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12:00 AM
No, something belongs, but I don't know how to wedge a symmetric thing of higher degree.
 
Oops, right. I meant $\sum_i \partial_i f \otimes (x_i \wedge x_{i_1} \wedge \cdots \wedge x_{i_{k-2}})$
 
OK, that makes sense.
But alternating sum on $i$?
 
Wondering if I should alternate. Because $df = \sum_i \partial_i f \,dx_i$, no sign.
This is like the exterior differential but there's an $\otimes$
 
OK.
Might need an unexpected sign in the next stage.
 
Let's just see what happens to $f \otimes x_2 \wedge \cdots \wedge x_k \in S W \otimes \wedge^{k-1} W$ under this weirdness. It first goes to $\sum_{i = 2}^k x_i f \otimes x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge x_k$, in $S^2 W \otimes \wedge^{k-2} W$. That goes to...
$\sum_{i = 2}^k \sum_j \partial_j (x_i f) \otimes x_j \wedge x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge x_k$, again in $SW \otimes \wedge^{k-1} W$. How do I simplify this?
 
12:05 AM
Huh?
Oh, right.
I'm slow.
Yeah, so here's why I thought some sign is needed.
 
I'm fast but I make 100 mistakes so we're even
 
Take the case of $f$ a quadratic monomial, of course.
Oh, wait, linear monomial.
So $f=x_k$.
 
Ah yeah, I missed a sign $(-1)^{i-1}$ in the very first thing.
One of the differentials has a sign (Koszul), the other doesn't (exterior)
At least so far
 
I think I'm going to need a martini for this computation. Keep going ;)
 
LOL
Yeah, just take $x_1 \otimes x_2 \wedge \cdots \wedge x_k$, what the hell.
Goes to $x_1 x_2 \otimes x_3 \wedge \cdots \wedge x_k - x_1 x_3 \otimes x_2 \wedge \cdots \wedge x_k + \cdots + (-1)^{k-1} x_1 x_k \otimes x_2 \wedge \cdots \wedge x_{k-1}$
 
12:10 AM
I think $x_k$ is better.
Oh, no, $x_\ell$.
Need the case $\ell\in\{2,\dots,k\}$ and the case $\notin$.
 
ted, i've noticed the same thing. (question from something fairly advanced with lots of very un-advanced confusion.) it's baffling. i could sometimes see it if it's chapter 1 material in a grad course, because who knows where people came from before that. but it's usually more advanced.
 
@TedShifrin Good point I think
 
@Balarka Probably not, but let's see.
@leslie There are also all sorts of high school (etc.) dilettantes trying to read things that they are not qualified to read.
 
i blame the internet for making it way too easy to do the equivalent of, hm, what's that? seems interesting. i think i'll get 20 graduate textbooks on it.
 
@BalarkaSen Goes to $x_2 \otimes x_1 \wedge x_3 \wedge \cdots \wedge x_k + x_1 \otimes x_2 \wedge x_3 \wedge \cdots \wedge x_k - x_3 \otimes x_1 \wedge x_2 \wedge x_4 \wedge \cdots \wedge x_k - \\ x_1 \otimes x_3 \wedge x_2 \wedge x_4 \wedge \cdots \wedge x_k + \cdots + (-1)^k x_k \otimes x_1 \wedge x_2 \wedge \cdots \wedge x_{k-1} + (-1)^k x_1 \otimes x_k \wedge x_2 \wedge \cdots \wedge x_{k-1}$
 
12:14 AM
And I think to some extent college students are taking things they shouldn't be. COVID hasn't helped.
 
Broke it lol
 
AGH @Balarka
 
@TedShifrin It was much nicer to have her home for her passing. She was as comfortable as we could make her and I was cradling her head and petting her neck and ear and her mom was petting her snout. She was licking her mom's hand, which always calmed her.
 
I'm glad it was peaceful and comfortable closure ... Y'all did the right thing for sure. @robjohn
 
We are sad and really miss her, but she is no longer suffering.
 
12:15 AM
Yeah, of course, you will be sad, but you saved her months of pain.
 
