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I am reading Richard Schoen's classical example on the multiplicity of solutions of yamabe problem. He says on $S^1(T)\times S^{n-1}$, there exists a critical number $T_0$ such that if $T\leq T_0$, then there is only metric, which is the product metric, whose scalar curvature is $n(n-1)$.

When $T\in (T_0,2T_0]$, there exist two metrics, one is product metric and the other is a minimizer for yamabe problem. He also said at this time the product metric has morse index 2. I have no idea of why you can say a metric has morse index? Is it infinite dimensional morse index? why 2?

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I assume you mean these notes: http://www.math.jhu.edu/~js/Math646/schoen.totalscalar.pdf?

In general it does not make sense to say that a "metric has morse index 2" without any context. Before proceeding further, you should make sure that you understand classical Morse theory for functions on finite dimensional manifolds. If you don't, I would recommend reading Milnor's famous book: http://www.amazon.com/Morse-Theory-Annals-Mathematic-Studies/dp/0691080089. It's very accessible.

What Schoen means here is that he is considering the Yamabe functional as a function on a conformal class (in this case the conformal class containing the product metric $S^1(T)\times S^{n-1}$). Here, the Yamabe functional is $$ Y_g(u) := E({u^\frac{4}{n-2}g}) = \frac{\int_M R_{u^\frac{4}{n-2}g}dV_{u^\frac{4}{n-2}g}}{\left(Vol(M,{u^\frac{4}{n-2}g}) \right)^{\frac{n-2}{n}}} $$

Then you can ask for critical points and for the Morse index.

The critical points of the Yamabe functional Y(u correspond to constant scalar curvature metrics in the conformal class. See, p. 39 of http://projecteuclid.org/euclid.bams/1183553962 (which is a great survey of the Yamabe problem). Essentially this amounts to computing that the derivative of the Yamabe functional vanishes $$ DY_g(u)v : = \frac{d}{dt} E((u+tv)^{\frac{4}{n-2}}g) = 0 $$ if and only if $u^{\frac{4}{n-2}}g$ has constant scalar curvature.

At a critical point, it make sense to talk about the Hessian of the Yamabe functional, i.e. $$ D^2Y_g(u)\cdot(v,w) : = \frac{d^2}{dsdt} E((u+sv+tw)^{\frac{4}{n-2}}g). $$

It's not hard to see that this is essentially the same as linearizing the scalar curvature equation. The reason for this is that the first derivative gives the constant scalar curvature equation, while the second derivative linearizes this equation. One may check that (see Schoen (1.8)), up to constants (which I won't try to get right here) the Hessian is given by $$ D^2Y_g(u)\cdot(v,w) = C\int_M ((n-1)\Delta_{u^{\frac{4}{n-2}}g}v + R_{u^{\frac{4}{n-2}}g}v)w dV_{u^{\frac{4}{n-2}}g}. $$ In particular, the "negative Morse directions" correspond to eigenfunctions of $\Delta_{u^{\frac{4}{n-2}}g}$ with eigenvalue less than $\frac{R_{u^{\frac{4}{n-2}}g}}{n-1}$.

In the $S^1(T)\times S^{n-1}$ discussion, Schoen does not make use of this fact/computation. He explicitly linearizes the equation for scalar curvature given in this case to find destabilizing directions (i.e. an eigenfunction of the linearized operator with negative eigenvalue). So, for example, if the product has "Morse index 2," then there are two such eigendirections for the linearized operator $L$.

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  • $\begingroup$ Thank you very much. I understand your answer for all except the last paragraph. I think you mean Schoen linearize $\frac{d^2u}{dt^2}-\frac{(n-2)^2}{4}u+\frac{n(n-2)}{4}u^{\frac{n+2}{n-2}}=0$ at $u_0$ to be $L=\frac{d^2}{dt^2}+(n-2)$. He found out it has two negative eigendirections $L$. $\endgroup$
    – Slm2004
    May 6, 2014 at 1:59
  • $\begingroup$ No problem! Yes, that's exactly what I meant, sorry if it wasn't clear. $\endgroup$ May 6, 2014 at 14:02

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