Correction at 4:40: transpose and inverse are not the same in SL_2(Z) (transpose=inverse is the defining feature for orthogonal matrices). Edit: again at 6:45. This is matrix _congruence,_ not matrix _similarity._ Edit: and again at 9:22. I'll stop pointing out individual instances (and each case isn't hard to fix) but I think this video might need a complete reworking - the notions of similarity and congruence are completely different, and it's pretty important to keep track of how differently they behave. Love your work overall! Just trying to help it be as useful, high quality, and misconception-free as possible.
He also completely neglected to mention why we don’t worry about the other reduced form 2x^2+2xy+3y^2. (It’s because this represents the other primes, 3 and 7 mod 20.) You don’t get to choose which form you’re represented by in that lemma, you have to eliminate the others by manual computation.
that's true, and the fact that the transpose equals the inverse is necessary to prove the last theorem. the step he skipped which involved Q'(1,0)=Q(p,q) requires the transpose to be the inverse. in that way, we see that T*T-1=I, or (p q r s)*(p r q s) = (p²+q² pr+qs p²+q² pr+qs) = (1 0 1 0). so we get that p²+q²=1 and pr+qs=0. accordingly, when you go from the matrix AQ to AQ', you have that (p q)*(p r q s)=(p²+q² pr+qs)=(1 0), on the left side, and (p q r s)*(p q)=(p²+q² pr+qs)=(1 0) on the right side.
@@beatrizkarwai6763 I actually think that part is fine - just plugging things into the definition, we get Q'(1,0) = (T*(1,0))^t A_Q (T*(1,0)) = (p,r)^t A_Q (p,r) = Q(p,r). So the only problem is that he swapped q and r.
The proof at 36:12 is wrong. The theorem only tells you that p is properly representable by SOME quadratic form with discriminant -20 iff (-5 / p) = 1, but that quadratic form doesn't need to be x² + 5y². Thus you only have an implication here meaning that if p is properly representable by x² + 5y² then p must be 1, 3, 7 or 9 mod 20. You then eliminate the cases 3 and 7 mod 20, but you haven't shown that p is actually representable by x² + 5y² if p = 1 or 9 mod 20. Discriminant -20 quadratic forms have a form class group of size 2, meaning that there are two different kinds of quadratic forms of discriminant -20 which represent different sets of integers. This is related to the fact that Z[-sqrt(5)] is not UFD.
Yay!! I think Michael Penn is teaching what is needed to prove the "hard" part of the if and only if statement in the classification of natural numbers N that can be expressed as the sum of 3 integer squares. He has a great playlist classifying all natural numbers of the form N=(x^2)+(y^2) (sum of two squares) and of the form N=(x^2)+(y^2)+(z^2) (sum of three squares) and proving Lagrange's Four Square Theorem: All natural numbers N can be expressed as N=(x^2)+(y^2)^(z^2)+(w^2), for x,y,z,w in Z. As long as you know a little modular arithmetic, the entire classification is quite manageable and easy to follow, except for one item, namely proving that n=(x^2)+(y^2)+(z^2) iff n CANNOT be represented in the form n=(4^a)*(8k+7) (Legendre's three square theorem from 1798). Michael Penn proves the backward direction by showing that if n CAN be written in that form, then n CANNOT be written as the sum of 3 squares (straightforward enough using induction) but he does not prove the forward direction, saying that it is "hard" and takes a lot of work beyond the scope of an elementary number theory course. I tried to understand the answer and it seems to be caught up in the notion of TERNARY quadratic forms, which share many properties with the BINARY quadratic forms discussed in this video, including the same definitions for "represented by", "equivalence", and "positive definite". There exist reduced representations of positive definite ternary forms as well. Most critically, it (apparently) can be proven that each positive definite ternary quadratic form with determinant 1 is equivalent to a sum of three squares. With that result we likely can prove the required result and to complete the proof of Legendre's three square theorem. I don't quite follow all the details, but I think this must be where we are heading.
Some serious confusion here. The transpose of an element of SL_2 is *NOT* it's inverse, as claimed here. This is easily seen by the counter example: T = ((2, 0), (0, 1/2)). This is its own transpose (it's diagonal) but it's inverse is trivially ((1/2, 0), (0, 2)). So what's happening in this video? Should we be liming ourselves to the case that T is in SO_2(Z)? Or should all the inverses be replaced with transposes? Either way, a lot of the logic breaks down. I fully the understand the odd slip in presenting a lecture, but when the main point is as fundamentally flawed as in here, I really do think that corrections need to be posted.