Sorry for your loss, @robjohn
 
There was no hope, with two forms of cancer and the decubitus ulcer, there was no way back.
 
@Balarka: I'm losted. So you stuck with your $f=x_1$?
 
@BalarkaSen thanks.
 
Yeah, it gives me something awful.
 
12:17 AM
@TedShifrin yes, there was no doubt, but it is still sad and we miss her greatly.
 
$\wedge$_$\wedge$
 
time to walk the other dog. BBL
 
See ya, @robjohn. Have a nice walk.
 
Take care, @robjohn
 
OK, let's do a small value of $k$ (which is what I meant the first time).
So we don't have infinitely many terms.
2
 
12:18 AM
Haha yeah
Chuckled out loud and it's the crack of dawn here
 
i liked seeing it. it made me turn on chatjax.
it also made me wonder if i was having a stroke.
 
The ducks did it.
 
This isn't even the worst thing I have MathJax'd here
Remember the Lie brackets?
 
I can always refer to a page of a my thesis (which was a variant of a horrendous computation Chern had done in a paper).
 
duck pond was something of a bust today. she complained about having to walk around, then saw that someone brought their pet rabbit to the park. that was all she wanted to look at. then she asked where the dead heron from last week was.
 
12:20 AM
Memories of the dead heron. She still is not walking decently?
 
she is almost normal on flat surfaces but her foot really juts out at a weird angle when she's walking on grass. i don't understand how it could make too big a difference, but it does.
 
Confidence?
Unsure footing, I mean.
 
Found it
Feb 28 '18 at 20:43, by Balarka Sen
$\gamma^X_p[0, t] \cup \gamma^Y_{\gamma^X_p(t)}[0, t] \cup \gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}[0, -t] \cup \gamma^Y_{\gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}(-t)}[0, -t]$
@leslietownes
 
Not Lie brackets, @Balarka. Multi-iterated flows.
 
Lol
 
12:23 AM
But, granted, for that I need a triple martini after this one.
 
one for every index
 
If you throw in more indices, I'll be drunk before you're done.
OK, back to your computation. Can we do $k=2$ or $3$?
 
$k = 3$ seems fine, yeah.
 
I still wonder if your $f$ is one of the indices in the wedge ... what happens there?
 
Here is what I think is correct. Let's call $d_K : W \otimes \wedge^{k-1} W \to S^2 W \otimes \wedge^{k-2} W$ the Koszul differential and $d_E : S^2 W \otimes \wedge^{k-2} W \to W \otimes \wedge^{k-1} W$ is the exterior-like differential. Then $d_E d_K - d_K d_E$ is identity, or some multiple of it
Last I did the computation I got some multiple which messed things up because characteristic
 
12:27 AM
balarka, i'm not even religious, but i'll pray for you.
 
Well, not $d_E d_K - d_K d_E$ where $d_K$ and $d_E$ are as above, but shifted a grade. They're prism operators wrt each other
I think
 
it seems like you're trying to do some kind of russell's paradox. the gamma of all gammas and supserscripts and subscripts that aren't superscripts or subscripts of themselves.
 
Yes, I know how chain homotopy works. All these computations (even $d^2=0$ or $\partial^2=0$) are a nightmare without careful bookkeeping.
 
Yeah
 
Forget that FB memory, @leslie.
 
12:30 AM
Oh and have you seen this @leslietownes
May 10 '20 at 21:56, by Balarka Sen
Here is the confusing calculation, if you care (I will do this with general $n$): We use all the identifications and notations from earlier. The map $f : \Bbb R^n \to \Bbb R$ gives derivative $Tf : T\Bbb R^n \to T\Bbb R$, $Tf(x, u) = (x, Df(x)u)$. Taking derivative again, and using $T^{(2)} \Bbb R^n = \Bbb R^n \times T_0 \Bbb R^n \times T_0 \Bbb R^n \times T_0T_0 \Bbb R^n$, I get $Tf(x, u_1, u_2, v) = (x, Df_x(u_1), Df_x(u_2), D^2f_x(v, u_2))$. Now, using

$$T^{(3)} \Bbb R^n = (\Bbb R^n \times T_0\Bbb R^n) \times (T_0 \Bbb R^n \times T_0T_0\Bbb R^n) \times (T_0 \Bbb R^n \times T_0 T_0 \Bbb
 
Oh, did you ever read Bill Pohl's paper?
I still think you're not doing this right.
 
