# Determining whether a space is really three or two dimensional? [closed]

A space purports to be three dimensional with the metric

$$dl^2=dx^2+dy^2+dz^2-\left(\frac{3}{13}dx+\frac{4}{13}dy+\frac{12}{13}dz\right)^2$$

How can I show that it actually represents a two dimensional space?

I tried diagonalizing it to see if it had a zero eigenvalue, for then it would imply that there exists a basis in which the representation of the metric tensor is really a 2X2 matrix i.e to show that there exists such a coordinate transformation which makes it a 2X2 matrix.

## closed as off-topic by Jim, ACuriousMind♦, Kyle Kanos, Danu, JamalSMar 10 '15 at 13:04

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• I edited to make the equation more readable, however as it stands this will likely be closed as a homework question, and is still somewhat confusing – Sean Mar 9 '15 at 18:36
• Zero is an eigenvalue! – MBN Mar 9 '15 at 21:20

Expand your line element and obtain the metric $g_{ij}$.

It is of the form $$g_{ij}=\delta_{ij}-n_in_j$$ where $n=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$ and so $n_in^i=1$ What you have now is a projection operator (because $g_{ij}g_{jk} = \delta_{ik}$, check it symbolically) which does this:

It takes any 3D vector $v$ and gives you its vector component along the plane perpendicular to the unit vector $n$, so it spans the 2D plane orthogonal to $n$

Proof:$$g_{ij}v^j=(\delta_{ij}-n_in_j)v^j=v_i-n_i(n_jv^j)$$ and the RHS is simply the vector $v$ minus its component along $n$, so you get the component orthogonal to $n$.

So this metric projects a 3D vector onto a plane.

In GR, these kinds of "degenerate metrics" are generally used when splitting space-time as a 3+1 foliation for solving initial value problems.

It is not a space-time because it is not Lorentzian. It is actually Riemannian. This exercise may be from a general relativity book, but is in fact a geometry question. So I take it that the question is to show that it represents a two dimensional space. But since it is in the general relativity tag one can be smart and guess the following.

Consider the vector $n^i=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$. It is a unit vector in a three dimensional Euclidian space. The given metric can be written as

$$g_{ij}=\delta_{ij}-n_in_j$$

where $\delta_{ij}$ is the usual Euclidan metric.

This shows that the given metric is the induced metric on the orthogonal to $n^i$ subspace.

• But the n^i space is still three dimensional! – MQRG Mar 10 '15 at 3:22
• n^i is not a space it is a vector. The subspace which is orthogonal to that vector is two dimensional and that is the space you are looking for. – MBN Mar 10 '15 at 7:31
• But you mentioned n^i subspace!, which is the 3D euclidean space , right? – MQRG Mar 10 '15 at 8:13
• Thanks for the clarification. But you mentioned n^i subspace! All I meant was that n^i is a vector in 3D. What is meant by a subspace being orthogonal to a vecor? Can you give a mathematical relation? – MQRG Mar 10 '15 at 8:25
• I wrote the orthogonal to n^i subspace (i.e. the subspace orthogonal to the vector), not n^i subspace. – MBN Mar 10 '15 at 8:57

Having a zero column in a diagonalization is bad (since the metric would be degenerate), but also bad would be if somehow it looked like $$dl^2=dx^2+dy^2+dz^2$$ or $$dl^2=-dx^2-dy^2-dz^2.$$ S you also want to avoid the metric being positive definite or negative definite. For more dimensions you'd also want to worry about having two spatial directions and two time directions!

So you want your metric to have a signature +--- or -+++ for a real spacetime, and for a lower number of dimensions, +-- or -++.

• Well, I guess that is why one has to show that it is actually a two dimensional space and in some basis looks like this! – MQRG Mar 10 '15 at 3:23
• @MariaQuadeer I tried making my answer very general, partly because of the homework policy and partly because I didn't find your question clear enough. I can't even figure out why you think zero isn't an eigenvalue. I'd expand the metric, write it as a symmetric matrix, and find the eigenvalues/eigenvectors. – Timaeus Mar 10 '15 at 4:05
• Yeah, thanks for the remark! Zero is an eigenvalue, I checked it again. – MQRG Mar 10 '15 at 8:26