Is there another way to write the LHS of the formula in the middle of the page here (i.imgur.com/CfIsUMD.png), in terms of the intersection, union, complement, and given symbols, without the element symbol? I don't think the element symbol was previously used in this way in the course I'm following and I'm not certain what the formula means.

@TedShifrin, what did Mike mean by $T+B$ being "equivalent" to $I+T^{-1}B$?

Best guess is that there is an iso (in this case T) such that composition with T gives T+B.

Right, that's what he meant.

@yh05 ask your question directly

I don't see how definition 4.33 and 4.1 coincides when A and x are real numbers. In definition 4.1, x has to be a limit point of E. Where else in definition 4.33, x can be either a limit point of E or not.

I

No, it's a limit point. You have neighborhoods of $x$ that have points other than $x$.

If $x$ is not a limit point, then $V\cap E$ will be just $x$ and so $t\ne x$ means it's vacuous.

If you have an isolated point of the domain, you can't talk about limits as you approach the point.

No, you have your order wrong. You start with a small open set in the range. Then what you say won’t work.

Do the case with $f(0)\ne f(1)$.

Now your artificial game won't work.

A constant function is not a good test.

I have somewhat of a very basic question. Let's say I have some contour integral involving the complex variable $z$. Now if I change my integration variable to $s=z+n$ where $n$ is some real number. The new contour in the $s$ plane that I need to integrate over, that would just be the old contour shifted along the real line by $n$, right?

Yes.

Doesn't my example show that definition 4.33 does not coincide with definition 4.1?

So I should add some assumptions to definition 4.33.

I don't know. I'm done with it.

Idle 3am thought: I wonder if there is a natural number $n>1$ such that $\vert\{G\mid G\text{ is a group},\vert G\vert=n\}\vert=n$

@Rithaniel oeis.org/A000001/b000001.txt suggests no

@xcodeking you just have to remember that others are struggling too, and that the fact that you're in grad school means you're probably alright at what you do

Well, by grad school I mean thesis year of masters

Can anyone help me?

I have a doubt.

@xcodeking sure I'm on my masters too, but the "advice" remains the same lol

Is there any other way to find the value of $\sqrt{11\sqrt{11+\sqrt{11+...4 times}}}$?

I did it by solving one by one sq root

A stupid combinatorics thing I've been trying to figure out: How many length-$k$ ordered sequences of $n$ elements be produced such that all $n$ elements occur at least once? (Here, $k\geq n$.)

Okay, I'm pretty sure I figured it out: You take the number of distinct partitions of ${1,2,\dots,k}$ into $n$ nonempty subsets (i.e., the number of element-index mappings irrespective of element order), and then multiply by $n!$ to account for the possible element orderings. Therefore, the total number is $n!S(n,k)$, where $S(n,k)$ is the Stirling partition number. Does this reasoning look correct?

Or rather, $S(k,n)$ if I use the correct argument order.

Is moduli space just a fancy name for the space of possible geometric solutions or something? I see it mentioned very often in the strings community and find it difficult to comprehend

@MikeMiller thanks! That's the second time I've seen that power series trick show up in functional analysis.

@xcodeking I will explain by way of example. Suppose you want to talk about the "space of all triangles in $\Bbb R^2$". How would you specify a triangle? One way would be to give all of its vertices. So as a naive attempt you might say $$U_{\text{Tri}} = \{(P,Q,R) \in (\Bbb R^2)^3 \mid P,Q,R \text{ do not all lie on a line}\}.$$

You can define that condition as an inequality in terms of the cross product if you really want. Anyway, this is not quite suitable.

When you draw triangles, you will find that you are getting 6 of each triangle --- depending on the way you order the vertices.

So one might say like $\mathcal{Tri} = U_{\text{Tri}}/S_3$. You have quotiented your naive space of triangles by the action of a group which measures the ambiguity --- you can permute the 3 vertices and get the same triangle, so you quotient by that ambiguity, and set your triples of vertices to be equivalent if they differ by a permutation like this.

