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\markboth{E. Kurniadi \& K. Parmikanti} %Jika lebih dari dua penulis, tuliskan sebagai Nama Penulis Pertama dkk.
{The Lie Algebra $\mathfrak{su}(3)$ Representation with Respect to Its Basis}

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\title{THE LIE ALGEBRA $\mathfrak{su}(3)$ REPRESENTATION WITH RESPECT TO ITS BASIS}

\author{EDI KURNIADI$^{a}$\footnote{penulis korespondensi}, KANKAN PARMIKANTI$^{b}$, %NAMA PENULIS KETIGA$^{c}$\\ (ditulis dengan huruf besar, \textbf{TANPA GELAR})
}

\address{$^{a}$ Department of Mathematics of Universitas Padjadjaran, Sumedang, Jawa Barat, Indonesia,\\
$^{b}$ Department of Mathematics of Universitas Padjadjaran, Sumedang, Jawa Barat, Indonesia.\\
%$^{c}$ Instansi Penulis Ketiga.\\
email : \email{edi.kurniadi@unpad.ac.id}}

\maketitle
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\begin{abstract}
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Accepted ..... \quad Revised ..... \quad Published ..... %tanggal-tanggal tersebut \textbf{dikosongkan} saja
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\bigskip

\textbf{Abstrak}. %Dalam bahasa Indonesia
 Aljabar Lie berdimensi delapan dari semua matriks anti-Hermit $3\times 3$ dengan \textit{trace}-nya sama dengan nol dinotasikan oleh $\mathfrak{su}(3)$ dan grup Lie-nya  dinotasikan oleh $\mathrm{SU}(3)$. Tujuan penelitian ini adalah menentukan semua representasi dari $\mathfrak{su}(3)$ melalui basisnya yang direalisasikan dalam ruang polinom homogen kompleks tiga variabel $\mathbb{P}_1$ berderajat tiga. Langkah pertama adalah mengkonstruksi representasi $\mathrm{SU}(3)$ pada $\mathbb{P}_1$ dan langkah kedua menentukan turunan dari representasi $\mathrm{SU}(3)$. Hasil yang diperoleh adalah delapan rumus eksplisit dari representasi $\mathfrak{su}(3)\curvearrowright \mathbb{P}_1$.   

\bigskip

\textbf{Abstract}. % Dalam bahasa Inggris
\textit{The eight-dimensional Lie algebra of $3\times 3$ anti-Hermitian matrices with \changed{its} traces equal to zero is  denoted by $\mathfrak{su}(3)$ whose Lie group is denoted by $\mathrm{SU}(3)$. The research aims to provide all representations of $\mathfrak{su}(3)$ with respect to its basis which is realized on the three complex variables homogeneous polynomials  $\mathbb{P}_1$ of degree three. The first step is to construct representations of $\mathrm{SU}(3)$ on the space $\mathcal{H}$ and the second step is to find all derived representations of $\mathrm{SU}(3)$. The obtained results are eight explicit formulas of representations $\mathfrak{su}(3)\curvearrowright\mathbb{P}_1$.
}  

\end{abstract}

\keywords{Anti-Hermitian Matrices, Derived Representation, $\mathrm{SU}(3)$, $\mathfrak{su}(3)$}

\section{Introduction}
Representation theory of Lie groups and Lie algebras are widely applied in both mathematics and physics (\cite{Mat}, \cite{Dragt}, \cite{Dai}). In mathematics, Lie groups and Lie algebras contribute in many branches of mathematics such as in biological mathematics and finance mathematics (\cite{Taka}, \cite{Her}). Moreover, representations of Lie algebras are vast studied by many researchers (\cite{Hu}-\cite{Rabi}). In addition, matrix Lie groups can be used as a model for understanding representation theory. 

The set of all $n\times n$ matrices whose  their adjoints are equal to their inverse and their determinant are equals one is denoted by $\mathrm{SU}(n)$.  In other words, $\mathrm{SU}(n)$ is the collections of $n\times n$ unitary matrices with determinant \changed{equal to} one. We write it as $\mathrm{SU}(n)=\{S\in \mathrm{U}(n) ; \mathrm{det}(S)=1\}$. The dimension of $\mathrm{SU}(n)$ is $n^2-1$. In case of $n=2$, the unitary-irreducible representation of $\mathrm{SU}(2)\curvearrowright \mathbb{P}_n$ where $\mathbb{P}_n$ is homogeneous polynomials of two-complex variables can be found in (\cite{Bern}, \cite{Brian}). 

