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I Bases for SU(3) Adjoint representation

  1. Aug 13, 2016 #1
    What are the bases for the adjoint representation for SU(3)?
     
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  3. Aug 13, 2016 #2

    fresh_42

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  4. Aug 14, 2016 #3
    I thought the raising and lowering operators may be involved since they are the eigenstates corresponding to the roots. My hang up is that adj(x)(y) = [x,y] where x,y are 4 x 4 matrices but adj(x) is 8 x 8 which doesn't seem to work. New to group theory ... sorry!
     
  5. Aug 14, 2016 #4

    fresh_42

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    I'm no physicist so I will give you an answer from the mathematical point of view.

    We are talking about unitary transformations here, i.e. complex linear operators.
    ##G = SU(3) = \{ u : \mathbb{C}^3 \rightarrow \mathbb{C}^3 \;|\; \overline{u}^{\dagger} = u^{-1} \text{ (unitary) and } \det(u) = 1\}##
    These are complex ##(3\times 3)-##matrices. It's Lie Algebra (tangent space) is
    ##\mathfrak{g} = \mathfrak{su}(3) = \{ X : \mathbb{C}^3 \rightarrow \mathbb{C}^3 \;|\; \overline{X}^{\dagger} = -X \text{ (skew-hermitian) and } trace(X) = 0\}##

    If we count degrees of freedom here, we get ##2\cdot 9 =18## real matrix entries (##9## complex coordinates), ##9## conditions by ## \overline{X}^{\dagger} = -X## and ##1## condition by ##trace(X) = 0##, i.e.$$\dim_{\mathbb{R}}G = \dim_{\mathbb{R}}SU(3) = \dim_{\mathbb{R}}\mathfrak{su}(3)=\dim_{\mathbb{R}}\mathfrak{g}=18-9-1=8$$

    In general a representation ##(G,V,\varphi)## of a group ##G## on a vector space ##V## is a group homomorphism (mapping which respects group multiplication) ##\varphi \,:\, G \longrightarrow GL(V)##.

    Now to the adjoint representation of ##G##.
    This is a representation ##(G,\mathfrak{g},Ad)##, i.e. ##Ad\,:\, G \longrightarrow GL(\mathfrak{g})## defined by conjugation
    $$Ad\,:\, u \longmapsto (X \mapsto u\,X\,u^{-1})$$
    Simultaneously there are representations of (the tangent space) ##\mathfrak{g}## on vector spaces ##V##.
    These are Lie Algebra homomorphisms (mappings which respect Lie multiplication, esp. the Jacobi identity) ## \mathfrak{g} \longrightarrow \mathfrak{gl}(V) ##.

    There is also an adjoint representation of ##\mathfrak{g}##.
    This is a representation ##(\mathfrak{g},\mathfrak{g},ad)##, i.e. ##ad\,:\, \mathfrak{g} \longrightarrow \mathfrak{gl}(\mathfrak{g})## defined by left multiplication
    $$ad\,:\, X \longmapsto (Y \mapsto ad(X)(Y) = [X,Y])$$

    Why are both representations, one of a group and one of a Lie algebra, called adjoint?
    Firstly, groups and Lie algebras are different objects. So there won't be confusion as long as we keep (Lie) groups and Lie Algebras (tangent spaces) apart. Secondly, they are closely related.

    As you might know, the exponential mapping can be applied here and it basically maps tangent vectors to group elements:
    $$\exp\,:\, \mathfrak{g} \longrightarrow G \; ; \; X \mapsto u = \exp(X) $$
    (The details aren't that easy, but for our purposes it will do. This remark also holds for the following.)

    Now both adjoint representations are closely related by the exponential function via
    $$\exp (Ad(u)(X)) = \exp (u\,X\, u^{-1}) = u\,\exp(X)\, u^{-1} \; \text{ and } \: Ad(\exp(X))= \exp(ad(X))$$

    This all is the mathematical environment of representations, (Lie) groups and Lie algebras. Of course there is a lot of work to do to proof all this and to exactly formulate it. However, what is important here to note, is the fact, that among all these objects of linear mappings one has to be precise which of them is meant:
    • ##G= SU(3)## as a group of unitary transformations
    • ##\mathfrak{g}=\mathfrak{su}(3)## as a Lie algebra of skew-hermitian matrices
    • ##V=\mathfrak{g}## as a representation space of ##G##
    • ##V=\mathfrak{g}## as a representation space of ##\mathfrak{g}##
    • ##GL(V)## as the representation of ##G## itself
    • ##\mathfrak{gl}(V)## as the representation of ##\mathfrak{g}## itself
    and likewise with the vector spaces:
    • ##\mathbb{C}^3## for the definition of ##SU(3)## and ##\mathfrak{su}(3)##
    • ##\mathfrak{su}(3)## as tangent space of ##SU(3)##
    • ##\mathfrak{su}(3)## as representation of ##SU(3)## via ##Ad##
    • ##\mathfrak{su}(3)## as representation of ##\mathfrak{su}(3)## via ##ad##
    • ##\mathfrak{gl}(\mathfrak{g})## as tangent space of ##GL(\mathfrak{g})##
    Usually it is clear by the context. But I think it does no harm to know about those traps.

