"In this case, the spinor bundle $S_M$ is associated with the $L_s$-lift $P_h$ of $L^h X$ : $$S_M = S_h = (P_h \times V) / L_s$$"

Blublublu

Enough bundles for tonight

so...the WEC is enough for what I want

It usually is

quoting HE, Sanchez

some good physics here boi

$$\int_{\mathscr H(t)\cap\mathscr U}T^{ab}t_{;a}\, d\sigma_b\le -\int_{\partial_1\mathscr U}T^{ab}t_{;a}\, d\sigma_b+P\int^t\int_{\mathscr H(t')\cap\mathscr U}T^{ab}t_{;a}\, d\sigma_b \, dt'$$

@Slereah hmm, how does one get the integrals closer

closer to what

to the other integral

$$\int^t\int_{\mathscr H(t')\cap\mathscr U}$$

No idea

Not sure what most of those symbols are

I literally mean the spacing of the integrals looks off

Oh

There's a backspace symbol IIRC

$$\int^t\!\!\int_{\mathscr H(t')\cap\mathscr U}$$

Better?

$$\int^t\!\!\!\!\int_{\mathscr H(t')\cap\mathscr U}$$

@Slereah ah right the exclamation mark

that's how one does dirac notation, right?

it is one possible way, yes

ugh how is surface measure defined in GR

@Slereah hmm, do you know if the "usual" induced metric $g_{\mu\nu}+n_\mu n_\nu$ is just the pullback metric?

Do you mean with $g$ the Riemannian metric?

and $n$ a timelike vector

$g$ is the metric of spacetime and $n$ is the normal to the spacelike hypersurface

Oh

I do not know

hmm, this is a bit of an abuse of notation (what else is new)

@Slereah Consider the embedding $i:\Sigma\to M$

If $h=g+n\otimes n$, then $h|T\Sigma=i^*g$

I'd believe it

Just gotta prove that $$g(v,w) + n \otimes n (v, w) - g(df(v), df(w)) = 0$$

What's the formula for $n$, is it just $df$

You got like $n \otimes n(v, w) = n(v) \cdot n(w)$ right?

Wait no, the second metric is the wrong space

Or is it

It's too late for diff geom