btw this series is incredibly illuminating even for people like me who already studied these basic group theory concepts from a more "visual" point in addition to traditional text books.
You are very quickly becoming one of my favorite math channels, I absolutely love every video you put out. I feel I learn something new every time. I can't wait till you get to GA.
Thanks! You are one of my favorite commenters today 🥰 We are currently publishing the linear algebra series on Patreon. After that, we will move to tensor algebra and then geometric algebra. Those videos will arrive on RUclips somewhere in the spring of 2024 if all goes according to plan.
Conjugacy is so interesting! It feels mysterious how this behavior only appears in non-commutative systems. To transform the system and do operations is to do the same in the original perspective along as you switch each of those operations with their conjugates.
Now that this series is covering topics that weren’t in my undergrad abstract algebra class, I’m starting to get really into. Conjugacy is such a cool idea :)
Yeah, and it's also an idea that comes back in many different areas of math, such as matrix similarity. We will talk about this again in the series on linear algebra.
Thank you for the video! I was always a bit confused as to the role commutativity played in algebra. This explanation really helped clear up some of my doubts without being bogged down in too much detail! I love your channel! Looking forward to more videos :)
Most interesting. Congugacy was where I lost the thread in group theory years ago. It was given as stuff to memorize without motivation which never works for me. Thank you
I'm not sure if you are aware but there is a great, obscure book "Bypasses - simple approach to complexity" by Z.A. Melzak that takes conjugacy application to the next (meta) level.
I'm honestly not sure. I've been trying to understand adjoints for some time now, but I can't wrap my mind around them yet. They have many applications and are very flexible & generic, so I hope to figure them out some day.
I don't know. I'm not sure if the individual parts of a quaternion satisfy the requirements for a group. Only group elements can be divided into conjugacy classes.
@@AllAnglesMath each of imaginary component has unit circle and each circle is orthogonal to others two. and each equivalence relation the value on a cross of two unit circles at the axis, no?
@@DeathSugarThe quaternions do not form a group under multiplication. The only mathematical structure which is a group under both addition and multiplication is the trivial ring. However, the nonzero quaternions do form a group under multiplication. The basis elements of the algebra anti-commute, so you could say this group is "anti-Abelian," although I believe this terminology does not actually exist in mathematics research, because it is not particularly useful. In any case, here is some exploration: to start with: we need understand how to invert a nonzero quaternion. As it happens, you do it equivalently as with the complex numbers: for some nonzero quaternion q, q^(-1) = q*/|q|^2, where q* is the quaternion conjugate, and |q| is the Euclidean absolute value. Incidentally, this means i^(-1) = -i, and the same holds of j and k. Hence, the conjugacy class of i is the set {qiq^(-1) in H\{0} : q in H\{0}}. Now, q = Re(q) + I(q)i + J(q)j + K(q)k, so q* = Re(q) - I(q)i - J(q)j - K(q)k, and |q|^2 is real, so it commutes with all quaternions. Therefore, qiq^(-1) = (Re(q) + I(q)i + J(q)j + K(q)k)i(Re(q) - I(q)i - J(q)j - K(q)k)/|q|^2 = ((Re(q) + I(q)i + J(q)j + K(q)k)(Re - I(q)i + J(q)j + K(q)k)/|q|^2)i = (q(q - 2I(q)i)/|q|^2)i. Analogous results hold for j and k. You can simplify this down to ((q^2 - 2I(q)qi)/|q|^2)i = ((q/|q|)^2 - 2I(q/|q|)(q/|q|))i. Let ρ = q/|q|, where ρ denotes a unit quaternion. Hence, one has (ρ^2 - 2I(ρ)ρ)i for all ρ. You can solve the equations ρ^2 - 2I(ρ)ρ = k and ρ^2 - 2I(ρ)ρ = j, and if they do have solutions, then this means i, j, k all have the smae conjugacy class. This is most likely the case.
Wowwww!! What an amazing video! So cleanly presented, well argued, and I feel like it finally made some really important concepts click for me! I've always struggled with the group theory concept of "conjugacy/conjugates" and the intuitive meaning of "conjugacy classes" of group elements, and this video was incredibly insightful in all ways. I'm deeply grateful! In particular, I had no idea that if two elements are "conjugate to each other" (by the action(?) of some other group element), they... am I understanding right that in some sense they can be seen to represent "the same action" but just "seen from a different basis" so to speak? Does the set of all conjugacy classes of a given group (i.e. equivalence classes under the equivalence relation of "X is conjugate to Y under sandwiching with some element g"), actually represent in some sense the set of "actually/essentially/truly distinct group actions", once we account for the confounding effects of "switching perspectives"/"changing basis"? Hmmmm... I still have a lot to learn, but the idea is fascinating... and seems like it could be a road to something really powerful!