Why is T transpose equal to T inverse for matrices in SL_2(Z)? Taking your later V+ example, we can try multiplying V+ x V+T [[1, 1], [0, 1]] x [[1, 0], [1, 1]] and we get [[2, 1], [1, 1]] which is clearly not the identity.
The statement of the final theorem is unclear. From the proof, it seems that you mean "THERE EXISTS a quadratic form of discriminant d that properly represents n iff ...", but it could also be read "ANY quadratic form of discriminant d properly represents n iff ..."
Noticed that myself too. The proof is wrong since the previous theorem only tells us that SOME quadratic form properly represents p iff (-5 / p) = 1. Sometimes there are different classes of quadratic forms with a given discriminant which properly represent different set of integers. In the case of d = -20 there are 2 such classes and x² + 5y² happens to be in the class which represents primes which are 1 or 9 mod 20.
I studied these a ton earlier this year because I've been going through branches of math in preparation for my Fundamentals of Engineering exam. Now, some of you may be asking "what does advanced linear algebra have to do with engineering?" Answer is not much past this exam, buuuut I have a huge math itch to scratch and love getting to the bottom of things. That being said, if anyone wants to get into these anymore than just this video, looking to understand where these surfaces come from and how they can be derived or thought of (projective geometry), I can share some resources. The awesome thing that came about from this was how these "forms" can be extended to n-dimensions. Quadratic forms are of degree n=2, linear forms of degree n=1, but then once you get higher and higher into the multi-term/multidimensional forms like cubics (e.g. Axyz + Bxy^2 + Cz^3 + ...) and quartics (e.g. Axyzw + Bx^2w^2 + Cx + ...), mathematicians have only started to look into these. Lots of potential research in this kind of field. Edit: linear forms := n=1, not 3
Correction at 4:40: transpose and inverse are not the same in SL_2(Z) (transpose=inverse is the defining feature for orthogonal matrices).
Edit: again at 6:45. This is matrix _congruence,_ not matrix _similarity._
Edit: and again at 9:22. I'll stop pointing out individual instances (and each case isn't hard to fix) but I think this video might need a complete reworking - the notions of similarity and congruence are completely different, and it's pretty important to keep track of how differently they behave.
Love your work overall! Just trying to help it be as useful, high quality, and misconception-free as possible.
He also completely neglected to mention why we don’t worry about the other reduced form 2x^2+2xy+3y^2. (It’s because this represents the other primes, 3 and 7 mod 20.) You don’t get to choose which form you’re represented by in that lemma, you have to eliminate the others by manual computation.
@@noahtaul thanks for catching this - I actually stopped watching after the first ten minutes haha
Yeah... For inverse matrix you switch the main diagonal and add a minus to the other elements and divide by the determinant which is 1 in this case.
that's true, and the fact that the transpose equals the inverse is necessary to prove the last theorem. the step he skipped which involved Q'(1,0)=Q(p,q) requires the transpose to be the inverse. in that way, we see that T*T-1=I, or (p q r s)*(p r q s) = (p²+q² pr+qs p²+q² pr+qs) = (1 0 1 0). so we get that p²+q²=1 and pr+qs=0. accordingly, when you go from the matrix AQ to AQ', you have that (p q)*(p r q s)=(p²+q² pr+qs)=(1 0), on the left side, and (p q r s)*(p q)=(p²+q² pr+qs)=(1 0) on the right side.
@@beatrizkarwai6763 I actually think that part is fine - just plugging things into the definition, we get Q'(1,0) = (T*(1,0))^t A_Q (T*(1,0)) = (p,r)^t A_Q (p,r) = Q(p,r). So the only problem is that he swapped q and r.
The proof at 36:12 is wrong. The theorem only tells you that p is properly representable by SOME quadratic form with discriminant -20 iff (-5 / p) = 1, but that quadratic form doesn't need to be x² + 5y². Thus you only have an implication here meaning that if p is properly representable by x² + 5y² then p must be 1, 3, 7 or 9 mod 20. You then eliminate the cases 3 and 7 mod 20, but you haven't shown that p is actually representable by x² + 5y² if p = 1 or 9 mod 20.
Discriminant -20 quadratic forms have a form class group of size 2, meaning that there are two different kinds of quadratic forms of discriminant -20 which represent different sets of integers. This is related to the fact that Z[-sqrt(5)] is not UFD.