In this research, we shall take for case $n=3$ in which the representation of the matrix Lie group $\mathrm{SU}(3)$ can be applied to the field of particle theory \cite{Brian}. Different from previous results, we give \changed{explicit} formulas for representations of $\mathrm{SU}(3)$. The derived representations of $\mathrm{SU}(3)$ arise to the notion of representation of the Lie algebra of $\mathrm{SU}(3)$ which is denoted by $\mathfrak{su}(3)$. 

It is well known that the Lie algebra $\mathfrak{su}(3)$ consists of $3\times 3$ anti-Hermitian matrices whose traces of them equals zero, namely, $\mathfrak{su}(3)=\{P\in \mathrm{M}(3,\mathbb{C}) ; P^*+P=O, \mathrm{tr}(P)=0\}$. Roughly speaking, it is also well known that  $\mathfrak{su}(3)$ contains all smooth left-invariant vector fields of $\mathrm{SU}(3)$. Furthermore, the representations of the Lie algebra $\mathfrak{su}(3)$ is important since it gives the best model in studying representation of matrix Lie groups. The obtained result of representation of the Lie algebra $\mathfrak{su}(3)$ in this research can be developed to the general case $\mathfrak{su}(n) \curvearrowright \mathbb{P}_n$ where $\mathbb{P}_n$ is homogeneous polynomials of $n$-complex variables. 

We organize this paper as follows: In part 1, we introduce the motivation and state of art of our research, in part 2, we deliver some of the relevant \changed{backgrounds}: The Lie group $\mathrm{SU}(3)$, the Lie algebra $\mathfrak{su}(3)$ and its basis, representation of matrix Lie groups, and derived representations, part 3, we state the main result, particularly, in Propotition \ref{prop2} equipped with complete proofs. 

