# Laplacian and Einstein manifold

1. Jan 31, 2013

### Messenger

I am researching a hypothesis and looking for anyone who is familiar with differential topology (specifically Einstein manifolds). I have access to the Besse book Einstein Manifolds but am also looking for any open questions in differential topology that I am not aware of. I am attempting to develop a solid proof link between the Laplacian and Einstein manifolds (listed in the book as not found yet, free meal up for grabs!). I have seen somewhere that there is a discrepancy concerning the physical representation of Minkowski space, but there were no details listed and can't find the reference anymore. If anyone happens to know a paper detailing this it would be appreciated.

The math is pretty basic, sort of a gauge theory of the Laplacian. In graphical form (as the equations might be confusing without them). So whether function goes to all zero scalars or some arbitrary scalar values will give the same Laplacian $\nabla^2(C-f)=\nabla^2\Phi=0$
Attempting to equate this to $4\Lambda-R=0$

If we flip the integrable function f about the x axis, there is no distinction between the derivatives of the following plots:

so for two dimensional gradients easier to see:

2. Jan 31, 2013

### fzero

The offer of a free meal was for a new example of a compact Ricci-flat manifold (also called Calabi-Yau). Such new examples were provided by intersections in products of (weighted) projective spaces, more generally studied as toric varieties. See http://www.scholarpedia.org/article/Calabi-Yau_manifold#Examples_of_compact_Calabi-Yau_manifolds for relevant references.

As for a link between the Laplacian and Einstein manifolds, I would think that any serious study would take you into Ricci flows. You can look at http://en.wikipedia.org/wiki/Ricci_flow for an obvious 2d example and references.

This is kind of bizarre. Perhaps you are misremembering something, as Minkowski space is pretty trivial from a differential geometry/topology point of view.

It's not possible to say anything meaningful without defining the quantities you're using. You have some harmonic functions $f,\Phi$. Are they somehow related to the metric on some manifold? Perhaps whatever you have in mind could be compared with the situation of Kaehler manifolds, where the metric is actually determined by a scalar function (locally in general, globally if the manifold is also complex).

3. Jan 31, 2013

### Messenger

Thanks!

That would be something I would be interested in looking at later on down the road

It wouldn't be the first time I had remembered something incorrectly :)

The Kaehler manifold isn't something I had considered yet and will look more into that. The basics of it are fairly simple. Take for instance the classical Newtonian gravitational potential $\Phi$. With no source functions present it is zero everywhere, but towards a source function it increases:

but the gradient of $\Phi$ isn't unique, since we could just as easily "flip the integral over" and use a scalar function that decreases from some value $\frac{d(C-f_3)}{dx}=\frac{d\Phi}{dx}$:

What interests me is comparing the weak field approximation $1-2\Phi$ to a normalized version of the decreasing scalar function:

I am trying to determine if I can link a scalar constant with a constant multiple of the metric. Approximating the last plot with the 00 component of a normalized Einstein manifold $R/\Lambda=g$.

4. Jan 31, 2013

### fzero

You should still try to more precisely state what it is that you want to do. Are you considering metrics which are asymptotically flat, i.e.,

$$g_{ab} = \delta_{ab} + h_{ab},~~~~\lim_{r\rightarrow\infty} h_{ab} \rightarrow \frac{C}{r}.$$

You'd seem to need something like this for a weak-field approximation to make sense. Now you can derive conditions such that g will be Einstein to linear order in h. The analysis should be similar to that in Ch 12 in Besse, which leads to a linearized Einstein equation for h, involving the Lichnerowicz Laplacian.

It's probably an important point that the difference between your functions f and $\Phi$ are the boundary conditions. As above, the weak-field approximation is only compatible with boundary conditions such that the harmonic function vanishes sufficiently fast at infinity (like $\Phi$ does). Perhaps how this applies to what you're doing will be clearer when you make a mathematically precise definition of your problem.

5. Jan 31, 2013

### Messenger

It would be something like
$$R=\Lambda g_{ab}$$
$$\frac{R}{\Lambda}=g_{ab}=diag(1,-1,-1,-1)$$
$$g_{ab}-\frac{R}{\Lambda}=diag(1-1,-1+1,-1+1,-1+1)=\delta_{ab}-h_{ab}$$
$$\lim_{r\rightarrow\infty} h_{00} \rightarrow \frac{R_{00}}{\Lambda_{00}}=\frac{f_3}{C}=1$$
$$\lim_{r\rightarrow 0} h_{00} \rightarrow \frac{R_{00}}{\Lambda_{00}}=\frac{f_3}{C}=0$$

Last edited: Jan 31, 2013
6. Feb 1, 2013

### fzero

You should probably be consistent with indices if you're going to use them. I realize that most of Besse is written with index-free notation, but you can't have one side of the equation be index free if you're using indices on the other side.

If $g_{ab}=\text{diag}(1,-1,-1,-1)$, then $R_{ab}=0$. This is a flat Minkowski metric. It is still technically Einstein with $\Lambda=0$.

