- #1

- 135

- 4

^{-1}D(g)U produces a block diagonal matrix for all g in G? For example, I am trying to figure out how the matrix (7) on page 4 of this document is obtained.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- A
- Thread starter nigelscott
- Start date

- #1

- 135

- 4

- #2

fresh_42

Mentor

- 15,542

- 13,640

I would study the Jordan Chevalley decomposition, but I don't know the algorithmic procedure.

- #3

- 17,333

- 7,205

Caveat: two eigenvectors in different irreps may have the same eigenvalue. Thus, you should pick eigenvectors that correspond to non-degenerate eigenvalues.

- #4

- 135

- 4

- #5

- 17,333

- 7,205

Also note that you just need to act with the generators of the group on your vectors until they do not produce new linearly independent ones.

I was planning to write down a common physical example which is the coupling of two spin-1/2 particles, but I am on my mobile. I might do it later.

Also note that there are some nice results about irreps, such as Shur’s lemma, that can allow you to do things more easily.

- #6

fresh_42

Mentor

- 15,542

- 13,640

- #7

Cryo

Gold Member

- 173

- 73

Basically lets say you have a group ##G##. A vector space ##V## and a (possibly reducible) representation of ##G## over ##V##,

##\mathbf{M}: G \to Hom(V,V)##

i.e. ##\forall g \in G## you have ##\mathbf{M}\left(g\right): V\to V##

Assuming you have the character tables for all irreducible representations of this group, with ##\chi^{(i)} : g\to\mathbb{C}## being the character if i-th irreducible representation, you can define:

##\mathbf{P}_i : V \to V##

such that ##\mathbf{P}_i = \frac{\#i}{\#G}\sum_{g\in G} \chi^{(i)}\left(g^{-1}\right)\mathbf{M}(g)##

where ##\#G## is the number of elements in ##G## and ##\#i## is the dimension of i-th irrep.

Now remember the great orthogonality theorem. Given irreducible representations ##D^{(i,j)}_{\alpha,\beta}##, we have:

##\frac{1}{\#G}\sum_{g\in G} D^{(i)}\left(g^{-1}\right)_{\alpha,\beta} D^{(j)}\left(g\right)_{\mu,\nu}=\frac{1}{\#i} \delta_{i,j}\delta_{\alpha,\mu}\delta_{\beta,\nu}##

Taking the trace over ##\alpha,\beta## we get:

##\frac{1}{\#G}\sum_{g\in G} \chi^{(i)}\left(g^{-1}\right) D^{(j)}\left(g\right)_{\mu,\nu}=\frac{1}{\#i} \delta_{i,j}\delta_{\mu,\nu}##

So now I can drop the matrix indices and write:

##\frac{1}{\#G}\sum_{g\in G} \chi^{(i)}\left(g^{-1}\right) \mathbf{D}^{(j)}\left(g\right)=\frac{1}{\#i} \delta_{i,j}\mathbf{Id}##

Lets get back to your representation. We know that there will be a way of decomposing ##M## into ireducible representations. Lets say ##\mathbf{M}=\mathbf{D}^{(1)}\oplus \mathbf{D}^{(3)}##

Then

##\mathbf{P}_1 = \frac{\#1}{\#G}\sum_{g\in G} \chi^{(1)}\left(g^{-1}\right)\mathbf{D}^{(1)}(g) \oplus \frac{\#1}{\#G}\sum_{g\in G} \chi^{(1)}\left(g^{-1}\right)\mathbf{D}^{(3)}(g) = \mathbf{Id}\oplus\mathbf{0}##

So ##\mathbf{P}_1## projects all vectors into the first irrep, i.e. it send the vectors from different irreps to zero, whilst keeping the vectors from the 1-st irrep untouched.

Now simply find all the eigenvectors of ##\mathbf{P_1}## (numerically). This will allow you to span all the 1-st irrep copies in your representation. Do it for all other projection operators for the irreps that occur in your representation (use characters to test this), i.e. find eigenvectors of all ##\mathbf{P}_i##. Finally, represent ##\mathbf{M}## in that eigenvector basis for all ##g\in G## (i.e. ##\mathbf{U}## that you wanted consists of these eigenvectors). This will be block-diagonal.

