@@kemsekov6331 I know.. Even just "Quaternions make my head spin" is classic. A math geek with a sense of humour. Too bad most people don't get it. BTW When I was teaching Boolean algebra, one of my engineers wore a t-shirt to clas that said only 1 out of 10 people get this. The other half don't get binary.
@@dacomputernerd4096: I was maybe not paying close attention to these details, but off the top of my head, none were noticed. I will have to rewatch though.
@@curtiswfranks Regardless of whether or not your assumption is correct, I believe the point they were making is that *proving it* would become incredibly difficult, rather than simply noticing the pattern to begin with.
@@isospectral3537 I know, I just called it the "1 dimensional sphere" bc someone not versed in differential geometry or topology might not understand what an n-sphere is
@@drpeyam Would there be a number system when you get a circle? It feels like you're going up '2 steps' here, like we're missing something in the middle between the two points and a whole sphere.
Have always loved quaternions from the first time I ran into them so this video was a bit like binge eating comfort brainfood. 7:31 I loved this moment, Dr. Peyam. **does little jump from excitement** "How cool is that?" That is exactly how I react to fun insights into mathematics.
Okay but did it take you "3 tries!" Or 3!=6 tried? Haha Thanks that's a very beautiful result! Direct connection to linear algebra with your layout. Is this where the matrix formulation originated?
This is similar to what we have seen in an optional course our school offered for students in their 11-12th year. It was about matrices and vectors: the expression of the quaternion can be seen as a vector in a 4D space, so it has 4 components. We also have seen the proof of the result of i•j=k using the definition of the vectorial product, since {i, j, k,} are the fundamental unit vectors that they are orthogonal between one another.
@@theodorostsilikis4025 bijectors are in the exterior product of a vector space. Here the vector space he is talking about is an exterior product over the reals^4 mod by a certain easy ideal. Bivectors are vectors.
If you wanted to do more than one calculation, then you could quickly show uv = uxv - u.v (where u, v are imaginary quaternions bi+cj+dk and uxv, u.v are the normal vector and scalar products in E^3), so everything in the future would not involve 16 terms. In particular (a+u)² = (a² - ||u||²) + (2a)u = -1 a=0, ||u||=1 is immediate. Quaternions are cute
@@cooltaylor1015 Quaternions are “4 dimensional numbers” a+bi+cj+dk where a,b,c,d are real numbers and i,j,k are symbols that satisfy i²=j²=k²=ijk=-1. u,v are arbitrary imaginary quaternions bi+cj+dk and fi+gj+hk, and x is the cross written between 3D points to mean the vector product (aka cross product). These appeared in the original comment because they are what I was talking about. Hope that helps.
@@cooltaylor1015 𝓾 and 𝓿 are normal vectors, where 𝓲, 𝓳, 𝓴, are the imaginary parts of a quaternion. × is cross product (𝓾×𝓿), similarly, dot is the dot product (𝓾•𝓿).
The geometric interpretation is interesting. A unit quaternion with 0 as the first element represents a 180 deg rotation. The general form is (0,b,c,d), where the point (b,c,d) on the unit sphere defines the axis of rotation. Any 180 deg rotation, repeated, will produce a 360 deg rotation which is (-1,0,0,0), thus as long as (b,c,d) is on the unit sphere, (0,b,c,d)^2 = (-1,0,0,0) always holds. The other interesting point is that quaternion rotations are equivalent to their own negative, so (1,0,0,0) and (-1,0,0,0) are the same in some sense.
I think you'd really enjoy Clifford algebra/geometric algebra. It generalizes complex numbers, quaternions and in general vectors, with multivectors and a product. It would be really cool if you could also look into Geometric Calculus (the use if geometric algebra for calculus) it's a powerful framework and a very general one (maybe even more than differential form, but I'm unsure about this). I think it's been used in partial diffential equations and computer vision and modeling. Thank you Dr. Peyam for the awesome videos!
Ah, how I love quaternionms! In 2018 I even calculated a formula for a quaternion to the power of another quaternion as a fun exercise, the answer is actually quite neat, it's a scaled triple rotation involving the original imaginary axis plus the axis given by the croos product of the vector parts
I will try get in touch with you so you can do a video on the Octonions and Sedenions as a compilation of quaternion pairs in each octet. We are just finishing a paper about creating the Octonions and sedenions as composite fractals of the quaternions and it appears to predict all three generations of quarks. Anyway this is a great video. I have numerous published papers in this area and this is by far the best explanation of the quaternion manifold I have ever seen. I must admit I see my model clearer now too. Thank you. I hope your are teaching this at a University.
When doing a 2D game in Unity it needs Q's but used weirdly. It's hard to get the rotation correct when 'z' must change but not 'x' or 'y'. It's even harder since sometimes degrees are used and sometimes radians.
So with complex numbers, we get a 0-sphere (two points equidistant from a central point). If we used triternian (?) numbers, we'd get a 1-sphere (circle). Here quaternions yield a 2-sphere (common sphere). So, octonians must yield a 6-sphere (the surface of a 7-dimensional ball). In general, using n-dimensional numbers yields a (n-2)-sphere. Super cool.
This means that x²+1 can be factorised in infinitely many distinct ways, not just as (x+i)(x-i)=(x+j)(x-j)=(x+k)(x-k). Why does the usual unique factorisation of polynomials fail? Lack of commutativity would be my guess.
An important remark is that the property of fields that a polynomial of degree n can not have more than n roots within a field (a field is a division ring with a commutative multiplication, for example the complex numbers) doesn't necessarily hold when you look for roots in a division field with a non-commutative multiplication (you can see it as a field where you omit commutativity of multiplication, for example the quaternions). In fields, you can prove than polynomials of degree n can not have more than n roots by using euclidean division ; but in the case of quaternions, the fact that there can be more roots prove there can not be an euclidean division for polynomials.
I did this same problem using a similar 4x4 layout but kept the signs all positive and preserved the order of the imaginary products, so I had terms in the sum like *(ij)* _bc_ and, in the next row, *(ji)* _cb._ This was done knowing in advance that any terms involving the six possible products of two different imaginary units were all going to cancel out. It meant that any term in my sum which had these two imaginary products in parentheses just got crossed out and whatever was left had to equal -1.
That's exactly correct. Quaternions actually predate the cross product iirc. Deeper insight is found in Geometric Algebra aka Clifford Algebra. There, you have scalars (0D), vectors (1D), bivectors (2D, think of them as oriented plane segments) and trivectors (3D, an oriented volume) and that's it for the 3D geometric algebra. Quaternions are then isomorphic to the even subalgebra, which just contains scalars and bivectors (and sums of them of course). The corr concept in geomtric algebra is the geometric product (aka clifford product), which induces the quaternion product onto the even subalgebra. The cross product on the other hand is the dual of the bivector part of the geometric product of two vectors, but in 3D you can identify vectors with bivectors (they are duals of each other), so the two products are deeply linked.
okay, but "b^2+c^2+d^2 = 1" just says that it is normalized quaternion and 0 is that quaternion represents rotation of pi around any axis, thus squared it gives rotation of 2*pi (full circle around any axis) which is -1 as quaternion
Is there a version of complex ( or quaternion) math, where the axes are completely symmetrical, instead of having one odd dimension (the real axis), where multiplication implies only scale change, but all the other axes imply rotation, as well as scale change?
It simply says that q is a pure unit quaternion of S^2. In other words, there are infinitely many imaginary units, not just i, j and k. Also, it says that the equation x^2+1+0 has infinitely many solutions, and hence the FTA is not true for skew fields.
Yes it was a strange ending to an otherwise excellent video. He embeds a spherical surface, S2, in R3 and then implies it has imaginary axes, which it cannot. It is a kind of argument from analogy which is a bit odd but it's not so hard to tell what he's trying to demonstrate there.
And now the arc from the Simplectic group to the double cover of the SO(3), and why we need half angles in the quarternion formulars for describing rotations.
My introduction to quaternions. Are you saying that the quaternion solution to q^2 = -1 is a three dimensional unit sphere and a point in a 4 dimensional space?
I think it was Hilbert in the 1800s, that proved that no, there is no algebra spanned by 1,I,j. I don’t know the reason why. Quaternions, & then octonians are the next 2 algebras, with additional properties lost as you go along that sequence.
3vector system is not close. This means sometimes you have a 3vector equation where the solution is a 3x3 matrix. Quaternion is a subset of a 4x4 matrix with a unique property that you can do algebra with the members and stay within the quaternion subset.
The closest you can get is what you have likely already thought of: the quaternions minus any one of the three imaginary terms. Suppose that you force it to be the case that d = 0. Then, (a + ib + jc)(a + ib + jc) = -1 a (a + ib + jc) + ib (a + ib + jc) + jc(a + ib + jc) = -1 This will leave you with a² - b² - c² = -1. From this it follows that, a = 0 and b² + c² =1, which is the kind of solution you were after. As @kazedcat has pointed out this form, taken as a system of numbers by itself, is not a good system since it is not closed under the operation of multiplication. Since it doesn't have closure it's not really anything more than a limited case of the quaternions and you are stuck using quaternions to do the trick in one dimension lower.
In a sense, those two points really are just the sphere in the one dimensional imaginary number line. (A sphere being the set of points equidistant from a center) It was a sphere all along hiding in plain sight!
Wait i don't have even a small idea of what is going on with j and k, but according to that arrow scheme on the right you can't commutate with these numbers: ki=j, but ik=-j Did I mess up something or this is just how it goes?
Math does not work this way... It would be broken if it did! If ki=j than ik must also equal j! If that isnt true, you simply arent using math, but a "math like" system of numbers.
