Is Veeky Forums as smart as a Ph.D. math student at a top 10 university?

>Is Veeky Forums as smart as a Ph.D. math student at a top 10 university?

You should already know the answer to this.

I guess it depends on how far you want to go with it.

To approach something like this, you definitely need a firm grasp on multivariable analysis, multi-linear algebra, point set topology and differential topology- particularly flows, forms and Lie groups.

You need to know a few ODE and PDE facts regarding solution spaces and solution existence, too. But it's not important to get acquainted with ODEs and PDEs that much before taking on this type of material.

Suppose there is such an embedding. Then the image is compact and thus contained in some open set. Taking the inverse image we would have a single coordinate patch for S^1 which is impossible. Is this correct? I am a 2nd year undergraduate.

So, basically, I'd recommend finishing your analysis book, then reading the first half of Munkres, Chapter 1-15 in Lee's Differential Topology, then attacking Do Carmo's Riem Geo.

Do you have 2 semesters of ODE's? That should be enough to get you through.

>Is this correct?

Mmm... no. This looks like a "proof" that no smooth embedding S^1 to R^n exists (which is not a true statement).

Hint: a correct proof will involve some kind of inequality pushing.

i'm a fucking physics undergrad and i could probably make a poorly argued solution to the first 3. god bless GR

oh i see. distance-preserving forces you to go in a straight line (otherwise the straight line through point A and B is shorter, contradicting distance-preservation). Also clearly the image is contained in the unit ball. But the only lines intersecting the unit balls in a single point are tangets.

>top 10
>University of Michigan

Top 10 for what, OP? Shitty American schools?

Top 10 American math grad school.

>clearly the image is contained in the unit ball.
Why? Any two antipodal points should have distance pi. Those would not be contained in a ball of unit size.

Go back to the mosque yuro.

Can a circle be divided by chords into equal area parts such that no two parts are congruent?

So a top 10 *program*, at a bottom-50% school, in a shit-tier-for-education country? Ranked only by graduate metrics that don't exist anywhere else?

Yeah, we read US News' rankings too, we just all accept them as BS off the bat

Why are you so angry? Did you not get into Umich?

My name is Mike too!

this isn't very related but since there might be some people who know about maths in this thread I thought I'd ask.

Is mathematics inherently biased in favour of the 3rd dimension, R^3 over say R^4 or R^14 ?

Because for example several times I've gotten the impression that there are quite significant things that exist or are valid in R^3 but not in most higher dimensions.

For example the vector cross product is only valid in 3 dimensions and 7 dimension.

so curl also doesn't exist in higher dimensions.

So aren't there a lot of ways in which R^3 is a lot more "rich" and has a lot more maths than higher dimensional spaces?

Isn't that spooky? Because pure mathematics doesn't have anything to do with the universe, yet it just so happens that there is a lot more pure mathematics in R^3 , which happens the spatial dimensions of our universe, than higher dimensional spaces.

I don't really have a good answer to your question, but this lecture on 4-dimensional topology/geometry is very interesting, although Mikhail Gromov is a bit hard to understand.

youtube.com/watch?v=u5DLpAqX4YA

Other mathematicians you might want to read and/or find lectures of if you're interested in this question are Thurston and Milnor.

These I could definitely do if I didn't already have diff. geom. HW to do.
Could probably do.
>8032826
Probably not without using a few resources to help me.

Why is this though? If you have two surfaces that are tangent along a curve, surely parallel transports along that curve will be different, no?

Is it because the cone is flat, i.e. developable? My intuition for parallel transport is pretty bad.

I must not understand something... why doesn't f(x,y) = (x,y,0) work?

you drop 3 and 4 altogether and you study more general things. you study for example R^n, as in properties that all of dimensions share. or you just study metric spaces which are more general

for curl and gradient, these generalize a LOT to differential forms

for cross product, this is just the orthogonal vector. why the hell would it be valid only on n = 3 and 7?

it's not spooky at all, there's not more mathematics in n=3 and nothing you talked about is "pure" mathematics

I did my PhD in applied mathematics at a top 3 university, so I suppose the answer to your question is yes. At least some of us are.

>what are exterior algebra

Physics student here, how the fuck does an oral exam in mathematics work? 2-3 hours? Do you sit there and talk/work out the proof live while this committee drills their eyeballs out?

>waggle e-penis

you present the results and work the proofs, yeah.

why would they drill their eyeballs out?

>so curl also doesn't exist in higher dimensions.
Yes it does. In terms of differential forms.

3-dimensions is not exceptionally special. It is just that vector calculus was developed explicitly for it b/c physics.

