# Is it true that for any compact Riemannian manifold, the sectional curvature is bounded?

My attempt at a proof of this statement goes something like this: Let $(M,g)$ be a Riemannian $n$-manifold, then consider the vector bundle $\mathrm{T}M\oplus\mathrm{T}M$, then sectional curvature defines a function $K:\mathcal{U}\to\mathbb{R}$ which is defined on an open subset $\mathcal{U}\subset\mathrm{T}M\oplus\mathrm{T}M$ and is continuous, since it is smooth in coordinates. Note that we can define a $\operatorname{Gr}_2\mathbb{R}^n$-bundle on $M$ by considering a quotient topology on $\mathcal{U}$ in which we identify $(v_1,v_2)\in\mathrm{T}_xM\oplus\mathrm{T}_xM$ with $(w_1,w_2)\in\mathrm{T}_xM\oplus\mathrm{T}_xM$ precisely when they cut out the same plane in $\mathrm{T}_xM$. If we define the resulting quotient space to be $\mathcal{V}$, then we obtain a continuous map $K:\mathcal{V}\to\mathbb{R}$ and $\mathcal{V}$ is compact, so that sectional curvature is bounded.

However, I was talking to another student recently, and he claimed with great confidence that the sectional curvature on compact Riemannian manifolds need not be bounded, which makes me wonder if my (sketch of a) proof is incorrect.

• Do not worry, your proof is correct. – Moishe Kohan Mar 29 '17 at 4:22
• He might be thinking of things similar to definitions common in hyperbolic geometry, where one considers a hyperbolic metric on a compact 3-manifold to be a complete one on the interior with certain specified singular structure as you approach the boundary. So maybe he's thinking of complete metrics on the interior of a compact manifold. – user98602 Mar 29 '17 at 8:34

## 1 Answer

Yes: see Section 9.3, page 166, in "Geometry of manifolds" by R.L. Bishop and R.J. Crittenden, Academic Press, 1964.

That page is available on Google Books, and the book can also be downloaded on Elsevier's website.