@@cooltaylor1015 yeah but he said that if you follow the arrows in one direction it's positive, and in the other it's negative, and the circle went j->i->k->j.., soo
That's kind of my point. His explaination of how the variables work in relation to one another simply doesn't make sense within math as I understand it. It's my ignorance, I am sure, and not the teacher's error. Hell, maybe this isn't math at all. Among the reals, the closest I can come to making sense of that is if all 3 are equal to 0. But then, it's not really positive or negative, and signing it is meaningless. i is almost certainly imaginary. It seems to be implied that j and k might be as well. Perhaps imaginary numbers can somehow break otherwise basic rules of math, and I've just never heard about it before...
@@cooltaylor1015 As you move to higher dimensional numbers, you start to lose properties. In the case of quartorians, the commutative property is lost. Keep in mind that i, j and k are not variables; they are a part of the number.
Is it just a coincidence that the triplet multiplication for quaternions looks very similar to taking the cross product of the bases vectors of a right-handed coordinate system?
I have no idea what quaternions are. Could anyone explain their application and what i, j, and k represent? Edit while watching: Observation All i, j and k all seem to be somehow similar to the square root of minus 1, as if they all had the same magnitude but perhaps in different directions.
There are no shortage of videos on this topic. And yes they are all roots of -1 and each is orthogonal to the other two as you suspected. They are predominantly used in the compounding of rotations in three dimensions. After all, if you are looking for a way to combine two or more rotations in 3-D you quickly run into enormous trouble. Today we might do this using rotation matrices but when quaternions were developed this was not how it was done. There are no shortage of applications in theoretical physics. For instance Hamilton who developed this mathematics, used quaternions for describing the theory of optics. The TH-cam channel Kathy Loves Physics might prove helpful, she has a good history of the discovery and the topic. th-cam.com/video/CdwxpSInhvU/w-d-xo.html
Not enough information for a single solution to either of those equations... But, for example, if a=4, x and y could both equal √2 And b would be (√2³)×2
Without watching the video... q=i (unit imaginary number) I am quite ignorant of the complex part of mathematics... It's likely I am wrong. Here's my thought process. q²=-1 √q²=√-1 q=√-1 √-1=i Therfore q=i Though ive also been tole that i=√-2 . Or that i can be any imaginary number. Contrarywise, Ive been told that √-2=2i . I am not sure which, if any, is correct.
Well, reading the other comments, it seems I solved the wrong problem... I'm still not going to watch the video. But it isn't fair to put an equation in the thumbnail if solving it isn't the object of the video. On the other hand, I've never hears of 'quarternations' before. Perhaps it is an operation implied by 'q'. But if that were the case, using a different script for q so we didn't think it was a variable would have been nice...
Ok. Watched the video u til 4:53, when the simplified equation is put on the board. The whole i×j=k, j×k=i, k×i=j thing is confusing. Epecially when you said doing it the other way results in a negative... I will assume i is imaginary, even though that hasn't been stated thus far. I imaging the math will require that. j and k may be imaginary too... I would have preffered different values of i if that is the case.... Anyway, we have another equation on the board that seems simple enough for me to work with. Let's give it a shot. q²=(a²-b²-c²-d²)=-1 There is also stuff under that I can't tell if it is part of the equation, or further simplification. I'll guess the latter, because I'm ending up with an overly simple q=-1 and not enough information to solve for a, b, c, or d... Putting the rest of the stuff into the equation... q=(a²-b²-c²-d²)+2abi+2acj+2adk=-1+0i+0j+0k Simplifying this will not be easy... The right side is still -1, as everything else equals 0. But I feel like the variables are there for ease of simplification purposes... But since they are all multiplied by 0, how does that work? Algebra class was a long time ago... Lets come from another angle. If adding the variables on the left doesn't change the equation, they all together they must equal 0, and can be ignored. Simply chop them off both sides. I'm ending up with q=-1 again. But this time, at least one of a, b, or c is equal to 0. I can't prove it, but I guess a. I'm terrible at this. I probably did the wrong problem again. Better watch the rest of the video...
Assuming you mean j = √-1. This would be a bit messy in almost every case but it is possible of course. The least messy cases are going to be symmetric 2x2 matrices. Symmetric because all eigenvalues are real and 2x2 because the matrix is as small as possible without being the completely trivial 1x1 matrix. For some matrix A, you can use a spectral (aka eigenvalue) decomposition which goes like A = SES^(-1) And then raise E to the power of √-1. That part is easy because, in the spectral decomposition, E is _a diagonal matrix containing only eigenvalues_ so that matrix raised to any power is just the diagonal elements raised to that power. You then multiply out S E^j S^(-1) to evaluate A^j. If you want a tool online which will just give you the spectral decomposition of any square matrix and which would handle imaginary numbers as powers of diagonal matrices, wolframalpha.com is a good site.
