# Why are the fundamental and anti-fundamental representation in $\text{SL}(2,\mathbb{C})$ not equivalent?

I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $$\text{SL}(2,\mathbb{C})$$, $$2 \times 2$$ matrix with determinant of $$1$$, are not equivalent. This means that no similarity transformation can map one of them to the other.

My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.

Can anyone please explain the explanation of my professor, or perhaps give another explanation?

• I had a hard time following the notation, but the quoted claim is not correct. There is a unique irreducible representation of each dimension up to isomorphism. Nov 25, 2018 at 7:28
• Those notes are talking about the real representations of the group $SO^+(3, 1)$, which has a double cover by $SL(2, \mathbb{C})$. So you're really asking about the real representations of the real Lie group $SL(2, \mathbb{C})$. Perhaps this helps. Nov 26, 2018 at 10:57
• Crossposted from physics.stackexchange.com/q/443092/2451 Dec 1, 2018 at 12:39
• ...now math.stackexchange.com/q/3021290/11127 Math mods: Please merge. Jan 28, 2019 at 16:44

1. For $$G~:=~SL(2,\mathbb{C})~:=~\{g\in {\rm Mat}_{2\times 2}(\mathbb{C})\mid \det g = 1 \}\tag{1}$$ viewed as a complex Lie group, the finite dimensional linear representations should by definition be complex manifolds, which rule out complex conjugate representations in the first place, cf. e.g. this Math.SE post. In physics texts (like the one OP is linking to) the irreducible representations are labelled by an half integer $$j\in \frac{1}{2}\mathbb{N}_0,$$ and of complex dimension $$2j+1$$.

2. For the same group $$G~:=~SL(2,\mathbb{C})~\cong~ Spin(1,3,\mathbb{R})\tag{2}$$ viewed as a real Lie group, it is not hard to see that the complex conjugate representation $$\rho: G\to GL(2,\mathbb{C}), \qquad \rho(g)~=~\bar{g}, \qquad g~\in~ G, \tag{3}$$ of the defining representation (1) is not equivalent, i.e. there does not exist an element $$M\in GL(2,\mathbb{C})$$ such that $$\forall g\in G: Mg=\bar{g}M. \tag{4}$$

3. One complexification of $$G$$ is $$G_{\mathbb{C}}~\cong~Spin(1,3,\mathbb{C})\cong SL(2,\mathbb{C})\times SL(2,\mathbb{C}).$$ In the physics literature the irreducible representations are typically labelled by a pair of half integers $$j_L,j_R\in \frac{1}{2}\mathbb{N}_0$$, cf. e.g. this Phys.SE post. The inequivalent left and right Weyl spinor representations (which OP's link mentions) are labelled $$(1/2,0)$$ and $$(0,1/2)$$, respectively.

• I do not understand why the Lie group being a complex manifold rules out complex conjugate representations. Can you establish why that is? I'm sure it's something obvious Mar 8, 2022 at 1:40
• Hi Craig, Thanks for the feedback. Before I get technical, do you agree that e.g. the map $\mathbb{C}\ni z\mapsto\bar{z}\in \mathbb{C}$ is not a holomorphic map? Mar 8, 2022 at 8:08
• Yes, I agree complex conjugation is not holomorphic (hence not a differentiable map) Mar 8, 2022 at 15:26
• Just to be sure, $z\mapsto\bar{z}$ is real differentiable but not complex differentiable. Mar 8, 2022 at 15:37
• Yes sorry I meant to say complex differentiable! Mar 8, 2022 at 19:30

Looking at Lie algebras as your source does (and using the physics convention for elements of the algebra):

I will modify the regular notation somewhat so as to better fit in with physics standards. For a real matrix Lie algebra with its standard/defining/fundamental representation over a complex vector space given by left multiplication, its complex-conjugate representation is given by $$\overline{\pi}(X)=-X^{\ast}$$ where the star denotes complex conjugation of the entries. The dual representation is given by $$\pi^{d}(X)=-X^{t}$$ If the generators are hermitian, as is the case of the Pauli matrices, you can see these two representations are exactly the same, as $$-\sigma^{\ast}=-\sigma^{t}$$.

On page 75 of your pdf, they show that $$\epsilon (-\sigma_k^{\ast}) \epsilon^{-1}=\sigma_{k}$$ for $$k=1,2,3$$ with $$\epsilon=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$$ This basically states that the fundamental representation of $$\mathfrak{su}(2)$$ is self-dual, as the dual/complex-conjugate/antifundamental representation acts as the standard/fundamental representation when making the change of basis given by $$\epsilon\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}$$ (Note: in fact, all representations of $$\mathfrak{su}(2)$$, and therefore $$\text{SU}(2)$$ as it is simply-connected, are self-dual; see here for a rather more technical explanation, which at its core uses the above change of basis.)

However, for the case of $$\mathfrak{so}(3,1)\simeq\mathfrak{sl}(2,\mathbb{C})$$ (the latter viewed as a real six-dimensional Lie algebra), first note from your source's basis $$\{\sigma_k,i\sigma_k\}_{k=1,2,3}$$ that the generators are no longer hermitian, and therefore the dual and complex-conjugate representations do not coincide. With antifundamental we are then referring to the complex-conjugate representation.

What your notes are saying then is that, having the complex-conjugate representation, we should take a change of basis such that the spacial (hermitian) part of the representation acts as the standard representation. (Note the typo in equation 8.83; $$\sigma^{\ast}_{k}$$ should be $$\sigma_k$$). But by doing so, the boost part does not act as in the standard representation; it differs by a sign. Basically: you cannot make a change of basis such that the complex-conjugate representation becomes exactly the standard one, and so it is inequivalent to it.

Note that these two representations are still self dual, since transposition ignores the $$i$$ factor. For the antihermitian operators we have that $$-\left(i\sigma_k\right)^t=-i\sigma_k^\ast\neq -\left(i\sigma_k\right)^\ast$$ for $$k=1,2,3$$ so the above trick of changing the basis still works.