Nope, let me see
 
LOL
One thing at a time, dammit.
 
balarka weirdly that doesn't look quite as bad. it's still very bad.
 
FINE!
What should I do
 
12:31 AM
@leslie You're running out of town to make costume.
Settle the Koszul.
 
OK let me write those differentials down on a piece of paper and see what I get again
 
I'm too lazy to do any work. I'm a retired bum with no brains.
 
ted at the moment we're thinking a red dress or shapeless felt thing outfitted with some yellow 'seeds,' some sort of green hat, and (to make it scary) a cape from her black cat costume that's got spider webs on it.
 
No one will have any idea what it is, unless she wears a label saying "I'm a scary strawberry."
feline strawberry
But it sounds charming.
 
You all already planning for Halloween?
 
12:33 AM
Not charmin, but maybe...
 
Oh it is nearby isn't it
 
Yes, it is. I'm a bad gay boy — I never get into Halloween, but people around me do and Leslie's munchkin sure does.
 
balarka, yes, although even general purpose stores around here have had halloween stuff in for several weeks. this seems relatively recent to me.
 
I should dress up as a Koszul complex for Halloween
 
Nah, it's like all holidays. Things month(s) ahead.
With wagging arrows, @Balarka. That should be cute.
But a lot of work.
 
12:35 AM
i've always liked slasher film/spooky halloween, the silly party angle has less appeal. but we are going all in this year.
 
@TedShifrin I'll watch a movie or two, maybe. "The Resolution" is a good kind of scary movie if you're into it.
Yeah I don't like the fashion Halloween either
 
Hitchcock's Marnie and The Birds were always scary enough for me.
 
The Birds is excellent
 
my favorite hitchcock is shadow of a doubt, which is a very halloweeny movie. i think i've talked about this.
 
Marnie is a seriously good psychological drama.
 
12:37 AM
I haven't watched that, jotted.
 
the birds and shadow of a doubt are also filmed (or located) in sonoma county, which is a rarity for big feature films and an added plus for me.
suspicion is also great, maybe my favorite cary grant performance. we might watch that before halloween.
 
Boy, I sure derailed this quickly.
 
hahaha
 
I'm composing the differentials on paper
 
Modern music.
Maybe not quite my dad's sort of composition.
 
12:41 AM
Hahah
Something promising emerging
 
holds breath
 
Have to use your favorite fact about homogeneous polynomials now
 
For linear, it's not too deep.
But, yes, I'd expect Euler to be the key here.
 
Yeah.
I just went ahead and did it generally
 
OK. All power to you.
 
12:50 AM
Alright if everything checks out I'm getting that if $d_{i, j} : S^i W \otimes \wedge^j W \to S^{i-1} W \otimes \wedge^{j+1} W$ is the Koszul differential and $\partial_{i, j} : S^i W \otimes \wedge^j W \to S^{i+1} W \otimes \wedge^{j-1} W$ is the exterior-like differential, then $\partial_{i-1, j+1} \circ d_{i, j} + d_{i+1, j-1} \circ \partial_{i, j} = (i + j) \mathbf{1}$.
We'd like to prove $\ker(d_{i, j}) \subseteq \mathrm{im}(d_{i+1, j-1})$ using this, I suppose
 
So is there evidence in any reliable text (e.g., Eisenbud, which I no longer possess) that this exercise is correct?
 
I checked Eisenbud but his Koszul is some strange business, not exactly this exact sequence.
 
So, with your signs, we know that $d^2 = \partial ^2 = 0$?
 
Yeah.
It seems it is crucially important that $i+j$ is invertible in the base ring, but I dunno
At least for the above argument.
 
Yeah, this seems correct to me.
Is there some magic from all the $i+j$ not divisible by the prime?
 