- #8

- 135

- 4

- #9

Cryo

Gold Member

- 173

- 73

Finally, represent ##\mathbf{M}## in that eigenvector basis for all ##g\in G## (i.e. ##\mathbf{U}## that you wanted consists of these eigenvectors). This will be block-diagonal.

What bugs me is how to get block-diagonal matricies if your representation is ##\mathbf{D}_1\oplus \mathbf{D}_1##, or something similar, i.e. if you have two or more copies of the same representation. The projection operators will not touch it.

Does anyone know?

- #10

fresh_42

Mentor

- 15,542

- 13,640

How is it given, if not already in block form, i.e. how do you know, that the two spaces are invariant?What bugs me is how to get block-diagonal matricies if your representation is ##\mathbf{D}_1\oplus \mathbf{D}_1##, or something similar, i.e. if you have two or more copies of the same representation. The projection operators will not touch it.

Does anyone know?

If they are, find a basis for both and perform the change of basis on your matrices.

- #11

Cryo

Gold Member

- 173

- 73

How is it given, if not already in block form, i.e. how do you know, that the two spaces are invariant?

It is motivated by my earlier post (see above) , but it is not too important.

So let us say that for all elements ##g## of a finite group ##G## we have representation ##\mathbf{M}\left(g\right):\mathcal{V}\to\mathcal{V}##, where ##\mathcal{V}## is the vector space. Crucially, ##\mathbf{M}## are not block-diagonal in the initial basis. By construction. From character calculus we know that ##\mathbf{M}\left(g\right)=\mathbf{D}\left(g\right)\oplus\mathbf{D}\left(g\right)##, where ##\mathbf{D}\left(g\right)## is an irreducible representation of ##G##

If they are, find a basis for both and perform the change of basis on your matrices.

That's my question, how do I find such basis for this specific situation, i.e. where I know that representation is a direct sum of two copies of the same irrep, but I do not know in which basis? What is the procedure?

- #12

fresh_42

Mentor

- 15,542

- 13,640

- #13

Cryo

Gold Member

- 173

- 73

The matrix representation is:

##\mathbf{D}\left(Id\right)=\left(\begin{array}\\1 & 0\\ 0 & 1\end{array}\right)##

##\mathbf{D}\left(r_{120^\circ}\right)=\left(\begin{array}\\-\frac{1}{2} & -\frac{\sqrt{3}}{2}\\\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{array}\right)##

##\mathbf{D}\left(r_{240^\circ}\right)=\left(\begin{array}\\-\frac{1}{2} & \frac{\sqrt{3}}{2}\\\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{array}\right)##

##\mathbf{D}\left(m_{90^\circ}\right)=\left(\begin{array}\\-1 & 0\\\ 0 & 1\end{array}\right)##

##\mathbf{D}\left(m_{210^\circ}\right)=\left(\begin{array}\\-\frac{1}{2} & -\frac{\sqrt{3}}{2}\\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right)##

##\mathbf{D}\left(m_{330^\circ}\right)=\left(\begin{array}\\-\frac{1}{2} & \frac{\sqrt{3}}{2}\\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right)##

The characters are: ##\chi=2,\,-1,\,-1,\,0,\,0,\,0##, respectively.

Next I create a bigger representation ##\mathbf{M}=\mathbf{U}\,\left(\mathbf{D}\oplus\mathbf{D}\right)\,\mathbf{U}^\dagger##, where ##\mathbf{U}## is simply a unitary real-valued matrix.