\section{Preliminaries}
In this section, we shall briefly review some of the relevant materials which shall be used in main results: $\mathfrak{su}(3)$ notion and its basis, representation theory, derived representations. As mentioned before the Lie algebra $\mathfrak{su}(3)$ can be witten in the following form:
\begin{eqnarray}
\mathfrak{su}(3)=\{P\in \mathrm{M}(3,\mathbb{C}) ; P^*+P=O, \mathrm{tr}(P)=0\}.	
\end{eqnarray}
\begin{definition}\label{defi1}\cite{sun}
	%\cite{B} mengacu ke referensi no. 2 di Daftar Pustaka. \ref{th1} mengacu kepada nomor teorema ini.
	The Lie algebra $\mathfrak{su}(3)$ has the basis $\mathcal{A}=\{\tau_k=-\frac{i}{2}\zeta_k, \quad i^2=-1$ \mbox{and} $k=1,2,\ldots,8\}$ where $\zeta_k$ are given in the following $3\times 3$ Hermitian matrices: 
	\begin{eqnarray*}
		\zeta_1=\left(\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right),\,\, \zeta_2=\left(\begin{array}{crc}0&-i&0\\i&0&0\\0&0&0\end{array}\right), \,\, \zeta_3=\left(\begin{array}{crc}1&0&0\\0&-1&0\\0&0&0\end{array}\right), \,\, \zeta_4=\left(\begin{array}{ccc}0&0&1\\0&0&0\\1&0&0\end{array}\right),  \\
		\zeta_5=\left(\begin{array}{ccr}0&0&-i\\0&0&0\\i&0&0\end{array}\right), \,\, \zeta_6=\left(\begin{array}{ccc}0&0&0\\0&0&1\\0&1&0\end{array}\right), \,\, \zeta_7=\left(\begin{array}{ccr}0&0&0\\0&0&-i\\0&i&0\end{array}\right), \,\, \zeta_8=\frac{1}{\sqrt{3}}\left(\begin{array}{ccr}1&0&0\\0&1&0\\0&0&-2\end{array}\right). 
	\end{eqnarray*}
\end{definition}
We recall that \changed{the} matrix Lie group $H$ is a subgroup of $\mathrm{GL}_n(\mathbb{C})$ and closed subgroup as well. A representation of $H$ is given by the following definition
\begin{definition}\label{defi2}\cite{Brian}
	A finite complex representation of a matrix Lie group $H$ on a finite complex vector space $\mathcal{V}$ or $H \curvearrowright \mathcal{V}$ is a Lie group homomorphism given by
	\begin{equation}
		\Phi : H\rightarrow \mathrm{GL}(\mathcal{V}). 
		\end{equation}
	\end{definition}
\changed{In the context of representation theory}, Definition \ref{defi2} is equivalent to the following definition:
\begin{definition}\cite{Brian}
	A finite complex representation of a matrix Lie group $H$ on a finite complex vector space $\mathcal{V}$ or $H \curvearrowright \mathcal{V}$ is a linear action of $H$ on  $\mathcal{V}$. Namely, for all $x\in H$ we have
	\begin{equation}
		\Phi(g) : \changed{\mathcal{V}\ni v}\mapsto \Phi(g)v:=g.v\in \mathcal{V}. 
	\end{equation}
\end{definition}
\begin{proposition}
	Let $\Phi:H \curvearrowright \mathcal{V}$ be a representation of a matrix Lie group on the complex vector space $\mathcal{V}$. Then there exists a unique $\pi:\mathfrak{h}\curvearrowright\mathcal{V}$ 
	\begin{equation}
		\pi(A)=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta A})|_{\delta=0}
	\end{equation}
	where $\Phi(e^{\delta A})=e^{\pi(A)}$. 
	\end{proposition}
The representation $\pi:\mathfrak{h}\curvearrowright\mathcal{V}$ is named by \emph{derived representation} associated to $\Phi:H \curvearrowright \mathcal{V}$. Indeed, if $H$ is simply connected then the representation $\pi:\mathfrak{h}\curvearrowright\mathcal{V}$ can be obtained from the representation $\Phi:H \curvearrowright \mathcal{V}$. In our case, $\mathrm{SU}(3)$ is topologically simply connected. Therefore, we can obtain the $\pi:\mathfrak{su}(3)\curvearrowright\mathbb{P}_1$ from $\Phi:\mathrm{SU}(3) \curvearrowright \mathbb{P}_1$ where $\mathbb{P}_1$ is three complex variable homogeneous polynoms. 

\section{Results and Discussion}
Firstly, we compute all exponential matrices in the basis $\mathcal{A}=\{\tau_k=-\frac{i}{2}\zeta_k, \quad i^2=-1$ \mbox{and} $k=1,2,\ldots,8\}$ of $\mathfrak{su}(3)$ as written in Definition \ref{defi1}. All computations we state in the following proposition. 

\begin{proposition} \label{prop1}
Let $\mathcal{A}=\{\tau_k=-\frac{i}{2}\zeta_k, \quad i^2=-1$ \mbox{and} $k=1,2,\ldots,8\}$ be a basis of $\mathfrak{su}(3)$ as written in Definition \ref{defi1}. Then the exponential matrices of elements of $\mathcal{A}$ can be listed in the following forms: 	
\begin{eqnarray*}
	e^{\delta\tau_1}=\left(\begin{array}{ccc}\cos \frac{\delta}{2}&-i\sin \frac{\delta}{2}&0\\-i\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right),\,\, e^{\delta\tau_2}=\left(\begin{array}{crc}0\cos \frac{\delta}{2}&-\sin \frac{\delta}{2}&0\\\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right), \\
	e^{\delta\tau_3}=\left(\begin{array}{ccc}e^{-i/2}&0&0\\0 &e^{i/2}&0\\0&0&1\end{array}\right),\,\, e^{\delta\tau_4}=\left(\begin{array}{rcr}\cos \frac{\delta}{2}&0&-i\sin \frac{\delta}{2}\\0 &1 &0\\-i\sin \frac{\delta}{2}&0&\cos \frac{\delta}{2}\end{array}\right),\\
	e^{\delta\tau_5}=\left(\begin{array}{ccr}\cos \frac{\delta}{2}&0&-i\sin \frac{\delta}{2}\\0 &1&0\\\sin \frac{\delta}{2}&0&\cos \frac{\delta}{2}\end{array}\right),\,\, e^{\delta\tau_6}=\left(\begin{array}{crr}1&0&0\\0&\cos \frac{\delta}{2}&-i\sin \frac{\delta}{2}\\0&-i\sin \frac{\delta}{2}&\cos \frac{\delta}{2}\end{array}\right),\\
	e^{\delta\tau_7}=\left(\begin{array}{ccr}1&0&0\\0 &\cos \frac{\delta}{2}&-\sin \frac{\delta}{2}\\0&\sin \frac{\delta}{2}&\cos \frac{\delta}{2}\end{array}\right),\,\, e^{\delta\tau_8}=\left(\begin{array}{ccc}e^{\frac{-i}{2\sqrt{3}}}&0&0\\0&e^{\frac{-i}{2\sqrt{3}}}&0\\0&0&e^{\frac{i}{\sqrt{3}}}\end{array}\right). 	
\end{eqnarray*}
\end{proposition}