THE BIG QUESTION: OK, a complex number z = a+bi. a quaternion number q = a+bi+cj+dk. But isn't there an in-between number t = a+bi+cj? I was told that this number would be forbidden, unnatural or fake. But why? I was told that this is the number Hamilton started with to describe a 3d space numbers, but couldn't get anywhere until he rejected it in favor of quaternions. But doesn't this number describe a 3d space, while a quaternion describes a 4d space? It gets much more complicated when one gets to octonions and beyond, because there are lots more "missing numbers" in between. The fractal 3d program that has given us all those beautiful fractals is (as far as I know) based on "t numbers," not quaternions. Please, Dr Peyam, do a big video with a good in-depth explanation. Thanks.
...of arbitrary choice, with 2 adjustment operators for dimensional harmony. The fractal lines of the exploding singularity exhausts the 3 operators and is its' spectre.
Use 4D Caterion Cq^2 =(-a-bi^2-cj^2-dk^2)=+h^2=-1 Cq^2 =(+a-bi^2-cj^2-dk^2)=-h^2=+1 Cq^2 =(+ah^2-b^2-cj^2-dk^2)=+i^2=-1 Cq^2 =(-ah^2-b^2-cj^2-dk^2)=-i^2=+1 h^0=+1, h^1=0, h^2=-1, h^3=0 These are rotations about +h,-h,+i,-i.(Real and imaginary spaces. The particle problem in G.R.
Woah woah woah you said i*j = k ≠ -k = j*i so quaternion multiplication isn't commutative? Dang don't you lose a lot of algebraic properties due to that? Also that would prevent the quaternions from being a field or even a ring? I always thought that quaternions were relatively well-behaved...
Quaternions make my head spin. Luckily, I can use them to perfectly describe that rotation.
thats funny.
👏🏾👏🏾👏🏾
Lmao
@@kemsekov6331 I know.. Even just "Quaternions make my head spin" is classic. A math geek with a sense of humour. Too bad most people don't get it. BTW When I was teaching Boolean algebra, one of my engineers wore a t-shirt to clas that said only 1 out of 10 people get this. The other half don't get binary.
haha
Next, do octonions and sedonions. It's much harder, because of the loss of some properties.
It would not just be an (n-1)-sphere?
...onions make me cry. but i like them. :)
@@dacomputernerd4096: I was maybe not paying close attention to these details, but off the top of my head, none were noticed. I will have to rewatch though.
@@curtiswfranks Regardless of whether or not your assumption is correct, I believe the point they were making is that *proving it* would become incredibly difficult, rather than simply noticing the pattern to begin with.
Why is onions sad it makes all of us sad 😭
3 tries or 3!=6 tries?
I LOVE how much fun you're visibly having while doing this. You're in the right line of work.
Thank you!!
In a way, the ±i solution in the complex numbers is just the 1 dimensional sphere, while with the quaternions you get three!
Yep
It's the 0-sphere in the 1D number line. Like the usual sphere is the 2-sphere in 3D space.
@@isospectral3537 I know, I just called it the "1 dimensional sphere" bc someone not versed in differential geometry or topology might not understand what an n-sphere is
@@drpeyam Would there be a number system when you get a circle? It feels like you're going up '2 steps' here, like we're missing something in the middle between the two points and a whole sphere.
Have always loved quaternions from the first time I ran into them so this video was a bit like binge eating comfort brainfood.
7:31 I loved this moment, Dr. Peyam.
**does little jump from excitement**
"How cool is that?"
That is exactly how I react to fun insights into mathematics.
Okay but did it take you "3 tries!" Or 3!=6 tried? Haha
Thanks that's a very beautiful result! Direct connection to linear algebra with your layout. Is this where the matrix formulation originated?
@C&C Studios nah, I joined Ravin. Everything is not a contest, after you've survived long enough
4:24 I followed so much Dr.Peyam that I'm not sure whether you failed 3 or 6 times at recording that sequence
It would have been less ambiguous if Peyam had said "3!!"
@@dlevi67 so 24 times?
@@dlevi67 oh wait, 24=(3!)!, but 3!!=1•3=3 since double factorial is the product of all the integers 1 through n with the same parity as n. I see
@@andrewkarsten5268 😁!
Alright, now this was pretty interesting/elegant solution! Thanks for breaking it down, Dr. Peyam
4:18 3 tries are a lot... I mean 6 right?
Hehe
This is similar to what we have seen in an optional course our school offered for students in their 11-12th year. It was about matrices and vectors: the expression of the quaternion can be seen as a vector in a 4D space, so it has 4 components. We also have seen the proof of the result of i•j=k using the definition of the vectorial product, since {i, j, k,} are the fundamental unit vectors that they are orthogonal between one another.
i,j and k are not vectors.i,j are bivectors and k is a 4-vector in 4 dimentions.
@@theodorostsilikis4025 bijectors are in the exterior product of a vector space. Here the vector space he is talking about is an exterior product over the reals^4 mod by a certain easy ideal. Bivectors are vectors.
You blew my mind at the end with the geometric interpretation. Bless you.
