When we say that the universe is flat, this means that the sum of angles of any triangle will always be 180°.

I believe that "flat" refers to the bending of spacetime in a 5-dimensional space (or a forth space dimension). This would however contradict the fact of the su of angles, as for warped space, the sum is smaller than 180°. Or does the geometry to the universe as a whole and "small" fluctuations like galaxies (which bend space) are ignored?

So: Is it correct to describe the shape of the universe in relation to a 5-dimensional surrounding? Like you can see in this picture:enter image description here (Image Credit ESA)

In this picture, one space dimension is omitted (so the spheres actually should be circles), so our 3-dimensional Universe/Brane "floats" in a higher-dimensional hyperspace or bulk. In This image, space is locally warped by mass (or energy, but this is not shown here), but space as a whole is flat.

Is it correct to imagine the flat shape of the universe like this (theoretically with one space dimension more)?

Update This question was closed as a duplicate of the following: Question 1, Question 2, Question 3.

While they are somewhat related to my question, they still ask for a different question, namely the following:

Question 1: Is your 3-dimensional universe part of the surface of a 4-dimensional sphere [like in the ant-sphere analogy] that we cannot perceive or access?

Question 2: The correctness of the bend-sheet-analogy for GTR

Question 3: Could the universe be bend over a forth dimension to form a 4-dimensional shphere?

The essence of my question was: When we refer to the shape of the universe (being flat for example), do we mean the same curvature as in GTR?

  • $\begingroup$ Are you asking if, in the same way we can embed a 2-dimensional surface like a sphere in 3-dimensions, we can embed 4-dimensional spacetime in a higher dimensional space? $\endgroup$
    – Charlie
    Commented Jul 2, 2020 at 18:27
  • $\begingroup$ @Charlie I know we can do that (e.g. in Einstein's General Theory of Relativity). My question is whether it is appropriate to do so when describing the Universe's shape. Does the curvature in "flat" (which is zero curvature) mean the same as curvature in GTR? $\endgroup$
    – jng224
    Commented Jul 2, 2020 at 18:32
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    $\begingroup$ I'm not entirely clear what you mean, but if you're asking if the curvature in spacetime is somewhat analogous to that of 2D curvature as shown in the image you've linked the answer I believe is yes. If you want a more mathematical demonstration someone more experienced will have to provide that. $\endgroup$
    – Charlie
    Commented Jul 2, 2020 at 18:34
  • $\begingroup$ Kip Thorne calls the embedding space "The Bulk". Are you asking about local bulk around massive objects, or cosmological bulk & the whole universe? $\endgroup$
    – JEB
    Commented Jul 2, 2020 at 20:21
  • $\begingroup$ @JEB To be honest, I'm not completely sure what's the difference there, but in "my" theory, the entire universe had to be placed in a higher-dimensional bulk. $\endgroup$
    – jng224
    Commented Jul 2, 2020 at 20:26

2 Answers 2


No, your belief is not correct. We do not, at least in General Relativity (GR), embed our spacetime in a higher dimensional space (or like you said in 4 dimensions of space).

Although I agree that it is possible to imagine many curved surfaces as embedded in a higher dimension, it is not how we do GR. In fact, the picture you have is one of the most misleading to interpret the mathematics of GR.

So what is happening is that you are restricted to your 4 dimensional manifold and you do not know what is outside it, like an ant restricted on a sphere would just imagine it as a 2 dimensional space and would not know that it is embedded in a 3 dimensional space.

Now to deal with such problems, Gauss had found the correct mathematical machinery which was refined by Reimann. In fact, the result you state about the sums of angle of a triangle on a curved surface is derived without embedding the surface in a higher dimension. We manage to figure out whether the space is curved or not by staying in that space and not seeing it from outside (by embedding).

The mathematics begins with the Gauss-Bonnet theorem and then leading to Reimanian geometry. What we calculate is the intrinsic curvature. For example: imagine a cylinder, you might see it as curved but it isn't a curved surface. It has zero intrinsic curvature. To get to that purely mathematically you need to show that Reimann curvature tensor vanishes but you can also see that intuitively. On the other hand a sphere is curved.

The cylinder has an extrinsic curvature (which can be calculated by embedding it) but no intrinsic curvature whereas a sphere has an intrinsic curvature.

GR is formulated in the language of intrinsic curvature. There's certainly nothing wrong with studying, say, a 2-sphere embedded in 3-dimensional space. But it's not necessary and requiring that such a higher-dimensional space even exists is an undue constraint. It's quite marvelous to realize that a 2-sphere can simply exist in nothing more than 2-dimensions: the geometry is encoded on the surface.

