If you’re in a compact setting, the geodesic result is probably not a good way of looking at things.

]]>Thank you for that fine remark about “random walks look like geodesics on large-dimension negatively curved manifolds” … I wonder if you can provide a reference for it?

It dovetails nicely with a theorem which states that any complex manifold that has an immersion in a large-dimension Euclidean manifold—meaning, pretty much any state-space manifold that physicists ever do practical computations upon—has negative holomorphic bisectional curvature.

It is becoming clear (IMHO) that one reason for the exponentiating increase in quantum simulation capability seen in recent decades is derives purely from Riemann/Kahler geometry — the negative biholomorphic sectional curvature theorem helps ensure that quantum simulation codes “seldom get lost.”

@article{, Author = {S. I. Goldberg and S. Kobayashi}, Journal = {Journal of Differential Geometry}, Pages = {225–233}, Title = {Holomorphic Bisectional Curvature},Volume = 1, Year = 1967}

]]>I, for one, appreciate Scott Aaronson’s brilliant speculations, made without being held back by immersion in tangential topics. Starting 1968 when I hung out with Feynman at Caltech, I stated to many people my opinion that our universe is indistinguishable from a simulation running in a larger universe, but as if our was a least-action fixed point in an infinite tree of such embeddings.

The question was: how much computation, with what complexity, does the universe take to calculate (locally, modulo spooky action at a distance) its own next state, and what artifacts of discretization of space-time (planck) can be distinguished from the limit to the GR continuum?

]]>I think the answer is that in the presence of negative curvature, a random walk will not return to its starting point, even in 2 dimensions. On a large enough scale, it will look like a geodesic. ]]>

To calculate the sectional curvature, we start at any point, and we choose two independent vectors. We define a *section* as the two-dimensional surface swept by geodesics whose tangent vectors at the chosen point are linear combinations of the two vectors.

We define the sectional curvature to be positive or negative, according to whether the 2D intrinsic geometry of the sectional surface is spherical or hyperbolic.

It turns out that the sectional curvature and the (four-index) Riemann curvature tensor are geometrically equivalent … if we know the curvature of every section, then we know the full Riemann curvature tensor, and *vice versa*.

The precise relation is simple. Given a Riemann manifold with inner product ⟨,⟩, the sectional curvature function S(,) and the Riemann curvature tensor R(,,,) are related by

S(u,v)=R(u,v,u,v)/(⟨*u*,*u*⟩⟨*v*,*v*⟩-⟨*u*,*v*⟩⟨*v*,*u*⟩)

where the vectors *u* and *v* span the desired section.

This illustrates that from a strictly geometric point of view, the sectional curvature function is the “natural” way to define curvature.

This relation can be regarded as the *definition* of the Riemann curvature tensor. Particularly on complex manifolds, the resulting explicit expression for R(,,,) in terms of the Hermitian metric is very simple and natural … with no Christoffel symbols required … this is one of Kähler’s famous “long list of miracles” that occur on complex manifolds.

In contrast, the full Ricci curvature has Myers’s theorem. Relativity is also about the full Ricci curvature. The cosmological constant is basically the negative of the scalar curvature.

Sectional curvature is the easiest to understand: it’s a function on 2-planes, given by the induced Gauss curvature on a surface tangent to that 2-plane. This misses all the structure of curvature, such as the possibility of separating scalar, Ricci, and Weyl curvature. It does have the advantage over the tensor point of view that it’s clear what it means to compare it to a number.

]]>In what sense are we talking of curvature in this instance?

In the cosmology of a homogeneous, isotropic universe, when people talk about the universe being “positively curved, negatively curved, or flat”, what they are talking about is the Ricci scalar curvature of the homogeneous 3-manifolds, i.e. the 3-dimensional slices through spacetime on which everything looks the same, which can sensibly be thought of as slices of constant “cosmological time”.

For 2-dimensional manifolds, the Ricci scalar is just twice the Gaussian curvature. For 1-dimensional manifolds, the Ricci scalar is 0 everywhere.

The Ricci scalar is an *intrinsic* measure of curvature, i.e. it can be derived solely from measurements within the manifold, and it doesn’t require, or refer to, any embedding of the manifold in a larger space.

Though the usual way to get to the Ricci scalar is to start with the Riemann curvature tensor and contract it on two indices to get the Ricci curvature tensor, and then take the trace of that … the “Direct geometric interpretation” given in the Wikipedia article on the Ricci scalar is a nice shortcut to the essential point: if you construct a Taylor series for the *ratio* between the volume of a small n-sphere of radius *r* in your manifold (i.e. the set of points you get by travelling at most a distance *r* along geodesics in every possible direction) and the volume of an n-sphere of radius *r* in Euclidean space, the second order term contains the Ricci scalar. Specifically, S is -3(*n*+2) times the second derivative of the volume ratio wrt *r*, evaluated at *r*=0.

But sometimes it’s easier to use a similar formula for the *surface area* of these n-spheres; S is -3*n* times the second derivative of the surface area ratio wrt *r*, at *r*=0.

For example, in any 1-dimensional manifold, the volume is just the length of an interval, 2*r*, which is the same as the Euclidean case, so the ratio is exactly 1, so the Ricci scalar is 0.

For a sphere of radius *a*, you need to consider a circle of geodesic radius *r*, i.e. a circle that subtends an angle of *r*/*a* from centre to perimeter. The 1-dimensional “surface area” in this case is just the circumference of this circle, and the ratio of “surface areas” is simply (*a*/*r*) sin (*r*/*a*). The second derivative of that ratio evaluated at *r*=0 is just -1/(3 *a*^2 ), and the Ricci scalar is -6 times that, or 2/*a*^2. Which, as I mentioned earlier, is just twice the Gaussian curvature.

by “fat helix/torus” do you mean the 2-dimensional boundary of a thickening? then everything you say is correct (zero on top…); in particular, it isn’t uniformly positive, so Myers’s theorem doesn’t apply. There’s probably some relationship between the intrinsic curvature of the 2-manifold and the extrinsic curvature of the 1-manifold. ]]>

I didn’t say R2 was convex, I said the surface of revolution of a parabola. A parabola isn’t any old line, but one with an embedding in R2; similarly, its surface of revolution comes with an embedding in R3. The metric of that surface is interesting enough that you might say intrinsic things about it (like that it has positive curvature, without being compact), but convexity is about its embedding.

I’ll take that as a flimsy excuse to mention the intrinsic property of the distance function being convex. This turns out to be the same as negative curvature. Also, I’ll take your mention of algebraic geometry as a flimsy excuse to use the phrase “Atiyah class.”

I don’t know what you mean about the curvature of a solid helix or torus. As an open set of R3, a solid torus looks to me to be locally the same as R3, ie, flat. I suppose you could consider a solid torus to have the translation invariant metric, so that it’s flat, but the inclusion in R3 is distorted. But that’s at a different level than the kind of distortion of curvature.

I would tend to say that the curvature of a 1-manifold is the zero 2-form (which I’d say exists). You might say that the sectional curvature is a function on the space of infinitesimal 2-planes, but the set of functions on the empty set is not itself not empty, but a point.

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