If you wanted to do more than one calculation, then you could quickly show uv = uxv - u.v (where u, v are imaginary quaternions bi+cj+dk and uxv, u.v are the normal vector and scalar products in E^3), so everything in the future would not involve 16 terms. In particular (a+u)² = (a² - ||u||²) + (2a)u = -1 a=0, ||u||=1 is immediate.
Quaternions are cute
Nice!
Where are u, v, and x coming from? I still can't figure out how j and k are relevant to the equation...
Solve for x and y
x^2 +y^2 = a
x^3+y^3 = b
@@cooltaylor1015 Quaternions are “4 dimensional numbers” a+bi+cj+dk where a,b,c,d are real numbers and i,j,k are symbols that satisfy i²=j²=k²=ijk=-1.
u,v are arbitrary imaginary quaternions bi+cj+dk and fi+gj+hk, and x is the cross written between 3D points to mean the vector product (aka cross product). These appeared in the original comment because they are what I was talking about. Hope that helps.
@@cooltaylor1015 𝓾 and 𝓿 are normal vectors, where 𝓲, 𝓳, 𝓴, are the imaginary parts of a quaternion. × is cross product (𝓾×𝓿), similarly, dot is the dot product (𝓾•𝓿).
The geometric interpretation is interesting. A unit quaternion with 0 as the first element represents a 180 deg rotation. The general form is (0,b,c,d), where the point (b,c,d) on the unit sphere defines the axis of rotation. Any 180 deg rotation, repeated, will produce a 360 deg rotation which is (-1,0,0,0), thus as long as (b,c,d) is on the unit sphere, (0,b,c,d)^2 = (-1,0,0,0) always holds.
The other interesting point is that quaternion rotations are equivalent to their own negative, so (1,0,0,0) and (-1,0,0,0) are the same in some sense.
I think you'd really enjoy Clifford algebra/geometric algebra. It generalizes complex numbers, quaternions and in general vectors, with multivectors and a product. It would be really cool if you could also look into Geometric Calculus (the use if geometric algebra for calculus) it's a powerful framework and a very general one (maybe even more than differential form, but I'm unsure about this). I think it's been used in partial diffential equations and computer vision and modeling. Thank you Dr. Peyam for the awesome videos!
Ah, how I love quaternionms! In 2018 I even calculated a formula for a quaternion to the power of another quaternion as a fun exercise, the answer is actually quite neat, it's a scaled triple rotation involving the original imaginary axis plus the axis given by the croos product of the vector parts
I will try get in touch with you so you can do a video on the Octonions and Sedenions as a compilation of quaternion pairs in each octet. We are just finishing a paper about creating the Octonions and sedenions as composite fractals of the quaternions and it appears to predict all three generations of quarks. Anyway this is a great video. I have numerous published papers in this area and this is by far the best explanation of the quaternion manifold I have ever seen. I must admit I see my model clearer now too. Thank you.
I hope your are teaching this at a University.
When doing a 2D game in Unity it needs Q's but used weirdly.
It's hard to get the rotation correct when 'z' must change but not 'x' or 'y'.
It's even harder since sometimes degrees are used and sometimes radians.
So with complex numbers, we get a 0-sphere (two points equidistant from a central point). If we used triternian (?) numbers, we'd get a 1-sphere (circle). Here quaternions yield a 2-sphere (common sphere). So, octonians must yield a 6-sphere (the surface of a 7-dimensional ball). In general, using n-dimensional numbers yields a (n-2)-sphere. Super cool.
Agreed 😁
This means that x²+1 can be factorised in infinitely many distinct ways, not just as (x+i)(x-i)=(x+j)(x-j)=(x+k)(x-k). Why does the usual unique factorisation of polynomials fail? Lack of commutativity would be my guess.
Every unique factorization domain is commutative.
Seit 3 Jahre suche ich diese hübsche demonstration
endlich😇😇
danke du bist verdammt gut
You're so cool Dr Peyam!!
An important remark is that the property of fields that a polynomial of degree n can not have more than n roots within a field (a field is a division ring with a commutative multiplication, for example the complex numbers) doesn't necessarily hold when you look for roots in a division field with a non-commutative multiplication (you can see it as a field where you omit commutativity of multiplication, for example the quaternions). In fields, you can prove than polynomials of degree n can not have more than n roots by using euclidean division ; but in the case of quaternions, the fact that there can be more roots prove there can not be an euclidean division for polynomials.
nice job taking it to a whole new level!
Just amazing 👏
Yay! I've been hoping for quaternions!
Awesome stuff
very clear!
Are the quareternions isomorphic to the lorentz group?
No. It's more proper to say that quaternions of norm 1 are isomorphic to SU(2).
Amazing video as always!
I did this same problem using a similar 4x4 layout but kept the signs all positive and preserved the order of the imaginary products, so I had terms in the sum like *(ij)* _bc_ and, in the next row, *(ji)* _cb._
This was done knowing in advance that any terms involving the six possible products of two different imaginary units were all going to cancel out. It meant that any term in my sum which had these two imaginary products in parentheses just got crossed out and whatever was left had to equal -1.
