What does it mean by a flat universe?

[Happiness]

My understanding:
When we draw a triangle on a flat piece of paper and measure the angles using a protractor, the sum of the angles is $180^\circ$. So we conclude that the universe is locally flat. Suppose we draw a very big triangle that spans across galaxies (say, using lasers and mirrors) and measure its angles. If the sum of the angles exceeds $180^\circ$, then the universe is closed. If it equals $180^\circ$, then the universe is flat. If it is less than $180^\circ$, then the universe is open.

Confusion 1:
There seems to be a circular reasoning in the definition of "flatness".
Suppose Alice lives in a universe that is both locally and globally closed. Since Alice's universe is locally closed, she would measure the sum of the angles of a triangle she draws on a "flat" piece of paper to be more than $180^\circ$, say $290^\circ$. Since the paper she uses is "flat", she concludes that the sum of the angles of a triangle in Euclidean or flat geometry is $290^\circ$. Then, she draws a very big triangle that spans across galaxies and finds its sum of angles to be $290^\circ$. She concludes her universe is both locally and globally flat, when in fact, it is not.

That means that even though we measure the sum of angles to be $180^\circ$ locally, our universe may not be locally flat. To Alice, our universe is locally open, since $180^\circ<290^\circ$.

Confusion 2:
It seems that the sum of angles in a triangle is always $180^\circ$ regardless of the local geometry.
If the local geometry is curved, then when we measure the angles of a triangle on a "flat" piece of paper, don't the lines of the protractor we are using bend accordingly by the curved local geometry such that the angle measured is still the same as the one measured in a flat geometry? In other words, since the protractor we are using exists in the curved local geometry, its lines (or scale lines or graduations) are bent by the curved local geometry so that the angle measured is still the same.

 

[PeterDonis]

Suppose Alice lives in a universe that is both locally and globally closed.

There is no such thing as "locally closed". Locally, any manifold looks flat, i.e., Euclidean (or Minkowskian if we are talking about spacetime). Curvature is a concept that only applies on larger scales.

For example, consider the Earth's surface. It is globally closed--a 2-sphere. But locally, it looks flat. See below.

If the local geometry is curved

There's no such thing; see above. Again, take the example of the Earth's surface. If I draw a triangle on the sidewalk in front of my house, the sum of its angles is 180 degrees. But if I draw a triangle on the Earth whose vertices are the North Pole, the intersection of the prime meridian with the equator, and the intersection of the 90 degree West meridian with the equator, the sum of the angles is 270 degrees, showing that the Earth's surface, globally, has positive curvature.

It is true that there are degrees of "local". If I draw a triangle in my local neighborhood with sides a mile long, taking care to make each side a geodesic, a surveyor with accurate equipment will be able to see that the sum of its angles is a little larger than 180 degrees. But it still won't be 270 degrees; with a triangle that small on the Earth's surface, it's impossible to get that much of a change in the sum of the angles. So a more complete way of expressing what "local" means is that the smaller the triangle, the smaller the difference you can possibly get in the sum of its angles from 180 degrees; and in the limit of an infinitesimal triangle (which the triangle drawn on my sidewalk is a very good approximation to, given the size of the Earth), the difference goes to zero.

 

[Happiness]

If the local geometry is always flat, how can people postulate that a way for extra dimensions to exist is for them to be very curved up? If I draw one side of the triangle in one of those curved-up dimensions, wouldn't the sum of angles be different from $180^\circ$? And if so, the local geometry would then be curved.

 

[PeterDonis]

If the local geometry is always flat, how can people postulate that a way for extra dimensions to exist is for them to be very curved up?

"Very curved up" means that for a triangle to be "local" along one of those extra dimensions, it has to be very small, much smaller than a triangle drawn on surfaces which are not "very curved up". So, for example, while a triangle drawn on my sidewalk is small enough to be considered "local" (no detectable curvature--looks flat) on the Earth's surface, a triangle drawn in the "very curved up" dimensions might need to be smaller than an atomic nucleus in order to look flat.

 

[Happiness]

Locally, any manifold looks flat, i.e., Euclidean (or Minkowskian if we are talking about spacetime). Curvature is a concept that only applies on larger scales.

Alan Guth mentioned that not all spaces are locally flat. Only a special class of spaces is locally flat and these spaces are called Riemannian spaces. He said these @53:22.

 

[ChrisVer]

what does flatness have to do with closeness?
A cylinder of given axial length L and radius R is a flat and closed object.
A sphere of radius R is not flat but closed object.
A saddle surface is not flat but open...
A limitless-piece of paper is flat and open...

 

[PeterDonis]

Alan Guth mentioned that not all spaces are locally flat. Only a special class of spaces is locally flat and these spaces are called Riemannian spaces.

He means the same thing here by "Riemannian space" as I meant by "manifold" in post #2. Note also that "Riemannian space" is too restrictive; spacetime is technically not a Riemannian space because the metric is not positive definite. It's a pseudo-Riemannian space, but those are also included in "manifold" and are also locally flat (you just have to define "flat" in a way that allows the metric to be non-positive-definite).

 

[Happiness]

He means the same thing here by "Riemannian space" as I meant by "manifold" in post #2. Note also that "Riemannian space" is too restrictive; spacetime is technically not a Riemannian space because the metric is not positive definite. It's a pseudo-Riemannian space, but those are also included in "manifold" and are also locally flat (you just have to define "flat" in a way that allows the metric to be non-positive-definite).

Okay. But that would not be an accurate use of the word "manifold". I was misled into believing that all manifolds are locally flat until I watched Alan Guth.

 

[PeterDonis]

I was misled into believing that all manifolds are locally flat until I watched Alan Guth.

Does Guth use the word "manifold"? Or just the word "space" (which is how you quoted him in your last post)? All the mathematical definitions that I have seen of the term "manifold" include local flatness as a condition. "Space" is a more general term that does not imply that condition.

 

[ChrisVer]

I was misled into believing that all manifolds are locally flat until I watched Alan Guth.

There are manifolds which don't admit a metric at all...

 

[PeterDonis]

There are manifolds which don't admit a metric at all...

Can you give an example?

 

[Happiness]

All the mathematical definitions that I have seen of the term "manifold" include local flatness as a condition. "Space" is a more general term that does not imply that condition.

I see.

In that case, I guess, a paraphrase of my original question would be how do we know or prove that the 3D physical space in our universe is a 3D Riemannian space or a 3D manifold (if we ignore the time dimension for now)? If we can't prove it, then there seems to be a circular reasoning in the definition of "flatness".

 

[PeterDonis]

how do we know or prove that the 3D physical space in our universe is a 3D Riemannian space or a 3D manifold

Because it works like one. Locally, we can make various geometric objects (spheres, cubes, etc.) and verify that they satisfy the propositions of Euclidean geometry. That shows that the space of our universe is locally Euclidean, whatever its global properties might be.

Similarly, if we include time, we can set up experiments to show that these locally Euclidean geometric objects also satisfy the propositions of Special Relativity, i.e., of Minkowskian spacetime geometry. That shows that the spacetime of our universe is locally Minkowskian, whatever its global properties might be.

 

[ChrisVer]

Can you give an example?

sorry that was going for topological spaces and not manifolds