$\mathcal{Tri}$ is an example of a moduli space. It's a space of geometric objects. When you try to present a geometric object in practice, you will find you have done so in too many ways --- there are different ways of presenting the same geometric object. You then quotient by that, setting equal two different ways of presenting the same geometric object.

A more complicated example: Suppose you wanted to study the space of knots in $\Bbb R^3$. The way you would usually write down a knot is as an injective map $\gamma: S^1 \to \Bbb R^3$.

You could write $\mathcal C_K = \{\gamma: S^1 \to \Bbb R^3 \mid \gamma \text{ is injective}\}.$

But if I parameterize the same loop differently (moving along at a different speed), it still draws the same knot in 3-space.

Two ways of parameterizing the same loop differ as $\gamma_2(t) = \gamma_1(h(t))$, where $h: S^1 \to S^1$ is a map with nonzero derivative. We call such a thing a diffeomorphism; the group of all of these is written $\text{Diff}(S^1)$.

Then the space $\mathcal{Knot} = \mathcal C_K /\text{Diff}(S^1)$. The moduli space of knots is the space of parameterized knots, after you quotient by setting two different parameterizations equal to each other.

@anakhro Sure

@MikeMiller that makes sense, thanks! I need to go over the knots part again, but the triangles thing makes it easier to understand

$$\{a_n\} = \{ 1/2, 2/2, 1/3, 2/3, 3/3, 1/4, 2/4, 3/4, 4/4, \cdots\}$$ to write the general term for this sequence the book writes

for any natural number $j$ $$1+2+\cdots j = \frac{j(j+1)}{2}$$ for an index $n$ written as $n= \frac{j(j+1)}{2} +k $ where $1 \leq k \leq j+1$ $$ a_n= \frac{k}{j+1}$$

Can someone explain me what the book actually did?

@xcodeking The triangle thing might be better. Don't worry about knots

Another example is the Grassmannian, the space of k-dimensional subspaces of an n-dimensional vector space

How do you write a subspace down? By writing down a basis

So you would start with $V(k, n) = \{v_1, \cdots, v_k \mid \text{linearly independent}\}$. But a different basis for the same subspace still gives the same subspace.

so you need to quotient by $V(k,n)/GL_k$ --- any two bases are related by a (k-dimensional) change of basis matrix

Stuff like this

I dealt with it with sporadic intense depression, like (at least) a significant minority of grad students do

The healthy attitude is that this is part of research and dead ends are what eventually lead to ways out of those dead ends (or new projects that aren't stuck)

After all, resolving an issue you're stuck on ends up clarifying the story

In practice I definitely get that it can feel pretty crushing though

@MikeMiller "The healthy attitude is that this is part of research and dead ends are what eventually lead to ways out of those dead ends" -- That definitely helps. Thanks!

@Shaun

:D

Does a manifold formed by two transversal manifolds have any special properties?

That's an interesting question.

In other words, can any manifold be formed as the intersection of transverse manifolds?

Maybe just consider some of the easiest intersections and manifolds?

e.g. cylinders.

Choose wisely, and I think you'll get it!

Okay thanks for the hint I'll think about it

So if I may ask, is this a way to create new manifolds from old ones?

Of course the intersection of transverse submanifolds is an example of a way to create new manifolds from old ones.

But it's not the only way (and not the easiest sometimes).

hmm that is interesting, I'll have to look into it more

@anakhro is there a relevant term or theorem I should search for?

For what?

for creating manifolds from two transversal manifolds

it might be in my manifold book

Guillemin & Pollack should have the proof of the fact that the intersection of transverse submanifolds is again a submanifold.

If that's what you are asking about.

Yeah that helps

also curious if one is dealing with two transverse lorentzian manifolds. the intersection will again be a lorentzian manifold. I don't quite understand how the "new" lorentzian manifold differs from the two transverse ones that form it, if that makes sense.

and I wonder if this is used at all in physics, because of the importance of lorentzian manifolds as models of spacetime