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nigelscott
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What are the bases for the adjoint representation for SU(3)?
I'm not quite sure what you mean here. Usually capital letters denote groups, such as ##SU(3)## and ##SU(2)##, while lower-case letters denote their tangent spaces, such as ##\mathfrak{su}(3)## and ##\mathfrak{su}(2)##, i.e. vector spaces. (see my definitions at the beginning of post #4)nigelscott said:So the G-M is one possible basis for SU(3) and the Pauli matrices are the basis of SU(2).
nigelscott said:Given this I have another question concerning the statement adj(X)Y = [X,Y]. For SU(2) the adjoint generators are 3 x 3 matrices acting on a 3 x 1 column matrix with components iσ1, iσ2 and σ3. So how does one compute the commutator.
nigelscott said:What are the bases for the adjoint representation for SU(3)?
fresh_42 said:I haven't calculated whether the Gell-Mann matrices correspond to the adjoint representation but I'd be surprised if not.
https://en.wikipedia.org/wiki/Gell-Mann_matrices
is usually noted by ##\operatorname{Ad}## while the adjoint representation of Lie algebras is noted by ##\mathfrak{ad}## or ##\operatorname{ad}##. The formerDavideL said:As fres_42 explained, there is the adjoint rep of a Lie Grup, and of its Lie Algebra. One in this field usually looks at the latter, the adjoint representation of the Lie Algebra ##su(3)##.
The Lie Algebra ##su(3)## is a ## n^2 -1 | _{n = 3} = 8##-dimensional vector space. It can be thought as spanned by the 8 3x3 Gell-Mann matrices multiplied by the imaginary unit, so that the vectors in ##su(3)## are 3x3 skew-hermitian traceless matrices. So the Gell-Mann matrices are a base of ##su(3)## itself.
The adjoint rep of a Lie Group ...
No, it doesn't. If we denote the Lie group by ##G## and the Lie algebra by ##\mathfrak{g}##, then we have... maps the Lie Group in the space of automorphisms of the Lie Group itself.
This is a bit true in this case, but unfortunately very, very misleading overall. E.g. ##\mathfrak{ad}(0)=0## is far from being invertible. So the essential answer is: No it is wrong. ##\mathfrak{ad}## is basically the left multiplication of the Lie algebra, and as an algebra isn't a field, it usually doesn't have neither a one nor do the elements ##\mathfrak{ad}_X=\mathfrak{ad}(X) ## have a multiplicative inverse. The fact that the ##\mathfrak{ad}(X)## are bijective on ## \mathfrak{su}(3) ## for ##X\neq 0## is because this is a semisimple Lie algebra, i.e. we have ## \operatorname{ker}(\mathfrak{ad(g)}) = \mathfrak{z(g)}=\{\,0\,\} ## a trivial center and ##\operatorname{im}(\mathfrak{ad(g)})=[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}## which makes it surjective. But these two properties are far from generally true for Lie algebras.In other words, ##ad_X## for ##X \in su(3)## is a linear invertible function acting on ##su(3)## itself, ...
See above. The image is a subspace of the Lie algebra and as such a vector space again. But there is a difference between ##\operatorname{im}(\mathfrak{ad})## which are linear transformations of ##\mathfrak{g}## and ##\operatorname{im}(\mathfrak{ad}(X))## which is a single vector (aka generator) in ##\mathfrak{g}##.... ##ad_X (Y) = [X, Y] \in su(3), \quad X, Y \in su(3)##.
As from basic linear algebra, we can associate ##ad_X## to a matrix, such that, per definition, this matrix applied to the components of ##Y## returns the components of ##[X,Y]##, as vectors in ##su(3)##. Of course, the components of a vector in ##su(3)## are 8 numbers, because the dimension of ##su(3)## is 8. This matrix must then be 8x8.
The image of ##su(3)## via ##ad## is a vector space itself (I'm not sure about this actually)...
, of dimension equal to that of ##su(3)##. The base of the adjoint representation of ##su(3)## is then given by 8 8x8 matrices.
These 8 matrices are given simply by the structure constants of ##su(3)##: denote the GM matrices as ##\lambda_i##, then ## [\lambda_i, \lambda_j] = f_{ijk} \lambda_k ##, with ##i, j, k = 1, ..., 8##. Then the base of the adjoint representation of ##su(3)## is given by the 8 matrices ##t_i = ad_{\lambda_i}##, $$ (t_i)_{jk} = f_{ikj} $$ In this sense,
It would be great if someone could confirm or correct what I wrote =)
The SU(3) Adjoint representation is a mathematical concept used in the study of symmetry groups and their representations. SU(3) refers to the special unitary group of 3x3 complex matrices, and the adjoint representation is a way of representing this group through the use of matrices.
The bases for the SU(3) Adjoint representation are the Gell-Mann matrices, named after physicist Murray Gell-Mann who developed them as a way to describe the symmetries of the strong nuclear force. They consist of eight 3x3 matrices, each representing a different combination of the three colors of quarks.
The bases for the SU(3) Adjoint representation are simply another way of expressing the Gell-Mann matrices. The two are mathematically equivalent and can be interchanged in calculations and representations of SU(3).
The SU(3) Adjoint representation is used extensively in the study of quantum chromodynamics (QCD), which is the theory that describes the strong nuclear force. The Gell-Mann matrices, as the bases for this representation, are used to describe the interactions between quarks and the force carriers known as gluons.
Yes, there are many other representations of SU(3) that have been studied and used in physics. Examples include the fundamental representation, which describes the three colors of quarks, and the symmetric and antisymmetric representations. Each of these representations has its own set of bases, which can also be expressed in terms of the Gell-Mann matrices.