@Robert Can you upload code for the site?

@fermesomme, using stacks-website and stacks-tools here along with the manifolds repo.

I had to make one minor adjustment to the stacks-tools code because the tags/tags file in the manifolds repo was empty (the stacks one isn't). I'm unsure of how to populate it. But my first guess is that this is the problem.

I'm getting errors executing make tags.

Which specifically?

I just ran the code I merged from you on the server and it ran make tags perfectly.

Also, I could give you root access to my server so you can work on it?

I have to get back to my studying full time and can't devote much time to this :/ as much as I want to get it to work

This is a paste of the output.

@RobertCardona Does make tags populate the tags/tags file?

I see. It assumes it's being run on the server

Not that I'm aware of.

It might be that something is going wrong.

I will have some spare time this friday and saturday. Hopefully, I can work on it.

There are key assumptions going on in places (like having those stacks stylesheets) that will break functionality if they are removed. I spent hours tracking down a bug yesterday.

So what are the current issues that you are facing?

Well, I'm doing every step that made the original stacks project work on the manifolds project, but it's not working.

It's not immediate to me why.

I can see that the database was populated with tags

But when I go to a specific chapter, I can see nothing: example

I should qualify the statement about the database. It has been populated with 'something'

Can you paste the output of tags/tags somewhere?

It's absolutely empty.

Let me change the permissions, perhaps it's not writeable?

Didn't seem to do anything.

Anyway, I'll be stuck in the airport for about 6hrs this Friday. I can make use of the time trying to fix the issues if you can provide me with access to the server.

Will do. Just find a secure way of sending me your email, so I know it's you, and I'll send you the info.

Okay! Try gitk and you can find my email there.

gitk or git-gui in the repo.

ok! great!

Now, I've to leave. Talk to you soon.

See ya!

Site is working: here

Yes, via github

Or you can leave a comment to have someone else do it.

I will add the other problems for chapter one right now

so you can comment individually

@fermesomme, hopefully if you still want to play around with it on Friday, you can remove/rewrite all the text on the site, as it still looks like the stacks project. To me that's just cosmetic and I've gotten the part I wanted down, which allows people to check out individual problems and contribute without knowing git and just minimum latex.

@AlexClark, here are the exercises for the first chapter. You can click on a specific one, say the first and leave a comment/proof/hint, etc.

refresh?

I think the browser cached it. I had the same problem before.

It's not perfect, but it's something.

I'm out for the day. Have to catch up one my studies!

For some strange reason, all the LaTeX heavy chats are not being shown as desired. Just the code is propping up even after refreshing. Do I need to do something to fix this??

@Vishesh: See the starred message on the right about How to use LaTeX in chat.

@ Huy Thanks a lot . I should have been more observant.

does it work now @Vishesh?

Yeah it does...though the rendering is really slow, sure it is just my browser.

what browser are you using?

Chrome

worked fine for me when I was using Chrome

maybe your PC is a bit slow?

its fine now....yeah my laptop...might be also due the fact that there are a lot of windows open, some with MathJax.

@RobertCardona... just a trivial point to add to the discussion about the left multiplication differential. It seems to me that invoking Proposition 2 to write $$ g \dfrac{d}{dt}|_{t=0}(c) = gc'(0) $$ is just cosmetic.

I think so too. It's just be definition, correct?

I didn't understand why the were saying $\mathbb R$-linearity and that other theorem came into play.

@RobertCardona: Did you see my answer?

The $\mathbb{R}$ -linearity is to take $g$ out and allow $\dfrac{d}{dt}$ to operate on $c$ explicitly and Proposition 2 seems to be mentioned only to make sure that we identify the fact that $GL(n)$ is a open submanifold and hence a chart has come into play in the background

@Huy, yes, but I didn't get an aha moment out of it. I'm still thinking about it.

@Huy...yeah force of habit using the double dollar signs...Sorry

Then try writing down matrix multiplication - maybe in terms of "vectors".

@Vishesh, just checked it again, We're trying to show $\frac{d}{dt} \bigg \vert_{t = 0} g c = gc'(0)$. What you wrote down follows from the definition: $\frac{d}{dt} \bigg \vert_{t = 0} c = c'(0)$.

@Huy, I think that helps a bit.

Say I take the case when $n = 2$ and look at $g = \begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{pmatrix}$ and take $c(t) = \begin{pmatrix} c_{11}(t) & c_{12}(t) \\ c_{21}(t) & c_{22}(t)\end{pmatrix}$.

I multiply them in the standard way and then take the derivative with respect to $t$ which is just taking the derivative of each component?

I can then split them up by linearity?

I have trouble getting it to look like the form of Proposition 2.

Aargh...my LaTeX i pretty shoddy. I am not able to delete my previous comment or edit it either

Don't worry about it!

I suppose I'm having trouble seeing what $\frac{\partial}{\partial x^i} \bigg \vert_g$ is for $i = 1, 2, \ldots, n^2$ in the case of $T_g(\text{GL}_n(\mathbb R))$.

How specifically it's a matrix and of what form it is.

Is it just a matrix with 1 in only one entry and zero everywhere else? It can't be because it wouldn't be invertible anymore.

I dont think the representation is the issue. It is the fact that $GL(n, \mathbb{R})$ is not exactly the Euclidean space so there is a chart (just the inclusion map) and so Proposition 2 is in action allowing to write the derivative in terms of the basis vectors.

Why $\dfrac{\partial}{\partial x_1}|_g$?

I'm trying to calculate explicitly what $\frac{d}{dt} \bigg \vert_{t = 0} g \cdot c$ is and put it in terms of $\sum_{i = 1}^{n^2} \dot c^i(0) \frac{\partial}{\partial x^i} \bigg \vert_{c(0)}$ which I know to be $c'(0)$, so I can conclude that the result is correct.

