2
$\begingroup$

If we know how to write the matrix representation of the fundamental representation of SU(N), could we use them to derive the matrix representation of other representations of SU(N)? (adjoint, anti-symmetric, or symmetric, etc.)

For example, for SU(2), we know the tensor product of 2-dimensional fundamental representation of SU(2) gives the 3-dimensional adjoint representation of SU(2): $$ 2 \times 2 = 3_A + 1_S. $$ So if we know the explicit 2-dimensional matrix representation of SU(2) in terms of its three Lie algebra generators, $$\sigma_1 = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0& -i \\ i&0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} , \tag{1} $$ how do we DERIVE the 3-dimensional matrix representation of SU(2) (which we know is the adjoint representation of SU(2) and vector representation of SO(3))? Namely in terms of its three Lie algebra generators, $$ \boldsymbol{L}_x = \begin{bmatrix}0&0&0\\0&0&-1\\0&1&0\end{bmatrix}, \quad \boldsymbol{L}_y = \begin{bmatrix}0&0&1\\0&0&0\\-1&0&0\end{bmatrix}, \quad \boldsymbol{L}_z = \begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix}. \tag{2} $$ Namely, how do derive eq (2) from the given eq (1) for the Lie algebra matrix representations?

$\endgroup$
8
  • $\begingroup$ The Lie brackets of matrices are given by commutators, i.e., $[A,B]=AB-BA$. This is given. But then the adjoint representation follows directly by definition. For example, $[x,y]=z$, so that the matrix of $L_x$ has first column zero and second column $(0,0,1)$. $\endgroup$ Aug 1, 2021 at 16:44
  • $\begingroup$ Dear Dietrich Burde - thanks, but I need to have explicit derivations. I know Lie bracket. I also already know the answer of SU(2) case. How about you illuminate the new question on the SU(5) case? math.stackexchange.com/q/4214397/141334 $\endgroup$ Aug 1, 2021 at 16:47
  • 1
    $\begingroup$ The adjoint matrices are exactly given by the Lie brackets, so I don't understand your question. $ad(x)(y)=[x,y]$ by definition. In your example here, $L_x=ad(x)$. $\endgroup$ Aug 1, 2021 at 16:48
  • $\begingroup$ Indeed, so this adjoint rep is not a good example. How about the other question on SU(5)? there I am not asking the adjoint rep but the other reps. $\endgroup$ Aug 1, 2021 at 16:49
  • $\begingroup$ This question has been answered several times on the physics site, since it is a standard drill of the physics curriculum. The most excruciatingly explicit answer, to the point of overkill, is this one; but I could just jot down the condensed version any good physics theory course requires as an exercise, if you were interested. $\endgroup$ Aug 4, 2021 at 0:35

1 Answer 1

3
$\begingroup$

Question: "Namely, how do derive eq (2) from the given eq (1) for the Lie algebra matrix representations?"

Answer: Let $k$ be the real numbers and $K$ the complex numbers and let $G:=SU(2)$ be the special unitary group. It follows its complexification $G_K\cong SL(2,K)$ is the special linear group. The irreducible representations of $SL(2,K)$ are classified. Let $V:=K\{e_1,e_2\}$ be the standard representation of $SL(2,K)$, with dual $V^*:=K\{x_1,x_2\}$. There is a canonical map of $SL(2,K)$-modules

$$ \wedge: V\otimes_K V \rightarrow \wedge^2 V \cong T$$

and since $T$ is the trivial $SL(2,K)$-module it follows $V \cong V^*$ is an isomorphism of $SL(2,K)$-modules. The irreducible $SL(2,K)$-modules are the symmetric powers

$$Sym^d(V) \cong Sym^d(V^*)$$

for $d \geq 0$. "Using" the isomorphism $G_K \cong SL(2,K)$ it follows the irreducible $SL(2,K)$-modules $Sym^d(V),Sym^d(V^*)$ are irreducible $G$-modules with dimension given by

$$dim_K(Sym^d(V))=d+1, dim_k(Sym^d(V))=2(d+1).$$

There are precise relations between irreducible $G$-modules and the irreducible $SL(2,K)$-modules (similar for $SU(3)$ and $SL(3,K)$). The irreducible representations of $SL(n,K)$ have been classified using the "Schur-Weyl" construction. Given any "partition" $\lambda$, there is a Schur-Weyl functor $\mathbb{S}_{\lambda}$ and when you apply this functor to the standard representation

$$V(\lambda):=\mathbb{S}_{\lambda}(V)$$

of $SL(n,K)$ you get all finite dimensional irreducible $SL(n,K)$-modules.

You can find some information in

Hall, Brian - Lie Groups, Lie Algebras, and Representations: An Elementary Introduction (Graduate Texts in Mathematics, 222)

and

Fulton/Harris - Representation Theory: A First Course.

Your comment: "For example, for SU(2), we know the tensor product of 2-dimensional fundamental representation of SU(2) gives the 3-dimensional adjoint representation of SU(2):"

Answer: If $dim_k(V)=2$ it follows $dim_k(V^{\otimes_k n}) = 2^n$. You cannot tensor a 2-dimensional representation with itself and get an odd dimensional representation. The representation $V\otimes V$ may decompose into a direct sum $V_3 \oplus V_1$ with $dim_k(V_i)=i$: The tensor product of two irreducible representations is not necessarily irreducible.

Groups such as $SL(n,K)$ and $SU(n)$ are semi simple Lie groups and they have the property that any finite dimensional $k$ or $K$ representation $V$ decompose into a direct sum

$$V \cong \oplus_i V_i$$

where $V_i$ is an irreducible module. The irreducible modules are classified by their "highest weight vector" and "highest weight". Hence if you want to check if two irreducible modules are isomorphic you must calculate their weights and compare. See the above books for more details.

Example: In your case you have given a basis for $\mathfrak{g}:=\mathfrak{su}(2,k):=k\{\sigma_i\}$

and an explicit representation

$$\rho: \mathfrak{g} \rightarrow End_k(k^2)$$

This induce an explicit adjoint representation

$$ad: \mathfrak{g} \rightarrow End_k(\mathfrak{g})$$

defined by $ad(x)(y):=[x,y]$. Since $dim_k(\mathfrak{g})=3$ you will get explicit matrices $ad(\sigma_i)$ for $i=1,2,3$, and you must calculate these matrices in your given basis $B$. If $x:=\sigma_1$ you must calculate the vectors

$$v_i:=[x, \sigma_i]$$

\begin{align*} [v_i]_B:=\begin{pmatrix} v_{1i} \\ v_{2i} \\ v_{3i} \end{pmatrix} \end{align*}

and construct the matrix

\begin{align*} M_1:= \begin{pmatrix} 0 & v_{12} & v_{13} \\ 0 & v_{22} & v_{23} \\ 0 & v_{32} & v_{33} \end{pmatrix} \end{align*}

Do this for $\sigma_2, \sigma_3$, construct $M_2,M_3$ and compare to the matrices you write down above. You must first check that the basis you give above is a basis for $\mathfrak{su}(2,k)$ as $k$-vector space.

$\endgroup$
21
  • $\begingroup$ thanks this is great - I voted +1 - how about the other one math.stackexchange.com/q/4214397/141334 $\endgroup$ Aug 1, 2021 at 18:01
  • $\begingroup$ About this "If $dim_k(V)=2$ it follows $dim_k(V^{\otimes_k n}) =2^n$. You cannot tensor a 2-dimensional representation with itself and get an odd dimensional representation." --> but you can write $2×2=3_𝐴+1_𝑆$, and hopefully get the $3_𝐴$ out of $2×2$ $\endgroup$ Aug 1, 2021 at 18:04
  • $\begingroup$ @anniemariecœur - the same argument applies: You should study the relation between $SU(n)$-modules and $SL(n,K)$-modules. The irreducible $SL(n,K)$-modules are explicitly constructed in Fulton/Harris using $\mathbb{S}_{\lambda}$. $\endgroup$
    – hm2020
    Aug 1, 2021 at 18:04
  • $\begingroup$ Thanks - but do you know that really can lead to explicit matrix representations of the dimension I asked? or do you know any refs with the matrix given explicitly? $\endgroup$ Aug 1, 2021 at 18:05
  • $\begingroup$ "There are precise relations between irreducible 𝐺-modules and the irreducible 𝑆𝐿(2,𝐾)-modules" what exactly do you define modules here? $\endgroup$ Aug 1, 2021 at 22:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .