# Bryant Soliton is asymptotically cylindrical?

This is my first question in mathoverflow.

I'm now reading Brendle's paper http://arxiv.org/pdf/1203.0270.pdf.

I'm confused about how to check Condition (ii) of asymptotically cylindrical condition for Bryant soliton.

How to make use use of Hamilton's compactness Thm to prove it?

Thanks a lot.

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One method is to repeat in dimensions $n$ Bryant's analysis in which he proves the existence and uniqueness of the so called Bryant soliton in $3$-dimensions: http://www.math.duke.edu/~bryant/3DRotSymRicciSolitons.pdf, which if you're asking such a question I suppose you have already done (it is well known to exist, but not written down anywhere as far as I know... the ODE argument should follow basically identically as Bryant's). If you have good enough control on the asymptotics of the warping function, you should be able to prove that the higher dimensional Bryant solitons satisfy the desired properties.

An easier method (which uses a good deal of machinery) is as follows. You should note the similarity to Proposition 2.2 in Brendle's 3D Inventiones paper:

Notice that up to a subsequence, we can take the blowdown limit of the $\hat g_{(m)}(t)$ by Hamilton compactness (you should think through why you can use this!). Furthermore, by looking at $\mathscr{L}_V(\hat g_{(m)})=1$ for any $V$ coming from a rotation of $\mathbb{S}^{n-1}$, we see that up to passing to a further subsequence, the blowdown limit is rotationally symmetric. On the other hand, it splits off a line, by e.g. Morgan--Tian Theorem 5.35. Thus, it must be a shrinking cylinder. Thus, because any sequence has this limit, we have the desired result.

EDIT: Just to remark, one must construct the Bryant soliton and prove certain properties about it in order to make this argument work. In particular, you need to use nonnegative sectional curvature in the above argument (do you see where?). I don't want to give the impression that no ODE analysis is necessary. However, this argument does avoid some possibly annoying analysis of the ODE.

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You are sooo helpful. Thanks. –  Naruto Uzumaki Sep 5 '13 at 2:06
Glad to be of assistance! Hope my answer was clear. I added a remark at the end, because I think I was a bit hasty in my original answer. –  Otis Chodosh Sep 5 '13 at 3:56
This is to supplement Otis Chodosh's fine answer. From Robert Bryant's analysis of his steady Ricci soliton, one has $g=dr^{2}+\phi(r)^{2}g_{S^{n-1}}$, where $$C^{-1}r^{1/2}\leq\phi(r)\leq Cr^{1/2},\quad\phi^{\prime}(r)=O(r^{-1/2} ),\quad\phi^{\prime\prime}\left( r\right) =O(r^{-3/2}).$$ Consider a basepoint $x_{i}$ on the $(n-1)$-sphere with $\{r(x)=r_{i}\}$ and rescale so that this sphere has radius $1$; i.e., let $$g_{r_{i}}=\frac{dr^{2}}{\phi(r_{i})^{2}}+\left( \frac{\phi(r)}{\phi(r_{i} )}\right) ^{2}g_{S^{n-1}}=ds^{2}+\left( \frac{\phi(\phi(r_{i})s+r_{i})} {\phi(r_{i})}\right) ^{2}g_{S^{n-1}},$$ where $s=\phi(r_{i})^{-1}(r-r_{i})$. We have $$\frac{d}{ds}\left( \frac{\phi(\phi(r_{i})s+r_{i})}{\phi(r_{i})}\right) =\phi^{\prime}(\phi(r_{i})s+r_{i})=O((\phi(r_{i})s+r_{i})^{-1/2} )=O(r_{i}^{-1/2}),$$ provided $s=O(\phi(r_{i})^{-1}r_{i})=O(r_{i}^{1/2})$. Now choose any sequence $\{x_{i}\}$ so that $r_{i}=r(x_{i})\rightarrow\infty$. From this one obtains a pointed cylinder limit of $(g_{r_{i}},x_i)$ in the Cheeger-Gromov sense without having to pass to a subsequence.
In some more general situations, one can apply Hamilton's Cheeger-Gromov type compactness theorem for complete solutions to the Ricci flow with semi-global curvature bounds. Some form of dimension reduction (given a line or using the strong maximum principle) together with Hamilton's classification (reproved by Perelman) of $2$-dimensional $\kappa$-solutions is also useful here.
As an analogue (although extrinsic), consider a parabola $y=x^{2}$ in the plane. Dimension reduction of this curve yields two parallel lines as follows. Let $x_{i}>0$ and consider the basepoint $(x_{i},x_{i}^{2})$. Multiply the parabola by the constant $x_{i}^{-1}$ so that distance from $(x_{i},x_{i}^{2})$ to the $y$-axis becomes $1$. We get $x_{i}y=(x_{i}x)^{2}$, i.e., $y=x_{i}x^{2}$. Now translate this curve so that $(1,x_{i})$ goes to $(1,0)$. The curve becomes $y=x_{i} x^{2}-x_{i}$. As $x_{i}\rightarrow\infty$ this curve converges in compact subsets of the plane to the pair of vertical lines $x=\pm1$, i.e., $S^{0}\times\mathbb{R}$.