I Differential Geometry: Comparing Metric Tensors

[dsaun777]

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?

 

[Orodruin]

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.

 

[WWGD]

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?

 

[Infrared]

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'.

 

[WWGD]

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'.

I understand that it can be done and that tangent spaces are "logarithmic", i.e. $T_{(x,y)}M_1 \times M_2=T_xM_1 + T_yM_2$ ( 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.

 

[Orodruin]

don't know how to tex direct sum

\oplus $\oplus$

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.

I would argue that there is no "standard way" in physics. It completely depends on what you are attempting to achieve.

 

[lavinia]

@WWGD

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?

 

[dsaun777]

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.

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

 

[Orodruin]

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

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.

 

[Infrared]

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?

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.

 

[lavinia]

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.

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

 

[Orodruin]

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

Then you can find a coordinate transformation such that the new coordinate derivative is.

 

[lavinia]

Then you can find a coordinate transformation such that the new coordinate derivative is.

So the product metrics on the torus are flat. Most metrics are not product metrics.

Suppose one has two manifolds of positive sectional curvature?

 

[Infrared]

@lavinia Then the product has non-negative sectional curvature (it can sometimes be zero- take a tangent plane spanned by one tangent vector from each factor)

 

[lavinia]

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?

@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]

@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.

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]

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.

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.

@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.

 

[Infrared]

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.

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

 

[dsaun777]

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.

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

 

[lavinia]

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

What do you mean by a bimetric?

 

[dsaun777]

What do you mean by a bimetric?

I mean more than one, in this case 2, metrics on a single manifolds

 

[lavinia]

I mean more than one, in this case 2, metrics on a single manifolds

A metric is a smoothly varying choice of a non-degenerate bilinear form on each tangent plane. What would a bimetric be?

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

 

[WWGD]

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

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.

 

[Orodruin]

What would a bimetric be?

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]

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.

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

 

[WWGD]

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, there's this interesting one, (not originally mine). It uses the degree of a map, specifically, homeomor-
phisms have degree $\pm 1$ and satisfy $def(f \circ g)=deg(f) \cdot deg(g)$, so that for any homeomorphism f, we have : $deg(f \circ f) ==1$.
Assume then there is a homeo h $X^2 \rightarrow \mathbb R^3$. Then we have a homeo
from $X^4 \rightarrow \mathbb R^6$. Consider the homeo $h: X^4 \rightarrow X^4 : (a,b,c,d) \rightarrow (d,a,b,c)$ with $h \circ h : (a,b,c,d) \rightarrow (c,d,a,b)$ which gives us the map $h' \circ h' \mathbb R^6 \rightarrow \mathbb R^6 (a,b,c,d,e,f) \rightarrow$

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

Yes, I don't know if this applies to any other type of space though. And it seems like a novel proof.

 

[WWGD]

Well, clearly a 2-manifold cannot be even locally diffeomorphic to $\mathbb R^3$, I just thought it was a cool proof, did not use Invariance of dimension, etc.

 

[Infrared]

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]

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 $X\times X=\mathbb{R}^{2n+1}$.

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]

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.

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.

 

[lavinia]

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.

Not really my point. The underlying intuition was that the iterated mapping $h^2$ on $X^4$ is orientation preserving. So how is $X^4$ orientable in the first place? What does that mean?E.g.if $X$ is a non-orientable manifold then $X^4$ is also non-orientable. For instance the four fold Cartesian product of the real projective plane with itself is not orientable. In fact is $w_{i}$ is the generator of its first $Z_3$ cohomology then the first Stiefel-Whitney class of the four fouldCartesian product is $w_1+w_2+w_3+w_4$.

BTW: I am curious to see your Kunneth formula proof.

 

[Infrared]

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 $H_6(X^4,X^4-\text{point})$.

 

[lavinia]

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 $H_6(X^4,X^4-\text{point})$.

Right. Now I get it. Nice.

There is a theorem here that is implicitly assumed which is that the square of any homeomorphism of an orientable manifold is orientation preserving. This is clear for a compact manifold but for a non-compact manifold, how does one prove it?

 

[Infrared]

I found a MO 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...

 

[lavinia]

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...

yes.

 

[WWGD]

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.

Myb bad, thanks for pointing it out , for links and followup. I am editing as we speak.

 

[Infrared]

There is a theorem here that is implicitly assumed which is that the square of any homeomorphism of an orientable manifold is orientation preserving. This is clear for a compact manifold but for a non-compact manifold, how does one prove it?

I think the point is the same as in the closed case. If $X$ is orientable, then all of the local homology groups $H_n(X,X-\text{point})$ are naturally isomorphic (identify the preferred generators), so a homeomorphism $f:X\to X$ with $f(x_1)=x_2$ gives an isomorphism $H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),$ and then $f\circ f$ has to induce the identity, so $f\circ f$ preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of $X$.

 

[lavinia]

I think the point is the same as in the closed case. If $X$ is orientable, then all of the local homology groups $H_n(X,X-\text{point})$ are naturally isomorphic (identify the preferred generators), so a homeomorphism $f:X\to X$ with $f(x_1)=x_2$ gives an isomorphism $H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),$ and then $f\circ f$ has to induce the identity, so $f\circ f$ preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of $X$.

Right. Two oriented charts are connected by a path of overlapping oriented charts.