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## Main Question or Discussion Point

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?

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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?

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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.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?

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WWGD

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Infrared

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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'.

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WWGD

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I understand that it can be done and that tangent spaces are "logarithmic", i.e. ##T_{(x,y)}M_1

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'.

\times M_2=T_xM_1 + T_yM_2## ( don't know how to tex direct sum), 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. Thanks for the link.

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\oplus ##\oplus##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.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.

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So an example would be the torus where there are 2 types of curvatures, negative and positive?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.

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No. This is due to a single metric on the torus that has different scalar curvature at different points.So an example would be the torus where there are 2 types of curvatures, negative and positive?

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.

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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.

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?

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lavinia

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Yes but what if ##∂/∂θ## is not a unit vector field?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.

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Then you can find a coordinate transformation such that the new coordinate derivative is.Yes but what if ##∂/∂θ## is not a unit vector field?

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lavinia

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So the product metrics on the torus are flat. Most metrics are not product metrics.Then you can find a coordinate transformation such that the new coordinate derivative is.

Suppose one has two manifolds of positive sectional curvature?

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lavinia

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@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.

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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

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.

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lavinia

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While it is true that the tangent bundle of a Cartesian product is a Whitney sum of two sub bundles over the product factors, a manifold whose tangent bundle splits as a sum of two sub bundles may not be a Cartesian product. For instance the 3 sphere's tangent bundle is trivial.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

However, the tangent bundle does determine when a smooth closed manifold is cobordant to a Cartesian product.

Aside: Two manifolds are cobordant if their disjoint union is the boundary of a one higher dimensional manifold. For instance, a manifold is always cobordant to itself since the disjoint union with itself is the boundary of the cylinder ##M×[0,1]##. The relation of cobordism is an equivalence relation. These equivalence classes are closed under Cartesian product and thus form a ring. Thom's theorem says that this ring is actually a polynomial ring (over ##Z_2##) with a single generator in each dimension not of the form ##2^{n}-1##. So products of generators are represented by Cartesian products of manifolds.

A manifold's cobordism class is determined by the Stiefel-Whitney numbers of its tangent bundle. For instance, a manifold is a boundary all by itself if and only if all of its Stiefel-Whitney numbers are zero. While in general Stiefel-Whitney numbers are extremely difficult to calculate, they nevertheless determine when a manifold is cobordant to a Cartesian product of generators of the cobordism ring.

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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 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.

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How would you have two metrics for the single manifold of the torus then? What types of manifolds can you have bimetrics on?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.

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lavinia

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What do you mean by a bimetric?How would you have two metrics for the single manifold of the torus then? What types of manifolds can you have bimetrics on?

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I mean more than one, in this case 2, metrics on a single manifoldsWhat do you mean by a bimetric?

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lavinia

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A metric is a smoothly varying choice of a non-degenerate bilinear form on each tangent plane. What would a bimetric be?I mean more than one, in this case 2, metrics on a single manifolds

There are many metrics. Two metrics are different if the bilinear forms on the tangent planes are not the same.

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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-You can prove this (##\mathbb{R}^{2n+1}\ncong X\times X##) with algebraic topology (homology)- do you have a different argument in mind?

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.

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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).What would a bimetric be?

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lavinia

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If ##X## is a smooth manifold then wouldn't ##X^2## be even dimensional?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.

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