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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?

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  • $\begingroup$ possible duplicate of More than one time dimension $\endgroup$ – John Rennie Apr 18 '15 at 16:03
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    $\begingroup$ 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
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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.

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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.

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    $\begingroup$ 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

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