12:56 AM
I guess one can try to cook up a counterexample over $\Bbb F_p$. Maybe with $W = \Bbb F_p^2$. If it goes wrong it has to go wrong at $0 \to S^{p-2}\Bbb F_p^2 \otimes \wedge^2 \Bbb F_p^2 \to S^{p-1} \Bbb F_p^2 \otimes \wedge^1 \Bbb F_p \to S^p \Bbb F_p^2 \to 0$?
Because $p-2 + 2 = p-1 + 1 = p+ 0 = p$, not invertible
 
OK.
Start with $p=2$. :)
 
Haha yeah
 
So we get to use $(x+y)^2 = x^2+y^2$?
 
But that's $W\otimes W = S^2 W \oplus \wedge^2 W$, which I guess is always true?
@TedShifrin That's my guess
Oh no it is not true
How can it be true you need to divide by 2 to symmetrize
 
Good luck with that?
 
12:59 AM
$A = (A + A^T)/2 + (A - A^T)/2$
You get wrecked mod 2 I think
 
Yuppers.
 
C'est fantastique
Need a cigarette lol
 
Bad boy. You quit.
Well, not so fast. Just because that formula fails doesn't mean it's false.
Actually, it does.
Done.
 
Yeah
 
Gotta love false homework. It doesn't offend me at this level :)
 
1:08 AM
The ironic thing is that he corrected a bunch of things for this particular homework a week ago and sent a revised version, added this problem to make up for the wrong ones
And this is also wrong
 
LOL ... I'm glad I was never that sort of prof :)
Within reason it's a good thing at this level.
If you state, Prove or give a counterexample, then all's fair.
 
Yeah.
 
@Balarka if I have a hyperbolic manifold, does the automorphism group have compact isotropy subgroups?
 
That said, I have one or two embarrassingly false things in my algebra book.
@Lukas: I still don't see why your hypotheses make it hyperbolic. It's not just Riemann mapping, is it?
 
@LukasHeger The manifold has to be $\Bbb H^n/G$ so I guess it should follow from isotropy subgroups of $\Bbb H^n$ being compact?
 
1:10 AM
And I think you have compact isotropy for any Riemannian case.
But I might be stooopid, as usual.
@Balarka He's in $n=1$.
 
I'm wondering about the generalization for all $n$ as well
@TedShifrin $\Bbb R^n$ doesn't have compact isotropy groups, does it?
 
Doesn't it?
 
It does, right? It's just O(n)
 
I think Ted is right
 
1:12 AM
hmm
 
It should be true for any Lie group acting by isometries on a Riemannian manifold.
 
So any action by isometries is proper?
 
Myers-Steenrod or something, no?
 
Whoa ... That sounds curvature-dependent.
Isn't it elementary?
Oh, maybe not.
If the manifold is non-compact, then it looks false.
 
Myers-Steenrod is Isom(M) is a Lie group, I'm misremembering. But we need that.
 
1:16 AM
Oh, that's what you're talking about. I don't know this stuff.
 
Me neither, let me think
 
Lie's third theorem fails in infinite dimensions :((
This is the sad thing I learned today.
 
So Lukas is right. I get a (semi-?)direct product of translations with $O(n)$, because I can translate other points there from as far away as I want.
 
When you say isotropy group you mean the stabilizer, no? Stabilizer is still $O(n)$
 
But I don't see why hyperbolic gets you anything here. Positive curvature, yes.
Not in $E(n)$.
 
1:20 AM
what is $E(n)$?
 
Euclidean group.
The group of isometries.
 
Confused. Maybe I'm doing too much algebra. I thought isotropy group means all isometries fixing, say, the origin.
 
Oh, Lukas got me confused.
 
no I think the stabilizer should still be $O(n)$
 
Dammit.
The preimage of the group action is what I'm thinking about.
With regard to Lukas's question about properness.
Yes, we agree on stabilizer.
 
1:22 AM
So if you have a sequence of isometries $f_n$ with $f_n(p) \to q$, I want to prove this sequence has a subsequential limit isometry -- this is exactly what properness means.
 
Why was the properness question raised?
 
Properness implies compact stabilizers
So we're thinking if it's actually just proper
It should be
 
And that's false here ... there are very far-away points that map to the origin.
In what you just wrote, isn't there $p_n$ along with $f_n$?
 
