Lemma: Suppose $$U \subset R^n$$ is connected and open with the property that if $$[a,b],[b,c] \subset U$$ then $$[a,c] \subset U$$, that is if two sides of a triangle are in $$U$$ then so is the third side, then $$U$$ is convex.

Proof: The set of points that can be reached with a straight line from the point $$p$$ is both an open set and a closed set. Because $$U$$ is connected the set of points we can reach from some point $$p$$ must be the whole space. Therefore taking any two point other than $$p$$ and making two sides of a triangle by connecting them to $$p$$ we conclude there must be a straight path between those points by hypothesis. Therefore $$U$$ is convex.

Last time: Again take U to be an open connected subset of $$R^n$$ and let $$c:U \rightarrow R$$ be hessian convex ($$d^2c>0$$) and $$d^2c$$ is a complete Riemannian metric then $$U$$ is convex.

A projective manifold determines and affine manifold of one dimension higher, called the tautological line bundle. Whether or not that projective manifold is convex is equivelent to whether or not there is a certian kind of metric on the affine manifold.

Tautological Line Bundle: Let $$M^n$$ be a real projective manifold. There is a developing map $$dev: \tilde{M} \rightarrow S^n$$. The sphere of course is a subset of $$R^{n+1}-\{0\}$$ and there is a radial projection $$\Pi_{\xi} R^{n+1}-\{0\} \rightarrow S^{n}$$, then the preimage of a point in the sphere will be a ray coming out of the origin. This is a line bundle $$R^{+} \equiv (0,\infty) \rightarrow R^{n+1}-\{0\}$$ which we can line of as an affine manifold. This gives us a natural line bundle over the sphere.

Pull Back Construction: The pull back of this line bundle to $$\tilde{M}$$. The fiber product: $$\xi \tilde{M} \{(\tilde{m},x) \in \tilde{M} \times (R^{n+1}-\{0\}) : \Pi_{\xi}(x)=dev(\tilde{m})\}$$. The obvious map $$(\tilde{m},x) \rightarrow \tilde{m}$$ which is the $$R^{+}$$ bundle. We can quotient out by the fundamental group $$\Pi_{M}: \tilde{M} \rightarrow M$$ and define an action of $$\Pi$$, $$M$$ on $$\xi \tilde{M}$$ by $$\tau \circ (\tilde{m},x):=(\tau \tilde{m},p(\tau)(x))$$ where $$\rho$$ is the holonomy $$\Pi,M \rightarrow SL^{\pm}(n+1,R)$$. This action is bundle preserving which means a map $$\xi M = \xi \tilde{M}/(\Pi,M) \rightarrow M$$ that is line preserving.

Example: If $$dev:\tilde{M} \hookrightarrow S^{n}$$ is injective then the line bundle is all the rays comming through the image of this map. So the line bundle is embedded in euclidean space $$R^{n+1} - \{0\}$$.

Example: $$dim=1$$ with $$M=S^1$$ we can think of that

$P= \begin{pmatrix} 2&0 \\ 0&1/2 \end{pmatrix}$

The universal cover we can think of as the real line. This doubles the $$x$$ coordinate andf halves the $$y$$ coordinate. So $$\xi \tilde{M} / (\rho(\Pi,M) = <p(\alpha)>)$$ we get an open annulus/cylinder. Affine manifold, $$\xi S^{1} = S^{1} \times (0,\infty)$$ the projection of $$\Pi({(x,y) : xy=c})$$. These can be thought of as Vinberg hypersurfaces.

Tautological $$S^1$$-bundle: $$\xi , M = \xi M/ Z$$ with $$k \in Z$$ and $$[\tilde{m},x] \in \xi M$$ then $$k \cdot [\tilde{m},x] = [\tilde{m},2^k \cdot x]$$ since we apply the action of doubling $$k$$ times of course. We get $$S^1 \rightarrow \xi, M \rightarrow M$$.

Given compact projective $$n$$-manifold $$M$$ to get compact $$S^{1}$$-bundle $$S^1 \rightarrow \xi,M$$ is compact and affine. We can show it is convex using a complete Hiessan metric. We are going to show $$M^n$$ is properly convex.

Definition: Radial flow ($$\Phi_t \circ \Phi_s = \Phi_{t+s}$$) or $$\Phi: R^{n+1}-\{0\} \rightarrow R^{n+1}-\{0\}$$, $$\Phi_{t}(x)=e^{-t}x$$. Which pulls back to a flow $$\Phi_t: \xi \tilde{M} \rightarrow \xi \tilde{M}$$ $$\Phi_{t}(\tilde{m},x) = (\tilde{m},\Phi_t(x)=e^{-t} \cdot x)$$. Note: The word radial refers the fact that in euclidean space points are just moving radially away towards the origin.

Flow function: Smooth $$c:\xi M \rightarrow R^{+}$$ which is flow equivariant. Which means that the image of a point $$c(\Phi_t(\tilde{m},x)) = \Phi_{t}(c(\tilde{m},x))$$. Note: $$c$$ is uniquely determined by $$c^{-1}(1)$$ (multiplicative identity “zero” is $$1$$).

Theorem: Suppose $$M^{k}$$ is a closed projective manifold and $$c:\xi M \rightarrow R^{+}$$ is a complete convexity function. Which means that $$c$$ is Hiessian convex, ($$d^2c>0$$) and $$d^2c$$ is a complete Riemannian metric and $$c$$ is a flow function. Then $$M$$ is properly convex.

Proof: We have $$dev: \xi \tilde{M} \rightarrow R^{n+1}-\{0\}$$ let $$\Omega = dev(\xi \tilde{M}) \subset R^{n+1}$$. Since $$c$$ is a complete Hiessian metric this implies that $$\Omega$$ is convex and the developing map is injective. It remains to show that $$P(\Omega) \subset RP^n$$ is properly convex. This can be shown with and elementary geometric argument, the proof of which starts at the end of page 8 of the Deforming Convex Projective Manifold paper (Thm 3.4).

Previous Post: Lecture 14: Deforming Strictly Convex Projective Manifolds