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dsaun777
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Is there ever an instance in differential geometry where two different metric tensors describing two completely different spaces manifolds can be used together in one meaningful equation or relation?
Not two different manifolds. However, it is possible to have two different metrics on the same manifold. This would induce two different geometries on that manifold and you could study the relation between them. The application of this to spacetime is an active research topic known as bimetric gravity.dsaun777 said:Is there ever an instance in differential geometry where two different metric tensors describing two completely different spaces manifolds can be used together in one meaningful equation or relation?
I understand that it can be done and that tangent spaces are "logarithmic", i.e. ##T_{(x,y)}M_1Infrared said:Yes for the first question. If ##M_1## and ##M_2## are manifolds, you can identify ##T(M_1\times M_2)=TM_1\times TM_2##. If ##M_i## have metrics, then set ##\langle (v,w),(v',w')\rangle=\langle v,v'\rangle+\langle w,w'\rangle## to get a metric on ##M_1\times M_2##. That is, you're just adding the metrics of the factors.
For the second, I haven't seen a way to assign a metric to either factor, and I can't think of a reasonable approach to do so.
Edit: One thing you could do is fix a point ##y\in M_2## and identify ##M_1=M_1\times\{y\}\subset M_1\times M_2##. Then restricting the metric on ##M_1\times M_2## gives a metric on ##M_1##, but of course this depends on the choice of ##y##. I guess you could average over ##y## if ##M_2## is compact but this is getting a little far-fetched...
As for relating multiple metrics on the same manifold, you might be interested in reading about Ricci flow (https://en.wikipedia.org/wiki/Ricci_flow), which is a way of time evolving a given metric on a manifold to 'smooth it out'.
\oplus ##\oplus##WWGD said:don't know how to tex direct sum
I would argue that there is no "standard way" in physics. It completely depends on what you are attempting to achieve.WWGD said:just curious if this is the standard way in Physics, of which I know little , and whether it is an important issue therein; more of the Physics perspective.
So an example would be the torus where there are 2 types of curvatures, negative and positive?Orodruin said:Not two different manifolds. However, it is possible to have two different metrics on the same manifold. This would induce two different geometries on that manifold and you could study the relation between them. The application of this to spacetime is an active research topic known as bimetric gravity.
No. This is due to a single metric on the torus that has different scalar curvature at different points.dsaun777 said:So an example would be the torus where there are 2 types of curvatures, negative and positive?
lavinia said:
You might like to check if there are any special properties of the curvature tensor of a product metric. For instance in the case of the torus what can be said if one starts out with two different metrics on the circle?
Infrared said:Shouldn't any product metric on ##S^1\times S^1## have identically vanishing curvature? ##S^1## with any metric is locally isometric to ##\mathbb{R}## (with the usual metric), so ##S^1\times S^1## with a product metric should be locally isometric to ##\mathbb{R}^2## I think.
Then you can find a coordinate transformation such that the new coordinate derivative is.lavinia said:Yes but what if ##∂/∂θ## is not a unit vector field?
Orodruin said:Then you can find a coordinate transformation such that the new coordinate derivative is.
WWGD said:Is there a standard way of defining a metric on a product manifold from the factor metrics, e.g., for the 2-torus ##S^1 \times S^1 ##? Or, conversely, is there a way of projecting a metric in a product onto its factors?
Thanks, Lavinia. I had this idea a while back of trying to figure out when/if a space X was a product space, i.e., homeomorphic to ##A \times B ##. One way I guess is to see whether X is a trivial bundle over the product factors. There is a nice argument on why ##\mathbb R ^{2n+1}## is not a product space, though it has nothing to see with the bundle trick.lavinia said:@WWGD
I guess the point about curvature of product metrics was that for an arbitrary metric the metrics on the factors obtained from some choice of projection do not recover the metric on the original manifold. I guess one could have simply said that not all of the metrics on a Cartesian products are product metrics. But this is a simple way to see it.
WWGD said:Thanks, Lavinia. I had this idea a while back of trying to figure out when/if a space X was a product space, i.e., homeomorphic to ##A \times B ##. One way I guess is to see whether X is a trivial bundle over the product factors. There is a nice argument on why ##\mathbb R ^{2n+1}## is not a product space, though it has nothing to see with the bundle trick.
