# Tag Info

### Normalizer of solvable linear group is an algebraic group?

The normalizer of every connected closed subgroup of $\mathrm{GL}_n(\mathbf{R})$ is the stabilizer of its Lie algebra, so is Zariski-closed.

### What does it mean that the Hessian is proportional to the metric?

Since the proof that such a function implies the metric is a warped product metric is fairly simple, I include a complete copy below. 1 Start with $\nabla^2 f = \lambda g$. For any vector field it ...
Accepted

### Compactification of a product of manifolds

Yes, the quotient $C_M = \overline{M} \times \overline{\mathbb{R}} / \left\{\{x\} \times \overline{\mathbb{R}} : x \in \partial\overline{M}\right\}$ seems to do the job, where I mean that the points ...

### Lie group framing and framed bordism

Lie group framing is a reference to the group action. A Lie group $G$ acts on the left of $G$ by the map $$(g,h) \longmapsto gh.$$ Similarly there are actions on the right, and conjugation actions, ...
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### Bounded covariant derivative of curvature tensor

I did not find any reference the proves a theorem as strong as stated in the question. Greene-Wu and Stefan Peters indepdently proved the following using almost linear coordinates (defined and studied ...

### Convergence of metric spaces of increasing dimension

Theorem. For all integers $0<m<n$ $d_{GH}(\mathbb{S}^m,\mathbb{S}^n)\geq \frac{\pi}{4}$. This is Theorem B on page 7 in S. Lim, F. Mémoli, Z. Smith, The Gromov-Hausdorff distance between ...

### How do we calculate the gradient of this function defined using the Riemannian logarithm on a Riemannian manifold?

In order to differentiate something containing the logarithm, it's best if you can differentiate the logarithm itself. And because you work in a neighborhood where the exponential map is a ...
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### Convex hull of 3 points in Cartan-Hadamard manifolds

I believe that the idea described by Ian Agol works, and can be elaborated on as follows. The general fact we want to establish is that the convex hull of a finite collection $X$ of points in $M$ is ...

### string bordism group and framed bordism group for $d \leq 6$ and $d \geq 7$

The map $i : * = B\{e\} \to BString$ over $BF = BGL_1(S)$ is $7$-connected, so induces a $7$-connected map $S = M\{e\} \to MString$ of Thom spectra, by the Thom isomorphism and Hurewicz theorem. At ...
1 vote

### Convergence of metric spaces of increasing dimension

Gromov has such a general result in Gromov, Misha. Metric structures for Riemannian and non-Riemannian spaces. Based on the 1981 French original. With appendices by M. Katz, P. Pansu and S. Semmes. ...
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### Derive distributional inequalities from pointwise estimates

First note that the result is true if the support of $\varphi$ is disjoint from $E$; this follows from integration by parts. Next note that every algebraic set is a finite union of smooth submanifolds ...

### Manifolds with nonpositive radial curvature

Assume that $M$ is three-dimenional and $\exp_p\colon\mathbb{R}^3\to M$ is a diffeomorphism. Let $g_r$ be the Riemannian metric induced on $\mathbb{S}^{2}$ by the map $x\mapsto \exp_p(r\cdot x)$. Note ...
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### Decompositions of $\partial_i$ to the radial direction and rotations in higher dimensions

The formula you gave in the beginning is a special case of the vector triple product formula in $\mathbb{R}^3$ $$a\times (b\times c) = (a\cdot c) b - (a\cdot b) c$$ Let $v$ be an arbitrary vector ...
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### Examples of hyperbolic manifolds of dimension $\geq$ 3 with disjoint totally geodesic hypersurfaces

I do not have a self-contained reference, but the key is Long, D. D.; Reid, A. W., Constructing hyperbolic manifolds which bound geometrically, Math. Res. Lett. 8, No. 4, 443-455 (2001). ZBL0992.57023....
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### Is any globally-hyperbolic manifold conformally equivalent to one with complete slices?

This is the original Nomizu-Ozeki article: Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401. Applying their proof ...
1 vote
Accepted

### Curve length in the Sasaki metric

For the inequality: first note that if $Y$ is a vector field parallel along $\alpha$, then $$\tau_\alpha (X + Y)(\alpha(0)) - (X+Y)(\alpha(1)) = \tau_\alpha X(\alpha(0)) - X(\alpha(1))$$ and  \...

### How to tackle the smooth Poincaré conjecture

Fastforward $11+\frac12$ years, I thought I'd mention that I tried to do what my original post suggested, which is when I started my PhD and then when I completed my thesis it naturally spawned this ...

### Developable 3-manifolds in $\mathbb{R}^4$

In this paper, I showed that an open and dense subset of any flat ruled submanifold (and, in particular, of any flat hypersurface without planar points) is a union of cylindrical, conical, and tangent ...

### reference for reading Schoen Yau positive mass theorem proof II

Already found one that might be helpful so I share it here, notes for a short course by Schoen at Tsinghua unversity, https://web.math.ucsb.edu/~zhou/Math_GR_2012.pdf

### Torsion-free Cartan connections

In my Introduction to Cartan geometries, p.30, I define a near parallelogram, a 5-sided figure, of constant vector fields flows, say of two elements $A,B$ in the Lie algebra of the model, to be a ...
1 vote

### Calculation of the top Chern class of spinor bundle over $S^{2n}$

I will present an explicit calculation using Chern-Weil theory, which makes an amusing use of Legendre's duplication formula for the gamma function. The Chern character form of a vector bundle $E$ ...

### Manifolds with negative dimension – Definition, References

One place where negative dimensional manifolds appear naturally is complex cobordism $U^*$. Intuitively, elements of the abelian group $U^n(X)$ are represented by families of $(-n)$-dimensional ...

### Does Hermite-Einstein imply Kähler-Einstein?

If one interprets the OP's question literally, the answer is 'yes', but I imagine that the OP didn't literally mean what the OP wrote. First, interpret everything literally: Assume that $(M,g,\omega)$...
Here's a counterexample. Take a cone of unrolled angle strictly between $\frac\pi2$ and $\pi$. Smooth the vertex and take $\gamma$ to be a radial ray far from the vertex. Then you can make a piecewise ...