That is the form of the result in Proposition 2 that is referenced by the proof.

@RobertCardona: What's a basis for $\mathbb{R}^{m \times n}$?

'I can see how $\dot c^i$ is just the derivative of the $i$th entry in the matrix (assuming we've laid the matrix out in one line row by row). And I see how that comes up in $\frac{d}{dt} \bigg \vert_{t = 0} gc$, but I then don't know how to split it up appropriately to get it in terms of the $\frac{\partial}{\partial x^i} \bigg \vert_g$.

$n$ columns all zero except for a $1$ at a specific entry?

$(0, \ldots, 0, 1, 0, \ldots, 0)$

Hmm... I guess I am not doing a good job of explaining what I understood. To me it the only reason they seem to mention Proposition 2 is to let you know that you are not in the Euclidean space. Sort of like " this time its ok, but next one may not be so simple, so better understand what actually was done here". Crude way of putting it, I know :)

yes, @RobertCardona, and what is the tangent plane at a certain point of $GL_n(R)$?

I see, but $T_g(\text{GL}_n(\mathbb R))$ is still a vector space $\frac{\partial}{\partial x^i} \bigg \vert_g$ as a possible basis. Each of those elements must have some meaning, correct? Even if we're going through the hassle of looking at it in terms of coordinates.

@Huy, does my answer to Vishesh also answer your question? a linear combination of that basis?

Or "in terms" of that basis.

you need to be more specific

what's the tangent space $T_0(R^n)$?

It's isomorphic to $\mathbb R^n$.

and so that above space is also isomorphic to $\mathbb R^{n^2}$.

yes

So it's just a point in that space.

so you can just take the usual basis for $R^{n^2}$

@RobertCardona i.e. this

Something like $\frac{\partial}{\partial x^{ij}} \bigg \vert_p$ where $i, j$ go through rows and columsn and only take the derivative of that entry?

Going back to the explicit example above: I can calculate $$\frac{d}{dt}\bigg \vert_0 gc(t) = \begin{pmatrix} g_{11} c_{11}'(t) + g_{12}c_{21}'(t) & g_{11} c_{12}'(t) + g_{12} c_{22}'(t) \\ g_{21}c_{11}'(t) + g_{22}c_{21}'(t) & g_{21}c_{12}'(t) + g_{22}c_{22}'(t)\end{pmatrix}.$$

How do I then write that into a sum that looks like $g\sum_{i, j} c_{ij}'(t) \frac{\partial}{\partial x^{ij}} \bigg \vert_0$?

I think I might have it: the key is that we see each $c_{ij}'(t)$ twice in the sum which accounts for both entries above and below? For instance the $g_{11}c'_{11}(t)$ and the $g_{21}c_{11}'(t)$?

No, that doesn't seem to work.

Right now, the way I'm seeing the sum is: $$c_{11}'(t) \begin{pmatrix} 1& 0 \\ 0 & 0 \end{pmatrix} + c_{12}'(t) \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c_{21}'(t) \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + c_{22}'(t) \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$

When I multiply by $g$, I should get the result as above, correct?

I think this is the way I'm supposed to look at it. It uses linearity as well as the Proposition 2.

Thanks both of you for helping! If this is incorrect, let me know :P

@RobertCardona: Those four matrices are precisely your basis vectors.

They are precisely $\frac{\partial}{\partial x^i}$

I think that was the key! To notice that! Thanks!

I understand the importance of linear algebra. It comes up everywhere. But I've always been bad at making simple observations when matrices or determinants come into play.

For instance, on the subreddit there was a recent post about matrices of maximal rank and showing the set is open. The proof doesn't seem obvious to me at all, but I've come across it before and have seen other students come up with the same proof independently (it involves properties of the minors).

@RobertCardona: Probably because didn't have enough practice yet. It comes up everywhere, it'll become as simple as adding two integers. :P

I understand. I agree. It's just slow going for me :/

just work at your own pace and don't let others being better/faster frustrate you. there's always going to be people who are better/faster.

I do that, but it's more difficult when your in a class that uses these tools. One must keep up with the material and that can be a bit more stressful.

Well you noticed it, thats all that matters. Yeah practice is the key. I have the same issue. At any rate if you are slow, you will still have me for company.

I learn things much better on my own, at my own pace. I can stop and think about things for a long time without forcing it. I tend to internalize better that way.

But courses are good for pushing you further, which I think can also be good. I'm very split on this.

In any case, I'll get back to my reading.

Hoping to get through the second chapter of Lee's book on Riemannian manifolds.

And maybe work through a few more problems in Chapter 1 of Smooth Manifolds.

Did you do topological manifolds too?

I think Riemannian will not be very easy to do simultaneously.

requires a good grasp of smooth manifolds rather quickly, but you can do the first chapters probably.

I worked through the first few chapters (quickly, just reviewing). But I've already taken a year long course in algebraic topology, so I feel prettty good with it.

I've worked through a big chunk of Tu's book on manifolds, so I'm somewhat comfortable with it.

ok

I'd really like to take a course in algtop, but I don't have time for it :(

What area are you mainly interested in?

Well, where I come from shockingly I have never had either a course on manifolds or Algebraic topology, but the only course offered was Riemannian Geometry.

@RobertCardona: I used to be very interested in theoretical physics and still am but realized my knowledge in pure maths was too limited to do theoretical physics as rigorous as I wanted to, so I had to stop after I learned QM and GR and go back to diffgeo. Right now I'm doing some more geometric topology and Lie groups.