\begin{proof} 
We just compute exponential matrices for $\delta\tau_1$. The exponential matrices of other elements can be computed in similar ways.  We can observe that the matrix $\tau_1$ has three linear independent eigenvectors
\begin{equation}
s_1=\left(\begin{array}{c}0\\0\\1\end{array}\right),\quad 	
s_2=\left(\begin{array}{r}-1\\1\\0\end{array}\right), 
s_3=\left(\begin{array}{c}1\\1\\0\end{array}\right). 
\end{equation}
corresponding to the eigenvalues $\lambda_1=0, \lambda_2=i/2$, and $\lambda_3=-i/2$. By using \changed{Proposition} 2.3 in \cite{Brian}, $e^{\delta\tau_1}$ can be computed as $e^{\delta\tau_1}=Pe^DP^{-1}$ where P is an invertible matrix contained all linearly independent eigenvectors of $\tau_1$ and $D$ is the diagonal matrix contained all $\delta$-multiples of eigenvalues of $\tau_1$. In other words, we get:
\begin{equation*}
e^{\delta\tau_1}=Pe^DP^{-1}=\left(\begin{array}{crc}0&-1&1\\0 &1&1\\1&0&0\end{array}\right)\left(\begin{array}{ccc}1&0&0\\0 &e^{i/2}&0\\0&0&e^{-i/2}\end{array}\right)\left(\begin{array}{crc}0&-1&1\\0 &1&1\\1&0&0\end{array}\right)^{-1}.	
\end{equation*}
Then we have the exponential matrix of $\delta\tau_1$ or 
\begin{equation*}
	e^{\delta\tau_1}=\left(\begin{array}{ccc}\cos \frac{\delta}{2}&-i\sin \frac{\delta}{2}&0\\-i\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right).
	\end{equation*}
  as required. 
\end{proof}

Secondly, we let $\mathrm{SU}(3)$ acts on the space $\mathbb{P}_1$ by $\Phi$. Thirdly, derived representation of $\Phi$ is the representation $\pi$ of $\mathfrak{su}(3)$ realized on the space $\mathbb{P}_1$.  Now we are ready to state \changed{our} main result in the following proposition: 

