dg.differential geometry – Do exist constant curvature manifolds (hyperbolic or elliptic) with torsion?


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measure theory – Question about a proposition in Munkres’s Analysis on Manifolds

I am reading through Munkres’s Analysis on Manifolds, and I get stuck in a proof of the lemma 18.1, that is stated as following:

Lema 18.1 Let $A$ be open in $mathbb{R}^n$; let $g:Ato mathbb{R}^n$ be a function of class $C^1$. If the subset $E$ of $A$ has measure zero in $mathbb{R}^n$, then $g(E)$ has measure zero in $mathbb{R}^n$.

He made out its proof in three steps. The first and second step are mentioned in the third, where he actually prove the theorem. Let me add some pictures of the third step.

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(If you need pictures of the other two steps in order to solve the question above, let me know, please)

Note: A $delta$-neighborhood of a set $X$ is the union of all open cubes (in this case) with width $delta>0$ and centered at $xin X.$ The theorem 4.6 in that book states that every compact set $K$ that is contained in an open set $Usubset mathbb{R}^n$ has a $delta$-neighborhood contained in $U$.

So, the problem is here: When he covers the set $E_k$ by countably many cubes $D_i$ with certain properties, he asserts: Because $D_i$ has width less than $delta$, it is contained in $C_{k+1}$.

Why this is true? I mean, if each cube $D_i$ is centered at some point lying at $C_k$ it is clearly true, but we don’t know if this happens. I tried to give a proof that we can assume that each $D_i$ can be choosen in a way that is centered in $C_k$ but I couldn’t prove that.

Can you help me to justify that assertion on the book? Thanks in advance.

Improving regularity of the boundary of a convex set in Riemannian manifolds

Let $X$ be a geodesically complete Riemannian manifold (we may assume that $X$ is simply connected and negatively curved, although I don’t think it matters). Given a closed, convex subset $K subset X$, there is a result by Rolf Walter that the $epsilon$-neighbourhood of $K$ (denoted $K_{epsilon}$) has $C^{1,1}$-regular boundary, i.e. the topological boundary of $K_{epsilon}$ is a $C^{1,1}$-submanifold of $X$. I was wondering if the regularity can be improved if we allow for more deformations of the convex set. Specifically, I would like to know the following:

${bf Question:}$ Given a closed, convex subset $K subset X$ and $epsilon > 0$, does there exist a convex closed $K’$ such that $K subset K’ subset K_{epsilon}$, and $partial K’$ is a $C^2$-submanifold of $X$?}

Considering that Walter’s result goes back to the 70s (Rolf Walter, “Some analytical properties of geodesically convex sets”), this seems like a question whose answer should be known, but I’m struggling to find anything. All methods I could come up with to deform $K$ so that its boundary becomes more smooth while keeping the set convex rely on the second fundamental form, which requires that we make the boundary $C^2$ first.

I should empathize that I do not want to assume that $K$ is compact. (I’m fine assuming that there is a cocompact action by isometries on $K$ though.)

ag.algebraic geometry – Manifolds with a Kähler deformation

Let $X$ be a compact complex non-Kähler manifold, then what conditions do we need to make it has a Kähler deformation? that is to say it can be deformed to a Kähler manifold.

Obviously not all the compact complex manifolds can be deformed to Kähler ones, for example, the Hopf surface, but certainly, there exist some non-Kähler manifolds which can be deformed to Kähler ones. For example, Hironaka has provided an example that except the central fiber, all the other fibers are projective manifolds, and the central fiber is a non-Kähler Moishezon manifold, so, conversely, for this Moishezon manifold, we can say it has a Kähler (even projective) deformation.

Then, are there any other examples of non-Kähler manifolds which has a Kähler deformation? or even a projective deformation? For example does a $partialbarpartial$-manifolds with trivial canonical bundle has a Kähler deformation? Has anyone think about it before? And what’s the latest progress on it?

ag.algebraic geometry – Manifolds with $[,]:H^1(X,T_X)times H^1(X,T_X)rightarrow 0$

Let $X$ be a compact complex manifold, for arbitrary $phi_1,phi_2in H^1(X,T_X)$, if the Lie bracket $(,):H^1(X,T_X)times H^1(X,T_X)rightarrow H^2(X,T_X)$ always maps $phi_1,phi_2$ to zero, i.e.$(phi_1,phi_2)=0in H^2(X,T_X)$, then $X$ admits an unobstructed deformation?

As we know, if we assume $H^2(X,T_X)=0$, then of course the Lie bracket gives a map $(,):H^1(X,T_X)times H^1(X,T_X)rightarrow 0$, and for a Calabi-Yau manifold, by Tian-Todorov lemma and $partial barpartial$-lemma, the Lie bracket also maps $H^1(X,T_X)times H^1(X,T_X)$ to $0$, so, generally, if we assume that the Lie bracket always maps $H^1(X,T_X)times H^1(X,T_X)$ to $0$, then can we say $X$ must have an unobstructed deformation?

3 manifolds – Thurston Universe gates in Knots: which invariant is it?

Today I discovered this nice video of a lecture by Thurston:

https://youtu.be/daplYX6Oshc

in which he explains how a knot can be turned into a “fabric for universes”. For example, the unknot can be thought as a portal to Narnia, and when you pass again you switch back to the Earth. This forms in a sense a $mathbb{Z}/2mathbb{Z}$. Then he proceeds to explore what fabric one gets with the treefoil and you get an $S_3$. I am sure there is some real geometry behind but I can’t grasp how to translate a portal into something homotopic.

A way to formalize this would be the following. Take a knot $K$ in $mathbb{R}^3$. Fix a tubular neighborhood $N$ of $K$. For each point $x$ in the knot take the loop $L_x$ obtained as the sphere bundle of $N$ at $x$ (a small circle around $x$ that jumps into the portal). Then there exist a connected 3-manifold $M$ with a (finite?) cover $M to mathbb{R}^3 setminus K$ such that the “monodromy in small circles” around $L_x$ has order two for all $x$. Then we set the “group of universes” as the group of cover automorphisms.

I think this captures the previous idea in the following sense: to each locus of $mathbb{R}^3 setminus K$ we have $n$ counterimages that represent the different worlds. Some branch should be chosen to make distinguishing between worlds possible. The constraint on monodromy ensures that if you jump twice through the same portal (at least for the portals very close to the boundary) you get back.

Does such a manifold exist for all knots? Is this construction just some simplification of the fundamental group of the complement?

ag.algebraic geometry – Non-projective complex manifolds such that every meromorphic function is rational?

Let $X$ be a smooth compact complex manifold. Suppose every meromorphic function $f : X to mathbb{C}$ is rational. Does this imply that $X$ is projective?

Of course, this is a partial converse to Chow’s theorem, which I doubt is true. Perhaps there are Moishezon examples which give a negative answer.

smooth manifolds – Finetely connected orientable surface

Let $(M,g)$ be a finetely connected orientable complete Riemannian surface, that is, $M$ is homeomorphic to a compact orientable surface $Sigma$ minus $k geq 1$ points. Do you have references or a proof for the fact that $(M,g)$ is conformal to a compact orientable Riemann surface with $k geq 1$ disks deleted?

This is part of an argument in the paper “ On complete minimal surfaces with finite Morse index in three manifolds”, by Fischer-Colbrie.

riemannian geometry – Curvature of locally conformally-flat manifolds and pull-back metrics.

Suppose $(M^n, g)$ is a Riemannian manifold and $forall pin M$ there is a nbhd $p in U$ and a local diffeomorphism $phi:(U, g|_U) to (mathbb{R}^n, delta)$, where $delta$ is the flat metric on $mathbb{R}^n$, such that $g|_U = phi^{*}delta$.

Let $R_g$ be the curvature tensor of $g$, then for each $pin U$ and $X,Y,Z,W in T_pU$ we have

$R_g(X,Y,Z,W)|_p = R_{phi^*delta}(X,Y,Z,W)|_p = (phi^*R_delta)(X,Y,Z,W)|_p = R_delta(phi_*X,phi_*Y,phi_*Z,phi_*W)|_{phi(p)} = 0$

because $R_delta = 0$ since $(mathbb{R}^n,delta)$ is flat. This is what I think a locally-flat manifold is, is this logic correct? (As an aside: if we can do this in a nbhd of each point, isn’t the whole manifold $M$ flat then?)

Now suppose instead that we have $(M^n,g)$, $pin Usubset M$, and a local diffeomorphism $phi:(U, g|_U) to (mathbb{R}^n, delta)$ such that $(e^{2f} g)|_U = phi^*delta =:tilde{g}$ for some positive $fin C^infty(U)$. Then the curvature tensor of $tilde{g}$ satisfies (I don’t know how to do the Kulkarni-Nomizu product in latex so I just use wedge)

$R_{tilde{g}} = e^{2f}(R_{g} – (nabla^2 f)wedge g +(dfotimes df)wedge g – frac{1}{2}(gwedge g))$

and by the above, this should still vanish on $U$, right? This is what I think a locally conformally-flat manifold is.

If I have this correct then on $(U, g|_U)$ we should have $Ricci = Scal = 0$ since $R=0$. But now think about the round sphere $(mathbb{S}^n,mathring{g})$: it is locally conformally-flat by using the stereographic projection, and the pull-back of $delta$ is conformal to $mathring{g}$ so shouldn’t $R_{mathring{g}} = 0$?? Shouldn’t it have vanishing Ricci and scalar curvatures too? I know it certainly does not have those tensors vanishing but I can’t understand what’s going on here.

I’m asking a lot of vague questions and this reflects my real confusion but if I could try to summarize my question it would be: Why doesn’t the pullback metric under a conformal diffeomorphism have vanishing curvature tensor? Or does it?

Are there known examples of almost complex manifolds admitting neither a symplectic nor a complex structure?

I have seen the the example of $S^6$ being touted around here and there but it does not seem to be generally confirmed that there is no complex structure on it.