Don't think so. Properness of action of $G$ on $M$ means $G \times M \to M \times M$, $(g, m) \to (gm, m)$ is proper, not that $G \times M \to M$ is proper, remember
 
Oh, ugh, right. I haven't taught this in ages.
"Never mind."
 
1:24 AM
I only remember this because I gave a talk among my friends a few days ago on group actions on manifolds.
 
@TedShifrin the motivation for asking about the hyperbolic case was that the Riemannian manifold was that the Riemannian manifolds for which I have the complex analysis argument happen to have a hyperbolic metric
 
Time for me to go cook dinner?
You should look at Kobayashi's Transformation Groups, @Lukas.
I no longer own that, either.
 
Anyway, thanks for helping me out with Koszul, @TedShifrin. Couldn't have done it without you.
 
Ha ha ha. I did nothing.
@Lukas There's also Kobayashi's book on hyperbolic (complex) manifolds.
 
Not true, the crucial suggestion was to look at $k = 2, n = 2$ and eventually $p = 2$. That's what made all the difference!
 
1:27 AM
Well, I did encourage you to take small numbers, which is my standard sort of comment on here.
You're welcome, a @Balarka :P
 
It's something I should keep in mind
 
Actually all isometry groups have compact stabilizers
it's in Kobayashi
thanks @Ted
 
Yeah, I thought so.
LOL, you're welcome.
 
yeah Ted was right before we confused him
 
nods
 
1:28 AM
That's almost always the case
 
I need another martini and dinner. Take care, guys. Always fun chatting.
 
Take care!
Watch The Resolution someday
 
bye @Ted
 
It's a good one
 
1:29 AM
But curiously Kobayashi doesn't seem to mention whether the action is proper
 
Look carefully :P
You can always also check Kobayashi/Nomizu. Lots of that stuff is in there.
I copied some of their stuff when I taught this.
 
@Ted should I mention you if I point this generalization out in an answer?
@BalarkaSen sounds like an homological algebra movie
 
HAH
 
Apparently the statement of Myers-Steenrod sometimes include compact stabilizers as well
so you were right @Balarka
 
Ah OK cool
 
2:09 AM
@TedShifrin thanks. It was a beautiful sunset.
 
2:32 AM
:)
@LukasHeger Not necessary.
 
2:43 AM
@Ted: So just to nail the coffin since I'm writing the solution: For $M = \Bbb F_2^2$, the exact sequence $0 \to \wedge^2 M \to S^1 M \otimes \wedge^1 M \to S^2 M \to 0$. The first arrow maps the basis vector $x \wedge y$ to $xy \otimes 1$, that's that. The image of this does not contain $x \otimes y + y \otimes x$, which maps to by the second arrow $xy + yx = 2xy = 0$.
One second, first arrow wrong.
$x \wedge y$ maps to $x \otimes y + y \otimes x$, actually, no?
Bit confused
Oh, maybe I should look at $(x + y) \otimes (x + y) - (x \otimes x) - (y \otimes y)$ in the middle term. This isn't in the image of $\wedge^2 M \to S^1 M \otimes \wedge^1 M$, which is spanned by $x \otimes y + y \otimes x$
Still it maps to $(x + y)^2 - x^2 - y^2 = 2xy =0$?
That it?
Sounds good to me, let me know what you think
 
3:18 AM
@BalarkaSen what exactly is this exact sequence? How do you have a map $\wedge^2 M \to M \otimes M$ with no assumptions of char $0$ which allows for antisymmetrization?
 
