I doubt having isolated fixed point implies yes. What if I take R^2 and a diffeomorphism fixing the origin which sends x-axis to the parabola y = x^2? I feel like that should be cookupable

Arnaud Mortier's comment's suggestion should be false

I don't understand your example.

needs three martinis to deal with today's chatroom

Oh I see, I can perturb the x-axis upwards to move it away from the fixed point....

Actually as long as dim F + dim N < dim M, you can perturb N off of the fixed set so that N has zero fixed points.

where F is the fixed point set.

True I guess

I see why thinking in terms of the diagonal is useful now

Intersections of things are always handy with the diagonal: You intersect the product $A\times B$ with the diagonal to see $A\cap B$.

Not that I see how that helps here.

Hi again Ted

in what sense are Z and N closed?

._.

What do you mean "in what sense"? They are closed subsets of $\Bbb R$.

every limit point of N is a point of N ?

hmm

There are no limit points. So that is true, vacuously.

Or, directly, the complement is visibly open.

that was my comfusing because i could make any ball of radius <1 so no limit points

thanks Ted :D

I don't understand that vacuously statement

i mean why when something does nto occur, we assume the defition is obeyed

like for the case of the empty set

it has nothing in it, so every point in the empty set is an interior point

since there is nothing, one could as well said every point is isolated no ?

@TedShifrin one last thing :D can you tell me what dense is ?

@MikeMiller @TedShifrin @BalarkaSen I was just given the following counterexample: $X:=\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=x^2 \}\cup\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=2x^2 \}$ in which case $T_p X=0$ yet $V=\{(x,y)\mid y=0\}$.

I read the defintion but got nothing to associated with

@Arrow for who ?

@Kasmir: Dense means the closure is everything.

@KasmirKhaan for a question I asked about earlier in this room about the equivalence of two conditions on certain subsets of Euclidean space.

@TedShifrin hmm, so the closure of Q is what?

@Arrow: So they've taken the usual absolute value non-manifold and bent it to have the tangent lines of the two branches be the same.

because I know Q is dense in Rr

@Kasmir: You tell me.

But why is $T_pX=\{0\}$, @Arrow?

Oh, tangent vectors of curves.

Yes

So the only differentiable curve has to stop there, yes.

This is strange to me... Both conditions seem so natural...

OK, nice counterexample :)

You need more differentiable structure to make (2) imply (1).

@Arrow This does not answer your original question about tangent sets, does it? dim T_p X is 0, not 1

I don't see that 1 implies 2 either though

I haven't read your new one

Well, you certainly would want the dimension of $T_pX$ to be the topological dimension, so this is a bad notion.

when we have a bounded set , is the sup of that set, a limit point ?

it seems to be yes for me

Yes, @Kasmir (working in $\Bbb R$, of course).

:D

What's the closure of $\Bbb Q$ in $\Bbb R$?

well hmm

the limit Points of Q is R

because between two reals there is a rational

@BalarkaSen which original question? about characterizing embedded submanifolds?

and vice versa

@Arrow yes

so i say the closure of Q is R

You can approach any real number with a sequence of rationals, for example.

Or any open interval contains rational points.

We gonna start talking about sequences soon

I feel so far all we doing is old analysis but with more regous

rigour*

No, the first one. The fact that you have a curve with a point with a 0-dimensional tangent space is troubling.

I like the second one fine.

I don't follow - wouldn't a zero dimensional tangent space be exactly what you want at a "tip"? Like the tangent set at the point of a cone.

No. I don't like tangent spaces to drop dimensions.

But who the hell knows ...

I'm just saying that the drop in dimension reflects the "non-smoothness" at the origin of the counterexample given.

the Zariski tangent space goes up in dimension at singular points

By the way, you can do a much more natural version of that counterexample. Just do the usual cusp $y^2=x^3$.

Right @Balarka.

Well surely that shouldn't be a differentiable manifold

Oh, but for the cusp the projection to the normal is a homeomorphism. Never mind.

Interesting.

I think the dimension drop reflects that the point is singular and thus that the example should not be a manifold.

I'm used to thinking about singular points being discovered by tangent spaces' going UP in dimension.

If up, why not down :)

What happens to you if we take the nodal cubic $y^2=x^2(x+1)$?

Tangent set's a cone isn't it

So it looks like $y^2=x^2$ near the origin.

two transverse lines

yeah

It's not a vector space

So we get $T_0 X$ the union of two lines.

so it's out of the question

But that's a perfectly lovely singular curve :)

Yes, but not a differentiable manifold

Well, your example is far from that, too.

@Arrow The example you are given is not a differentiable submanifold either

@BalarkaSen @TedShifrin exactly - my point is that a "wrong dimensional" tangent set should reflect whether or not something is a manifold, no?

That is why I don't understand why @TedShifrin prefers the second notion to the first - the second fails to detect the singular point in the example I was given.

Oh, I see, so my cusp example fits your (2) ... but not (1).

Right.

Because I like the geometry of (2) in the smooth setting.

That raises the question of whether (1) implies (2) again

@Arrow Which brings us back to your previous question...

Yes... So is it possible to interpolate a convergent sequence by a curve within $X$..

Oh, what about turning the cusp upside down. Look at the union of two vertical semicircles at a common point.

No, never mind.

As soon as you can cut your finger on the graph the tangent set will have lower dimension (if it's a vector space) I think

Right, cusps make tangent vectors die.

@TedShifrin another thing which bugs me is how to get a local homeomorphism between a neighborhood of $p$ in the tangent space and a neighborhood of $p$ in $X$ given the first condition holds.

I bet we need something like Lipschitz to get a Lipschitz inverse function theorem.

I don't think it works just in the topological category.

Me neither

I think you atleast want locally flat submanifold

hmm Ted, the empty set is open and all the metric X is closed right?

Yeah, that's probably the definition of locally flat.

And vice versa, too, Kasmir.

can you give me something so i belive that?

><

Just the definitions.

Why is the empty set open?

Well, for every $x$ in the empty set, blah ... holds vacuously.

every point of it is interior

Right, cuz there ain't no points.

@BalarkaSen to what would you apply this theorem? The orthogonal projection onto the tangent space?

aha so we can "play"with the defintions

one question one might ask, in proving stuff

Yes, Kasmir. As with beginning linear algebra, here everything is about playing with the definitions.

@Arrow What theorem

we can use 2 things, either emoty set is closed and X is open, or vice versa

so we cant use both at same time

right?

Lipschitz inverse function theorem (sorry, misread the name of the author of the message)