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Lie group representations

by spookyfish
Tags: representations
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spookyfish
#1
May11-14, 07:51 PM
P: 53
The vectors [itex]\vec{\alpha}=\{\alpha_1,\ldots\alpha_m \}[/itex] are defined by
[tex]
[H_i,E_\alpha]=\alpha_i E_\alpha
[/tex]
they are also known to be the non-zero weights, called the roots, in the adjoint representation. My question is - is this connection (that the vectors [itex]\vec{\alpha} [/itex] defined by the commutation relations above in some representation, are also the roots of the adjoint representation) is true only when [itex]\vec{\alpha} [/itex] is in the defining representation, or is it true for any representation?
I hope my question is clear
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andrien
#2
May12-14, 02:49 AM
P: 1,020
Which other representation of roots? Are you talking about the basis of the roots like Cartan-weyl basis or Dynkin basis?
spookyfish
#3
May12-14, 09:11 AM
P: 53
No. Sorry, I am talking about the representation in which [itex]\vec{\alpha} [/itex] is defined. for example, in su(3), the defining representation has 3 weights (because the space is 3 dimensional) and the vectors [itex]\vec{\alpha} [/itex] are the difference between these weights. The vectors [itex]\vec{\alpha} [/itex] also coincide with the roots of the adjoint representation.
Now, suppose I wanted to consider a different arbitrary representation of su(3), not the 3-dimensional and not the adjoint 8-dimensional. It would have a different number of weights. Would the root vectors (same roots vectors, defined in the adjoint representation) still carry me between the different weights, or will they now not coincide with the vectors [itex]\vec{\alpha} [/itex] satisfying
[tex]
[H_i,E_\alpha]=\alpha_i E_\alpha
[/tex]

Bill_K
#4
May12-14, 09:50 AM
Sci Advisor
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P: 4,160
Lie group representations

The root vectors are not defined in the adjoint representation, or any representation, they are defined in the Lie Algebra, by the formula you gave,
[tex]
[H_i, E_\alpha]=\alpha_i E_\alpha
[/tex]
Quote Quote by spookyfish View Post
Would the root vectors (same roots vectors, defined in the adjoint representation) still carry me between the different weights
Yes, this is easy to show. If m is an eikenket of H with weight m, namely H|m> = m |m>, then Eα|m> is an eigenket with weight m + α. This follows from the above commutator.
spookyfish
#5
May12-14, 10:03 AM
P: 53
I see. But since [itex]H_i [/itex] and [itex]E_\vec{\alpha} [/itex] are different in different representations, aren't the root vectors [itex]\vec{\alpha} [/itex] also different?

then, if they are, either they are equal to the weights in the adjoint representation, or not.
Bill_K
#6
May12-14, 10:10 AM
Sci Advisor
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P: 4,160
Quote Quote by spookyfish View Post
I see. But since [itex]H_i [/itex] and [itex]E_\vec{\alpha} [/itex] are different in different representations, aren't the root vectors [itex]\vec{\alpha} [/itex] also different?
H and Eα are elements of the Lie algebra. In any representation there will be matrices that correspond to them, but the commutators of those matrices (and hence the root vectors) had better be the same, or else it would not be a representation!

Note that the root vectors are always ℓ-dimensional vectors, where ℓ is the rank of the group. They don't depend on the dimensionality of the representation.
spookyfish
#7
May12-14, 10:30 AM
P: 53
I see. Thank you


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