The map is the Koszul differential, $S^0 M \otimes \wedge^2 M \to S^1 M \otimes \wedge^1 M$, $\alpha \otimes \beta_1 \wedge \beta_2 = \alpha \beta_1 \otimes \beta_2 - \alpha \beta_2 \otimes \beta_1$
Of course, $\alpha$ is a constant here
It's not an exact sequence, just a complex
 
oh okay
 
The big discovery of the day is that the Koszul complex is not exact unless you have some characteristic hypothesis
When I say the Koszul complex I mean the classical Koszul, I don't quite know what it has to do with that regular sequence thing
 
the Koszul complex I know just has exterior powers, no symmetric ones
just an algebraic version of the complex defining de Rham cohomology
 
yeah IDK that stuff
This is like the Lie algebra cohomology complex if you prefer
Or equivariant cohomology, which is obtained from $\Omega^*(M) \otimes S^* \mathfrak{g}$ by taking $G$-invariants
 
3:28 AM
$M \otimes M \to \mathrm{Sym}^2M$ is the usual quotient map?
 
yeah
 
Then I don't see how the sequence you wrote can fail to be exact, at least when $M$ is a finite-dimensional vector space. The fact that $M \otimes M \to \mathrm{Sym}^2 M$ is surjective and has kernel generated by $x \otimes y - y \otimes x$ is just the definition of $\mathrm{Sym}^2M$. So the only thing that could go wrong beyond that is that $\wedge^2 M \to M \otimes M$ is not injective. But that's impossible by dimension counting
I must be missing something
 
How is $(x + y) \otimes (x + y) - (x \otimes x) - (y \otimes y)$ in span of $x \otimes y - y \otimes x$?
The former is also in the kernel of $M^{\otimes 2} \to S^2 M$
$M = \Bbb F_2^2$
 
Be careful. Write $x\otimes y$ as a sum of symmetric and skew.
To Lukas.
 
What is your definition of $S^2M$? The definition I know is literally $M \otimes M / \langle x\otimes y -y\otimes x\rangle$
 
3:42 AM
Mine is not a quotient.
If we’re splitting, we need submodules.
 
if it's not a quotient then what is the map $M \otimes M \to S^2M$?
 
Do you agree or not that $(x + y) \otimes (x + y) - (x \otimes x) - (y \otimes y)$ is in the kernel of that
 
@BalarkaSen can't you multiply that out and get $x \otimes y + y\otimes x$? I feel like I'm being really dumb right now
 
$x \otimes y = y \otimes x$ because $S^2 M$ consists of symmetric tensors, right?
So $2(x \otimes y) = 0$ in char 2
Oh one moment
Something's wrong
 
I thought $x \otimes y + y\otimes x$ is an element of $M \otimes M$ here
 
3:45 AM
Yeah something bizarre is up but I can't focus anymore because it's too early in the morning
I'll think about it and get back to you tomorrow
 
@BalarkaSen if you work with symmetric tensors, not with $M \otimes M / \langle x\otimes y -y\otimes x\rangle$, then I don't see the map $M \otimes M \to S^2 M$
 
$M$ is a free module in my case. $S^2 M$ is degree 2 homogeneous polynomials
So you just send a tensor to the corresponding homogeneous polynomial
It seems right that $0 \to \wedge^2 M \to M \otimes M \to S^2 M \to 0$ should be exact, but it sounds wrong that you can symmetrize a tensor in characteristic 2
Don't really have the brain power to think this through today
 
Lukas, in char 2 your two quotients (sym and alternating) are identical.
Isn’t that a hoot.
Some things are amiss in char 2.
 
@TedShifrin but why? alternating is quotient by $x \otimes x$, symmetric is quotient by $x \otimes y - y \otimes x$
 
boycotts char 2
 
3:51 AM
how do you go from $x \otimes y - y\otimes x$ to $x \otimes x$ without dividing by two?
I'd agree if you use $x \otimes y + y \otimes x$ of course
 
that sounds like the right kind of fix
for me its x o y + y o x = 0
 
Which I do usually.
 
yeah I think you have to work with $x \otimes x$ in char $2$
in char $\neq 2$, these are of course equivalent
 
Hmm …
I’ll have to ponder.
 
@BalarkaSen I don't think that the exactness of $0 \to \wedge^2 M \to M \otimes M \to S^2 M \to 0$ directly gives you a way to symmetrize tensors. That would require you to have a splitting $S^2 M \to M \otimes M$
 
3:55 AM
sounds right
 
4:52 AM
@TedShifrin being ponderous again?
 
5:06 AM
continuing my quest for convex psqs
 
@robjohn If I can’t have a Torrey pine, I settle for Ponderosa.
 
pining for pines?
 