Seems like multiplication of i,j,k is like a cross product if i,j,k were vectors?
That's exactly correct. Quaternions actually predate the cross product iirc.
Deeper insight is found in Geometric Algebra aka Clifford Algebra. There, you have scalars (0D), vectors (1D), bivectors (2D, think of them as oriented plane segments) and trivectors (3D, an oriented volume) and that's it for the 3D geometric algebra. Quaternions are then isomorphic to the even subalgebra, which just contains scalars and bivectors (and sums of them of course). The corr concept in geomtric algebra is the geometric product (aka clifford product), which induces the quaternion product onto the even subalgebra. The cross product on the other hand is the dual of the bivector part of the geometric product of two vectors, but in 3D you can identify vectors with bivectors (they are duals of each other), so the two products are deeply linked.
You could also read "A history of vector analysis" by Michael J. Crowe; the connections are nicely explained there.
Beautiful!
This video ist great!
Damnit....you piqued my curiousity threshold haha
okay, but "b^2+c^2+d^2 = 1" just says that it is normalized quaternion and 0 is that quaternion represents rotation of pi around any axis, thus squared it gives rotation of 2*pi (full circle around any axis) which is -1 as quaternion
a, b, c, and d could all equal ✓¼
And, I suppose, infinitely many other things. We need more information to solve.
❤Love You Sir❤
4:20 But gladly the sixth try went perfectly! :)
I would guess that you had to try each of the permutations of k, j, and i before you got it right.
Is there a version of complex ( or quaternion) math, where the axes are completely symmetrical, instead of having one odd dimension (the real axis), where multiplication implies only scale change, but all the other axes imply rotation, as well as scale change?
Will there be an octonion
So what surface describes the solution set in *H* for the _cube_ roots (or the nth roots) of unity? 🤔
It simply says that q is a pure unit quaternion of S^2. In other words, there are infinitely many imaginary units, not just i, j and k. Also, it says that the equation x^2+1+0 has infinitely many solutions, and hence the FTA is not true for skew fields.
Ah ha! J and k ARE imaginary! I thought they might be!
Should have used different values of i to make that more clear...
I certainly wasn't expecting that. You should have put in some dramatic sound effect for the big reveal. 🙂
I love the geometric view at the end of the video.
But also, this isn't the square root of -1, it's the solutions to q² = -1, which is different.
@@angeldude101 i = √-1 is the definition of i. It defines the principal branch.
Yes it was a strange ending to an otherwise excellent video. He embeds a spherical surface, S2, in R3 and then implies it has imaginary axes, which it cannot. It is a kind of argument from analogy which is a bit odd but it's not so hard to tell what he's trying to demonstrate there.
And now the arc from the Simplectic group to the double cover of the SO(3), and why we need half angles in the quarternion formulars for describing rotations.
What is a term squared in quaternions?
Oh all of them are -1
My introduction to quaternions. Are you saying that the quaternion solution to q^2 = -1 is a three dimensional unit sphere and a point in a 4 dimensional space?
Yep
Is there any kind of complex like numbers spanned by 1,i,j and if not, why not?
I think it was Hilbert in the 1800s, that proved that no, there is no algebra spanned by 1,I,j. I don’t know the reason why. Quaternions, & then octonians are the next 2 algebras, with additional properties lost as you go along that sequence.
The thumbnail had me flashing back to my college days… 😂
Is it ‘a sphere’? Or is it the sum of all possible rotations around a central point by the three polar vertices of an octahedron?
Now do Octonians!
wow, that is amazing!
And now I won't be able to sleep tonight. Crazy result.
oh yeah another fun video by dr.peyam
😐 Ok, but why the two white lines in the thumbnail? 🤔
So cool!
Ty. ℍ³=1 has been bugging me for awhile.
Is there a number system that results with the solution to -1 being 0 + the unit circle? Maybe a 3 variable number system...
3vector system is not close. This means sometimes you have a 3vector equation where the solution is a 3x3 matrix. Quaternion is a subset of a 4x4 matrix with a unique property that you can do algebra with the members and stay within the quaternion subset.
The closest you can get is what you have likely already thought of: the quaternions minus any one of the three imaginary terms. Suppose that you force it to be the case that d = 0. Then,
(a + ib + jc)(a + ib + jc) = -1
a (a + ib + jc) + ib (a + ib + jc) + jc(a + ib + jc) = -1
This will leave you with a² - b² - c² = -1.
From this it follows that, a = 0 and b² + c² =1, which is the kind of solution you were after.
As @kazedcat has pointed out this form, taken as a system of numbers by itself, is not a good system since it is not closed under the operation of multiplication. Since it doesn't have closure it's not really anything more than a limited case of the quaternions and you are stuck using quaternions to do the trick in one dimension lower.
it took 3 factorial tries to get that right??
so cool!
In a sense, those two points really are just the sphere in the one
dimensional imaginary number line. (A sphere being the set of points equidistant from a center)
It was a sphere all along hiding in plain sight!
