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It is well-known that $\mathrm{SL}(2,\mathbb{C})=\mathrm{Spin}^+(3,1)$ is the double cover of the connected component containing the identity of the Lorentz group, see https://physics.stackexchange.com/questions/669776/sl2-mathbbc-double-cover-of-so1-3. The group $\mathrm{Spin}^+(3,1)$ is contained in the Clifford algebra $\mathrm{Cliff}_{3,1}$. The Clifford algebra $\mathrm{Cliff}_{3,1}$ acts on the exterior algebra $\Lambda\mathbb{C}^2$, see section 3.2 of https://arxiv.org/pdf/0904.1556.pdf. This gives a complex 4-dimensional representation of $\mathrm{Spin}^+(3,1)$, called the spin representation of $\mathrm{Spin}^+(3,1)=\mathrm{SL}(2,\mathbb{C})$.

The outer automorphism of the real Lie group $\mathrm{SL}(2,\mathbb{C})$ is $\mathbb{Z}/2$ generated by the complex conjugation exchanging the two semi-spin representations of $\mathrm{SL}(2,\mathbb{C})$.

My question: How does the spin representation of $\mathrm{SL}(2,\mathbb{C})$ decompose into the direct sum of two semi-spin representations and how does the complex conjugation exchange the two semi-spin representations of $\mathrm{SL}(2,\mathbb{C})$, thus fix the spin representation of $\mathrm{SL}(2,\mathbb{C})$?

Thank you!

Borromean
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    The outer automorphisms of ${\rm SL}(2,\Bbb C)$ are (a) trivial if we regard it as a complex Lie group, and (b) generated by complex conjugation if we regard it as a real Lie group. Remind us what it is exactly you're calling the spin representation of ${\rm SL}(2,\Bbb C)$. (Also, if you remember where you "heard" something, you should cite sources when possible.) – anon May 27 '22 at 15:56
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    In fact, even when you look at the Lie algebra $\mathfrak{sl}2(\mathbb C)$ as a (six-dimensional) real Lie algebra (which one might want to clarify in notation, like $R{\mathbb C \vert \mathbb R}$, cf. https://math.stackexchange.com/q/4177437/96384), then its Satake-Tits diagram is made up of two dots connected with an arrow, and its complexification is just two dots, which of course have a non-trivial diagram automorphism. In general, as soon as one looks at various real forms, questions like these become more intricate to answer than just looking at a Dynkin diagram. – Torsten Schoeneberg May 27 '22 at 17:42
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    In short, your first sentence is true a priori only for complex semisimple Lie algebras, and apparently your problem is not set in that case. – Torsten Schoeneberg May 27 '22 at 17:44
  • Thanks for your comments! I edited my question. Actually, I do not understand the spin representation. – Borromean May 28 '22 at 02:04
  • Remind us what it is exactly you're calling the spin representation of ${\rm SL}(2,\Bbb C)$. – anon May 28 '22 at 05:19
  • @runway44 I added my understanding of the spin representation of $\mathrm{SL}(2,\mathbb{C})$. – Borromean May 28 '22 at 05:58
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    ${\rm SL}(2,\Bbb C)$ is a set of matrices and ${\rm Spin}^+(3,1)$ is a subset of a Clifford algebra, so they're not technically equal as sets. Which means implicitly there is some way to reinterpret elements of one as elements of the other. How is this done? And how does the Clifford algebra $C\ell_{3,1}$ act on $\Lambda \Bbb C^2$? – anon May 28 '22 at 06:37
  • The new question is answered on the Lie algebra level in https://math.stackexchange.com/q/4389806/96384, https://math.stackexchange.com/q/4177437/96384, and https://math.stackexchange.com/a/4026224/96384. Since the real $SL_2 (\mathbb C)$ is the simply connected Lie group over the algebra, its rep theory is virtually identical to the Lie algebra one. All irreps are given by pairs of $\ge 0$ integers $(m,n)$; complex conjugation flips the pair; the semi-spin ones must be $(1,0)$ and $(0,1)$; the spin one is $(1,1)$ which, however, should be a tensor product (not sum) of the semi-spin ones. – Torsten Schoeneberg May 28 '22 at 16:02
  • Compare also https://math.stackexchange.com/q/3738143/96384. – Torsten Schoeneberg May 28 '22 at 16:02
  • @TorstenSchoeneberg Thanks for your comments! But I think that the spin representation is $(1,0)\oplus(0,1)$, not $(1,1)$ since the elements of a spin representation are called spinors and the Dirac spinor is the sum of two Weyl spinors. Also, $(1,1)$ is a vector representation, not a spinor spresentation. – Borromean May 29 '22 at 04:04
  • Ah, of course, right. Then it makes sense it's a sum, and it makes sense that conjugation flips the summands hence stabilizes the whole thing. – Torsten Schoeneberg May 29 '22 at 04:53

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