11. Representations of matrix groups
In the last part of this module, we will look at some introductory content into representations of matrix groups. Recall:
\(\operatorname{GL}_n(\mathbb F)\) — all \(n\times n\) matrices with non-vanishing determinant
\(\operatorname{O}_n=\{A\in\operatorname{GL}_n(\mathbb R)\colon AA^\top=I\}\)
\(\operatorname{U}_n=\{A\in\operatorname{GL}_n(\mathbb C)\colon AA^\dagger=I\}\) where \(A^\dagger=\bar A^\top\) is the hermitian conjugate, that is, transposition of the complex conjugation
\(\operatorname{SU}_n=\{A\in\operatorname{GL}_n(\mathbb C)\colon AA^\dagger=I\text{ and }\det A=1\}\)
\(\operatorname{Lor}=\{A\in\operatorname{GL}_n(\mathbb R)\colon A^\top\eta A=\eta\}\)
In the following we will focus on \(\operatorname{SU}_n\) and try to find some of their (irreducible) representations.
There are various ways allowing to construct such representations, for example using the theory of Lie groups and their relations to Lie algebras, but we will focus on the method based on the so-called transplantation of a function.
Let us focus on \(\operatorname{SU}_2\). Consider a set of monomials of degree \(n\) in the two variables \(x_1\) and \(x_2\):
\[x_1^n,x_1^{n-1}x_2,x_1^{n-2}x_2^2,\dotsc,x_1x_2^{n-1},x_2^n\]
Every element \(A\) of \(\operatorname{SU}_2\) can be written as
\[\begin{split}A=\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}\text{ where }a\bar a+b\bar b=1\text{ and }a,b\in\mathbb C\end{split}\]
Then
\[\begin{split}AA^\dagger=\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}\cdot\begin{pmatrix}\bar a&-b\\\bar b&a\end{pmatrix}=\begin{pmatrix}a\bar a+b\bar b&-ab+ab\\-\bar b\bar a+\bar a\bar b&b\bar b+a\bar a\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\end{split}\]
as expected.
Define:
\[\begin{split}\begin{pmatrix}y_1\\y_2\end{pmatrix}=A^{-1}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}\bar a&-b\\\bar b&a\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}\implies\begin{cases}y_1=\bar ax_1-bx_2\\y_2=\bar bx_1+ax_2\end{cases}\end{split}\]
Then every monomial \(y_1^ky_2^{n-k}\) can be expanded in the basis of monomials \(x_1^lx_2^{n-l}\). The coefficients of this expansion define an \(n\times n\) matrix \(M(A)\) that provides an \((n+1)\)-dimensional representation of \(\operatorname{SU}_2\):
\[\operatorname{SU}_2\ni A\mapsto M(A)\in\operatorname{GL}_{n+1}(\mathbb C)\]
Example
\(n=0\). The only monomial we get is \(1\) and \(M(A)=(1)\). This is the trivial, one-dimensional representation.
\(n=1\). There are only two monomials: \(y_1\) and \(y_2\). Then \(y_1=\bar ax_1-bx_2\), \(y_2=\bar bx_1+ax_2\) and
\[\begin{split}M(A)=\begin{pmatrix}\bar a&-b\\\bar b&a\end{pmatrix}\end{split}\]
This is a two-dimensional representation of \operatorname{SU}_2$.
\[\begin{split}\left.\begin{aligned}
y_1^2&=\bar a^2x_1^2-2\bar abx_1x_2+b^2x_2^2\\
y_1y_2&=\bar a\bar bx_1^2+(a\bar a-b\bar b)x_1x_2-abx_2^2\\
y_2^2&=\bar b^2x_1^2+2a\bar bx_1x_2+a^2x^2
\end{aligned}\right\}\implies M(A)=\begin{pmatrix}
\bar a^2&-2\bar ab&b^2\\
\bar a\bar b&a\bar a-b\bar b&-ab\\
\bar b^2&2a\bar b&a^2
\end{pmatrix}\end{split}\]
This is a three-dimensional representation of \(\operatorname{SU}_2\). It can be easily checked by checking that:
\[M(A)\cdot M(B)=M(A\cdot B)\leftarrow\text{homomorphism condition}\]
This way one can construct an \((n+1)\)-dimensional representation \(\theta_n\) of \(\operatorname{SU}_2\) for all \(n=0,1,2,\dotsc\)
the representations \(\theta_n\) are irreducible and mutually inequivalent (since they have different dimensoins)
it turns out that the list of representations \(\theta_n\) contains all irreducible representations
We can calculate characters of these representations:
\(\chi_{\theta_0}(A)=1\)
\(\chi_{\theta_1}(A)=\operatorname{Tr}\begin{pmatrix}\bar a&-b\\\bar b&a\end{pmatrix}=a+\bar a=\operatorname{Tr}(A)\)
\(\chi_{\theta_2}(A)=\bar a^2+a\bar a-b\bar b+a^2=(\bar a+a)^2-a\bar a-b\bar b=\left(\operatorname{Tr}(A)\right)^2-1\)
more complicated beyond \(n=2\).