  • 1
    $\begingroup$ Thanks for your answer, I think my understanding has gotten a little clearer. However, I have two (new) questions: Would it be wrong to embed spacetime to higher dimensions (not just unnecessary)? And if space is warped, does there not have to be any higher dimension in which it ia warped (similarly to a (more or less) 2D sheet of paper that we can bend in 3D space)? Should I open a new question for these? $\endgroup$
    – jng224
    Commented Jul 2, 2020 at 19:35
  • 3
    $\begingroup$ @Jonas 1) No that is not wrong fundamentally. There are many advantages in not embedding, in that it is not always possible to find an appropriate space in which the embedding can be done. You see you know about the thing in which you live so you formulate your mathematics in such a way that you get your result that too without going to a higher space. 2) by wrap I guee you mean curvature. Offcourse it can have but that is called extrinsic curvature. GR deals with intrinsic curvature. And do understand intrinsic curvature you need to understand parallel transport of a vector.. $\endgroup$
    – Shashaank
    Commented Jul 2, 2020 at 19:45
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    $\begingroup$ @Jonas I guess I was helpful, if so you can accept the answer. For your 2) question I gave a short answer, the long answer needs significant explanation and that too in terms of advanced mathematics which I cannot possibly do in a comment. I suggest you to learn the mathematics first as it will be easier to understand then. Probably after that you can get the point yourself and wouldn't require to ask a question. Because these things are best understood in the language of Reimannian geometry. But still your question is an important one. Hope that was helpful... $\endgroup$
    – Shashaank
    Commented Jul 2, 2020 at 19:51
  • $\begingroup$ as a mathematical trick to compute quantities, sometimes it can be useful to embed in higher dimensional space (one example of this is sometimes called the embedding space formalism), this can allow you to exploit symmetries or use a more convenient coordinate system $\endgroup$
    – 4xion
    Commented Jul 3, 2020 at 4:37

There is no need for a higher-dimensional space in which to embed the spatial manifold. The Riemann curvature is a measure of the intrinsic curvature of the surface -- it is independent of and does not require any embedding.

The Riemann tensor is the fundamental quantity that describes the intrinsic curvature of surfaces. A nice way to visualize how it "measures" curvature intrinsically (without reference to an embedding space), is to examine how a single vector, $V^\mu$, ends up when it is parallel transported along two different curves, $C$ and $C'$. The following picture is from Nakahara 7.3: enter image description here

Starting a $p$, parallel transport of $V^\mu(p)$ to $q$ a distance $\epsilon$ away along $C$ gives $V^\mu_C(q) = V_0^\mu - V_0^\kappa \Gamma^\mu_{\nu \kappa}(p)\epsilon^\nu$. Then, along from $q$ a distance $\delta$ to $r$ gives $$V^\mu_C(r) = V^\mu_C(q)-V^\kappa_C(q)\Gamma^\mu_{\nu \kappa}(q)\delta^\nu$$ which we can write as $$V^\mu_C(r) \simeq V_0^\mu - V_0^\kappa\Gamma^\mu_{\nu \kappa}(p)\epsilon^\nu - V_0^\kappa\Gamma^\mu_{\nu \kappa}(p)\delta^\nu - V_0^\kappa[\partial_\lambda \Gamma^\mu_{\nu \kappa}(p)-\Gamma^\rho_{\lambda \kappa}(p)\Gamma^\mu_{\nu \rho}(p)]\epsilon^\lambda\delta^\nu$$ where we've kept terms up to second-order in $\epsilon$ and $\delta$.

You can do the same exercise along the other curve, $C'$. Then, when you take the difference of the vectors at the point $r$ you get $$V^\mu_{C'}(r) - V^\mu_C(r) = V_0^\kappa[\partial_\lambda \Gamma^\mu_{\nu \kappa}(p) - \partial_\nu \Gamma^\mu_{\lambda \kappa}(p) - \Gamma^\rho_{\lambda \kappa}(p)\Gamma^\mu_{\nu \rho}(p) + \Gamma^\rho_{\nu \kappa}(p)\Gamma^\mu_{\lambda \rho}(p)]\epsilon^\lambda \delta^\nu = V_0^\kappa R^\mu_{\kappa \lambda \nu}\epsilon^\lambda \delta^\nu,$$ where $R^\mu_{\kappa \lambda \nu}$ is the Riemann tensor. And so we can think of the Riemann curvature as arising from the fact that the orientation of a vector undergoing parallel transport depends on the path taken on curved surfaces. Importantly, there is no reference to any embedding space.

Of course, it is often helpful to visualize spaces with positive spatial curvature as spheres existing in a higher-dimensional space, but that's because we humans are used to seeing things this way. Part of the fun of differential geometry is learning to do away with these perceptual habits and understanding surfaces in terms of their intrinsic geometries.

  • $\begingroup$ Thanks for the answer although I have to admit that I could not follow along with the math part (might be coming back to that in, like a few years when I have sufficient mathematical skill :D). Do you mean (in your last paragraph) that it is possible, but unnecessary, to use more dimensions or that it is wrong? $\endgroup$
    – jng224
    Commented Jul 2, 2020 at 19:37
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    $\begingroup$ @Jonas There's certainly nothing wrong with studying, say, a 2-sphere embedded in 3-dimensional space. But it's not necessary and requiring that such a higher-dimensional space even exists is an undue constraint. It's quite marvelous to realize that a 2-sphere can simply exist in nothing more than 2-dimensions: the geometry is encoded on the surface. $\endgroup$
    – bapowell
    Commented Jul 2, 2020 at 20:08

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