WWGD said:There is a nice argument on why ##\mathbb R ^{2n+1}## is not a product space, though it has nothing to see with the bundle trick.
How would you have two metrics for the single manifold of the torus then? What types of manifolds can you have bimetrics on?Orodruin said:No. This is due to a single metric on the torus that has different scalar curvature at different points.
Also note that what you are talking about is a particular metric on the torus, probably the one induced by the standard embedding in Euclidean space. However, there are other metrics where the torus is flat.
dsaun777 said:How would you have two metrics for the single manifold of the torus then? What types of manifolds can you have bimetrics on?
I mean more than one, in this case 2, metrics on a single manifoldslavinia said:What do you mean by a bimetric?
dsaun777 said:I mean more than one, in this case 2, metrics on a single manifolds
EDIT : Yes, there's this interesting one, (not originally mine, but can't find the original source. It may ned a bit tidying up). It uses the degree of a map, specifically, homeomor-Infrared said:You can prove this (##\mathbb{R}^{2n+1}\ncong X\times X##) with algebraic topology (homology)- do you have a different argument in mind?
There are theories of bimetric gravity in physics. Of course, that is just about having two different metrics on a single manifold (both dynamical in the sense that they have their own equations of motion).lavinia said:What would a bimetric be?
WWGD said:EDIT : Yes, there's this interesting one, (not originally mine, but can't find the original source. It may ned a bit tidying up). It uses the degree of a map, specifically, homeomor-
phisms have degree ## \pm 1## and satisfy ##deg(f \circ g)=deg(f) \cdot deg(g)##, so that ##deg(f \circ f) ==1 ## for all homeomorphisms.
Assume then there is a homeo h ## X^2 \rightarrow \mathbb R^3 ##. Then we have a homeo
from ## X^4 \rightarrow \mathbb R^6 ## and 4-ples (a,b,c,d) correspond to 6 -ples :(n,m,o,p,q,r) . Consider the
homeo ##s:(a,b,c,d) \rightarrow (d,a,b,c) ## so that ## s \circ s : (a,b,c,d) \rightarrow (c,d,a,b)## and then ##deg s \circ s =1 ##. But this implies that the corresponding composition ## s' \circ s' : X^6 \rightarrow X^6 : (a,b,c,d,e,f) \rightarrow (o,p,q,r,n,m)## is an orientation-preserving homeo . But it is not, since its determinant is -1, which is a contradiction.
Yes, there's this interesting one, (not originally mine). It uses the degree of a map, specifically, homeomor-Infrared said:You can prove this (##\mathbb{R}^{2n+1}\ncong X\times X##) with algebraic topology (homology)- do you have a different argument in mind?
Yes, I don't know if this applies to any other type of space though. And it seems like a novel proof.lavinia said:If ##X## is a smooth manifold then wouldn't ##X^2## be even dimensional?
Infrared said:It is a cool proof, and it doesn't assume that ##X## is a (even topological) manifold, so is stronger than an argument from invariance of domain, etc.
A more mundane proof would be to remove a point and use a Kunneth formula on ##X\times X=\mathbb{R}^{2n+1}##.
lavinia said:If it does not assume that ##X## is a topological manifold then how does one make the argument about orientation reversal?
For instance, how does ##f## possibly have degree ##-1## for an arbitrary topological space? Suppose that ##X## is a non-orientable manifold.
Infrared said:The proof does not consider the degree of a map ##X\to X##, but of a map ##X^4\to X^4##. This makes sense because ##X^4\cong\mathbb{R}^6##. I think @WWGD had a typo of ##X^6## for ##\mathbb{R}^6## in his post.
Infrared said:Sorry I must be missing something. ##\mathbb{R}^6## is orientable, and ##X^4## is homeomorphic to ##\mathbb{R}^6##, so it is also orientable. If you're asking what definition of orientable works here, I think you can use the usual definition of being able to consistently choose generators for the local homology groups ##H_6(X^4,X^4-\text{point})##.
yes.Infrared said:I found a SO thread (https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space) that has both the Kunneth argument (first answer) and @WWGD's orientation solution (link in second answer)
I worry that we've hijacked the thread...