@Semiclassical Hey. You don't wanna miss the inequality above. :-)

@MikeMiller $M$ be a smooth manifold, $x_0$ a point on $M$ and $\vec{v}$ a tangent vector on $M$ at $x_0$. Given any smooth function $f$ on $M$, taking directional derivative of $f$ along $\vec{v}$ at $x_0$ spits out a real number. I'm using the word smooth very informally, as I don't know the definition yet, and the only context in which I know what directional derivative of a smooth function over a manifold means, is when the manifold is $\Bbb R^n$.

However, speaking out of intuition, I can bet this map $C(M) \to \Bbb R$ is linear and satisfies the Leibniz rule, as it does so for $M = \Bbb R^n$ too (I expect this to hold as smooth manifolds are by definition locally diffeomorphic to $\Bbb R^n$). Analogizing this, pick an affine variety $V$. $x \in V$ be a point in the variety. Define a tangent vector to be a linear function $k[V]_x \to k$ from the stalk at $x$ satisfying the Leibniz rule.

Collection of these tangent "vectors" gives you the tangent space. But I am most unhappy with this definition, as it doesn't reveal any good geometric/sheafy information about the variety around $x$ - at least not in this formulation.

I have no idea if that ends up working, but it's definitely close to something that does.

Anyway this is a cut-paste definition obtained from copying line-by-line what happens for the analytic case.

Really, @MikeMiller? I'd actually not be surprised if it turns out to be something dumb.

Have you tried any examples? What does it give you for $x^2-y^2$?

at 0

What I really want to do is to look at the stalk of the structure sheaf at $x$, define these hypothetical tangent (co)vectors there, and dualize them. That seems to be the correct way to go about it.

OK, no, let me look.

yeah, something like that should work fine

now that we land onto the world of mathematics from the land of general nonsense, I have no idea what to do. how in the world can I even find out what the linear functions $A_x \to k$ satisfying leibniz rule are, where $A = k[x, y]/(x^2 - y^2)$?

by hand

...

actually, be more explicit and tell me what it means to satisfy leibniz's rule here

$v : k[V]_x \to k$ is a linear function such that $v(f g) = f v(g) + g v(f)$, where $f, g \in k[V]_x$ are germs at $x$.

what does $fv(g)$ mean?

i guess you mean to say $f(x)$. fine.

grr, my bad. i mean $v(fg)(x) = f(x)v(g) + g(x)v(f)$

: - )

anyway that wasn't a learning exercise or whatever i just wanted to clarify what you meant.

right

...and a very e x p a n d e d smiley.

at least it's better than . ` - )

Hello

Hi

@BalarkaSen this is not a terribly hard exercise... there's no need to ellipsis at me

I got to ask, how do you guys study for long periods of time?

with a pen and a paper and lots of amphetamines?

Ineffectively.

@BalarkaSen haha, I mean, after some periods I get headaches

hmm. if $A_x$ is the stalk at $x$, note that leibniz implies $v(fg) = 0$ whenever one of $f, g \in A_x$ are in the maximal ideal of $A_x$ consisting of the germs vanishing at $x$.

meh, that's not relevant and is obviously a true fact.

@Nickolas if you get headaches you stop and do something else

@MikeMiller weird, no idea how to do this. either it's too hard, or probably i am sleepy.

@Balarka: If it's either of those it's the latter.

No need to torcher yourself @Nickolas

@Rigor sure, no need to

ok, I am going to get some sleep. will think about this tomorrow.

Later pal

i guess there's some nasty computations involved

anyway, g'night all

bye

No, there's not. Good night.

Too many people have this "no pain, no gain" attitude.

@Chris'ssistheartist I wont call that a factorization

@IWantToRemainAnonymous Right.

@IWantToRemainAnonymous Does it seem straightforward to you? $$\int _0^1\int _0^1\int _0^1\frac{1}{1+x y+x z+y z} \ dx \ dy \ dz \le 2 \log^3(2)$$

@Chris'ssistheartist Not that much

@IWantToRemainAnonymous OK. Just curious how it seems to the others. I created it a few hours ago.

it seems like it's really hard to find somebody to talk to about nonstandard analysis :3

Hence the "non" :P

@KevinDriscoll I am a bit confused what $\frac{\partial^2 u}{\partial t^2}$ and $\frac{\partial^2 u}{\partial x^2}$ stand for in the wave equation.

To use an analogy, when I studied Newtonian mechanics, we have the derivative of a function x(t) where first order is velocity and second order is acceleration. I get those. I want a similar understanding of the meaning of the derivatives for the wave equation.

So I was asking if you could explain this difference. Please :)