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Which one of the isomorphism is correct? $$SL(2,\mathbb{C})\cong \operatorname{Spin}(1,3,\mathbb{R})?$$ $$SL(2,\mathbb{C})\cong \operatorname{Spin}(3,1,\mathbb{R})?$$

If $SL(2,\mathbb{C})\cong \operatorname{Spin}(1,3;\mathbb{R})$, then what would another interpretations of the isomorphism $\operatorname{Spin}(3,1,\mathbb{R}) \cong ?$

Vice versa.

See also https://math.stackexchange.com/a/3027018/141334

Bernard
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annie marie cœur
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    Is not $Spin(1,3, \mathbb R) \simeq Spin(3,1, \mathbb R)$? – Torsten Schoeneberg May 18 '21 at 21:21
  • The invariant metrics are different. The former is $(+1,-1,-1,-1)$, the latter is $(-1,+1,+1,+1)$. – annie marie cœur May 18 '21 at 23:06
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    That does not mean the Lie groups cannot be isomorphic. – Torsten Schoeneberg May 19 '21 at 04:13
  • If a matrix $A$ preserves a quadratic form $q$, then it also preserves the quadratic form $-q$. Negating a quadratic form reverses its signature. Thus, the groups $O(n,1)$ and $O(1,n)$ are literally the same as subgroups of the general linear group. – Moishe Kohan May 20 '21 at 20:39
  • @anniemarieheart This question currently has a net score of -2, and two close votes. My guess is that the downvoters and close voters are concerned that this question lacks context. Your question might be better received if you added some details---why do you think that one or the other (or both) is correct? or incorrect? Do you have any thoughts on this problem? For other ways in which you can add context, please see the meta question How to ask a good question. – Xander Henderson May 22 '21 at 16:12

1 Answers1

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Indeed, we always have $$ {\rm Spin}(p,q,\Bbb R)\cong {\rm Spin}(q,p,\Bbb R), $$ see for example wikipedia, or nLab.

For the orthogonal groups we have the isomorphism $$ f\colon O(p,q)\rightarrow O(q,p),\; A \mapsto \phi\circ A\circ \phi^{-1}, $$ where $\phi\colon \Bbb R^{p+q}\rightarrow \Bbb R^{q+p}$ is the so-called anti-orthogonal map.

Dietrich Burde
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