Thank you for your very positive and enthusiastic comment. You are right that conjugate elements are "the same", just seen from a different "basis"'. The wikipedia page for "conjugacy class" literally says that 2 such elements "cannot be distinguished by using only the group structure", and I think that's what they mean. We will look at matrix similarity in the next series, which is basically the exact same concept for matrices. There, we will see that 2 matrices are similar/conjugate when they represent exactly the same underlying linear transformation, but just seen in 2 different bases.
@@AllAnglesMath Similarity, equivalence, symmetry implies duality! Subgroups are dual to subfields -- the Galois correspondence. Lie groups are dual to Lie algebras. Domains are dual to co-domains. "Always two there are" -- Yoda. Real is dual to imaginary -- complex numbers are dual. Conjugacy is duality!
You always mention that the cosets other than the subgroup itself aren't subgroups because they don't contain the identity. I think there's a more "basic" sense in which they're not groups, and the lack of identity can be "derived" from it: The cosets aren't subgroups simply because they are not closed under the group operation. If they were, you could multiply an element by itself until you obtained the identity (and past that, the element's inverse too); at least for finite groups/cosets.
Good point. Perhaps this is indeed a more "basic" reason. However, the lack of an identity is easier to check visually, so it works better for educational purposes.
Subgroups are dual to subfields -- the Galois correspondence. Lie groups are dual to Lie algebras. Domains are dual to co-domains. "Always two there are" -- Yoda. Real is dual to imaginary -- complex numbers are dual. Conjugacy is duality!
Lie groups are dual to Lie algebras. Domains are dual to co-domains. "Always two there are" -- Yoda. Real is dual to imaginary -- complex numbers are dual.
btw this series is incredibly illuminating even for people like me who already studied these basic group theory concepts from a more "visual" point in addition to traditional text books.
Very glad to hear that we can offer additional value with our slightly different way of looking at things. Thanks for the positive comment!
You are very quickly becoming one of my favorite math channels, I absolutely love every video you put out. I feel I learn something new every time. I can't wait till you get to GA.
Thanks! You are one of my favorite commenters today 🥰
We are currently publishing the linear algebra series on Patreon. After that, we will move to tensor algebra and then geometric algebra. Those videos will arrive on RUclips somewhere in the spring of 2024 if all goes according to plan.
Conjugacy is so interesting! It feels mysterious how this behavior only appears in non-commutative systems. To transform the system and do operations is to do the same in the original perspective along as you switch each of those operations with their conjugates.
This is incredibly well made, thank you. I have never understood groups so well until now.
Thank you so much for this positive comment!
veryveryveryveryveryvery good stuff thank you. please keep going with this.
We intend to keep going as long as possible.
First learned about conjugates from Rubik's cube algorithm theory, guess it makes sense that it comes from group theory.
Now that this series is covering topics that weren’t in my undergrad abstract algebra class, I’m starting to get really into. Conjugacy is such a cool idea :)
Yeah, and it's also an idea that comes back in many different areas of math, such as matrix similarity. We will talk about this again in the series on linear algebra.
20:30 Connecting it to matrix diagonalization was so insightful. Thank You!
We will talk about this much more in the upcoming series on linear algebra.
This is easily the best explanation of conjugacy that I've ever seen. Thanks :)
Thank you so much for this comment!
Thank you for the video! I was always a bit confused as to the role commutativity played in algebra. This explanation really helped clear up some of my doubts without being bogged down in too much detail! I love your channel! Looking forward to more videos :)
Most interesting. Congugacy was where I lost the thread in group theory years ago. It was given as stuff to memorize without motivation which never works for me.
Thank you
It is interesting how many concepts in mathematics and even physics arise naturally in group theory
I'm not sure if you are aware but there is a great, obscure book "Bypasses - simple approach to complexity" by Z.A. Melzak that takes conjugacy application to the next (meta) level.
Thanks for the tip, I'll see if I can get my hands on a copy.
@@AllAnglesMath in case of problem with obtaining a book - library genesis (libgen)
Wish I had this series in college
Thank you!
Are conjugates an example of adjoint pairs from category theory?
I'm honestly not sure. I've been trying to understand adjoints for some time now, but I can't wrap my mind around them yet. They have many applications and are very flexible & generic, so I hope to figure them out some day.
So does it mean that quaternion's imaginary parts fall into different conjugacy class and that makes them into weird equivalence relation?
I don't know. I'm not sure if the individual parts of a quaternion satisfy the requirements for a group. Only group elements can be divided into conjugacy classes.
@@AllAnglesMath each of imaginary component has unit circle and each circle is orthogonal to others two. and each equivalence relation the value on a cross of two unit circles at the axis, no?