11.1. Beyond \(\operatorname{SU}_2\)
We can use a similar construction for all \(\operatorname{SU}_n\) by considering homogeneous monomials in \(n\) variables: \(x_1,x_2,\dotsc,x_n\).
Importantly, for every \(n\) there always exists an \(n\)-dimensional irreducible representation of \(\operatorname{SU}_n\), called the fundamental representation.
Even more importantly, all finite-dimensional representations of \(\operatorname{SU}_n\) can be generated from the fundamental representation using the so-called tensor method.
11.2. Tensor method
Given two representations \(R_1\) and \(R_2\) of a group \(G\), let \(V_1\) and \(V_2\) be their representation spaces, with \(\dim V_1=n\) and \(\dim V_2=m\). Denote by \(V_1\otimes V_2\) the space of pairs: \(V_1\otimes V_2=\{(v_1,v_2)\colon v_1\in V_1,v_2\in V_2\}\).
Then the group \(G\) acts on \(V_1\otimes V_2\) by
\[ g.(v_1,v_2)=(g.v_1,g.v_2)\]
This action defines a representation that is the tensor product of represnetations \(V_1\) and \(V_2\). We will denote this representation \(V_1\otimes V_2\).
Example
For \(\operatorname{SU}_2\) we can strudy the representation \(\theta_1\otimes\theta_1\): If \(e_1,e_2\) are the basis vectors of \(\theta_1\) and \(M_{\theta_1}(A)=\begin{pmatrix}\bar a&\bar b\\-b&a\end{pmatrix}\), then
\[\begin{split}\begin{align}
M_{\theta_1}(A)e_1&=\begin{pmatrix}\bar a&\bar b\\-b&a\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}\bar a\\-b\end{pmatrix}=\bar ae-1-be_2\\
M_{\theta_1}(A)e_2&=\begin{pmatrix}\bar a&\bar b\\-b&a\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}\bar b\\a\end{pmatrix}=\bar be_1+ae_2
\end{align}\end{split}\]
Therefore
\[\begin{split}\begin{align}
A.(e_1,e_2)&=(\bar ae_1-be_2,\bar ae_1-be_2)=\bar a^2e_1,e_1)-\bar ab(e_1,e_2)-\bar ab(e_2,e_1)+b^2(e_2,e_2)\\
A.(e_1,e_2)&=(\bar ae_1,be_2,\bar be_1+ae_2)=\bar a\bar b(e_1,e_1)+\bar aa(e_1,e_2)-b\bar b(e_2,e_1)-ab(e_2,e_2)\\
A.(e_2,e_1)&=(\bar be_1+ae_2,\bar ae_1-be_2)=\bar a\bar b(e_1,e_1)-b\bar b(e_1,e_2)+a\bar a(e_2,e_1)-ab(e_2,e_2)\\
A.(e_2,e_2)&=(\bar be_1+ae_2,\bar be_1+ae_2)=\bar b^2(e_1,e_1)+a\bar b(e_1,e_2)+a\bar b(e_2,e_1)+a^2(e_2,e_2)
\end{align}\end{split}\]
Now, the basis of \(\theta_1\otimes\theta_1\) is:
\[(e_1,e_1),(e_1,e_2),(e_2,e_1),(e_2,e_2)\]
and we get the following 4×4 matrix
\[\begin{split}
M_{\theta_1\otimes\theta_1}=\begin{pmatrix}
\bar a^2&\bar a\bar b&\bar a\bar b&\bar b^2\\
-\bar ab&\bar aa&-b\bar b&a\bar b\\
-\bar ab&\bar aa&-b\bar b&a\bar b\\
b^2&-ab&-ab&a^2
\end{pmatrix}
\end{split}\]
It provides a four-dimensional representation of \(\operatorname{SU}_2\). This representation is reducible. Notice that
\[\begin{split}\begin{align}
\chi_{\theta_1\otimes\chi_1}(A)&=\bar a^2+\bar aa+\bar aa+a^2=\bar a+2a\bar a+a^2\\
&=\left(\operatorname{Tr}(A)\right)^2=\left(\operatorname{Tr}(A)\right)^2-1+1\\
&=\chi_{\theta_0}(A)+\chi_{\theta_2}(A)
\end{align}\end{split}\]
It means that the tensor representation \(\theta_1\otimes\theta_1\) decomposes as:
\[\theta_1\otimes\theta_1=\theta_0\oplus\theta_2\]
11.3. General story
We will use Young diagrams to study tensor representations of \(\operatorname{SU}_n\).
