There is an embedding of $\mathfrak{su}(2)$ in $\mathfrak{su}(3)$ given by sending the standard basis $$ \begin{pmatrix}i/2&0\\0&-i/2\end{pmatrix},\quad \begin{pmatrix}0&-1/2\\1/2&0\end{pmatrix},\quad\begin{pmatrix}0&-i/2\\-i/2&0\end{pmatrix} $$ of $\mathfrak{su}(2)$ to $$\begin{pmatrix}i\\&0\\&&-i\end{pmatrix},\quad\frac{1}{\sqrt{2}}\begin{pmatrix}&-1\\1&&-1\\&1\end{pmatrix},\quad-\frac{i}{\sqrt{2}}\begin{pmatrix}&1\\1&&1\\&1\end{pmatrix}. $$ in $\mathfrak{su}(3)$.

Since $\mathrm{SU}(2)$ is simply connected, this lifts to a Lie group homomorphism $$ \varphi:\mathrm{SU}(2)\longrightarrow\mathrm{SU}(3). $$

Question. Is there an explicit formula for this map?

We can express any matrix in $\mathrm{SU}(2)$ as $$ \begin{pmatrix}\alpha&-\bar{\beta}\\ \beta&\bar{\alpha}\end{pmatrix} $$ where $\alpha,\beta\in\mathbb{C}$ and $|\alpha|^2+|\beta|^2=1$. Are there explicit expressions for the coordinates of $\varphi(\begin{smallmatrix}\alpha&-\bar{\beta}\\ \beta&\bar{\alpha}\end{smallmatrix})$ in terms of $\alpha,\beta$?

Using the exponential map, we can find special cases, such as $$ \varphi\begin{pmatrix}\alpha&0\\0&\bar{\alpha}\end{pmatrix}=\begin{pmatrix}\alpha^2\\&1\\&&\bar{\alpha}^{2}\end{pmatrix} $$ but I am having some difficulties finding the general formula.

  • $\begingroup$ Have you tried $\begin{pmatrix}i/2&0&0\\0&-i/2&0\\0&0&1\end{pmatrix}$,$\begin{pmatrix}0&-1/2&0\\1/2&0&0\\0&0&1\end{pmatrix}$,$\begin{pmatrix}0&-i/2&0\\-i/2&0&0\\0&0&1\end{pmatrix}$ with proper normalization? $\endgroup$ – mike Aug 22 '19 at 23:10
  • 1
    $\begingroup$ @mike This is a different embedding than the one in the question. Unfortunately, they are not conjugate. $\endgroup$ – Simon Parker Aug 23 '19 at 0:27

I don't know where you got the Lie algebra homomorphism from, but there is just one reasonable possibility what it can be. If you look at your formula, you see that three matrics in $\mathfrak{su}(3)$ that you are writing out don't have a joint eigenvektor, so they defined a complex irreducible representation of $\mathfrak{su}(2)$ and there is just one such representation in complex dimension 3. The simplest realization of this is via the adjoint representation on $\mathfrak{sl}(2,\mathbb C)$. This can be endowed with the Hermitian inner product defined by $(X,Y)\mapsto \frac12tr(X^*Y)$ and an orthonormal basis for this is given by $\left \{\begin{pmatrix} 0 & \sqrt2\\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\begin{pmatrix} 0 & 0\\ \sqrt2 & 0 \end{pmatrix}\right\}$. If I got it right, your matrices just describe the adjoint action of $\mathfrak{su}(2)$ in this basis.

The corresponding group homomorphism sends $A\in SU(2)$ to conjugation by $A$ as a linear map on $\mathfrak{sl}(2,\mathbb C)$. So one just has to determine the matrix representation of this with respect to the above basis. Working this out, you see that (if I got everything right) $$\phi\begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha\end{pmatrix}= \begin{pmatrix}\alpha^2 & \sqrt2 \alpha\bar\beta & -\bar\beta^2 \\ -\sqrt2 \alpha\beta & |\alpha|^2-|\beta|^2 & -\sqrt2\bar\alpha\bar\beta \\ -\beta^2 & \sqrt2 \bar\alpha\beta & \bar\alpha^2 \end{pmatrix}.$$

  • 2
    $\begingroup$ +1 for the key observation that the mapping has to come from the irreducible 3-dimensional representation of $SU(2)$. Conjugation then leaves a lot of extra room, but that's just details. $\endgroup$ – Jyrki Lahtonen Aug 24 '19 at 13:14
  • 2
    $\begingroup$ Another way of thinking about this rep is it's the usual double cover $SU(2)\rightarrow SO(3)$ together with the simple observation that $SO(3)\subseteq SU(3)$. $\endgroup$ – Jason DeVito Aug 24 '19 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.