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In this script (Link) regarding GUTs it is stated that the unique 3-dimensional complex representation of $\operatorname{SU}(2)$ up to isomorphism is given by Sym$^2\mathbb{C}^2$, the symmetric tensors of rank 2. I was wondering however whether this is indeed true. The author asserts that this representation can also be understood as the Adjoint representation on $\mathbb{C} \otimes \mathfrak{su}(2) \simeq \mathfrak{sl}(2,\mathbb{C})$. The adjoint representation of $\mathfrak{sl}(2,\mathbb{C})$ is however the special linear tensors of rank 2 and not just the symmetric tensors of rank 2. Am I missing something or is the statement wrong?

Samuel Adrian Antz
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gwinship
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If $$V_L~\cong~\mathbb{C}^2\tag{1}$$ denotes the fundamental/defining/left Weyl/spin-$(\frac{1}{2},0)$ representation of the Lie group $\operatorname{SL}(2,\mathbb{C})$, then $$\begin{align} \{M\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid M^t=M\} ~\cong~& {\rm Sym}^2V_L\cr ~\cong~& \mathfrak{sl}(2,\mathbb{C})\cr ~\cong~&\{M\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}M=0\}\cr ~\cong~&\mathfrak{su}(2)\otimes \mathbb{C} \end{align}\tag{2}$$ is the adjoint/spin-$(1,0)$ representation, cf. e.g this Phys.SE post.

Qmechanic
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  • I understand that this should yield the spin-(1,0) representation. The source of my confusion was regarding the isomorphism between Sym$^2V_L$ and $\mathfrak{sl}(2,\mathbb{C}$. Are they only isomorphic as complex linear spaces? I'd assume there is no isomorphism between $Sym^2V_L$ and $\mathfrak{sl}(2,\mathbb{C})$ represented as the symmetric matrices and as traceless matrices such that it is invariant under the action of some matrix from SL(2,$\mathbb{C}$) in the fundamental representation. I hope this clarifies my point of confusion. – gwinship Aug 04 '22 at 08:05
  • They are equivalent representations. – Qmechanic Aug 04 '22 at 08:45
  • Thanks, I only thought about the vector space isomorphism. – gwinship Aug 04 '22 at 09:00