@@DeathSugarThe quaternions do not form a group under multiplication. The only mathematical structure which is a group under both addition and multiplication is the trivial ring.
However, the nonzero quaternions do form a group under multiplication. The basis elements of the algebra anti-commute, so you could say this group is "anti-Abelian," although I believe this terminology does not actually exist in mathematics research, because it is not particularly useful.
In any case, here is some exploration: to start with: we need understand how to invert a nonzero quaternion. As it happens, you do it equivalently as with the complex numbers: for some nonzero quaternion q, q^(-1) = q*/|q|^2, where q* is the quaternion conjugate, and |q| is the Euclidean absolute value. Incidentally, this means i^(-1) = -i, and the same holds of j and k. Hence, the conjugacy class of i is the set {qiq^(-1) in H\{0} : q in H\{0}}. Now, q = Re(q) + I(q)i + J(q)j + K(q)k, so q* = Re(q) - I(q)i - J(q)j - K(q)k, and |q|^2 is real, so it commutes with all quaternions. Therefore, qiq^(-1) = (Re(q) + I(q)i + J(q)j + K(q)k)i(Re(q) - I(q)i - J(q)j - K(q)k)/|q|^2 = ((Re(q) + I(q)i + J(q)j + K(q)k)(Re - I(q)i + J(q)j + K(q)k)/|q|^2)i = (q(q - 2I(q)i)/|q|^2)i. Analogous results hold for j and k. You can simplify this down to ((q^2 - 2I(q)qi)/|q|^2)i = ((q/|q|)^2 - 2I(q/|q|)(q/|q|))i. Let ρ = q/|q|, where ρ denotes a unit quaternion. Hence, one has (ρ^2 - 2I(ρ)ρ)i for all ρ. You can solve the equations ρ^2 - 2I(ρ)ρ = k and ρ^2 - 2I(ρ)ρ = j, and if they do have solutions, then this means i, j, k all have the smae conjugacy class. This is most likely the case.
Wowwww!! What an amazing video! So cleanly presented, well argued, and I feel like it finally made some really important concepts click for me!
I've always struggled with the group theory concept of "conjugacy/conjugates" and the intuitive meaning of "conjugacy classes" of group elements, and this video was incredibly insightful in all ways. I'm deeply grateful!
In particular, I had no idea that if two elements are "conjugate to each other" (by the action(?) of some other group element), they... am I understanding right that in some sense they can be seen to represent "the same action" but just "seen from a different basis" so to speak? Does the set of all conjugacy classes of a given group (i.e. equivalence classes under the equivalence relation of "X is conjugate to Y under sandwiching with some element g"), actually represent in some sense the set of "actually/essentially/truly distinct group actions", once we account for the confounding effects of "switching perspectives"/"changing basis"? Hmmmm...
I still have a lot to learn, but the idea is fascinating... and seems like it could be a road to something really powerful!
Thank you for your very positive and enthusiastic comment.
You are right that conjugate elements are "the same", just seen from a different "basis"'. The wikipedia page for "conjugacy class" literally says that 2 such elements "cannot be distinguished by using only the group structure", and I think that's what they mean.
We will look at matrix similarity in the next series, which is basically the exact same concept for matrices. There, we will see that 2 matrices are similar/conjugate when they represent exactly the same underlying linear transformation, but just seen in 2 different bases.
@@AllAnglesMath Similarity, equivalence, symmetry implies duality!
Subgroups are dual to subfields -- the Galois correspondence.
Lie groups are dual to Lie algebras.
Domains are dual to co-domains.
"Always two there are" -- Yoda.
Real is dual to imaginary -- complex numbers are dual.
Conjugacy is duality!
You always mention that the cosets other than the subgroup itself aren't subgroups because they don't contain the identity.
I think there's a more "basic" sense in which they're not groups, and the lack of identity can be "derived" from it: The cosets aren't subgroups simply because they are not closed under the group operation. If they were, you could multiply an element by itself until you obtained the identity (and past that, the element's inverse too); at least for finite groups/cosets.
Good point. Perhaps this is indeed a more "basic" reason. However, the lack of an identity is easier to check visually, so it works better for educational purposes.
@@AllAnglesMathIndeed. The lack of identity element is immediately evident and straightforward!
Subgroups are dual to subfields -- the Galois correspondence.
Lie groups are dual to Lie algebras.
Domains are dual to co-domains.
"Always two there are" -- Yoda.
Real is dual to imaginary -- complex numbers are dual.
Conjugacy is duality!
Lie groups are dual to Lie algebras.
Domains are dual to co-domains.
"Always two there are" -- Yoda.
Real is dual to imaginary -- complex numbers are dual.