Metric tensor after constructing a quotient space.

center o bass

Suppose we have some two-dimensional Riemannian manifold $M^2$ with a metric tensor $g$. Initially it is always locally possible to transform away the off-diagonal elements of $g$. Suppose now by choosing the appropriate equivalence relation and with a corresponding surjection we construct the quotient space $M^1\times S^1$ by $q:M^2 \to M^1\times S^1$. Now assuming the the metric tensor respect the equivalence relation there will be uniqe components, in say coordinate basis, $\tilde g_{ij}$ on $M^1\times S^1$ such that basis, $\tilde g_{ij} \circ q =g_{ij}$.

Now to my question: Is there any reason why we might not be able to transform away the off-diagonal elements of $\tilde g_{ij}$?

Related Differential Geometry News on Phys.org

jgens

Gold Member
You need to explain what M1 is here and what the quotient map q:M2→M1 X S1 is so people will actually know what you are asking.

center o bass

You need to explain what M1 is here and what the quotient map q:M2→M1 X S1 is so people will actually know what you are asking.
The idea was to be general. I think of $M^2$ as a any manifold such that it is possible to make it into $M^1 \times S^1$, where $S^1$ is a circle, by an appropriate surjection (equivalence relation) on $M^2$. In physics this is called the Kaluza-Klein mechanism or compactification.

I guess one could just consider $M^2$ to be the plane $\mathbb{R}^2$ as a special case and $q$ to be the usual surjection that makes it into a cylinder. I don't know if the fact that $g$ is taken to be Lorentzian plays any role in the argument (see below). So to repeat my question: as the components of the metric tensor can be considered maps from the set $M^2$ to $\mathbb{R}^2$; then after constructing the quotient space $g_{ij}$ will 'pass the quotient' (see the theorem below) in the sence that there exists a uniqe $\tilde g_{ij}$ on $M^1 \times S^1$ such that $g_{ij} = \tilde g_{ij} \circ q$. Is there now any reason why the off-diagonal components could not be transformed away with an appropriate coordinate transformation?

(Some background on KK theory: In Kaluza-Klein theory one assumes a five-dimensional manifold with a Lorentzian $(M^5,g)$ where the $g_{\mu 5}$ for $\mu = 1,2,3,4$ plays the role of the electromagnetic vector potential (in coordinate basis) $A_\mu$. An explanation of why we do not see the fifth dimension is that the five-dimensional manifold $M^5$ was 'compactified' in the early universe such that $M^5 \to M^4\times S^1$ where $M^4$ is the spacetime from general relativity and the circle is taken to be 'small'. I've read several places that after this 'compactication' the off-diagonal components of the metric tensor can not be transformed away, but this is never explained. Hence my question.)

Theorem (Passing to the Quotient). Suppose $q:X \to Y$ is a quotient map, $Z$ is a topological space, and $f:X \to Z$ is any continuous map that is constant on the fibers of $q$(i.e., if $q(x) = q(x')$ then $f (x) = f (x')$. Then there exists a unique continuous map $\tilde f: Y \to Z$ such that $f = \tilde f \circ q$.

Last edited:

jgens

Gold Member
If the resulting $\tilde{g}_{ij}$ are non-degenerate and symmetric, then it should be possible to choose local coordinates where the off-diagonal terms vanish. If you cannot guarantee that condition, then (for the most part) you are out of luck. It's just a matter of linear algebra.

center o bass

If the resulting $\tilde{g}_{ij}$ are non-degenerate and symmetric, then it should be possible to choose local coordinates where the off-diagonal terms vanish. If you cannot guarantee that condition, then (for the most part) you are out of luck. It's just a matter of linear algebra.
So you can not think of any reason why $g$, being a tensor field on the initial manifold, breaks up into something else when the quotient space is constructed? In KK theory it is stated that it breaks up into a vector field $A_\mu$ a scalar field $g_{55}$ and a four-dimensional tensor field.

jgens

Gold Member
So you can not think of any reason why $g$, being a tensor field on the initial manifold, breaks up into something else when the quotient space is constructed?
Having no knowledge how the quotient space is constructed or how exactly these new $\tilde{g}_{ij}$ are introduced I see no reason why they ought to be symmetric. So I would not be surprised if there were no local coordinates where the off-diagonal terms vanish. Someone who knows about this particular construction can hopefully shed some additional light on the matter for you.

center o bass

If the resulting $\tilde{g}_{ij}$ are non-degenerate and symmetric, then it should be possible to choose local coordinates where the off-diagonal terms vanish. If you cannot guarantee that condition, then (for the most part) you are out of luck. It's just a matter of linear algebra.
I have posted on this question earlier (https://www.physicsforums.com/showthread.php?t=728572) where I got an answer from fzero which involved topology. This was my motivating for reading up on topology the last few weeks. Perhaps if you read his answer (and my question) it would become a little more clear what my problem is here. Only if you have the time of course.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving