On Wednesday, I asked several of my students what tools they use to collaborate online on their problem sets. Several of them mentioned Discord. I am currently trying to set up a Discord channel for my class. I imagine I am not the only one in this situation, so I am writing up my progress as I go here . If you have relevant knowledge, please leave an answer to this question or edit mine!
If you have questions to discuss, let’s do that in the comment thread here. And please promote this on twitter and wherever else math teacher’s gather!
I did a dry run with 6 of my students this afternoon. We spent 10 minutes introducing each other to the system, broke into two groups of 3 and spent 10 minutes solving a math problem (problem 19.2 here) and 30 minutes debriefing. Most students found it awkward but workable. Here are things I/we saw as good:
- Connection was reliable; much less dropping out than the video conferencing tools they report their other courses are using.
- The presence of the chat stream with integrated graphics and LaTeX encouraged people to write things down, just like we encourage students to write on blackboards when doing group work in class. This made it easier for me to jump in and out of conversations.
- It was pretty easy for me to jump back and forth from one group to the other. (Not sure how I would do with 4 or more groups, though.)
- We really did solve a nontrivial problem and have a nontrivial conversation about it!
Things we didn’t like
- Both groups needed, at one point, to draw an equation or commutative diagram like this one:
One group did the LaTeX, the other draw a commutative diagram in MS Paint and dropped it into the thread. Both expressed frustration about how much slower this was than drawing a picture in person. - Some students felt that it was awkward not seeing the people they were talking to.
![\mathbb{Q}(\cos \tfrac{2 \pi}{7} ) \cong \mathbb{Q}[x]/f(x) \mathbb{Q}[x] \cong \mathbb{Q}(\cos \tfrac{4 \pi}{7}).](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BQ%7D%28%5Ccos+%5Ctfrac%7B2+%5Cpi%7D%7B7%7D+%29+%5Ccong+%5Cmathbb%7BQ%7D%5Bx%5D%2Ff%28x%29+%5Cmathbb%7BQ%7D%5Bx%5D+%5Ccong+%5Cmathbb%7BQ%7D%28%5Ccos+%5Ctfrac%7B4+%5Cpi%7D%7B7%7D%29.&bg=ffffff&fg=444444&s=0&c=20201002)
-folds just like the one for maps between open subsets of 
.” I think most students probably benefit from seeing things done carefully for a term first.
to 
. Will politely told me that I was the third team to get there — he and Eric Naslund already had the result, as did Fedor Petrov. But I think there might be some expository benefit in writing up the three arguments, to see how they are all really the same trick underneath. 
be a prime power and let 
be the cyclic group of order 
. Let 
be a set which does not contain any three term arithmetic progression, except for the trivial progressions 
. Then ![\displaystyle{|A| \leq 3 | \{(m^1, m^2, \ldots, m^n) \in [0,q-1]^n : \sum m^i \leq n(q-1)/3 \}|.}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B%7CA%7C+%5Cleq+3+%7C+%5C%7B%28m%5E1%2C+m%5E2%2C+%5Cldots%2C+m%5En%29+%5Cin+%5B0%2Cq-1%5D%5En+%3A+%5Csum+m%5Ei+%5Cleq+n%28q-1%29%2F3+%5C%7D%7C.%7D&bg=ffffff&fg=444444&s=0&c=20201002)
. Until recently, all people could prove was bounds like 
, and this is still the case if 
be a list of 
triples in 
such that 
, but 
if 
are not all equal. Then the same bound applies to 
. Once the problem is cast this way, if 
, so our hypotheses are that 
but 
if 
on the set 
which is isomorphic to 
by
. For example, 
and 
.
and 
to have degree 
, then 
is homogenous of degree 
.
.
, we have
.
if and only if 
in 
variables, 
, 
and 
, where 
and 
. Consider the polynomial
.
is a polynomial in 
variables. 
is a polynomial on 
. We identify this domain with 
. Then 
if and only if 
in the group 
.
. In this section, “degree” always has this meaning, not the standard one. The degree of 
and the degree of 
.
, 
or 
has lowest degree. We see that we can write
, 
and 
are monomials of degree 
.
of degree 
, but this is straightforward.
be the polynomial obtained from 
by 
where 
with 
if 
and 
if 
. So 
, but 
. In 
. The polynomial 
nonzero monomials and 
. Replacing 
. Now rerun the argument with 
. Our new bound is three times the number of monomials 
are 
. 
has degree 
. Identify ![[0,p-1]^r](https://s0.wp.com/latex.php?latex=%5B0%2Cp-1%5D%5Er&bg=ffffff&fg=444444&s=0&c=20201002)
with ![[0,q-1]](https://s0.wp.com/latex.php?latex=%5B0%2Cq-1%5D&bg=ffffff&fg=444444&s=0&c=20201002)
by sending 
to 
. We can thus think of ![[0,p-1]^{rn}](https://s0.wp.com/latex.php?latex=%5B0%2Cp-1%5D%5E%7Brn%7D&bg=ffffff&fg=444444&s=0&c=20201002)
as ![[0,q-1]^n](https://s0.wp.com/latex.php?latex=%5B0%2Cq-1%5D%5En&bg=ffffff&fg=444444&s=0&c=20201002)
. We get the bound ![N \leq 3 | \{(m^1, m^2, \ldots, m^n) \in [0,q-1]^n : \sum m^i \leq n(q-1)/3 \}|](https://s0.wp.com/latex.php?latex=N+%5Cleq+3+%7C+%5C%7B%28m%5E1%2C+m%5E2%2C+%5Cldots%2C+m%5En%29+%5Cin+%5B0%2Cq-1%5D%5En+%3A+%5Csum+m%5Ei+%5Cleq+n%28q-1%29%2F3+%5C%7D%7C&bg=ffffff&fg=444444&s=0&c=20201002)
, just as in the prime case.
is a well defined function 
when 
. Moreover, using Lucas again,
by
.
. 
by 
just as before. So 
if and only if 
gives a sum of terms of the form 
. Considering such a term to have “degree” 
, we see that 
. 
, 
or 
, has least degree. We obtain
and 
are products of binomial coefficients and, taking 
, we have 
, 
and 
.![N \leq 3 | \{(m^1, m^2, \ldots, m^n) \in [0,q-1]^n : \sum m^f \leq n(q-1)/3 \}|](https://s0.wp.com/latex.php?latex=N+%5Cleq+3+%7C+%5C%7B%28m%5E1%2C+m%5E2%2C+%5Cldots%2C+m%5En%29+%5Cin+%5B0%2Cq-1%5D%5En+%3A+%5Csum+m%5Ef+%5Cleq+n%28q-1%29%2F3+%5C%7D%7C&bg=ffffff&fg=444444&s=0&c=20201002)
, exactly as before.
. Let 
be the ring of functions 
with pointwise operations, and let ![\mathbb{F}_p[\Gamma]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BF%7D_p%5B%5CGamma%5D&bg=ffffff&fg=444444&s=0&c=20201002)
be the group ring of 
. We think of 
.
, 
, …, 
be generators for 
the functions annihilated by the operators 
where 
. For example, 
is the functions 
which obey 
for any 
, 
and 
. We think of 
as polynomials of degree 
, and the dimension of 
.
by 
and 
otherwise. Define 
by 
. 
, 
and 
for the generators of the three factors in 
.
, then 
as a function on 
.
for 
.
. 
.
be a one-sided ideal in 
. In our case, this is the ideal generated by 
. Then we obtain a bound ![N \leq 3 \dim \mathbb{F}_p[\Gamma]/I](https://s0.wp.com/latex.php?latex=N+%5Cleq+3+%5Cdim+%5Cmathbb%7BF%7D_p%5B%5CGamma%5D%2FI&bg=ffffff&fg=444444&s=0&c=20201002)
for sum free sets in 
variables 
. The formulas to translate between our variables are a mess: For example, my 
is their 
. However, we both agree on what it means to be a polynomial of degree 
.
, as 
. We just don’t know it is that what we’ve found! 
be an abelian group. A subset 
of 
with 
, 
, 
, other than the trivial solutions 
can have size at most 
, which is 
. This was the first upper bound better than 
, and has set off a storm of activity on related questions.
in 
, and we see that it is free of arithmetic progressions if we have 
if and only if 
. So, if 
, then this is the same as a set free of three term arithmetic progressions, but the colored version allows us the freedom to set the three coordinates separately. 
are treated separately, if 
is odd, we may as well replace 
by 
and just require that 
if and only if 
such that
and
.
? We suspect it is the same number as in Ellenberg-Giswijt, but we don’t know!
. 
, almost all the 
, almost all 
zeroes, 
ones and 
twos. The number of such 
, where 
is the entropy 
.
be the probability distribution on 
which maximizes entropy, subject to the constraint that the expected value 
is 
. Then almost all 
copies of 
. I’ll call 
the EG-distribution.
-symmetric probability distribution on 
such that 
was the marginal probability that the first coordinate of 
was 
, 
and 
with probability 
and 
, 
and 
with probability 
, then the resulting distribution on each of the three coordinates is the EG-distribution 
, and we can realize the growth rate of the EG bound for 
does not exist, then we can lower the upper bound! So, here is our result:
be the corresponding marginal distribution, with 
the probabilty that the first coordinate of 
for such a 
. Then
and
on 
, then the marginal distribution is the EG distribution. The problem is about the distributions at the boundary; it seems hard to show that it is always beneficial to perturb inwards.
, there is more than one distribution on 
for some function 
. 
, whereas the bound for sum-free is still 
. But, as you will see, I am chasing much rougher bounds here.
would you vote to convict?
and (2) there is no way this question would have a precise round answer. As you will see, I was quite wrong.
. Can we explain this?
.
. Assuming that 
, and it certainly should be, almost all of the contribution to that sum will come from terms where 
. In that case, 
. And we’ll roughly care about 
such terms. So the odds of acquittal are roughly 
.
to be as small as possible. For 
be as small as possible.
:
; that’s clearly wrong and our approximation that 
, the minimum is where 
.
, 
, 
(since 
is clearly wrong), 
. Holy cow, 
is actually right!
values of 
and the other only occurs for one value. I should have realized two things (1) the bell curve is tightly peaked, so it is really only the 
by the largest it can be.
since it is the largest of the 
entries in a row of Pascal’s triangle which sums to 
.
and 
. We further use the convenient trick of replacing a 
with a 
, up to bounded error to get that the odds of failure are bounded between 
and 
. 
. Since 
, one of the two exponentials in the first case is larger than 
members, 
, so our upper bound predicts only a one percent probability of failure. More accurately computations give 
. So the whole conversation deals with the overly detailed analysis of an unlikely consequence of a bizarre hypothetical event. Fortunately, this is not a problem in the study of Talmud!
Secret Blogging Seminar 