##\mathbf{U}=\exp\left(\mathbf{P}\right)##

##\mathbf{P}=\left(\begin{array}\\ 0 & -1 & 0 & 0 \\ 1 & 0 & -1 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 1 & 1& 0 \end{array}\right)##

Clearly identy will look as 4d identity, but (e.g.):

##\mathbf{M}\left(m_{210^\circ}\right)\approx\left(\begin{array}\\

0.3221 & 0.5709 & 0.1493 & -0.7403 \\

0.5709 & -0.5176 & -0.5786 & -0.2674 \\

0.1493 & -0.5786 & 0.7693 & -0.2261 \\

-0.7403 & -0.2674 & -0.2261 & -0.5738 \\

\end{array}\right)##

The trace of this is still zero, so I know this corresponds to reflection, but as you can see it is not at all block-diagonal. Same goes for other matricies. Now, if I was given these matricies (##\mathbf{M}\left(Id\right),\,\mathbf{M}\left(r_{120^\circ}\right),\,\dots ##) and told that they are a representation of D3 group, I could work out that this representation is isomorphic to ##\mathbf{D}\oplus\mathbf{D}## using characters alone. But then, how would I find the basis in which this representation is block-diagonal? How would I find ##\mathbf{U}## if I was not given it?

- #14

fresh_42

Mentor

- 15,542

- 13,640

- #15

- 17,333

- 7,205

This is not necessarily true. It may be written in a form which is rotated away from the block form and still be a ##\mathbf{D}\oplus\mathbf{D}## representation, i.e., it may be written in a basis that mixes the irreps. This would not mean that it is not a ##\mathbf{D}\oplus\mathbf{D}## representation.

- #16

fresh_42

Mentor

- 15,542

- 13,640

But he wrote the ##\mathbf{D}## as ##2\times 2## matrices and ##\mathbf{M}## as ##4\times 4##!This is not necessarily true. It may be written in a form which is rotated away from the block form and still be a ##\mathbf{D}\oplus\mathbf{D}## representation, i.e., it may be written in a basis that mixes the irreps. This would not mean that it is not a ##\mathbf{D}\oplus\mathbf{D}## representation.

There are no second diagonals possible in the given situation.

In case ##\mathbf{M}## is given anyhow, plus the abstract knowledge that the representation splits, in which case the information cannot arose from given block matrices, then my formerly given statement applies: look for invariants and start with all semisimple parts.

- #17

Cryo

Gold Member

- 173

- 73

This is not necessarily true. It may be written in a form which is rotated away from the block form and still be a D⊕D\mathbf{D}\oplus\mathbf{D} representation

Precisely. The rules of the 'game' is that you are given ##\mathbf{M}\left(g\right)##'s for each ##g\in G## and you are told what ##G## is, and ##G## is finite. Then

How would I find ##\mathbf{U}## if I was not given it?

- #18

Cryo

Gold Member

- 173

- 73

look for invariants and start with all semisimple parts.

Can you please explain it in more detail, or maybe give a reference to it? I have 6 4-by-4 matricies, and I know that they reps of D3 and that they are isomorphic to direct sum of two copies of the two-dimensional irrep of that group. What do I do to block-diagonalize all ##\mathbf{M}##'s?

Last edited:

- #19

fresh_42

Mentor

- 15,542

- 13,640

- #20

Cryo

Gold Member

- 173

- 73

I'd be glad to get any help you can provide

- #21

fresh_42

Mentor

- 15,542

- 13,640

J.E. Humphreys: Linear Algebraic Groups

https://www.amazon.com/dp/0387901086/?tag=pfamazon01-20

is an excellent book and good to read. The keyword is Jordan decomposition (chapter VI ff.).

The difficulty is, that books are full of fancy theorems about properties of e.g. tori, but don't give an explicit algorithm. The way leads indeed over characters and tori (maximal diagonalizable subgroups).

I think it is along the lines:

Write all elements ##g=g_s+g_n## additive ##(*)## as a sum of semisimple (diagonalizable) and nilpotent endomorphisms, then pass over to ##g_u=1+g_s^{-1}g_n## such that ##g=g_sg_u## is the multiplicative decomposition into a semisimple and unipotent part. ##g_s=p(g)\, , \,g_n=q(g)## for polynomials without constant term.

Then all semisimple parts are simultaneously diagonalizable, which yields the basis we look for.

(*) The additive part can be found in his other book:

J.E. Humphreys: Introduction to Lie Algebras and Representation Theory

https://www.amazon.com/dp/0387900535/?tag=pfamazon01-20

if not in any linear algebra book (using the characteristic polynomial of ##g## and its linear factors.)

Share: