Tensor Representation of Lie Group

This article is one of Lie Group & Representation contents.

Tensor representation is most common way to represent group action.

(Another representation is spinor representation of $SO(N)$)

SO(N) tensor

Vector

$N\times N$ rotation matrix defined as $$R^T R=I$$ and $$\det R=1$$

This matrix rotate $N$-dim vector $$V^i \rightarrow V'^i = R^{ij} V^j$$ with $i,j=1,2,\cdots, N$.

Tensor

Then $2$-tensor represent as $$T^{ij}\rightarrow (\sum^N_{k=1} \sum^N_{l=1}) T'^{ij}=R^{ik}R^{jl}T^{kl}$$  

We can reduce $N^2$-dim $2$-tensor into $\frac{1}{2}N(N+1)$-dim symmetric tensor, $\frac{1}{2}N(N-1)$-dim antisymmetric tensor.

$A^{ij}$ is tensor, that means $A^{ij}\rightarrow  A'^{ij} \equiv R^{ik} R^{jl} A^{kl}$

$$A^{ij}\equiv T^{ij} - T^{ji},\ A^{ij}=-A^{ji}$$

$S^{ij}$ is tensor, that means $S^{ij}\rightarrow  S'^{ij} \equiv R^{ik} R^{jl} S^{kl}$

$$S^{ij}\equiv T^{ij} + T^{ji},\ S^{ij}=S^{ji}$$

In symmteric tensor, trace component is scalar. $S^{kk}=\delta^{kl}S^{kl}$

$$\tilde{S}^{ij}=S^{ij}-\delta^{ij}(S^{kk}/N)$$

(For $SO(2N)$, this also reduce by self-dual.)

Invariant Symbol

Trace: $$\delta^{ij} R^{ik} R^{jl} = \delta^{kl}$$

Antisymmetric simbol: $\epsilon^{\cdots l \cdots m \cdots}=-\epsilon^{\cdots m \cdots l \cdots}$ and $\epsilon^{12\cdots N}=1$ $$\epsilon^{ijk\cdots n} R^{ip} R^{jq} R^{kr} \cdots R^{ns}=\epsilon^{pqr\cdots s}$$

Dual Tensor

Given antisymmetric tensor $A^{ij}$, define another antisymmetric tensor $$B^{k\cdots n} = \epsilon^{ijk\cdots n} A^{ij}$$

Adjoint Representation

$$[J_a, J_b]=i\epsilon_{abc} J_c$$

$$(J_a)_{bc} = -i\epsilon_{abc}$$

algebra equation become jacobi identity.

SU(N) tensor

Vector and Tensor

$$U^\dagger U = 1$$

Eleminates phase factor $e^{i\alpha}$: $$\det U=1$$

Vector is $$\psi^i \rightarrow \psi'^i = U^{ij} \psi^i$$

Tensor is $$\varphi^{i_1 i_2 \cdots i_m} \rightarrow \varphi'^{i_1i_2\cdots i_m} = U^{i_1j_1}U^{i_2j_2} \cdots U^{i_mj_m} \varphi^{j_1j_2\cdots j_m}$$

Up and Down

Above is not complete of tensor.

Let's define $$\psi_i \equiv \psi^{i*}$$

Then trace is $$\zeta^\dagger \psi = \sum^N_{j=1} \zeta^{j*}\psi^j = \zeta_j \psi^j$$

Also, rotation is $$\psi^i \rightarrow \psi'^i = U^i_j \psi^j$$ and $$\psi_i \rightarrow \psi'_i = \psi_j (U^\dagger)^j_i$$

Then trace is $$\zeta_i \psi^i \rightarrow \zeta'_i \psi'^i = \zeta_j (U^\dagger)^j_i U^i_k  \psi^k = \zeta_j (U^\dagger U)^j_k \psi^k = \zeta_j \delta^j_k \psi^k = \zeta_j \psi^j$$

Irreducible Tensor Representation

$$\varphi^{i_1i_2\cdots i_m}_{j_1j_2\cdots j_n}$$ with $(i_1i_2\cdots i_m)$, $(j_1j_2\cdots j_n)$ are symmteric or antisymmetric (traceless).

(This is only true in $SU(N)$. In general case, we have to use highest weight representation)

Completly Symmetric Tensor

$$S_{(i_1 i_2 \cdots i_k)}$$ This has $nHm$-dimension.

For example, let $S_{ij}=T_{ij}+T_{ji}$, when $T_{ij}=F_iF_j$, $F$ is fundamental representation.

Then highest weight state is $T_{11}$, in Dynkin Coefficient(Chevally Basis), $$[1,0,\cdots,0]+[1,0,\cdots,0]=[2,0,\cdots,0]$$

In general, completly symmetric $k$-tensor is $[k,0,\cdots,0]$ representation in Dynkin Coefficient(Chevally Basis).

Completly Antisymmteric Tensor

$$S_{[i_1 i_2 \cdots i_k]}$$ This has $nCm$-dimension.

For example, let $S_{ij}=T_{ij}-T_{ji}$, when $T_{ij}=F_iF_j$, $F$ is fundamental representation.

Then highest weight state is $T_{12}$, in Dynkin Coefficient(Chevally Basis), $$[1,0,\cdots,0]+[-1,1,\cdots,0]=[0,1,\cdots,0]$$

In general, completly symmetric $k$-tensor is $[k,0,\cdots,0]$ representation in Dynkin Coefficient(Chevally Basis).

Antisymmetric Symbol (Levi-Civita symbol)

$$\epsilon_{i_1i_2\cdots i_N}U^{i_1}_{j_1}U^{i_2}_{j_2}\cdots U^{i_N}_{j_N} = \epsilon_{j_1j_2\cdots j_N}$$

This symbol can use to raise and lower indices for antisymmetric tensor to another antisymmetric tensor.

For example, in $SU(4)$, $\varphi_{kpq}\equiv \epsilon_{ijpq}\varphi_k^{ij}$ is tensor, because $$\varphi_{kpq}\equiv \epsilon_{ijpq}\varphi_k^{ij} \rightarrow \epsilon_{ijpq}U^i_lU^j_m(U^\dagger)^n_k\varphi^{lm}_n=(U^\dagger)^s_p(U^\dagger)^t_q((U^\dagger)^n_k \epsilon_{lmst}\varphi^{lm}_n)=(U^\dagger)^s_p(U^\dagger)^t_q(U^\dagger)^n_k \varphi_{nst}$$

This is largest completly antisymmetric tensor, with 1 component(dimension).

$SU(N)$ tensors

$[n_1,n_2,\cdots,n_{N-1}]$

Each $n_i$ are number of symmetric components (horizontal)

Each $i$ are number of antisymmetric (vertical)

In Young Tableaux, this have $n_i$ numbers of hight $i$ blocks.

Also, this is highest weight representation, it is just $$[n_1,n_2,\cdots,n_{N-1}]$$

(In Dynkin Coefficient of Chevally Basis.)

Young Tableaux

Symmetric Group $S_n$ has 'Young Diagram' representation. we use this at index of tensor. 

For $SU(3)$, representation is $$A^{i_1\cdots i_n}_{j_1\cdots j_m}$$ where $(i_1\cdots i_n)$ and $(j_1\cdots j_m)$ are symmetric, traceless.

We can raise all the lower indices with $\epsilon$ tensors to get $$a^{i_1\cdots i_n k_1l_1\cdots k_ml_m}=\epsilon^{j_1k_1l_1}\cdots\epsilon^{j_mk_ml_m}A^{i_1\cdots i_n}_{j_1\cdots j_m}$$

Then antisymmetric in each pair, $k_i \leftrightarrow l_i$, and symmetric in the exchange of pairs, $k_i,l_i \leftrightarrow k_j,l_j$.

Symmetrize in the components in the rows, then antisymmetrize in the components in the columns. Also has a property the is the analog of traclelessness. $$\epsilon_{i_1k_1l_1}a^{i_1\cdots i_n k_1l_2 \cdots k_ml_m}=0$$

For $SU(3)$, the tensors corresponding to Young tableaux with more than three boxes in any column vanish beause no tensor can be completly antisymmetric in four or more indices which take on only three valuees.

Clebsch-Gordan decomposition

Dimension

Hook length formula shows dimension of representation with Young Tableaux of $S_n$.

For $SU(N)$, $$D=\prod_{\mbox{boxes}}\frac{N+d_\mbox{box}}{h_\mbox{box}}$$ where $d_\mbox{box}$ is assigned as shown here: 

and $h_\mbox{box}$ are called the hook lengths and are given by one plus the number of boxes to the right of the box plus the number of boxes below the box. For example, 

Fundamental Representation

Highest weight: $[1,0,\cdots,0]$

Tensor Form: $T^i$

Group element act $$T\rightarrow UT$$

Just vector.

Adjoint Representation

Adjoint representation is generator number's dimensional representation.

$$\mbox{ad}_X Y=[X,Y]$$

(Also in matrix element, it say just same as structure constant.)

This is similar to embedding finite group to symmetric group $S_n$ for generator.

So in adjoint representation, each weight is same as root system.

In other view, it is just group action with conjugation.

$$\mbox{Ad}_{e^{iX}}(Y)=e^{iX}Ye^{-iX}$$

For $SU(N)$, highest weight: $[1,0,\cdots,0,1]$

Tensor Form: $T^i_j$ (traceless)

Group element act $$T\rightarrow U^{-1}TU$$

For $SO(2N)$ and $SO(2N+1)$, highest weight: $[0,1,0,\cdots,0]$

Tensor Form: $T^{[ij]}$ (antisymmetric)

For $Sp(2N)$, highest weight: $[2,0,\cdots,0]$

Tensor Form: $T^{(ij)}$ (symmetric)

Reference

Zee - Group Theory in a Nutshell for Physicists

Georgi - Lie Algebras in Particle Physics

Hall - Lie Groups, Lie Algebras, and Representations

Matthew Foster - 8. Highest Weight Representation

The Algebra su(N), the group SU(N), and Chern-Simons theory

$S_k$ and Tensor Representations

Wikipedia - Compact Group