Incomplete 3.dim Lorentz manifolds

Reading this paper where closed 3-dim. Lorentz manifolds with noncompact isometry groups are studied, I wonder if all of them are geodesically complete. One class of 3-dim. closed Lorentz manifolds with noncompact isometry group are locally homogenuous modelled on $ \mathbb{R}^3, Heis$ or $ \widetilde{PSL}(2,\mathbb{R})$ (with certain Lorentz metrics).

By a result of Zeghib where he classified the lorentzian model geometries in 3 dimensions and also studied which of them are complete, we know that the above examples are all complete.

My question is are there examples of closed 3-dim. Lorentz manifolds with noncompact isometry group which are not geodesically complete?

My first idea was to look at some of the metrics which are given kind of explicity in the above work and study the corresponding geodesic equation in order to see if there are solutions which only exist for a finite time. But I’m getting nowhere with this (which may be due to my lacking PDE skills…).

Flag manifolds as incidence correspondences

Let $ G$ be a reductive group, $ B$ a Borel and $ P_j$ the maximal parabolics, indexed by the vertices $ j$ of the Dynkin diagram. Then $ B = \bigcap_j P_j$ , so the flag manifold $ G/B$ injects into $ \prod_j G/P_j$ .

In the classical types, this is a useful concrete description of $ G/B$ : In type $ A_{n-1}$ , $ GL_n/P_j$ is the Grassmannian of $ j$ -planes in $ n$ -space, so a point of $ \prod_j GL_n/P_j$ is a sequence of subspaces $ (V_1, V_2, \cdots, V_{n-1})$ , with $ \dim V_j = j$ . The subspace $ G/B$ is the flags $ V_1 \subset V_2 \subset \cdots \subset V_{n-1}$ .

Similarly, in types $ B$ and $ C$ , the $ G/P_j$ are isotropic flags, and we get the standard descriptions of the type $ B$ /$ C$ flag manifolds as complete isotropic flags.

In type $ D_n$ , the $ n-2$ vertices on the long leg of the diagram correspond to isotropic subspaces of dimensions $ 1 \leq j \leq n-2$ , and the two extra vertices correspond to the two types of isotropic $ n$ -plane: We get that $ G/B$ is flags $ V_1 \subset V_2 \subset \cdots \subset V_{n-2} \subset V_n^+, \ V_n^-$ , where $ V_j$ is an isotropic $ j$ plane and $ V_n^+$ and $ V_n^-$ are of the $ +$ type and the $ -$ type. This isn’t the standard description of complete isotropic flags, but it is easy to convert to one: Given $ (V_n^+, V_n^-)$ , we can recover $ V_{n-1}$ as $ V_n^+ \cap V_n^-$ and, given $ (V_{n-1}, V_n^+)$ , we can recover $ V_n^-$ as the other isotropic $ n$ -plane between $ V_{n-1}$ and $ V_{n-1}^{\perp}$ . Note that the condition $ V_{n-2} \subseteq V_n^+,\ V_n^-$ together with the types of $ V_n^{\pm}$ actually forces $ \dim (V_n^+ \cap V_n^-) = n-1$ .

I note that, in each of these cases, I only need to impose conditions on pairs $ (V_i, V_j)$ where there is an edge $ (i,j)$ in the Dynkin diagram. This leads me to my question:

For two Dynkin vertices $ a$ , $ b$ , let $ \pi_{ab}$ be the projection of $ \prod G/P_j$ onto $ G/P_a \times G/P_b$ . Is it still right in the exceptional types that $ G/B = \bigcap \pi_{ab}^{-1}\left( \pi_{ab}( G/B) \right) \subset \prod G/P_j$ where the intersection runs over edges $ (a,b)$ of the Dynkin diagram? Does this question make sense, and is the answer yes, for Kac-Moody groups?

Motivation: I’ve committed to teach a course on Bruhat orders and Schuberty things next Fall, and I’m thinking about how to tell combinatorialists about flag manifolds in other types without getting into the whole construction of reductive/Kac-Moody groups.

Reference/Known results on the singular behaviour of the fibres of a holomorphic map between compact Kähler manifolds

I have been interested in the following situation of late: Let $ X$ and $ Y$ be compact Kähler manifolds with $ \dim_{\mathbb{C}}(Y) < \dim_{\mathbb{C}}(X)$ and let $ f : X \to Y$ be a surjective holomorphic map with connected fibres. Let $ S = \{ s_1, …, s_k \}$ denote the critical values of $ f$ , which is a subvariety of $ Y$ .

I cannot find a detailed account of how bad the singular behaviour of the fibres of $ f$ can be. For example, do the fibres contain $ (-1)$ curves (i.e., curves with self-intersection number $ -1$ ) or $ (-2)$ curves?

If anyone can provide references where I can get a better understanding of this, that would be tremendously appreciated.

Gradient flows on Hilbert manifolds

I would like to know if gradient flows of Morse-Bott functions on a Riemannian manifold always converge towards a unique critical point, provided that the flow line is bounded.

To be more precise, a Morse-Bott function on a Riemannian manifold $ (M,g)$ is a smooth map $ f:M \longrightarrow \mathbb R$ such that the set of critical points $ Crit f \subset M$ is a submanifold and $ T_x Crit f = ker Hess f_x$ for all $ x \in Crit f$ . It might be more convenient to express this by using the linear map $ A_x:T_xM \longrightarrow T_xM$ given by $ g_x(A_xv,w) = Hess f_x(v,w)$ for $ v,w \in T_xM$ . I am interested in the behaviour of maps $ x:\mathbb R_+ \longrightarrow M$ satisfying $ \dot x(t) = -\nabla f(x(t))$ . Does $ lim_{t \to \infty} x(t)$ exist?

An easy computation shows that $ df_{x(s)} \longrightarrow 0$ for $ s \to \infty$ . So if $ x$ is bounded (e.g. if $ M$ is compact) $ x$ has to “converge” towards $ Crit f$ . Unfortunately, this does not rule out that $ x$ has several limits in $ Crit f$ . (This can not happen for Morse functions, since in this case the critical points are isolated).

  1. The simplest case is the case where $ M$ is finite dimensional. As far as I can see, the answer is already given by Austin and Braam in their paper “Morse-Bott theory and equivariant cohomology”, 1995. In particular Theorem A.9 is important, stating that the stable manifold of a connected component $ C \subset Crit f$ , given by $ W^s(C) = \{x \in M | \phi_t(x) \to C\}$ where $ \phi$ is the flow of the negative gradient, is indeed a submanifold.

  2. The second case is the case where $ M$ is a Hilbert manifold (infinite dimensional) endowed with a complete metric on the tangent bundle. Although I could only find a paper dealing with this question in the Morse case (see Abbondandolo, Majer: “A Morse complex for infinite dimensional manifolds”, Appendix C), I assume that it should be not to hard to extend the result to the Morse-Bott case. The important point here is that the operator $ A_x$ mentioned above is symmetric and defined on the whole tangent space, so it has to be bounded.

  3. The last case is the one I am particularly interested in. We start again with a Hilbert manifold $ M$ , but now endowed with an incomplete Riemannian metric (in my case, I have a Sobolev space $ W^{1,2}$ which is only endowed with the $ L^2$ metric). Let $ H$ denote the complete Hilbert space where $ T_xM \subset H$ is dense. Then the operator $ A$ is unfortunately of the form $ A_x:T_xM \subset H \longrightarrow H$ . Now $ A_x$ is unbounded, but at least still symmetric and even essentially self-adjoint. However, here I am failing to apply the results from case 2, which strongly used the boundedness of $ A$ . I would be very content if a similar result could be shown at least for the case, where $ Crit f$ is finite dimensional.

A remark: In the infinite dimensional case, one might have problems with the flow, since it might not be defined for all times. However, I start with a flow line which is already defined for all $ t$ . Furthermore, I am only interested in the case where the flow line really “converges” towards $ Crit f$ since in my set up, this has already been proven with completely different methods.

I would also be thankful for counterexamples where the gradient flow does not have a unique limit. But I expect those examples not to contain Morse-Bott functions.

Hodge theory, conformal manifolds and Fredholm modules-understanding the proof of one Lemma

I would like to understand the proof of Lemma 1, page 339 in this book. Very briefly, the context is as follows: we have even dimensional oriented conformal manifold with the Hodge star operator chosen and we would like to construct a Fredholm module based on this data. In order to do so, we need an auxiliary operator, denoted by $ S$ which has the property that the graph of $ S$ coincides with the image of exterior derivative. There are several points which are not clear for me in the argument:

Q1. Why any differential form in $ \mathcal{H}_0^+$ which is orthogonal to harmonic forms may be written as $ \frac{1+\gamma}{2} d\alpha$ for some $ \alpha$ ?

