Representation Theory of Finite Groups: We build the character tables for S4 and A4 from scratch. As an application, we use irreducible characters to decompose a tensor product.
This came up in my exam today and thanks to your video I was able to do it. Everyone of your videos that I've watched has been useful and I'm really grateful that you took the time to make them.
Thank you sooo much for this perfectly clear and logical explanation! Now I absolutely understand the formulas in our lecture notes which before didn't make any sense to me. Very glad I found you on TH-cam, keep it up.
That's a big story by itself. In differential geometry, these correspond to spaces of functions on homogeneous spaces or sections of homogeneous vector bundles. When the group is compact, the Peter-Weyl theorem and Frobenius reciprocity allow us to do a good deal of Fourier analysis. Hope that helps! If not, PM me with more questions (more space to respond that way).
I really appreciate your content. Have you made videos on the applications of representation theory and character theory for Differential Geometry, Fourier Analysis, and PDEs? Thank you for your time.
@nahaymath The first step in the theory for finite groups is to develop the theory of irreducible representations and show that each irreducible occurs as a summand of the group algebra. This severely restricts the dimensions of irreducibles. Finite groups always have a faithful representation - the left or right action of G on the group algebra CG. Irreducible faithfuls may or may not occur. If G is abelian, an irreducible faithful exists if and only if G is cyclic. - Bob
I've wondered why the golden ratio shows up in the character table of AT, do you have any idea? A high-level reason why we might expect that, not a walkthrough of how to find the A5 character table
Great question. If you look at the table, there are two things to note about where the golden ratio appears. First, it appears in each of the three dimensional irreducibles, which come from the symmetry group of the icosahedron. These two three dimensionals are conjugate by a transposition in S5. Then it appears in the column for (12345), which rotates the icosahedron about a vertex by 72 degrees. Oriented correctly, the matrix for the rotation is [ cos(72) -sin(72) 0 \ sin(72) cos(72) 0 \ 0 0 1]. The character value is the trace, which is 1 + 2 cos(72). Finally, five fold symmetries and the Golden Ratio are a rabbit hole all its own. (I have a video on how to calculate cos(72), sin(72) mechanically. Numberphile should have one on the relation of five-fold symmetries and the Golden Ratio.)
Hello Math Dr. Bob. A question; at about 10:45, how is it you select the sgn representation of disjoint 2-cycles together with the identity element such that it becomes a normal subgroup? I am asking this assuming one can not see the group structure and only characters? According to Wikipedia: "All normal subgroups can be recognized from its character table ... Each normal subgroup of G is the intersection of the kernels of some of the irreducible characters of G." Now I see that you picked two linear characters on the representation sgn, and I also know that 3 + 1 = 4 and that divides 24 the order of the group, but would this mean that every union of every linear representation of elements of conjugacy classes which divide the order of the group necessarily give a subgroup with order the sum of the elements of the conjugacy class. If not, what is the criteria? Thanks Dr. Bob.
+Brian Droncheff It actually arises as the kernel of pi2. In this case, we are taking the trace of the 2x2 identity element, which is 2. So dimension is important for finding kernels in the character table.
1. Show directly by conjugating by any generating set, say (12), (1234). 2. Normal subgroups are unions of conjugacy classes. So use the character table to find Ker of each rep. In this case, Ker of a rep is all entries =dimension.
This came up in my exam today and thanks to your video I was able to do it. Everyone of your videos that I've watched has been useful and I'm really grateful that you took the time to make them.
Thank you sooo much for this perfectly clear and logical explanation! Now I absolutely understand the formulas in our lecture notes which before didn't make any sense to me. Very glad I found you on TH-cam, keep it up.
Your welcome!
Thank you for very nice and clear explanation!
Thank you Doctor Bob. Your walking through examples really helped.
That's a big story by itself. In differential geometry, these correspond to spaces of functions on homogeneous spaces or sections of homogeneous vector bundles. When the group is compact, the Peter-Weyl theorem and Frobenius reciprocity allow us to do a good deal of Fourier analysis.
Hope that helps! If not, PM me with more questions (more space to respond that way).
I really appreciate your content. Have you made videos on the applications of representation theory and character theory for Differential Geometry, Fourier Analysis, and PDEs? Thank you for your time.
Thank you very much. This is very clear and helpful.
Your welcome! I actually have to come back to these to use myself.
@nahaymath The first step in the theory for finite groups is to develop the theory of irreducible representations and show that each irreducible occurs as a summand of the group algebra. This severely restricts the dimensions of irreducibles.
Finite groups always have a faithful representation - the left or right action of G on the group algebra CG. Irreducible faithfuls may or may not occur. If G is abelian, an irreducible faithful exists if and only if G is cyclic. - Bob
Thank you for the very detailed description. May I ask how one would find the kernel of the characters in S4?
Beautiful video 🙂🙏🏻🙏🏻... Thank you so much sir 🙂🙏🏻🙏🏻...
Dude, that's awesome. You should make more videos!
I'm not saying I'm done, but it feels like it. What's in place definitely runs by itself.
Youre a godsend.
I've wondered why the golden ratio shows up in the character table of AT, do you have any idea? A high-level reason why we might expect that, not a walkthrough of how to find the A5 character table
Great question. If you look at the table, there are two things to note about where the golden ratio appears. First, it appears in each of the three dimensional irreducibles, which come from the symmetry group of the icosahedron. These two three dimensionals are conjugate by a transposition in S5. Then it appears in the column for (12345), which rotates the icosahedron about a vertex by 72 degrees. Oriented correctly, the matrix for the rotation is
[ cos(72) -sin(72) 0 \ sin(72) cos(72) 0 \ 0 0 1].
The character value is the trace, which is 1 + 2 cos(72). Finally, five fold symmetries and the Golden Ratio are a rabbit hole all its own. (I have a video on how to calculate cos(72), sin(72) mechanically. Numberphile should have one on the relation of five-fold symmetries and the Golden Ratio.)
Hi Dr Bob, can you explain how the induced reps work? I understand how to compute with F reciprocity but don t understand why we do any of it. Tom
Thank you.
Thanks, That's useful, but do you have a note for that theory?
Hello Math Dr. Bob. A question; at about 10:45, how is it you select the sgn representation of disjoint 2-cycles together with the identity element such that it becomes a normal subgroup? I am asking this assuming one can not see the group structure and only characters? According to Wikipedia: "All normal subgroups can be recognized from its character table ... Each normal subgroup of G is the intersection of the kernels of some of the irreducible characters of G." Now I see that you picked two linear characters on the representation sgn, and I also know that 3 + 1 = 4 and that divides 24 the order of the group, but would this mean that every union of every linear representation of elements of conjugacy classes which divide the order of the group necessarily give a subgroup with order the sum of the elements of the conjugacy class. If not, what is the criteria?
Thanks Dr. Bob.
+Brian Droncheff It actually arises as the kernel of pi2. In this case, we are taking the trace of the 2x2 identity element, which is 2. So dimension is important for finding kernels in the character table.
Sir how to prove A4 is normal in S4? Plz
1. Show directly by conjugating by any generating set, say (12), (1234). 2. Normal subgroups are unions of conjugacy classes. So use the character table to find Ker of each rep. In this case, Ker of a rep is all entries =dimension.
MathDoctorBob thankyou sir😊