Mathematics - 2025-02-22

leslie townes

00:37

@Thorgott minor nitpick, but i think where you have I you want to use \Xi

Ben Steffan

00:50

$\overline{\underline{\Xi}}$

it's a mille feuille

Thorgott

I just typo'd the $s$, but I appreciate creative suggestions

Ben

02:45

Hello. I was wondering how to improve my post to reopen it and avoid downvoting? math.stackexchange.com/questions/5037038/…

leslie townes

03:03

ben this is a toughie because the thing about infinite decimals is asked so often it may be hard to ask a question that is both sufficiently detailed to have an answer, and also not be a duplicate of another question

all i have to offer is that it is very easy to conjure up ideas of lining up decimals next to one another in ordered ways that allow for things like "1 and then infinitely many things and then a 2 and then infinitly many things dn then a 5" but none of that has any obvious correspondence to what people do when they are talking about decimal representation of real numbers

so its very easy to conjure up strings of symbols or verbal statements without there automatically being a real number decimal somehow "just corresponding" to it

Ben

06:07

Thanks! :)

ChoMedit

09:08

Hi. could someone give me an advice on improving my post?

https://math.stackexchange.com/questions/5037890/when-does-a-proper-transform-of-real-embedded-submanifold-of-complex-manifold

one potato two potato

10:13

peep

Thomas Finley

11:44

0

Q: Find the Moment of Inertia (MI) of a triangular lamina about its base?

Show that the moment of inertia of a triangular lamina $ABC$ about the base BC is $Mh^2/6$ where $h$ is the height of $\Delta ABC$ and $M$ is it's mass. I tried to solve this problem as follows: First we divide the triangle into 2 right angled triangles by constructing a perpendicular from verte...

I need some help with this :/

youthdoo

16:03

Hello, I don't know where to ask, but how many reputation do you need to be able to click on the score of a post and see how many upvotes/downvotes exactly are casted?

I viewed the table of privileges but cannot find this.

Ben Steffan

math.stackexchange.com/help/privileges/established-user

1k, in other words

psie

Consider a finite-dimensional vector space $X$. Let $X_1$ be a subspace of $X$. Then there's a projection $P$ in $X$ with range $X_1$. I get the existence of a such a $P$. Let $\{u_1,\ldots,u_n\}$ be a basis of $X$ such that $\{u_1,\ldots,u_k\}$ is a basis for $X_1$ (here $k<n$). Then put $$P(c_1u_1+\cdots+c_nu_n)=c_1u_1+\cdots+c_ku_k.$$ Rudin then claims that if $0 < \mathrm{dim} X1 < \mathrm{dim} X$, there are infinitely many projections in $X$ with range $X_1$. Why?

Ben Steffan

@psie that is not true for any ground field (obviously)

psie

@BenSteffan by ground field, do you mean e.g. $\mathbb R$?

Ben Steffan

yes

psie

16:08

hmm, ok

Ben Steffan

so you should first tell us what ground fields are being considered

(it is true for some ground fields but not others; I worded that badly)

psie

Ben Steffan

that's no help whatsoever

find out what rudin means by "vector space"

psie

ok

It seems like his definition of vector space is a subspace of Euclidean space :)

Ben Steffan

shudders

alright, so the ground field is $\mathbb{R}$

psie

16:14

yes

Ben Steffan

good, then the statement is true, and you can get your infinitely many projections by taking the suggested $P$ and scaling in one coordinate :)

psie

@BenSteffan ok 👍 so for instance, if $u_1=e_1$ and $u_2=e_2$ (the standard basis vectors), then $P_1(c_1e_1+c_2e_2)=c_1e_1$ and $P_2(c_1v_1+c_2e_2)=c_1v_1$ where $v_1=2e_1$, are different projections, right?

Ben Steffan

that's a strange way to write it but yes

actually no, that's the same projection

psie

well, they do have the same range, I suspect

Ben Steffan

they're the same map

since you replaced $e_1$ with $v_1$ both in the input and output

try $P_2(c_1 e_1 + c_2 e_2) = 2 c_1 e_1$

psie

16:24

true :) ok

Thorgott

that's not a projection

psie

@Thorgott right, we're running into trouble with $P^2=P$, or?

Ben Steffan

oh, is that a condition

Thorgott

yes

psie

yeah, should have included that. Rudin's definition is $P^2=P$ (in other words, $P(Px)=Px$).