    The Gell-Mann matrices are a basis of ##\mathfrak{su}(3)## as ##8-## dimensional representation space for ##SU(3)##.

    Eigenvectors of a unitary ##(3\times 3)-##transformation matrix ##u## under the adjoint representation ##Ad## of ##SU(3)## on the vector space ##\mathfrak{su}(3)## of skew-hermitian ##(3\times 3)-##matrices, its tangent space, are therefore skew-hermitian matrices ##X_u \in \mathfrak{su}(3)## with ##Ad(u)(X_u) = \lambda_u \cdot X_u## for some scalar ##\lambda_u##.
    The tangent space ##\mathfrak{su}(3)## of ##SU(3)## is spanned by the ##8## Gell-Mann matrices (as one possible basis).

    The Wikipedia entry on special unitary groups (and related) is a bit more "physical".
     
    Last edited: Aug 14, 2016
  6. Aug 20, 2016 #5
    Thanks for your reply. I'm no mathematician so your explanation is a little over my head. So the G-M is one possible basis for SU(3) and the Pauli matrices are the basis of SU(2). 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.
     
  7. Aug 21, 2016 #6

    fresh_42

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    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)

    This distinction is important, although both consist of ##(3\times 3)-## matrices, resp. ##(2\times 2)-## matrices.

    Groups in this context are written multiplicative, representing the consecutive application of two transformations.
    Their neutral element, the transformation that does nothing at all, is the identity (matrix).
    And all elements ##A## have a (multiplicative) inverse ##A^{-1}##, i.e. ##\det (A) \neq 0##.

    Vector spaces are written additive, representing the addition of vectors, which in this case are matrices. Every matrix is viewed as a vector.
    Therefore you must not mix up the coordinates, in which the matrices are written, with the coordinates that might write them as vectors.
    Their neutral element, the vector that does nothing at all, is zero, the zero matrix.

    Now, we usually speak of generators in the case of groups, which do not have a basis. Zero is missing here, so how would one define linear independency? Generators ##g_{\iota}## here are elements, such that every ##G \ni g = \prod_{\iota} g^{r_{\iota}}_{\iota}##.

    Their tangent spaces, as vector spaces do have a basis. The zero matrix (here) is the zero of these vector spaces. However, the identity matrix is usually not part of these vector spaces, at least it doesn't have to be. And the inverse matrices don't have to exist either.

    With respect to the definitions I gave in post #4 you should be able to decide whether the Gell-Mann and Pauli matrices are unitary with determinant ##1## or whether they are skew-hermitian.

    A commutator (in the case ##X## and ##Y## are matrices / transformations / operators) can be computed by ##ad(X)(Y) = [X,Y]=XY-YX##.
    (Mathematically this should be stated with more caution, but it will do in the context here.)

    Now, for a brief look on the Pauli matrices. They are a little bit mean for they play multiple roles.

    1. The Pauli matrices ##\sigma_i## are hermitian.

    Thus they cannot be a basis for ##\mathfrak{su}(2)##.

    However, together with ##\sigma_0 = 1## (as a vector!), they form a basis of the complex vector space ##\mathbb{M}_{\mathbb{C}}(2,\mathbb{C})=\mathfrak{gl}(2,\mathbb{C})=\{ (2 \times 2) \text{ complex matrices }\}##.

    In addition and again together with ##\sigma_0## they form a basis of the real vector space ##\{A \in \mathbb{M}_{\mathbb{R}}(2,\mathbb{C})\;\vert \; \overline{A}^{\dagger}=A\}## of complex hermitian matrices. (Remember that ##\mathbb{C}## is also a two-dimensional real vector space.)

    2. The Pauli matrices ##\sigma_i## are unitary.

    Thus they might be a candidate to generate ##SU(2)##. Unfortunately, ##\det \sigma_i = -1 \; (i=1,2,3)##, so they do not belong to ##SU(2)## either.