\begin{proposition} \label{prop2}
Let $\mathcal{A}=\{\tau_k=-\frac{i}{2}\zeta_k, \quad i^2=-1$ \mbox{and} $k=1,2,\ldots,8\}$ be a basis for $\mathfrak{su}(3)$ as written in Definition \ref{defi1} and let 
	\begin{equation}
		\begin{split}
		\mathcal{B}=\{v_1=x^3, v_2=y^3, v_3=z^3, v_4=x^2y, v_5=x^2z, v_6=xy^2, v_7=xz^2,\\
		               \changed{v_8=y^2z, v_9=yz^2, v_{10}=xyz} \} 
		               \end{split}
	\end{equation}
be a basis for three-complex homogeneous polynomial space $\mathbb{P}_1$ of degree 3. 
Then the representations of $\pi:\mathfrak{su}(3)\curvearrowright\mathbb{P}_1$ with respect to the basis of  $\mathcal{A}$ are of the forms:
\begin{eqnarray}
	\hat{\tau}_1v_1&=\pi(\tau_1)v_1&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_1})v_1|_{\delta=0}=\frac{3}{2}iv_4\\
	\hat{\tau}_2v_2&=\pi(\tau_2)v_2&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_2})v_2|_{\delta=0}=-\frac{3}{2}v_6\\
	\hat{\tau}_3v_3&=\pi(\tau_3)v_3&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_3})v_3|_{\delta=0}=0\\
	\hat{\tau}_4v_4&=\pi(\tau_4)v_4&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_4})v_4|_{\delta=0}=iv_{10}\\
	\hat{\tau}_5v_5&=\pi(\tau_5)v_5&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_5})v_5|_{\delta=0}=v_{10}-\frac{1}{2}v_1\\
	\hat{\tau}_6v_6&=\pi(\tau_6)v_6&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_6})v_6|_{\delta=0}=iv_{10}\\
	\hat{\tau}_7v_7&=\pi(\tau_7)v_7&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_7})v_7|_{\delta=0}=-v_{10}\\
	\hat{\tau}_8v_{10}&=\pi(\tau_8)v_{10}&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_8})v_{10}|_{\delta=0}=0.
	\end{eqnarray}
\end{proposition}

\begin{proof} 
Let $x$ be any element of $\mathrm{SU}(3)$ and $v$ be element of $\mathbb{P}_1$. The map $[\Phi(x)v](z)=v(x^{-1}z), z\in \mathbb{C}^3$ is a actually a representation. To see that, let us compute:
\begin{eqnarray*}
	\Phi(x)[\Phi(y)v](z)&=[\Phi(y)v](x^{-1}z)\\
	                    &=v(y^{-1}x^{-1}z)\\
	                    &=\Phi(xy)v(z). 
\end{eqnarray*}
We just compute for the cases $\hat{\tau}_1v_1$, $\hat{\tau}_2v_2$, and the last $\hat{\tau}_2v_3$. Another case can be computed in the similar ways.  Firstly,  from Proposition \ref{prop1} we have 
\begin{equation*}
	e^{\delta\tau_1}=\left(\begin{array}{ccc}\cos \frac{\delta}{2}&-i\sin \frac{\delta}{2}&0\\-i\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right). 
\end{equation*}
By direct computation, we have the inverse of $e^{\delta\tau_1}$ is of the form
\begin{equation*}
	e^{\delta\tau_1}=\left(\begin{array}{ccc}\cos \frac{\delta}{2}&i\sin \frac{\delta}{2}&0\\i\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right). 
\end{equation*}
In addition, we can compute the representations $\Phi(e^{\delta \tau_1})v_1$ as follows:
\begin{eqnarray*}
	\Phi(e^{\delta \tau_1})v_1(z)&&=v_1(e^{-\delta \tau_1}z)\\
	                             &&=v_1\left[\left(\begin{array}{ccc}\cos \frac{\delta}{2}&i\sin \frac{\delta}{2}&0\\i\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)\right]\\
	                             &&=v_1\left[\left(\begin{array}{c}x\cos \frac{\delta}{2}+iy\sin \frac{\delta}{2}\\xi\sin \frac{\delta}{2}+y\cos \frac{\delta}{2}\\z\end{array}\right)\right]\\
	                             &&=(x\cos \frac{\delta}{2}+iy\sin \frac{\delta}{2})^3. 
	\end{eqnarray*}
The derived representation of $\Phi(e^{\delta \tau_1})v_1$ is given by $\hat{\tau}_1v_1$ which is nothing but the representation $\pi$ of $\mathfrak{su}(3)$ represented on $\mathbb{P}_1$. By direct computations, then we get an explicit formula in the following form:
\begin{eqnarray*}
\hat{\tau}_1v_1&=\pi(\tau_1)v_1&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_1})v_1|_{\delta=0}\\
                               &&=\frac{\mathrm{d}}{\mathrm{d}\delta}(x\cos \frac{\delta}{2}+iy\sin \frac{\delta}{2})^3|_{\delta=0}\\
                               &&=3(x\cos \frac{\delta}{2}+iy\sin \frac{\delta}{2})^2(-x/2 \sin \frac{\delta}{2} + iy/2 \cos \frac{\delta}{2})|_{\delta=0}\\
                               &&=3(x)^2(iy/2)=\frac{3}{2}i x^2y\\
                               &&=\frac{3}{2}i v_4. 	
\end{eqnarray*}
Secondly, we can compute inverse of $e^{\delta \tau_2}$, namely we have:
\begin{equation*}
e^{-\delta \tau_2}=\left(\begin{array}{crc}0\cos \frac{\delta}{2}&\sin \frac{\delta}{2}&0\\-\sin \frac{\delta}{2} &\cos \frac{\delta}{2}&0\\0&0&1\end{array}\right).
\end{equation*}
We can observe that
\begin{eqnarray*}
	\Phi(e^{\delta \tau_2})v_2(z)&&=v_2(e^{-\delta \tau_2}z)\\
	&&=(-x\sin \frac{\delta}{2}+y\cos \frac{\delta}{2})^3. 
\end{eqnarray*}
Similarly we have
\begin{eqnarray*}
	\hat{\tau}_2v_2&=\pi(\tau_2)v_2&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_1})v_1|_{\delta=0}\\
	&&=\frac{\mathrm{d}}{\mathrm{d}\delta}(-x\sin \frac{\delta}{2}+y\cos \frac{\delta}{2})^3|_{\delta=0}\\
	&&=3(x\cos \frac{\delta}{2}+iy\sin \frac{\delta}{2})^2(-x/2 \sin \frac{\delta}{2} + iy/2 \cos \frac{\delta}{2})|_{\delta=0}\\
	&&=3(y)^2(-x/2)=-\frac{3}{2}i xy^2\\
	&&=-\frac{3}{2} v_6. 	
\end{eqnarray*}
Thirdly, since $e^{\delta \tau_2}$ is the diagonal matrix then the inverse of $e^{\delta \tau_2}$ is easy to consider. \changed{Then} we have \changed{the} representation $\hat{\tau}_3v_3=0$ as desired. 
\end{proof}



