Systoles and the generalized Geroch conjecture

Almost a year ago, I said here that I would write a sequence of posts on some simple facts and observations related to the systolic inequality but got distracted and didn’t manage to do much of that…

I was reminded last week as I heard professor Guth’s talk on systoles for the 4th time (Yes, the same talk! –in Toronto, Northwestern, India and here at the IAS). It’s interesting that I’m often thinking about different things each time I hear the same talk. This one is about the generalized Geroch conjecture.

Geroch conjecture: \mathbb{T}^n (the n-torus) does not admit a metric of positive scalar curvature.

The conjecture is proved by Schoen and Yau (1979).

Now, scalar curvature can be seen as a limit of volume of balls:

Definition: The scalar curvature of M at p is

\displaystyle \mbox{Sc}(p) = c_n \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^{n+2}}

where \mbox{Vol}_E is the Euclidian volume and c_n is a positive constant only depending on the dimension n.

Note that since our manifold does not have any cone points,

\displaystyle \lim_{r\rightarrow 0} \frac{\mbox{Vol}_E (B(\bar{0},r)) - \mbox{Vol}_g(B(p,r))}{r^n}

must vanish. Further more, the Riemannian structure on M forces the r^{n+1} term to vanish.

Since for this context we only care about is whether the scalar curvature is larger or smaller than 0, we can be even more simple-minded: M has positive scalar curvature at p all small enough balls around p has smaller volume than their Euclidean cousins (with a difference of order propositional to r^{n+2}). In light of this definition, we have:

Restatement of the Geroch conjecture: For all g on \mathbb{T}^n, there exists some point p s.t. \mbox{Sc}(p) \leq 0.

This is to say, small enough balls around some point $p$ are not small enough for it to have positive scalar curvature. What if instead we look at balls of a fixed radius instead of those infinitesimal balls? This naturally leads to

Generalized Geroch conjecture: For any (\mathbb{T}^n, g), for all r, there exists p s.t. \mbox{Vol}_g(B(p, r)) \geq \mbox{Vol}_E(B(\bar{0}, r)).

(For those r larger than the injectivity radius, we lift M to its universal cover so that all homotopically non-trival loops are ‘unfolded’)

Let’s take a look at the 2-torus to get a feel of the conjecture:

The flat torus, of course, has 0 r-scalar curvature at all points.

For the regular rotational torus, we take the ball around the saddle point of the gradient flow, the ball look like a saddle, as shown below.

To see that this has area larger than the analogous Euclidean ball, we can cut it along radial rays into thin triangles, each triangle can be ‘almost flattened’ to a Euclidean triangle, but we have a more triangles than in the Euclidean case.

What if we try to make the surface spherical for most of the area and having those negative scalar curvature points taking up a very small potion. One of my first attempts would be to connect a few spheres with cylinders:

We have a few parameters here: the number of balls n, the width of the connecting cylinders w, the length of the connecting cylinders l and the radius of each sphere R.

If cylinders are too long (longer than 2r), then we can just take the ball in the middle of the cylinder, the volume when lifted to universal cover would be equal to Euclidean.

If the width of cylinders are much smaller than r, then the ball around a point in the gluing line would have volume almost a full spherical ball plus a half Euclidean ball, which would obviously be larger than a full Euclidean ball.

Hence the more interesting case is to have very short, wide tubes and as a consequence, have many balls forming a loop. In this case, the ‘worst’ ball would be centered at the middle of the tube, it intersects the two spheres connected by the tube in something a bit larger than a spherical half-ball.

I haven’t figured out an estimate yet. i.e. can the advantage taken from the fact that spherical ball are smaller than Euclidean balls cancel out the ‘a bit larger than half’? I think that would be interesting to work out.

Finally, let’s say what does this has to do with systoles:

Theorem: Generalized Geroch conjecture \Rightarrow \mbox{Sys}(\mathbb{T}^n, g) \leq \frac{2}{\omega_n^{\frac{1}{n}}} \mbox{Vol}_g(\mathbb{T}^n)^{\frac{1}{n}} (which is the systolic inequality with a constant better than what we have so far)

Proof: Suppose not,

\mbox{Sys}(\mathbb{T}^n, g) > 2 (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}

Let r = (\frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n})^{\frac{1}{n}}, by the generalized Geroch conjecture we have some B(p, r) larger than the Euclidean ball. i.e.

\mbox{Vol}_g(B(p, r))>\omega_n r^n = \omega_n \frac{\mbox{Vol}_g(\mathbb{T}^n)}{\omega_n} = \mbox{Vol}_g(\mathbb{T}^n)

Since the systole is at least 2r, hence B(p, r) cannot contain any homotopically non-trival loop i.e. it does not “warp around” and get unfolded when passing to the universal cover. Hence volume of a ball with radius r cannot be larger than the volume of the whole manifold. Contradiction

3 thoughts on “Systoles and the generalized Geroch conjecture

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