I'm having some minor confusion about the proof of the Corlette-Donaldson Theorem found here (Theorem 3.14)


For completeness, the statement is as follows.

Theorem: Let $M$ be a closed manifold and $\rho: \pi_1(M)\to SL_n(\mathbb{C})$ be a semisimple representation. Then there is a $\rho$-equivariant harmonic map from $\tilde{M}\to SL_n(\mathbb{C})/SU_n(\mathbb{C})$.

(Look at the survey starting from page 27, for the notations and conventions). The proof uses the well-known heat flow method, pioneered by Eells and Sampson. Namely, one considers the equation $$\frac{\partial}{\partial t} u_t = \tau(u_t)$$ where $\tau(\cdot)$ denotes the tension fields, $t\in [0,\infty)$, and $u_0$ is some fixed $\rho$-equivariant map. Note that if $u_t$ is connected to $u_0$ along the heat flow then $u_t$ is necessarily $\rho$-equivariant as well. One notes you can solve this equation for all times and there is a limiting map that is necessarily harmonic.

Here is the source of my confusion.

Eells-Sampson prove that given a map $f$ from a compact manifold to some other manifold, there is a short-time solution to the heat equation. They then give some sufficient conditions for long time existence. In particular, if this solution remains bounded and the target manifold has negative curvature, then we get long time existence and convergence to a harmonic map. Hamilton proves a similar statement but for maps between compact manifolds with boundary.

However, Wentworth is working on some cover of $M$, not necessarily compact. To obtain long time existence, namely a solution, Wentworth quotes the papers of Eells-Sampson, and Hamilton. He then shows the solution remains bounded and seems to apply some results from Eells-Sampson. I am wondering, why is he able to use these results? There must be something about the $\rho$-equivariance that is making this possible, but I'm not seeing it. (A similar thing is done in, say, Donaldson's original paper, and the same issue arises).



I won't answer your question "why can he use these results" directly, but let me indicate the structure of the proof and hopefully this will alleviate some of your confusion.

For the long time existence argument, I don't think you need look at Eells-Sampson or Hamilton, or any other paper. The limiting convergence to a harmonic map is in the end an application of Arzela-Ascoli to a new sequence of maps $g_{n}\cdot f_{n}$ where $g_{n}\in {SL}_{n}(\mathbb{C});$ this sequence is cooked up to ensure for some $x\in \widetilde{X},$ we have $g_{n}\cdot f_{n}(x)=p$ is fixed, which will allow us to apply Arzela-Ascoli. Here, the $f_{n}: \widetilde{X}\rightarrow SL(n,\mathbb{C})/SU(n)$ is some sequence of maps along the flow. The non-compactness of the universal cover $\widetilde{X}$ is not an issue, because you only need to prove convergence on a compact fundamental domain for the action of the fundamental group, since the map is equivariant. Using Arzela-Ascoli and the previously obtained estimates in the paper, you always obtain a limiting harmonic map $f_{\infty}: \widetilde{X}\rightarrow SL(n,\mathbb{C})/SU(n).$ But, the key point is that these maps $g_{n}\cdot f_{n}$ are now $\rho_{n}:=g_{n}\circ \rho \circ g_{n}^{-1}$-equivariant. Since $SL(n, \mathbb{C})$ is non-compact, it's not clear that the $\rho_{n}$ sub-converge to some $\rho_{\infty}$ for which $f_{\infty}$ is $\rho_{\infty}$-equivariant. Moreover, even if you achieve this, it's not clear that $\rho_{\infty}$ is conjugate to $\rho,$ which is what needs to be proved. What he proves, using the semi-simple hypothesis, is that the $g_{n}$ sub-converge to some $g\in SL_{n}(\mathbb{C})$ and therefore $\rho_{\infty}=g\circ \rho \circ g^{-1}$ and $f_{\infty}$ is $\rho_{\infty}$-equivariant.

If $\rho$ is not reductive, what is happening? Well, it's a key fact that the orbit closure of $\rho$ under the conjugation action contains a unique reductive orbit. The limiting harmonic map you get $f_{\infty}$ will be equivariant for a new reductive representation $\rho_{\infty},$ which lies in the orbit closure of $\rho,$ but is not conjugate to $\rho.$

A helpful alternative viewpoint getting rid of the non-compactness is to see your maps $g_{n}\cdot f_{n}$ as sections of the bundles


which are now bundles over the compact Riemann surface $X.$ But of course now to apply Arzela-Ascoli, you'll need to understand the sense in which the $E_{n}$ converge to $E_{\infty}$ and use some version of Arzela-Ascoli which takes into account varying targets. This is perhaps a bit more conceptual than the proof I describe above, but it turns out to be technically harder to work out all the details, at least I assume, since I haven't seen a proof along these lines in any of the sources I know.

  • $\begingroup$ Okay, thanks. I'll have to think about this a bit. Also, another thing (which I didn't make clear I was also asking about): to get the short time solution do you just view the original map as a section of the bundle, and then apply Eells-Sampson? This would be legal as now the domain is compact. $\endgroup$ – user470881 May 17 '19 at 23:41
  • $\begingroup$ The short time existence is its own animal, which might be best formalized in terms of the bundle section point of view, but will always have a fundamental complexity. At heart it’s an implicit function theorem argument, which means it’s really about the contraction mapping theorem. I suggest you understand everything else before tackling short time existence, and then work in that direction. I’ll try to supply some references later, but its a bit late for that in Heidelberg tonight. $\endgroup$ – Andy Sanders May 17 '19 at 23:53
  • $\begingroup$ Cool, thanks. Looking forward to it. $\endgroup$ – user470881 May 18 '19 at 0:14
  • $\begingroup$ In this thesis of Baker arxiv.org/pdf/1104.4409.pdf, the theory of short time existence for evolution equations for sections of vector bundles seems to be treated in great detail. Of course, this would need to be souped up to deal with more general fiber bundles to apply directly as here we have a section of a fiber bundle, not a vector bundle, which we wish to evolve. Sorry for not being able to give you a more definitive answer. $\endgroup$ – Andy Sanders May 18 '19 at 17:43
  • $\begingroup$ Well thank you very much. I'll take a look at this. $\endgroup$ – user470881 May 18 '19 at 19:06

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