Exactly!
3! Or 6 tries?
You’ll never know 😉
The initialization "DK" also stands for "Dead Kennedys"
or Denmark
Never forget Kelvin's opinion of quaternions. Look it up...
Wait i don't have even a small idea of what is going on with j and k, but according to that arrow scheme on the right you can't commutate with these numbers: ki=j, but ik=-j
Did I mess up something or this is just how it goes?
Math does not work this way... It would be broken if it did!
If ki=j than ik must also equal j!
If that isnt true, you simply arent using math, but a "math like" system of numbers.
@@cooltaylor1015 yeah but he said that if you follow the arrows in one direction it's positive, and in the other it's negative, and the circle went j->i->k->j.., soo
That's kind of my point. His explaination of how the variables work in relation to one another simply doesn't make sense within math as I understand it.
It's my ignorance, I am sure, and not the teacher's error. Hell, maybe this isn't math at all.
Among the reals, the closest I can come to making sense of that is if all 3 are equal to 0. But then, it's not really positive or negative, and signing it is meaningless.
i is almost certainly imaginary. It seems to be implied that j and k might be as well.
Perhaps imaginary numbers can somehow break otherwise basic rules of math, and I've just never heard about it before...
@@cooltaylor1015 As you move to higher dimensional numbers, you start to lose properties. In the case of quartorians, the commutative property is lost. Keep in mind that i, j and k are not variables; they are a part of the number.
a cannot not be zero 6:00
wait "3!" tries, sooo, do you mean "3, like really 3" or just 6?
We’ll never know 😉
Is it just a coincidence that the triplet multiplication for quaternions looks very similar to taking the cross product of the bases vectors of a right-handed coordinate system?
I know that they are very different concepts. But I noticed they share the same pattern.
4:23 Haha, did it take 3 turns, or 3! turns?
I guess we’ll never know 😉
Great T-Shirt! :D
how u work with maths if nobody knows u ._. i feel sad
why don't 3d multiplications play nicely with spherical coordinates in the same way that complex numbers do with polar coordinates?
Well 3 > 2
@@drpeyam ok but i don't think that they play that nicely with hyperspherical coordiantes either so what then
2 < 3 < 4
technically we don't see anything
I have no idea what quaternions are. Could anyone explain their application and what i, j, and k represent?
Edit while watching:
Observation
All i, j and k all seem to be somehow similar to the square root of minus 1, as if they all had the same magnitude but perhaps in different directions.
There are no shortage of videos on this topic. And yes they are all roots of -1 and each is orthogonal to the other two as you suspected.
They are predominantly used in the compounding of rotations in three dimensions. After all, if you are looking for a way to combine two or more rotations in 3-D you quickly run into enormous trouble. Today we might do this using rotation matrices but when quaternions were developed this was not how it was done.
There are no shortage of applications in theoretical physics. For instance Hamilton who developed this mathematics, used quaternions for describing the theory of optics.
The TH-cam channel Kathy Loves Physics might prove helpful, she has a good history of the discovery and the topic.
th-cam.com/video/CdwxpSInhvU/w-d-xo.html
It took 3! times = 6?
Or was it just an exclamation?
Can’t completely know if it comes from a mathematician! 😂😂
I'm just going to assume that j²=-1 and k²=-1. I don't know why they do, but the rest makes sense if they do.
Yes that is correct
Solve for x and y
x^2 +y^2 = a
x^3+y^3 = b
Not enough information for a single solution to either of those equations...
But, for example, if a=4, x and y could both equal √2
And b would be (√2³)×2
@@cooltaylor1015 actually this problem has a general solution for 'a' and 'b'. But process is lengthy . There may be a simple technique to solve this.
@@cooltaylor1015 the solution is about two pages, and i have this solution. And looking for others, how they solve this problem
Check the solutions and verify them
Three factorial tries. I'm sure it wouldn't take me anymore than five tries.
6?
Without watching the video...
q=i (unit imaginary number)
I am quite ignorant of the complex part of mathematics... It's likely I am wrong.
Here's my thought process.
q²=-1
√q²=√-1
q=√-1
√-1=i
Therfore q=i
Though ive also been tole that i=√-2 . Or that i can be any imaginary number.
Contrarywise, Ive been told that √-2=2i . I am not sure which, if any, is correct.
Well, reading the other comments, it seems I solved the wrong problem...
I'm still not going to watch the video. But it isn't fair to put an equation in the thumbnail if solving it isn't the object of the video.
On the other hand, I've never hears of 'quarternations' before. Perhaps it is an operation implied by 'q'. But if that were the case, using a different script for q so we didn't think it was a variable would have been nice...
???
You should watch the video, q is standard notation
Ok. Watched the video u til 4:53, when the simplified equation is put on the board.
The whole i×j=k, j×k=i, k×i=j thing is confusing. Epecially when you said doing it the other way results in a negative... I will assume i is imaginary, even though that hasn't been stated thus far. I imaging the math will require that. j and k may be imaginary too... I would have preffered different values of i if that is the case....