@copper.hat I was working on something... oh, well
 
is the theory behind the product log function complicated? i like those neat solutions wolframalpha gives to differential equations
 
ah sorry, just figured, sorry for being insensitive.
 
5:13 AM
@shintuku You mean the Lambert W function? It's just an inverse, but it is useful.
@copper.hat insensitive? you were just quicker.
 
yeah the lambert W function, wolframalpha outputs closed form solutions to differential equations with it
 
@robjohn as usual i am overthinking...
 
I met Lambert in 1908. He was overrated.
You guys can start working on birch jokes for tomorrow.
 
went out for drinks in berkeley tonight. ended up at eureka! near the berkeley bart station. one of our party forgot her id so we couldn't get into tipper & reed.
 
I know none of these …
 
5:16 AM
neither did i...
the place has changed considerably since i worked downtown in the early 2000's
lots of boba places
 
hm, does the lambert w function have some use for pen and paper calculations or is it mostly used to provide computer results?
 
thankfully arinell's is still there
 
seems like its mostly discussed in numerical analysis books
 
because $xe^x$ appears a lot...
 
@TedShifrin I love birch beer. For a while, it was available in stores here, then it disappeared. I can still get it online.
@shintuku You will find it used in a number of answers on this site.
 
5:24 AM
the birch tree has some celtic symbolism
 
you can invert $x^2e^x$, $x^3e^x$, etc. with Lambert W.
 
cool!
 
I was going to replace my ponderosa with a white birch.
 
@TedShifrin we had two birch trees in our backyard. The drought took care of them.
 
Sigh.
 
5:27 AM
You're going to kill a tree and put in a crappier tree?
 
crappier?
 
You! Where did that nonsensical group theory question come from?
 
@TedShifrin it's a bounty by someone on mian
*main
100 pts
I couldn't do it
 
What does $\Bbb Z^*$ mean?
 
$\Bbb{Z}_p^*$ is what I meant
 
5:29 AM
Oh good grief.
 
basically, prove that it's cyclic, without developing ring or field theory
 
@shintuku: $\operatorname{W}(x)=\sum\limits_{n=1}^\infty\frac{(-n)^{n-1}}{n!}x^n$ for $|x|\lt\frac1e$
 
So I guess you shouldn't use distributivity
 
Yeah, the proofs I know use something.
 
@TedShifrin why not put in a fruit tree?
 
5:33 AM
at robjohn thanks, i'm attempting to apply it to a solution I have
 
I am not growing trees.
 
Put in a native fruit tree :X
Or bamboo :D
 
Here’s a group theory result which will lead to what you want. If $G$ is a finite abelian group, let $m$ be the largest order of its elements. Then the order of every element divides $m$.
 
if $f$ is continuous and $f(x)=f(mx+c)$ for all $x\in R$ where m and c are fixed. Then, f is constant. Is the result true?
 
This is trickier than it seems without theorems on the structure of f.g. Abelian groups.
No @Koro
 
5:40 AM
I forgot to add that m is a constant such that $|m|\ne 1$.
and nothing is known about c
 
Oh, you forgot .
 
:'(
 
Then I don’t know.
 
I know that if we had f is continuous and $f(x)=f(mx)$ for all x, where $|m|\ne 1$, then f is a constant.
 
5:42 AM
@Koro do you have an example of such a continuous function?
 
Yes, consider any constant function. f(x)=5 for all $x\in R$.
@AbstractSpacecraft This can be proven. But the problem is if we have $f(mx+c)$ (that is this c is added extra).
that's what I am trying to analyse.
 
Take an arbitrary continuous function on $[0,1]$ with $f(0)=f(1)$. Use the rule $f(2x+1) = f(x)$ to define it everywhere else.
 
Koro should have said $m\ne 0,1,-1$.
 
5:57 AM
Prof. Ted, that f(2x+1)=f(x) one is an exercise. I'll try that and try to generalize to mx+c.
m=0 is a trivial case. It makes f constant.
 
Always do a concrete case to see what’s happening.
 
So I didn't say $m\ne 0$.
 
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