Our universe is often described as having 3 space-like dimensions and 1 time-like dimension.

Can hypothetical universe exist with more than space- and time-like dimensions?

If so how would these dimensions appear like?

  • $\begingroup$ possible duplicate of More than one time dimension $\endgroup$ – John Rennie Apr 18 '15 at 16:03
  • 5
    $\begingroup$ not a duplicate of this question - the question here asks about other types of dimensions than just time and space, while the proposed duplicate asks (and gets answers only) about more time and/or space dimensions than we seem to observe. $\endgroup$ – Martin Apr 19 '15 at 10:54

Our model for spacetime is that of a manifold, which is the mathematical term for something that looks like $\mathbb{R}^n$ in any zoomed-in patch, and where all these patches are stitched together in a sensible way. On our manifold we have $n$ coordinates -- real numbers that describe each point and vary smoothly from point to point.

We also add to our model the notion of angles and sizes, and this is accomplished via a metric $g$, which gives us an inner product between vectors. For example, if you have a direction vector $\vec{v}_1$ and another $\vec{v}_2$, the angle between them is $g(\vec{v}_1, \vec{v}_2)$. If $\vec{v}$ is the tangent vector along some path, then $g(\vec{v}, \vec{v})$ gives something like the squared infinitesimal distance along the path (so square rooting and integrating gives you the total distance).

Now we take $g$ to have some basic properties.

  • It must act linearly on its arguments, so for example $g(\vec{v}_1+\vec{v}_2, \vec{v}_3) = g(\vec{v}_1, \vec{v}_3) + g(\vec{v}_2, \vec{v}_3)$. Without this property, the angle between two physical directions would depend on how you choose to write down the formula. You can therefore represent $g$ as an $n \times n$ matrix, where the scalar value $g(\vec{v}_1, \vec{v}_2)$ is given by matrix multiplication of the row vector $\vec{v}_1$, the matrix $g$, and the column vector $\vec{v}_2$.
  • Furthermore, we require $g$ to be symmetric: $g(\vec{v}_1, \vec{v}_2) = g(\vec{v}_2, \vec{v}_1)$, always. Without this property, the angle between two directions would depend on which direction you write down first.
  • And in case it wasn't clear, $g$ should only return real numbers. (What would a complex angle even mean?) Since its inputs only consist of real numbers (since the coordinates themselves are real), this means $g$ as a matrix can only have real entries.

Now that we have a real, symmetric matrix, we can apply all sorts of standard linear algebra results to it. In particular, the eigenvalues of such a matrix must be real. Moreover, we can diagonalize $g$ at any point such that its eigenvalues become $0$ or $\pm1 $. Physically, this means we can change coordinates at a point such that the unit direction vectors at that point have squared length $0$ or $\pm1$.

The degenerate $0$ case is problematic, and is often a sign that your mathematical description is failing. In any event, the coordinate direction corresponding to eigenvalue $0$ would be null -- a direction in spacetime taken by something traveling at the speed of light.

This leaves the $\pm1$ cases. If the unit coordinate direction has squared length $+1$, we call the direction spacelike. If it is $-1$, we call the direction timelike. Null is the borderline case between the two, but again, using null coordinates is troublesome.

As a result of our reasonable physically motivated requirements on $g$, there is no room for other types of dimensions. If $g$ diagonalizes to having $s$ $+1$'s and $t$ $-1$'s, it corresponds to $s$ spacelike dimensions and $t$ timelike ones. In particular, by changing coordinates we can rescale any nonzero real numbers to $\pm1$, and complex numbers are disallowed entirely.


Over the real numbers, any non-degenerate quadratic form is determined (up to a change of basis) by its signature, which consists entirely of $1$s and $-1$s.

  • 3
    $\begingroup$ Okay, I understand how this answers the question. It doesn't elaborate on why there are no complex or imaginary signatures, but brevity isn't a fault. Nevertheless, I think it is easy to see how most people wouldn't understand that this answers the question. So maybe you could expand on this so that it caters to a non-expert level? More than just the technical way to say "two types of signatures means only two types of dimensions" $\endgroup$ – Jim Apr 18 '15 at 14:50
  • $\begingroup$ @ACuriousJim This was almost exactly how I was going to answer the question, but simply state the proviso that "if you confine yourself to real-valued co-ordinates" (then you have signatures. Beyond this, I can't really think of an intuitive explanation: even though I tend to agree with you. $\endgroup$ – WetSavannaAnimal Apr 19 '15 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.