Once in a while we need to re-organize our environment and our thinking. Sometimes it is good, sometimes it is a disaster. It is much like with the Law of Three:
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| “There is good, there is evil, and there is the specific situation that determines which is which.” |
So I reorganized my Notes, and added some new stuff and new exercises.
Here is the link to pdf.
29-07-25 18:00 I have a problem with my calculations. Will be silent until it will be resolved.
01-08-25 16:31 Problems solved. Can continue.
It's all so complicated. Right now I can only list the typos:
ReplyDeletein formula (33): γ1γ2γ3γ4 = diag(1, 1, −1m − 1)
on page 5: In [3] we were discussing the grpup
on page 6: Exercise 3.2 Vrify the last statement
in formula (39): x1^2
and the intriguing end of the text:
"Thus group G on Cl(2) induces a homomorphism of G into the group O(4, 2) of isometries of R2,2 as follows:"... how?
Thank you Anna!
DeleteOoh God! And I was so happy about what I have written! And it is a disaster. How good is to have a feedback!
I will have to write a road-map at the beginning, explaining what we are doing and why.
Typos fixed. File updated. Thanks!!!!
In Ex.2.2, I even cannot understand what is ρ((g, h)(g′, h′)); since function ρ should have two arguments, while (g, h)(g′, h′) seems to be a product of two real numbers, doesn't it?
ReplyDeleteChanged a little bit and added a line in the exercise. If you have still problems, please, let me know what makes my formulation so confusing. I really appreciate your feedback.
DeleteEx.2.2(1) is easy, but only given Exercise 2.1 is done.
ReplyDeleteIf we know that "group G, for our representation of Cl(2), can be identified with SL(2, R)", so elements a, g, h ∈ belongs to one and the same group SL(2, R), and conjugation of a∈Cl(2) in the form ρ(g, h) = g a h^-1 is just a combination of group operations and thus is also in the same group.
Ex.3.1 is straightforward. Done.
ReplyDeleteAimed at Ex.3.2, I obtained that in terms of gamma matrices (28)-(31) general element x has the form:
[0, 0, x1-x3, x2+x4
0, 0, x2-x4, -x1-x3
x1+x3, x2+x4, 0, 0
x2-x4, -x1+x3, 0, 0]
and its determinant = -x1^2 - x2^2 + x3^2 + x4^2
What else should be done to varify the statement (35)?
Well, this section starts with: "In [3] we were discussing the group Spin(2,2) assuming the form of generators given by Eqs. (1)-(2). The
Deletesame reasoning applies to the new gamma matrices"
Exercise (3.2) is meant you to go back to [3] and make sure that with the new set of gammas we are led indeed to the same form of the Spin(2,2) group as with old gammas, called e's in [3].
Ark, thank you for the prompt.
DeleteSo, I found that
(gamma1)(gamma2) = {{ε,0},{0,ε}}.
Then, we can show that the expression for ν from [3]
ν(x) = Ux^TU^-1
is the same as formula (10) ν(a) = −ωa^Tω,
since {{ε,0},{0,ε}}^-1 = - {{ε,0},{0,ε}}.
Finally, for g = {{A,0},{0,B}}, we can calculate
Δ(g) = ν(g)g and find that
Δ(g) = {{det(A)I2, 0},{0, det(B)I2}}.
This is precisely formula (7) from [3].
If det(A) = det(B) = 1, then Δ(g) is simply identity matrix.
Is that what was needed?
Ex.3.3 is done:) Note a curious relation: ω R^-1 = R^T ω
ReplyDeleteand, of course, the same for L.
This is almost the orthogonality condition, but with a help of omega. It seems that we get it for arbitrary matrices with one condition that its norm =1.
Page 4, p.2.1: "We will be interested in the representation ρ of the direct product G×G on Cl(2)"
ReplyDeleteboth G's should be the same, right?
Yes. Thanks. Will correct.
DeleteReturning to [3] https://ark-jadczyk.blogspot.com/2025/07/spin22.html,
ReplyDeleteI would like to ask about the Spin group:
the key definition of group Spin(2,2) stems from
Spin+(V,q) = {g ∈ Cl(V,q)+: Δ(g) = 1, g ∈ Γ(V.q)} (3)
And why do we need the last condition g ∈ Γ(V.q)?
Does it mean only that g is INVERTIBLE element of Cl(V,q), or I missed anything else?
In general an invertible element of Cl(V,q) will not have the property that g e_i g^{-1} will be a linear combination of the basis e_i in V. It will be just some element of the Clifford algebra. Γ(V.q) is the group of invertible elements that do have this property.
DeleteA priori it is not clear that Γ(V.q) has more than just identity element. But it is a good exercise to check that, for instance, the products e_i e_j are all in Γ(V.q) .
Ark, thank you for the answer.
Delete"It will be just some element of the Clifford algebra".
But I thought that any element of the Clifford algebra is a linear combination of the basis e_i, isn't it?
"to check that the products e_i e_j are all in Γ(V.q)":
DeleteIf g ∈ Γ(V.q), then by definition of group Γ,
g e_i g^{-1} and
g e_j g^{-1} are both in Γ.
For the product e_i e_j, we have
g e_i e_j g^{-1} =
g e_i g^{-1} g e_j g^{-1} = product of two elements of Γ, which should be in Γ by virtue of the group property.
"But I thought that any element of the Clifford algebra is a linear combination of the basis e_i, isn't it?"
DeleteIf e_i, i=1,..,4 is a basis in V=R^{2,2}, then Cl(2,2) is 2^4=16-dimensional. To have a basis in Cl(2,2) we need a6 basis elements. They are
1, e_i,
e_i e_j, i<j,
e_i e_j e_k, i<j<k,
e_1 e_2 e_3 e_4.
We need to distinguish between a basis in V and a basis in Cl(V,q).
V is only a 4D subspace of a 16D Cl.
"For the product e_i e_j, we have ..."
DeleteWhat I had in mind that if g=e_i e_j is always in Γ. It is easier to start with: every e_i is in Γ. To show this use the fundamental relation
e_i e_j + e_j ee_i = 2 eta_{ij}
thus e_ e_j = 2 eta_{ij} - e_j e_j
Then calculate e_i e_j e_i^{-1} to see that you get a linear combination of e_k's.
Therefore for each x=x^j e_j in V we have
e_i x e_i^{-1} is in V
Thus e_ is in Γ. And, since Γ is a group, also products of e_i are in Γ.
"We need to distinguish between a basis in V and a basis in Cl(V,q)".
DeleteYes, surely I know that, but I had in mind Cl(2) algebra at that moment, don't know why... Probably, because I have got a sore throat and cannot think properly :(
Thank you for the explanation about e_i's in Γ, got it.
Take care of your sore throat. And me know when you are well.
DeleteArk, thank you very much for your care and attention. It was a mild form of covid caught while we travelled in Kaliningrad/Koeningsberg. Now I am at home and feel myself much better. I will write some comments at https://ark-jadczyk.blogspot.com/2025/08/knowledge-protects-ignorance-endangers.html
Delete