We represent the fundamental representation as a single box:
\[\text{fundamental rep of }\operatorname{SU}_n \to \square\]
Other representations will be related to more general Young diagrams.
11.4. Rules for tensor products of Young diagrams
We only consider tensor products of a generic Young diagram \(\Gamma\) times the fundamental reprensetation.
Write down the diagram \(\Gamma\otimes\) a single box
\[\left\{\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\square \right\}\]
Attach the single box on the right of each row of \(\Gamma\)
\[\left\{\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\textcolor{blue}{\square}=\overset{\displaystyle\square\square\textcolor{blue}{\square}}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\textcolor{blue}\square}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square\textcolor{blue}{\square}}{\square\phantom\square}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\underset{\displaystyle\textcolor{blue}\square{\phantom\square}}{\square{\phantom\square}}}{\square\phantom\square}}
\right\}\]
Remove illegal Young diagrams (i.e. the ones where the number of boxes in rows is not non-increasing.
\[\left\{\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\square=\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}
\oplus
\textcolor{red}{
\overset{\text{illegal}}
{
\left.\overset{\displaystyle\square\square}{\underset{\displaystyle\square\square}{\square\phantom\square}}\!\!\!\!\!\!\!\!\right/}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\underset{\displaystyle\square{\phantom\square}}{\square{\phantom\square}}}{\square\phantom\square}}
\right\}\]
If there is any column with exactly \(n\) boxes, then remove this column. For example, assuming we work in \(\operatorname{SU}_4\), then:
\[\begin{split}\left\{
\begin{aligned}
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\square&=\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}
\oplus
\textcolor{red}{
\left.\overset{\displaystyle\square\square}{\underset{\displaystyle\square\square}{\square\phantom\square}}\!\!\!\!\!\!\!\!\right/}
\oplus\,\,\,\,
\textcolor{red}{
\left|\!\!\!\!
\textcolor{black}{
\overset{\displaystyle\square\square}{\underset{\displaystyle\underset{\displaystyle\square{\phantom\square}}{\square{\phantom\square}}}{\square\phantom\square}}}\right.}\\
&=\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}\oplus\square
\end{aligned}
\right\}\end{split}\]
Example
Consider the following tensor product of Young diagrams in \(\operatorname{SU}_3\):
\[\underset{\displaystyle\square{\phantom\square}}{\square\square}\otimes\square\otimes\square \]
First calculate
\[\begin{split}\begin{align}
\underset{\displaystyle\square{\phantom\square}}{\square\square}\otimes\square
&=\underset{\displaystyle\square\phantom{\square\square}}{\square\square\textcolor{blue}\square}
\oplus
\underset{\displaystyle\square\textcolor{blue}\square}{\square\square}
\oplus\,\,\,
\textcolor{red}{\left|\!\!\!\textcolor{black}{\overset{\displaystyle\square\square}{\underset{\displaystyle\textcolor{blue}\square{\phantom\square}}{\square\phantom\square}}}\right.}\\
&=\underset{\displaystyle\square\phantom{\square\square}}{\square\square\square}
\oplus
\underset{\displaystyle\square\square}{\square\square}
\oplus\square
\end{align}\end{split}\]
Then
\[\begin{split}\begin{align}
\underset{\displaystyle\square{\phantom\square}}{\square\square}\otimes\square\otimes\square&=
\left(\underset{\displaystyle\square\phantom{\square\square}}{\square\square\square}
\oplus
\underset{\displaystyle\square\square}{\square\square}
\oplus\square\right)\otimes\square\\
&=\underset{\displaystyle\square\phantom{\square\square\square}}{\square\square\square\textcolor{blue}\square}
\oplus
\underset{\displaystyle\square\textcolor{blue}\square\phantom{\square}}{\square\square\square}
\oplus\,\,\,
\textcolor{red}{
\left|\!\!\!\textcolor{black}{\overset{\displaystyle\square\square\square}{\underset{\displaystyle\textcolor{blue}\square\phantom{\square\square}}{\square\phantom{\square\square}}}}\right.}
\oplus
\textcolor{red}{
\left.
\textcolor{black}{\underset{\displaystyle\square\square\phantom{\square}}{\square\square\textcolor{blue}\square}
\oplus
\underset{\displaystyle\square\square\textcolor{blue}\square}{\square\square\phantom\square}}\!\!\!\!\!\!\!\!\!\!\right/}\,\,\,\,\,\,
\oplus\,\,\,
\textcolor{red}{\left|\!\!\!\textcolor{black}{
\overset{\displaystyle\square\square}{\underset{\displaystyle\textcolor{blue}\square\phantom\square}{\square\square}}}\right.}
\oplus\square\textcolor{blue}\square\oplus\underset{\displaystyle\textcolor{blue}\square}\square\\
&=\underset{\displaystyle\square\phantom{\square\square\square}}{\square\square\square\square}
\oplus
\underset{\displaystyle\square\textcolor{blue}\square\phantom{\square}}{\square\square\square}
\oplus
2\,\underset{\displaystyle\square\square\phantom\square}{\square\square\square}
\oplus
2\,\square\square+2\,\underset{\displaystyle\square}\square
\end{align}\end{split}\]
11.5. Dimensions
Each Young diagram corresponds to an irreducible representation of \(\operatorname{SU}_n\). There is a simple formula to find the dimension of each representation.
Let \(\Gamma\) be the Young diagram associated to an irreducible representation \(R\). Then
\[\dim(R)=\frac ND\]
To find \(N\) and \(D\), draw two copies of the Young diagram and and fill them up with numbers:
\[\overset{\displaystyle\boxed{n\phantom{\,+0}}\boxed{n+1}\boxed{n+2}}{\underset{\displaystyle\boxed{n-2}\phantom{\boxed{n+0}\boxed{n+0}}}{\boxed{n-1}\boxed{n\phantom{\,+0}}\phantom{\boxed{n+2}}}}\]
\[\overset{\displaystyle\boxed5\boxed3\boxed1}{\underset{\displaystyle\boxed1\phantom{\boxed0\boxed0}}{\boxed3\boxed1\phantom{\boxed0}}}\]
\[\dim\left(\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square\phantom{\square\square}}{\square\square\phantom{\square}}}\right) = \frac{n(n+1)(n+2)(n-1)n(n-2)}{5\cdot3\cdot1\cdot3\cdot1\cdot1}\]
Example
Let us check that the dimensions on both sides of the equation:
\[
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\square=\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}\oplus\square
\]
agree, assuming that these are representations of \(\operatorname{SU}_4\).
\[\begin{split}
\begin{align}
\dim\left(\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\right)&=\left\{\frac{\displaystyle\overset{\displaystyle\boxed4\boxed5}{\underset{\displaystyle\boxed2{\phantom{\boxed0}}}{\boxed3\phantom{\boxed0}}}}{\displaystyle\overset{\displaystyle\boxed4\boxed1}{\underset{\displaystyle\boxed1\phantom{\boxed0}}{\boxed2\phantom{\boxed0}}}}\right\}=\frac{4\!\!\!/\cdot5\cdot3\cdot2\!\!\!/}{4\!\!\!/\cdot2\!\!\!/\cdot1\cdot1}=15\\
\dim(\square)&=\left\{\frac{\displaystyle\boxed4}{\displaystyle\boxed1}\right\}=\frac41=4\\
\dim\left(\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}\right)&=\left\{\frac{\displaystyle\overset{\displaystyle\boxed4\boxed5\boxed6}{\underset{\displaystyle\boxed2\phantom{\boxed0\boxed0}}{\boxed3\phantom{\boxed0\boxed0}}}}{\displaystyle\overset{\displaystyle\boxed5\boxed2\boxed1}{\underset{\displaystyle\boxed1{\phantom{\boxed0\boxed0}}}{\boxed2\phantom{\boxed0\boxed0}}}}\right\}=\frac{4\cdot5\!\!\!/\cdot6\cdot3\cdot2\!\!\!/}{5\!\!\!/\cdot2\cdot1\cdot2\!\!\!/\cdot1}=36\\
\dim\left(\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}\right)&=\left\{\frac{\displaystyle\overset{\displaystyle\boxed4\boxed5}{\underset{\displaystyle\boxed2\phantom{\boxed0}}{\boxed3\boxed4}}}{\displaystyle\overset{\displaystyle\boxed4\boxed2}{\underset{\displaystyle\boxed1\phantom{\boxed0}}{\boxed3\boxed1}}}\right\}=\frac{4\!\!\!/\cdot5\cdot3\!\!\!/\cdot4\cdot2\!\!\!/}{4\!\!\!/\cdot2\!\!\!/\cdot3\!\!\!/\cdot1\cdot1}=20\\
\end{align}
\end{split}\]
We found:
\[\begin{split}
\begin{align}
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\phantom\square}}\otimes\square&=\overset{\displaystyle\square\square\square}{\underset{\displaystyle\square{\phantom{\square\square}}}{\square\phantom{\square\square}}}
\oplus
\overset{\displaystyle\square\square}{\underset{\displaystyle\square{\phantom\square}}{\square\square}}\oplus\square\\
\text{dim:}\qquad15\cdot 4&=36+20+4\qquad\text{✓}
\end{align}
\end{split}\]