    3. So why to consider Pauli matrices ##\sigma_i## at all?.

    Well, they have some very important properties.
    • ##\sigma^2_i = \sigma_0 = 1 \, ; \, \det(\sigma_i) = -1 \, ;\, tr(\sigma_i)= 0 \,;\, ## the eigenvalues of ##\sigma_i## are ##\pm 1##
    • $$\sigma_i \sigma_j = \delta_{ij} \sigma_0 + i \sum_{k=1}^{3} \epsilon_{ijk} \sigma_k$$
    • $$[\sigma_i , \sigma_j] = \sigma_i \sigma_j - \sigma_j \sigma_i = 2i \sum_{k=1}^{3} \epsilon_{ijk} \sigma_k$$
    • $$\{ \sigma_i , \sigma_j \} = \sigma_i \sigma_j + \sigma_j \sigma_i = 2 \delta_{ij} \sigma_0$$
    • ... and many more ...

    4. The ##i## multiples of Pauli matrices
    ##i\sigma_j \; (j=1,2,3)##
    are skew-hermitian.

    Thus ##i \cdot \sigma_j## (without ##\sigma_0=1\, !\,##) are indeed a basis for the real 3-dimensional Lie algebra ##\mathfrak{su}(2)=\mathfrak{su}(2,\mathbb{C})=\mathfrak{su}_\mathbb{R}(2,\mathbb{C})## of complex, skew-hermitian matrices with zero trace over the real numbers. (Again, remember complex numbers are a real vector space.)

    In addition ##\exp(-i\frac{\alpha}{2} \; (\vec{e}\cdot \vec{\sigma})) = \sigma_0 \, \cos(\frac{\alpha}{2}) -i \; (\vec{e}\cdot \vec{\sigma}) \sin(\frac{\alpha}{2})## with a unit vector ##\vec{e}## of ##\mathbb{R}^3## and the ##i\sigma_j## generate the rotation group ##SU(2)## which is isomorphic to the group of quaternions of norm ##1##, and is thus diffeomorphic to the ##3-##sphere.
    For further properties have a look on https://en.wikipedia.org/wiki/Special_unitary_group.

    If we write ##i \sigma_1 = X \;,\; i \sigma_2 = Y \;,\; i \sigma_3 = H\;## then we have ##[H,X]=-2i X \;,\; [H,Y]=2i Y \;,\; [X,Y]= -2H## which is basically the Lie algebra ##\mathfrak{sl}(2)## which is isomorphic to ##\mathfrak{su}(2)##. The representations of ##\mathfrak{sl}(2)## are completely classified and can be found in probably every textbook that defines a Lie algebra. Therefore all representations of ##\mathfrak{su}(2)## are also completely known.

    5. Now let's go physical.

    In this section I will suffer not one but several difficulties. Firstly, as mentioned before, my physical understanding here comes from what I read on PF, Wiki or alike. So we will be likely on a similar level. Secondly, the technical parts on what I've found unfortunately weren't in English, so I'll have to retype that stuff here. (There are also a lot of pdf on the internet, but I don't know whether and which have a copyright or not, which means I won't recommend them here. You may have a look on your own. Just google "generator of representation". But be prepared to find entire books on this matter.)
    Another difficulty for me is, that physicists tend to use the word generator for my personal sensation far too easy and too often for entirely different things. I might be wrong due to the lack of my understanding, but it's the impression I got. So I will translate and retype some brief concepts I found on Wikipedia without explicitly mention it in the following. (Its English version is less specific.)

    The Pauli matrices belong to the special case of angular momentum operators for ##l=1/2## (see above section 4).
    The latter operate on basis vectors ##v_m \; (m \in \{-l,-l+1,...,l-1,l \})## of an angular momentum ##l-##multiplet with quantum numbers ##m## as follows (##\hbar = 1##):
    $$(L_3).v_m = mv_m \\ (L_+).v_m=\sqrt{(l-m)(l+m+1)}v_{m+1} \\ (L_-).v_m=\sqrt{(l+m)(l-m+1)}v_{m-1}$$
    Here the ##L_i## are defined as ##L_3 = \frac{1}{2}\sigma_3 \;,\; L_+= \frac{1}{2}(\sigma_1 + i \sigma_2) \;,\; L_- = \frac{1}{2} (\sigma_1 - i \sigma_2)##. The ##L_i## which I have from Wiki and the ##X,Y,H## from the previous section are all elements of ##\mathfrak{sl}(2)##, i.e. there are even two sets of basis vectors, and therefore the preferable choice if it comes to representations. (The "##+##" in ##L_+## indicates the maximal and the "##-##" in ##L_-## the minimal root of ##\mathfrak{su}(2)##.)
    The ##v_m## are the eigenvectors (of ##L_3##, the Cartan subalgebra) in the representation space and ##m## its eigenvalues.

    ##2l+1## is a natural number and for a given ##m## there are ##2l+1## different quantum numbers ##-l,-l+1,\dots ,l-1,l##. For ##l=1/2## the angular momentum operators apply to the components of the linear combinations of ##v_{\frac{1}{2}}## and ##v_{- \frac{1}{2}}## by multiplication of the ##L_i## which are defined via Pauli matrices.
    (I'm on thin ice here, so take it with care and perhaps you want have a look at the following page: https://en.wikipedia.org/wiki/Angular_momentum_operator)

    Finally I'll get physical one more time and add some "translations" between Pauli matrices, linear combinations according to standard basis vectors, and according to eigenvectors (again from Wikipedia and I hope it makes more sense to you than it does to me).
    $$\sigma_1 = \sigma_x = \begin{pmatrix}0&1\\ 1&0\end{pmatrix}=|0 \rangle \langle 1| + |1\rangle \langle 0|=|+\rangle \langle +| - |-\rangle \langle -| \\ \sigma_2 = \sigma_y = \begin{pmatrix}0&-i\\ i&0\end{pmatrix}=i (|1\rangle \langle 0|+|0\rangle \langle 1|)=|\phi^+ \rangle \langle \phi^+ |-|\phi^-\rangle \langle \phi^-| \\ \sigma_3 = \sigma_z = \begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=|0\rangle \langle 0|-|1\rangle \langle 1|=|0\rangle \langle 0|-|1\rangle \langle 1|$$

    Here are (with the vectors meant in ##\mathbb{C}^2##)

    $$ |0 \rangle \doteq \begin{pmatrix} 1 \\ 0 \end{pmatrix} \; , \; |1 \rangle \doteq \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ |+ \rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \; , \; |- \rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \\ |\phi^+ \rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix} \; , \; |\phi^- \rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \end{pmatrix} $$

    The matrices ##\frac{\hbar}{2}\sigma_i## are operators for the components of spin 1/2 systems.
    The exponential equation above (section 4) describes the change of spin states under a rotation by ##\alpha## with a rotation axis ##\vec{e}##. If we set ##\alpha = 2\pi## then the state becomes his negation and only another rotation by ##2\pi## gets us back to were we started from. Therefore it's a half-spin-system.
     
    Last edited: Aug 21, 2016
  8. Aug 22, 2016 #7
    I thank you for your patience in answering my questions. My background is in semiconductor physics so this group theory stuff is a little to abstract for me. I think at this point an actual computation would help me immensely. My basic challenge is that in the adjoint representation of SU(2) the generators are 3 x 3 matrices as determined by (Ta)bc = -iεabc. There are the bases -iσi etc. of the vector space and the statement ad(x)y = [x,y]. I just can't figure out how all these things play together. I looked a thttp://mathworld.wolfram.com/AdjointRepresentation.html but the intermediate steps are missing.
     
  9. Jul 9, 2018 #8
    The post is quite old, but I've run into this topic now so I'll try to bring it back to life rather than opening a new one.

    I may be totally mistaken about the following, but here's my poor's man understanding of the matter.

    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 maps the Lie Group in the space of automorphisms of the Lie Group itself. In other words, ##ad_X## for ##X \in su(3)## is a linear invertible function acting on ##su(3)## itself, ##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 =)
     
  10. Jul 13, 2018 #9

    fresh_42

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    is usually noted by ##\operatorname{Ad}## while the adjoint representation of Lie algebras is noted by ##\mathfrak{ad}## or ##\operatorname{ad}##. The former
    No, it doesn't. If we denote the Lie group by ##G## and the Lie algebra by ##\mathfrak{g}##, then we have
    \begin{align*}
    \operatorname{Ad}\, : \,&G \longrightarrow GL(\mathfrak{g})\\
    &g \longmapsto (\,Y \mapsto (\operatorname{Ad}(g))(Y)=gYg^{-1}\,)\\[6pt]
    \mathfrak{ad}\, : \,&\mathfrak{g}\longrightarrow \mathfrak{gl(g)}\\
    &X \longmapsto (\,Y \mapsto (\operatorname{\mathfrak{ad}}(X))(Y)=[X,Y]=XY-YX\,)
    \end{align*}
    so that the adjoint transformation of the Lie group is an automorphism (in this case a linear one) of the Lie algebra, not the Lie group itself. This is expressed in the word representation which usually refers to a vector space or a module; not necessarily, but in 999 of 1000 cases here on PF.
    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.
    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}##.

    ...
     
    Last edited: Jul 13, 2018
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