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10:00 PM

I think so?

Is a geodesic of the spacetime a geodesic of the hypersurface?

There are spacetimes which are geodesically complete but have incomplete curves, but since those are accelerated curves, I don't think that matters for a spacelike surface

0celouvskyopoulo7

10:12 PM

@Slereah No

Slereah

yeah probably not

0celouvskyopoulo7

Are geodesics in $\Bbb R^3$ geodesics on the sphere?

Slereah

You can foliate Minkowski space with hyperboloid

I think it's probably true, though

0celouvskyopoulo7

So I have a symmetric guy $K_{AB}$ and a metric $\gamma_{AB}$. I want a function $f$ and a one-form $\Phi_A$ with $K_{AB}=f\gamma_{AB}+\Phi_A\Phi_B$

Surely this must be possible

Slereah

what is a symmetric guy

0celouvskyopoulo7

10:14 PM

symmetric tensor

Slereah

Decompose the tensor into the diagonal and the off diagonal terms, mb

should be easier to show

0celouvskyopoulo7

eh, $\Phi_A\Phi_B$ has both diagonal and off terms

wtf why is this so hard

Slereah

Maybe just consider this at a single point

is the dimension of the space of symmetric matrices equal to that of $f$ and $\Phi$

I think it may fall short

0celouvskyopoulo7

Wait, this is a two dimensional manifold.

I can probably just write everything down.

Slereah

In 2D that may work

0celouvskyopoulo7

10:24 PM

In 2D there are 3 dimensions of symmetric matrices.

Slereah

yeah

Hence why it may work

0celouvskyopoulo7

$f\gamma$ generates one

I guess the dyads generate two?

Slereah

I think so yeah

0celouvskyopoulo7

Linear algebra is awful.

@Slereah why?

Slereah

In 4D I think that would be $10$ versus $5$

Gut feeling?

I don't have a better reason than "it is described by three numbers"

"Causal structure | an infestation of closed timelike curves"

0celouvskyopoulo7

10:30 PM

where is ACM when you need him

Slereah

At this hour, probably in bed

The Germans are big on punctuality

0celouvskyopoulo7

what's "harmonic" in French?

harmonique?

Slereah

yes

0celouvskyopoulo7

So this is probably stupid

Suppose I just pick any old $\Phi$

And I suppose $K_{AB}-\Phi_A\Phi_B=f\gamma_{AB}$

And I then have $f=\gamma^{AB}(K_{AB}-\Phi_A\Phi_B)/2$

That probably doesn't work...

Slereah

Well it's three equations,

$$a = f \alpha + k^2$$
$$b = f \beta + p^2$$
$$c = f \gamma + pk$$

0celouvskyopoulo7

10:45 PM

How about this

Slereah

Gotta check if this has a solution for all values

Diagonalize the metric, find the eigenvalues?

I dunno

0celouvskyopoulo7

$$\begin{pmatrix}K_{11} & K_{12}\\ K_{12} & K_{22}\end{pmatrix}=\begin{pmatrix} f+\Phi_1^2 & \Phi_1\Phi_2\\ \Phi_1\Phi_2 & f+\Phi_2^2\end{pmatrix} $$

Slereah

Gauss coordinates?

0celouvskyopoulo7

I already diagonalized the metric

just set it = identity for now

$$K_{11}K_{22}-K_{12}^2=f^2+f(1-K_{11})+f(1-K_{22})$$

not great

Slereah

Consider $K_{11} = K_{12} = a, K_{22} = b$, then $\Phi_1 = a/\Phi_2$, so $f + \Phi_1^2 = a$ and $f + \Phi_2^2 = b$, so $\Phi_2^2 - \Phi_1^2 = b-a$, or $\Phi_2^2 - 1/\Phi_2^2 = b-a$, $\Phi_2^4 - 1 = (b-a) \Phi_2^2$

for $\Phi_2^2 = x$, does $x^2 + (a-b) \Phi_2^2- 1 = 0$ have a positive root

for all values of $a$ and $b$

I assume the matrix has to be real here

The determinant is $(a - b)^2 + 4$, so... yeah I guess it has a positive root

Dang it

@ACuriousMind Do you know if it's a true property or not?

So that I don't waste time trying to find counterexamples

you could prove it with three different values for each components of $K$, but I don't know how to solve cubic roots off the top of my head

0celouvskyopoulo7

10:59 PM

Q: Decomposing symmetric tensor field into sum of metric and tensor product

Consider a symmetric tensor field $K_{ij}$ on a 2-dimensional Riemannian manifold with metric $g_{ij}$. How does one show that there exists a (smooth) function $f$ and one-form $\phi$ such that $$K_{ij}=fg_{ij}+\phi_i\phi_j?$$ I tried raising the $j$ index (so the metric becomes the identity) and...