal group of $\hat{\mathbb{C}}$ is not compact seems to be crucial. So is there an example of bubbling when $\Sigma$ is not $\mathbb{C}$, especially in genus bigger than $2$?



Bubbling example for harmonic maps
Yes. The genus of $\Sigma$ is not really relevant. Here's an example: Let $f$ and $g$ be two meromorphic functions on $\Sigma$, where $g$ is nonconstant, and consider the sequence of maps $u\_n: \Sigma\to N^4 = \mathbb{CP}^1\times\mathbb{CP}^1$ given by
$$
u\_n(p) = \bigl([f(p)],[n\,g(p)]\bigr).
$$
(Here, $N$ is given the product metric and the metric on $\mathbb{CP}^1$ is the standard metric of constant sectional curvature $1$.) As $n$ goes to $\infty$, the energy densities of these (holomorphic and, hence, harmonic) maps stay bounded away from the zero divisor of $g$, but go to infinity in a neighborhood. In the limit, one has $u^\infty(p) = \bigl([f(p)],[\infty]\bigr)$, and the energy of $u^\infty$ is essentially the degree of $f$, while the energy of $u\_n$ is essentially the degree of $f$ plus the degree of $g$. The number of 'bubbles' is the number of points in the zero divisor of $g$, and this can be arbitrarily large.