Hodge theory ensures us that such $ \omega$ lies in the image of $ d+d^*$ and the claim should somehow follow from the fact that $ d^*=-*d*$ but I don’t see how since there is no obvious commutation relation between the grading $ \gamma$ and $ d$ .

Q2. I don’t see why $ \|\frac{1+\gamma}{2} d \alpha \|_2 =\| \frac{1-\gamma}{2}d\alpha \|_2$ .

In fact after simple computation this is equivalent to the fact that $ \langle d \alpha, \gamma d\alpha \rangle = -\langle \gamma d \alpha, d \alpha \rangle$ . But how to get this minus sign? I thought that $ \gamma$ should be self adjoint.

Q3. Even if we have this equality, does it really imply that $ d\alpha$ is determined uniquely?

Leaving this specific context for a moment, we can imagine two vectors in $ \mathbb{R}^2$ say $ \xi_1=(1,1)$ and $ \xi_2=(-1,1)$ and a projection $ P(x,y):=(0,y)$ Then $ P\xi_1=P\xi_2$ and $ \| P \xi_1 \|= \| (I-P)\xi_1 \|$ but $ (I-P)\xi_1 \neq (I-P)\xi_2$ . In our context the role of $ P$ is played by $ \frac{1-\gamma}{2}$ (which indeed is a projection).

I suspect that this uniqueness is needed in order to define $ S$ in such a way that $ S (\frac{1-\gamma}{2} d\alpha):=\frac{1+\gamma}{2} d \alpha$ . But even so I still don’t see why the graph of $ S$ should be equal to the image of $ d$ (image or closure of image?) since I can’t get rid of the harmonic part in the expression $ \omega+S \omega$ for $ \omega \in \mathcal{H}_0^{-}$ (which is mapped to 0 by $ S$ ).

I realize that these are technical questions and can be asked separetly, nevertheless I think that the whole context may be here important therefore I decded to ask all my questions in one post-I hope that this will be fine.

Examples of certain compact Kaehler manifolds

A Kaehler manifold is a complex manifold which has a Kaehler metric and Ricci curvature tensor $ R_{ij}$ . The Ricci curvature tensor is a Hermitian matrix having real eigenvalues. My question is: Is there a compact Kaehler manifold $ M$ of complex dimension $ n$ (with $ n>1$ ) such that the following conditions are satisfied:

(1) At most points of some region $ U$ of $ M$ , the Ricci curvature tensor is neither positive nor negative. At these points, the Ricci tensor has positive and negative eigenvalues. Also, the Kaehler or Riemannian volume of this region is small.

(2) Outside the region mentioned in (1), the Ricci tensor is positive (definite) and the Riemannian volume of this region (the region outside $ U$ ) is comparatively large.

(3) The canonical line bundle $ K$ of $ M$ is neither positive nor negative. (Note that the curvature of $ K$ is given by the Ricci curvature).

In complex dimension $ n=1$ , the canonical line bundle of a compact Riemann surface is either positive, negative or trivial. So, I am asking examples of complex dimension greater than 1.

Critical dimensions D for “smooth manifolds iff triangulable manifolds”

I am aware that at least for lower dimensions,

“smooth manifolds iff triangulable manifolds”

at least for dimensions below a certain critical dimensions D.

My question is that for

  • For orientable manifolds, in which lower dimensions $ \leq$ D, such that “smooth manifolds iff triangulable manifolds” is true, but above that dimensions > D is false? (So that “smooth manifolds may not be triangulable manifolds.”)


  1. orientable SO-manifolds with SO(D) co/bordism structure.

  2. orientable Spin-manifolds with Spin(D) co/bordism structure.

  • For non-orientable manifolds, in which lower dimensions $ \leq$ D, such that “smooth manifolds iff triangulable manifolds” is true, but above that dimensions > D is false? (So that “smooth manifolds may not be triangulable manifolds.”)


  1. non-orientable O-manifolds with O(D) co/bordism structure.

What are their critical dimensions D?

In this post, we learn:

“All orientable 5-dimensional manifolds are triangulable.” “In 6 dimensions, there are non-triangulable orientable manifolds.”

Are these referred to topological manifolds? Or smooth manifolds?

However, I heard an alternative view from a geometry topologist that (maybe standard viewpoint to some of you, but I apologize for my background ignorance):

“All smooth manifolds are uniquely triangulable. No critical dimensions D constraint or orientability constraints.”

p.s. Your correct answers are more important. References (the proofs) are secondary but very helpful.

Many thanks! Really appreciate your holiday time answer and inspiration!