Thorgott

16:33

it's also equivalent to any other reasonable definition of projection

Ben Steffan

fine, then take $u_{k + 1}, \ldots, u_n$ and map them into $X_1$ in a way of your choosing

psie

16:50

ok, thank you!

leslie townes

psie as a side note, i have no idea why rudin or anybody would make such a remark. maybe stick a pin in it, see if he ever uses it later, and if he does, report back on when and how. :)

in an inner product space context you have a concept of orthogonality which gives you a kind of uniqueness for projections (one to one correspondence between subspaces and orthogonal projections). very common for folks to work in that context.

psie

19:02

@leslietownes will do :)

Thorgott

20:19

0

Q: Gluing fiber bundle trivializations along a collar?

The following is discussed after Definition 2.6 in The Homotopy Type of the Cobordism Category by Galatius-Madsen-Tillmann-Weiss. Let $X$ be a smooth manifold without boundary and $a_0,a_1,\varepsilon\colon X\rightarrow\mathbb{R}$ smooth functions such that $a_0(x)\le a_1(x)$ and $\varepsilon(x)>...

differential-topology fiber-bundles

Balarka Sen

20:54

@Thorgott I am a bit unclear on notations, but it appears to me that the map $q := (\pi, f) : W[a_0, a_1] \to X$ satisfies the property that (a) $q$ is submersion onto $X$ restricted to the interior of $W[a_0, a_1]$, (b) $q$ is submersion onto $X$ restricted to the boundary of $W[a_0, a_1]$. Is this accurate?

Thorgott

$(\pi,f)$ is the projection to $X\times(a_0-\varepsilon,a_1+\varepsilon)$, the projection to $X$ is $\pi$

the map $\pi$ is a submersion everywhere, the map $(\pi,f)$ is a submersion "near the boundary", though I'm not sure if it's a submersion when restricted to the boundary exactly

Balarka Sen

I think if you can show that you are in good shape. If $q : (M, \partial M) \to N$ is a proper map and $q|M^\circ : M^\circ \to N$ and $q|\partial M : \partial M \to N$ are both surjective submersions, then $q$ is a fiber bundle.

Thorgott

but I know already that $W[a_0,a_1]\rightarrow X$ is a fiber bundle, the issue is what happens "past the boundary"

Balarka Sen

Don't you want to know whether $(\pi, f) : W[a_0, a_1] \to X \times [a_0, a_1]$ is a fiber bundle or not?

Thorgott

ah, that's your point

then the issue is that we don't know about $(\pi,f)\colon W[a_0,a_1]\rightarrow X\times[a_0,a_1]$ being submersive away from the boundary

all we know that $(\pi,f)$ is submersive near the boundary and that $\pi$ is submersive

Balarka Sen

21:10

Am I right in saying that $(\pi, f)$ maps the interior of $W[a_0, a_1]$ to the interior of $X \times [a_0, a_1]$?

Thorgott

yeah, that's by definition

you can just imagine this stuff as actual intervals to be honest

cause you can always rescale

Balarka Sen

So the only way that $(\pi, f)$, restricted to the interior of $W[a_0, a_1]$, can fail to be submersive is if $f$ has a critical point in the interior of $W[a_0, a_1]$?

This situation seems possible to me. Take $X = \{pt\}$ and $W \subset X \times (-1-\epsilon, 1+\epsilon) \times \Bbb R$ be a circle embedded in a square.

It satisfies all the conditions you are asking for, yet $(\pi, f) : W[-1, 1] \to X \times [-1, 1]$ is not a fiber bundle. It's the standard height function on a circle.

Balarka Sen

21:49

@Thorgott The trivializations of $(\pi, f) : (\pi, f)^{-1}(X \times (a_i-\epsilon, a_i+\epsilon)) \to X \times (a_i-\epsilon, a_i+\epsilon)$ (for $i = 0, 1$) and $\pi : W[a_0, a_1] \to X$ are both obtained from the following fashion:

In the first case, choose local vector fields $\partial/\partial x_1, \cdots, \partial/\partial x_n$ and $\partial/\partial t$, lift them up by the submersivity property of $(\pi, f)$ and then flow along them to generate a trivializing chart. In the second case, use only $\partial/\partial x_1, \cdots, \partial/\partial x_n$, lift them up by the submersivity p…

Essentially following the proof of Ehresmann's theorem. This suggests to me that the trivializations are compatible over $X$. That is, in fact, $\pi : (\pi, f)^{-1}(X \times (a_i- \epsilon, a_i+\epsilon)) \to X$ is a fiber bundle and the trivializations are compatible with that of $\pi : W[a_0, a_1] \to X$.

So they glue to give a fiber bundle $\pi : W \to X$.

And this does not seem to me to have much to do with $(\pi, f) : W \times [a_0, a_1] \to X \times [a_0, a_1]$ being a fiber bundle. As the example above shows, that can be false.

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