Are Vectors Defined by Commutation Relations Always Roots in Any Representation?

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Discussion Overview

The discussion centers around the relationship between vectors defined by commutation relations in Lie algebras and their identification as roots in various representations. Participants explore whether these vectors, denoted as \(\vec{\alpha}\), are consistently roots across different representations, particularly in the context of the Lie algebra su(3).

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions whether the vectors \(\vec{\alpha}\) defined by the commutation relations are roots in any representation or only in the defining representation.
  • Another participant seeks clarification on what is meant by "other representation of roots," suggesting possible bases like the Cartan-Weyl or Dynkin basis.
  • A participant explains that in the defining representation of su(3), the vectors \(\vec{\alpha}\) are the differences between weights and coincide with the roots of the adjoint representation.
  • There is a discussion about whether root vectors defined in the adjoint representation would still correspond to different weights in arbitrary representations of su(3).
  • One participant asserts that root vectors are defined in the Lie algebra and that the commutation relations imply they should remain consistent across representations.
  • Another participant expresses concern that since \(H_i\) and \(E_{\vec{\alpha}}\) differ in various representations, the root vectors \(\vec{\alpha}\) might also differ.
  • A later reply emphasizes that while matrices corresponding to \(H\) and \(E_{\vec{\alpha}}\) vary in different representations, the commutators—and thus the root vectors—should remain the same.
  • It is noted that root vectors are always \(ℓ\)-dimensional, where \(ℓ\) is the rank of the group, and do not depend on the dimensionality of the representation.

Areas of Agreement / Disagreement

Participants express differing views on whether the root vectors \(\vec{\alpha}\) are consistent across representations. Some argue for their consistency based on the properties of Lie algebras, while others question how differences in representations might affect these vectors. The discussion remains unresolved.

Contextual Notes

Participants highlight that the definitions and properties of root vectors depend on the context of the Lie algebra and the specific representations being considered. There are unresolved aspects regarding the implications of different representations on the nature of the root vectors.

spookyfish
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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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Which other representation of roots? Are you talking about the basis of the roots like Cartan-weyl basis or Dynkin basis?
 
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]
 
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]
spookyfish said:
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.
 
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.
 
spookyfish said:
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.
 
I see. Thank you
 

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