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22:00
i dont see a calabi yau manifold in the night sky
Howdy @robjohn. Any thoughts on this? I don't have the definitive answer to his follow-up.
I have not, thankfully
I've switched out of the math finance program into a certification program for data science
Actually, that seems to be a very hot area. Sadly, I know nothing.
i see commutative diagrams in the night sky
22:03
Well, a Balarka, we agree that counting is for the birds :P
Actually, @robjohn (if you're thinking about it), I'm now confused. Since he's applying the derivative only to the vector field $X$ itself, that looks a lot like the integrated form of $\mathscr L_X X = 0$. Hmm.
Ah, OK, let's try this. Let $Z_i$, $i \in \Bbb N$ be iid $\text{Unif}\{-1, 1\}$. Define $X_0 = 0$, $X_n = \sum_{i = 1}^n Z_i$.
$X_{\lfloor nt\rfloor}/\sqrt{n}$ should converge to $B_t$
I run the motion for a long time and scale like CLT tells me to
Clear that $X_{\lfloor nt \rfloor}/\sqrt{n} \sim N(0, t)$, the law of $B_t$, by CLT.
just construct brownian motion by proving that there exists a unique stochastic process with x, y, and z properties
if it's good enough for determinants...
Independent normal increments is also a obvious
I guess sample continuity is the messy business again
didn't we just talk about the determinant thing yesterday
still don't know a non-constructive way of getting existence
You don't grant our argument based on existence of tensor product?
22:17
proof by arrows
how is the double dual functor defined?
also used by the English to prove their claims on France circa the 1300s, incidentally
lol
brilliant
@orientablesurface it's just the composition of the dual functor with itself
oh alright
thanks
22:19
hang on, I'm still slow today---dual functor's contravariant, right?
oh yeah of course it is
$V \to W$ becomes $(W \to \Bbb F) \to (V \to \Bbb F)$ by left composition
I didn't see a way of making that argument work without explicitly writing down wedge products at some point
contracontra is co but coco is
@Thorgott: So what? I just skew symmetrize the tensor product.
that's effectively the same as writing down the Leibniz formula
for the record, the fact that the identity functor is covariant and the dual functor is contravariant is the trivial, but perhaps unsatisfying, reason that they are not naturally isomorphic
2
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22:23
there is a generalized version of natural transformations that allows you to compare covariant with contravariant functors and even that does not make identity and dual isomorphic
that is to say, their not being isomorphic is more than just a formality issue
why are we discussing functors at all? an isomorphism V -> W^* is choice of a nondegenerate pairing V x W -> R. There is no natural nondegenerate form on a vector space, so there is no natural isomorphism. When W = V^*, there is: it's evaluation
Forget category garbage
Nobody cares about this bullshit
define "natural"
non-categoricists = owned
what you are not
"natural" = "even me, an idiot, would think of it"
done
22:28
I wish I was that cool
@TedShifrin Just got back from a long trip. Let me see what's going on.
Welcome back. But you were here in the room yesterday, so I didn't know :)
Yeah, I think his result is actually correct. Just because you're applying to the vector field $X$ itself. But now I cannot figure out what's wrong with my counterexample.
@TedShifrin I'm sorry. I am not familiar with the $\Phi$s and $T$s. I would need to do a bit of research simply to understand the question.
I hate his $T_p$ notation. It's just the usual space derivative $D$. $\Phi$ is the flow of the vector field $X$.
These people who love the tangent functor. Yuck. Write $Df(p)$ or $Df_p$ for goodness sake.
22:55
Anyhow, all repaired now.
23:40
@TedShifrin Glad to hear it, I was having a bit of supper before taking my dogs for a walk.
23:59
Is there a particular choice of coordinates for describing the Schwarzschild spacetime that is favored for numerical computation/simulation?
Looking at en.wikipedia.org/wiki/… it seems like Lemaitre coordinates are best-suited for that purpose.
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