Given the flat metric, this equation is ill-defined. Even if we just had

$$0=\text{diag}(1-1,-1+1,-1+1,-1+1)=\delta_{ab}-h_{ab},$$

then it just means that $h_{ab}=\delta_{ab}$, which would be another constant, flat metric.

These are inconsistent with $h_{ab}$ being a constant. Also, your problem with indices has turned up, since $\Lambda$ is a scalar, so it has no indices.

Anyway, looking back at your graphs, you seem to want a source-free solution to the Einstein equations that is an increasing function of the radius. This is impossible, as the vacuum equations admit the Schwarzschild solution as the only spherically symmetric solution. Technically this is due to Birkoff's theorem, but on physical grounds it is the statement that the Newtonian potential must decrease as we move away from a point-mass at the origin.

7. Feb 1, 2013

### Messenger

True, I should probably write:
$$R_{ab}-\frac{1}{2}Rg_{ab}=\Lambda g_{ab}$$
$$R_{ab}=0$$ when $$g_{ab}=diag(1,-1,-1,-1)$$
$$-\frac{1}{2}Rg_{ab}=\Lambda g_{ab}=diag(\Lambda,-\Lambda,-\Lambda,-\Lambda)$$
$$-\frac{1}{2}Rg_{00}=\Lambda g_{00}$$
$$-\frac{1}{2}R\frac{1}{\Lambda}g_{00}=g_{00}=1$$
I do understand about Birkhoff's Theorem, but it is curious to me that the Laplacian/Poisson can handle this. I wonder why the field equations can't be written such that a spherical Schwarzschild solution works for either a positive mass density, a negative mass density, or even an energy density that decreases from something like a QFT value. I don't understand why the symmetry would break down. Why would differential topology care about absolute energy levels and not just the derivatives?

I had assumed from $$g_{ab} = \delta_{ab} + h_{ab},~~~~\lim_{r\rightarrow\infty} h_{ab} \rightarrow \frac{C}{r}.$$ that you were considering $$h_{ab}$$ as a perturbation and not as a constant.

Last edited: Feb 1, 2013
8. Feb 1, 2013

### fzero

If $R_{ab}=0$, then $R = g^{ab} R_{ab} =0$, so $\Lambda=0$. In your last equation you are trying to divide by zero.

Well you now seem to be including a cosmological constant, so that changes the solutions slightly. With a cosmological constant $\Lambda$ and vanishing stress tensor, we have the de Sitter-Schwarzschild solution (in your signature)

$$g = \text{diag} (f(r), - 1/f(r) , r^2, r^2\sin^2\theta),$$

$$f(r) = 1 - \frac{r_s}{r} - \frac{\Lambda r^2}{3}.$$

For $\Lambda>0$, the corresponding potential does begin to increase at large enough r, where the effect of the CC dominates the source at the origin. This space has positive curvature asymptotically. For $\Lambda=0$, we have the ordinary Schwarzschild black hole. For $\Lambda<0$, we call it anti-de Sitter-Schwarzschild and the space is asymptotically negatively curved.

Birkoff's theorems tell us that these are the only solutions and that they are static. Our source has delta function support at the origin (because the stress tensor is zero), and the parameter $r_s$ is a constant (by Birkoff). Physical considerations like the weak energy condition would suggest that $r_s<0$ is unphysical, but there's no mathematical reason not to consider the possiblity.

9. Feb 1, 2013

### Messenger

Thanks, then that wouldn't work. I was under the impression that the Ricci scalar can't vanish if there is a cosmological constant present, even if the Ricci tensor does. I need to make sure why I misunderstood that part.

I do know that the small curvature from WMAP implies a very small $\Lambda$ but I was wondering whether there was a geometric method to account for this by resetting the energy measurement point. However, it seems I need to make sure I understand all the cases of when the Ricci scalar is and isn't zero first, as you pointed out. Going to take a look at Gaussian surfaces.

10. Feb 1, 2013

### fzero

Take the trace of the Einstein equation with CC:

$$R_{ab} - \frac{1}{2} R g_{ab} +\Lambda g_{ab}=0,$$

$$\frac{2-n}{2} R + n \Lambda=0,$$

which allows us to rewrite

$$R_{ab} = \frac{2n\Lambda}{n-2} g_{ab} = k g_{ab}.$$

This last condition is the definition of an Einstein metric and from a mathematical perspective, we don't necessarily care that $k$ is even related to the CC. When $k=0$, we have a special case of Ricci-flat metrics, of which Minkowski space is a simple example.

It's impossible for the Ricci scalar to be nonzero if the Ricci tensor is zero, since the former is the trace of the latter. We can also see that $\Lambda\neq 0$ does in fact mean that $R\neq 0$.

The problem with your previous post is that, in an intermediate step, you were taking $g_{ab}$ as the Minkowski metric, which is necessarily flat, so not a solution when $\Lambda\neq 0$. I probably helped a bit to confuse you by introducing perturbations around a flat metric in post #4. In your OP you were already considering a nonzero CC, for which flat space is not a solution.