22:40 it seems that applying U is more useful
23:00 if you apply U then you have to check that -b/2 is between -c and c
Yay!! I think Michael Penn is teaching what is needed to prove the "hard" part of the if and only if statement in the classification of natural numbers N that can be expressed as the sum of 3 integer squares. He has a great playlist classifying all natural numbers of the form N=(x^2)+(y^2) (sum of two squares) and of the form N=(x^2)+(y^2)+(z^2) (sum of three squares) and proving Lagrange's Four Square Theorem: All natural numbers N can be expressed as N=(x^2)+(y^2)^(z^2)+(w^2), for x,y,z,w in Z. As long as you know a little modular arithmetic, the entire classification is quite manageable and easy to follow, except for one item, namely proving that n=(x^2)+(y^2)+(z^2) iff n CANNOT be represented in the form n=(4^a)*(8k+7) (Legendre's three square theorem from 1798). Michael Penn proves the backward direction by showing that if n CAN be written in that form, then n CANNOT be written as the sum of 3 squares (straightforward enough using induction) but he does not prove the forward direction, saying that it is "hard" and takes a lot of work beyond the scope of an elementary number theory course.
I tried to understand the answer and it seems to be caught up in the notion of TERNARY quadratic forms, which share many properties with the BINARY quadratic forms discussed in this video, including the same definitions for "represented by", "equivalence", and "positive definite". There exist reduced representations of positive definite ternary forms as well. Most critically, it (apparently) can be proven that each positive definite ternary quadratic form with determinant 1 is equivalent to a sum of three squares. With that result we likely can prove the required result and to complete the proof of Legendre's three square theorem. I don't quite follow all the details, but I think this must be where we are heading.
That escalated quickly!
Interesting! Do you plan to get into class field theory at some point to figure out which numbers can be represented by a particular quadratic form?
If you are reading this. You are 15:40
We described the procedure to apply U or V(+/-) to modify the matrix - but did we prove that this procedure will eventually stop?
Some serious confusion here. The transpose of an element of SL_2 is *NOT* it's inverse, as claimed here. This is easily seen by the counter example: T = ((2, 0), (0, 1/2)). This is its own transpose (it's diagonal) but it's inverse is trivially ((1/2, 0), (0, 2)). So what's happening in this video? Should we be liming ourselves to the case that T is in SO_2(Z)? Or should all the inverses be replaced with transposes? Either way, a lot of the logic breaks down.
I fully the understand the odd slip in presenting a lecture, but when the main point is as fundamentally flawed as in here, I really do think that corrections need to be posted.
I can't believe how many linear algebra theory mistakes are in this video.
That application of quadratic reciprocity at the end was satisfying.
7:55 For a moment I understood, "if these two things are equal, I will never post a video again." Luckily for everyone, it was not the case. :)
Why is T transpose equal to T inverse for matrices in SL_2(Z)? Taking your later V+ example, we can try multiplying V+ x V+T
[[1, 1], [0, 1]] x [[1, 0], [1, 1]]
and we get
[[2, 1], [1, 1]]
which is clearly not the identity.
This video should be remade.
39:05
I guess there’s no good place to stop (though I can’t blame anyone, this video was kind of unexpected!)
32:00 why is Q(1,0) n and not n+c?
He forgot to write the y^2 in the Q(x, y) = nx^2 + bxy +cy^2.
I c
This man be like a super maths saiyan
I would have said "There are integers p and r, such that Q(p,r)=n", because Q'(1,0)=Q(p,r) and not Q(p,q).
I don t think you have done zeikmondys and lifting the exponent theorem
What
The statement of the final theorem is unclear. From the proof, it seems that you mean "THERE EXISTS a quadratic form of discriminant d that properly represents n iff ...", but it could also be read "ANY quadratic form of discriminant d properly represents n iff ..."
Noticed that myself too. The proof is wrong since the previous theorem only tells us that SOME quadratic form properly represents p iff (-5 / p) = 1. Sometimes there are different classes of quadratic forms with a given discriminant which properly represent different set of integers. In the case of d = -20 there are 2 such classes and x² + 5y² happens to be in the class which represents primes which are 1 or 9 mod 20.
I studied these a ton earlier this year because I've been going through branches of math in preparation for my Fundamentals of Engineering exam. Now, some of you may be asking "what does advanced linear algebra have to do with engineering?" Answer is not much past this exam, buuuut I have a huge math itch to scratch and love getting to the bottom of things.
That being said, if anyone wants to get into these anymore than just this video, looking to understand where these surfaces come from and how they can be derived or thought of (projective geometry), I can share some resources.
The awesome thing that came about from this was how these "forms" can be extended to n-dimensions. Quadratic forms are of degree n=2, linear forms of degree n=1, but then once you get higher and higher into the multi-term/multidimensional forms like cubics (e.g. Axyz + Bxy^2 + Cz^3 + ...) and quartics (e.g. Axyzw + Bx^2w^2 + Cx + ...), mathematicians have only started to look into these. Lots of potential research in this kind of field.
Edit: linear forms := n=1, not 3