In some way's there are very special properties about R^3. In others, no. As other people have mentioned, without to much effort you can define cross products in any dimension (some people have made it seem crazy and it's not). Still, there are some very special things about 3-space (R^3 really isn't any different than R^n). One thing that comes to mind is the Poincare conjecture. It was proven in dimesions greater than 4 much before it was done in 3. This is true for a great number of reasons but one big one is that in higher dimensions, you can use some powerful techniques from topology to just classify the types of maps that exist between things. Trouble is that in 3 dims, many of the arguments break down (for somewhat complicated reasons) and you basically have to contruct the fucking map from a given thing to the 3 sphere (proving poincare). But Hamilton figured out that you could basically heat flow (if you havent had pde's look up the heat equation on wiki and there's a good picture) a given manifold metric (way of measuring distances) until you wound up with a metric that gave it constant curvature (a sphere). Now the cool thing that made this type of argument possible in 3 dims is a special property about the way we measure curvature. There is an end all be all measure of a manifolds curvature called the Reimann tensor. It is a 4 tensor (think of a vector as a 1 tensor and a matrix as a 2 tensor ...) in all dimensions higher that 4. It's a piece of shit to work with. But, if you trace 2 of it's dimensions you get the Ricci tensor which is only a 2 tensor (for tracing, think about how if you multiply a matix and a vector you get a vector, thats sort of like tracing, it's maxrix multiplying something (the metric) down one or more of you dimensions to lower the dimensionality). The ricci tensor is much much easier to work with (since it's basically a matrix) and you can prove lots of good shit about it. T

Turns out in 3D, the Reimann and Ricci tensors are equivalent. SO you can completly describe an objects curvature by its RIcci tensor and not give a fuck about Reimann. This is one of many things that made proof of Poincare in 3d, using the sort of heat flow argument possible. Really, the important this is that you only need a matrix to describe an incredibly important property of manifolds in 3d, and this is not true in higher dimensions, but it doesn't get any worse than 4. I left out a millennium problems worth of details, which happen to be the cool bits. But, I think that this is a good and interesting question, so I thought that I'd give an example from Reimannian geometry of why 3d is special.

Sorry, *special propeties about 3-manifolds, not R^3, fuck R^3

It turns out that there's a very easy way to calculate the Levi-Civita connection on an embedded manifold M that only depends on M's tangent space as a subspace of the ambient manifold.

en.wikipedia.org/wiki/Riemannian_manifold#Riemannian_manifolds_as_metric_spaces

>you drop 3 and 4 altogether and you study more general things.
Not necessarily. There's some very spooky stuff in dimension 4. Check out the smooth Poincare conjecture.

Define 'smart'

>Yeah, we read US News' rankings too, we just all accept them as BS off the bat

Retard alert.

90% of Veeky Forumsentists cant even into basic math and probability

>Is Veeky Forums as smart as a Ph.D. math student at a top 10 university?
>University of Michigan

THinks Michigan is a Top Ten university....you're obviously not as smart as you think you are.

Michigan is ranked #9 for math, dumbass. You're obviously as not as smart as you think you are.

Does it matter enough to argue with the guy? Anyone actually in the math community is aware that Michigan is a very strong department.

>University of Michigan
>top 10

don't make me lel

topuniversities.com/university-rankings/university-subject-rankings/2016/mathematics#sorting=rank region= country= faculty= stars=false search=

Try 17th

>applied math
>smart

Pick one

I feel like this question is a waste of time. Parallel transport is an intrinsic aspect of the geometry and is entirely independent of the embedding. The Christoffel symbol (the thingy responsible for parallel transport) can be expressed in terms of the metric tensor of [math]S^2[/math] and that shows that it's intrinsic.

idk why I said metric tensor of [math]S^2[/math] I of course meant [math]S^1[/math] lol

I pick:

>smart

>It turns out that there's a very easy way to calculate the Levi-Civita connection on an embedded manifold M that only depends on M's tangent space as a subspace of the ambient manifold.

what does this have to do with the "hint"? how does the cone help?

Kek. I'm an undergrad and this is easier than my class's exam.

When I get home I'll post it if the thread is still alive. We had 1 hour to solve it.

>babby's first differential geometry

Well, if you can show that the connection in both spaces is the same along the path c, it should be fairly obvious that the parallel transport along c will also be the same (why? stare at the parallel transport equation [math]\nabla_{c'} V=0[/math]).

Once you've shown what's desired in the hint, ask whether or not there exists a local isometry from C-{just the tip} to E^2, that is, whether or not you can unroll the cone and preserve the metric.

Is there a way to see the connections are the same without laborious calculations? If you need to do the calculation to find the connection on the sphere anyway, then again on the cone, what's the point? Is it any easier than just solving the transport equation on the sphere?

I've seen this before of course, though a long time ago. In retrospect I am wondering what the point is, and I can see a down-side of giving wrong intuition. It seems like the implication is that two surfaces tangent along a curve have the same parallel transport along that curve, and that's wrong.

... googling a bit, I find this

math.stackexchange.com/questions/80270/how-does-parallel-transport-work-on-the-sphere

where "the best answer" demonstrates my concern well.

Any ranking that puts Princeton at #8 and not at #1 is trash. The US News ranking is more accurate.