Anyway, we have another equation on the board that seems simple enough for me to work with. Let's give it a shot.
q²=(a²-b²-c²-d²)=-1
There is also stuff under that I can't tell if it is part of the equation, or further simplification. I'll guess the latter, because I'm ending up with an overly simple q=-1 and not enough information to solve for a, b, c, or d...
Putting the rest of the stuff into the equation...
q=(a²-b²-c²-d²)+2abi+2acj+2adk=-1+0i+0j+0k
Simplifying this will not be easy... The right side is still -1, as everything else equals 0. But I feel like the variables are there for ease of simplification purposes... But since they are all multiplied by 0, how does that work? Algebra class was a long time ago... Lets come from another angle.
If adding the variables on the left doesn't change the equation, they all together they must equal 0, and can be ignored. Simply chop them off both sides.
I'm ending up with q=-1 again. But this time, at least one of a, b, or c is equal to 0. I can't prove it, but I guess a.
I'm terrible at this. I probably did the wrong problem again. Better watch the rest of the video...
Oops, q² in the equation, not q. So q=i . So I'm definitely still wrong...
You opened my "i"-s.
Okay that was so bad I laughed.
Maths is all fun and games until someone loses an i.
4:19 six times? 😉
Can't wait for you to find the jth power of a matrix
Hahahaha omg
linear algebra
Define j. Or is it just any variable...
@@cooltaylor1015 I _think_ they mean to imply j² = -1.
Assuming you mean j = √-1.
This would be a bit messy in almost every case but it is possible of course. The least messy cases are going to be symmetric 2x2 matrices. Symmetric because all eigenvalues are real and 2x2 because the matrix is as small as possible without being the completely trivial 1x1 matrix.
For some matrix A, you can use a spectral (aka eigenvalue) decomposition which goes like
A = SES^(-1)
And then raise E to the power of √-1.
That part is easy because, in the spectral decomposition, E is _a diagonal matrix containing only eigenvalues_ so that matrix raised to any power is just the diagonal elements raised to that power.
You then multiply out S E^j S^(-1) to evaluate A^j.
If you want a tool online which will just give you the spectral decomposition of any square matrix and which would handle imaginary numbers as powers of diagonal matrices, wolframalpha.com is a good site.
THE BIG QUESTION:
OK, a complex number z = a+bi. a quaternion number q = a+bi+cj+dk. But isn't there an in-between number t = a+bi+cj? I was told that this number would be forbidden, unnatural or fake. But why?
I was told that this is the number Hamilton started with to describe a 3d space numbers, but couldn't get anywhere until he rejected it in favor of quaternions. But doesn't this number describe a 3d space, while a quaternion describes a 4d space?
It gets much more complicated when one gets to octonions and beyond, because there are lots more "missing numbers" in between.
The fractal 3d program that has given us all those beautiful fractals is (as far as I know) based on "t numbers," not quaternions.
Please, Dr Peyam, do a big video with a good in-depth explanation.
Thanks.
No family between complex and quaternion (i,j) ? Where is the circle ? 😺
this reflects my monetary situation. irreversible.
So it took six tries, then?
Poincares' crown has a kingdome.
...of arbitrary choice, with 2 adjustment operators for dimensional harmony. The fractal lines of the exploding singularity exhausts the 3 operators and is its' spectre.
Use 4D Caterion
Cq^2 =(-a-bi^2-cj^2-dk^2)=+h^2=-1
Cq^2 =(+a-bi^2-cj^2-dk^2)=-h^2=+1
Cq^2 =(+ah^2-b^2-cj^2-dk^2)=+i^2=-1
Cq^2 =(-ah^2-b^2-cj^2-dk^2)=-i^2=+1
h^0=+1, h^1=0, h^2=-1, h^3=0
These are rotations about +h,-h,+i,-i.(Real and imaginary spaces.
The particle problem in G.R.
Naice naice naice
4:23 is that 3 times or 3 factorial? LOL
I guess we’ll never know 😉
+cek+cek+bcek-bcek
what about amk and ahk?
?
@@drpeyam CE and BCE are other names for AD and BC.
So... 2cek? The bcek-bcek cancels out
3:25 Gesundheit
Quaternions are non-commutative. Octernions, however, are not even associative.
That thumbail 🤣
Ca ressemble un peu a l'Espace de Minkowsky . P^4 ( X Y Z t )
So... in quaternions the multiplication is not commutative. Disappointing (but not surprising since neither it is in vectors in 3 dimensions or more).
Somehow I understood what he said 😂
Woah woah woah you said i*j = k ≠ -k = j*i so quaternion multiplication isn't commutative? Dang don't you lose a lot of algebraic properties due to that? Also that would prevent the quaternions from being a field or even a ring? I always thought that quaternions were relatively well-behaved...
Yes, that’s the point! We lose commutativity! And worse, for Octonions we lose associativity
And for the sedenions we lose sanity.