\section{Conclusion}
We concluded that the representations of $\pi:\mathfrak{su}(3)\curvearrowright\mathbb{P}_1$ with respect to the basis of  $\mathcal{A}$ are of the forms:
\begin{eqnarray}
	\hat{\tau}_1v_1&=\pi(\tau_1)v_1&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_1})v_1|_{\delta=0}=\frac{3}{2}iv_4\\
	\hat{\tau}_2v_2&=\pi(\tau_2)v_2&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_2})v_2|_{\delta=0}=-\frac{3}{2}v_6\\
	\hat{\tau}_3v_3&=\pi(\tau_3)v_3&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_3})v_3|_{\delta=0}=0\\
	\hat{\tau}_4v_4&=\pi(\tau_4)v_4&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_4})v_4|_{\delta=0}=iv_{10}\\
	\hat{\tau}_5v_5&=\pi(\tau_5)v_5&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_5})v_5|_{\delta=0}=v_{10}-\frac{1}{2}v_1\\
	\hat{\tau}_6v_6&=\pi(\tau_6)v_6&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_6})v_6|_{\delta=0}=iv_{10}\\
	\hat{\tau}_7v_7&=\pi(\tau_7)v_7&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_7})v_7|_{\delta=0}=-v_{10}\\
	\hat{\tau}_8v_{10}&=\pi(\tau_8)v_{10}&=\frac{\mathrm{d}}{\mathrm{d}\delta}\Phi(e^{\delta \tau_8})v_{10}|_{\delta=0}=0.
\end{eqnarray}
We can generalize to the case representations of $\pi:\mathfrak{su}(N)\curvearrowright\mathbb{P}_k$ where $\mathbb{P}_k$ is $N$ complex variables of degree $N$. 
\section{Acknowledgement}
Authors would like to thank reviewers for valuable comments and suggestions for this article. 

\newpage

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