Timeline for Correspondence between real forms and real structures on complex Lie groups
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24 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 6, 2016 at 7:56 | comment | added | nfdc23 | It can indeed be confusing to sort things out on the purely analytic side (where certain ideas from the algebraic theory break down). Look at D.2 and especially D.3 (which are all about the link between compact Lie groups and reductive complex Lie groups) in Appendix D of the article on reductive group schemes in smf4.emath.fr/en/Publications/PanoramasSyntheses/2014/42-43/… | |
| Oct 6, 2016 at 6:47 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 24, 2015 at 5:09 | comment | added | Mikhail Borovoi | Explanation: by an anti-holomorphic map I mean a semi-algebraic anti-holomorphic map, i.e., given by polynomials in $\bar g_{i,j}$. | |
| May 24, 2015 at 2:19 | comment | added | Mikhail Borovoi | Serre proves the existence of descent (the map from (B) to (A) ) also for quasi-projective varieties (not necessarily affine) and for all algebraic groups (not necessarily affine), see Corollary 2 in Serre's book. For not quasi-projective varieties descent is not always possible, even for $\mathbb{C}/\mathbb{R}$. | |
| May 24, 2015 at 2:05 | comment | added | Mikhail Borovoi | This is called "Galois descent", a reference is Serre's book "Algebraic Groups and Class Fields", Ch. V-20, Prop. 12 (page 142 of Russian edition). This is a reference for any Galois extension $k_1/k$. You should take $k=\mathbb{R}$, $k_1=\mathbb{C}$. | |
| May 24, 2015 at 1:45 | comment | added | Mikhail Borovoi | This is a bijection between the isomorphism classes of: (A) affine real algebraic groups $G_0$ (resp. affine real varieties) and (B) affine complex algebraic groups $G$ (resp. varieties) endowed with a anti-holomorphic involutive automorphism $S$. The map from (A) to (B): $G_0\mapsto G_0\times_R C$. The map from (B) to (A): you take the ring ($\mathbb{C}$-algebra) of regular functions $R=\mathbb{C}[G]$ on $G$, consider the ring ($\mathbb{R}$-algebra) of $S$-invariants $R^S$, and set $G_0={\rm Spec}\,R^S$. | |
| May 23, 2015 at 6:43 | comment | added | Sergei Akbarov | @MikhailBorovoi, excuse me, I did not understand, for which class of groups this is a one-to-one correspondence? And is this a folklore, or there exist references? | |
| May 23, 2015 at 0:03 | comment | added | Mikhail Borovoi | 2. Conversely, if $G_0$ is a compact Lie group, then it is a real algebraic group, i.e., it is the set of real zeros in ${\rm GL}(n,\mathbb{R})$ of a finite set polynomials with real coefficients in the $n^2$ matrix elements $g_{i,j}$. Let $G$ denote the complex Lie group of complex zeros of these polynomials in ${\rm GL}(n,\mathbb{C})$, it has a canonical real structure $S\colon g\mapsto \bar g$ (the complex conjugation on the matix elements). This is the desired complexification of a real algebraic group $G_0$. | |
| May 22, 2015 at 23:50 | comment | added | Mikhail Borovoi | 1. If $G$ is a complex Lie group, and $S\colon G\to G$ is an anti-holomorphic involutive (i.e., $S^2=1$) automorphism, then the subgroup of fixed points $G^S$ of $S$ in $G$ is the desired real Lie group (the real form corresponding to the real structure $S$). | |
| May 22, 2015 at 14:56 | comment | added | Sergei Akbarov | @YCor: Yes, I thought the same, an involutive anti-linear Lie algebra automorphism. So this is not a one-to-one correspondence?... I think, at least for reductive groups (i.e. for complexifications of compact real Lie groups) this must be so... That's true? | |
| May 22, 2015 at 11:07 | comment | added | YCor | I guess you mean that $dS$ is an involutive $\mathbf{C}$-skew-linear Lie algebra automorphism. Also in your definition both $GL_n(\mathbf{R})$ and $GL_n^+(\mathbf{R})$ are real forms on $GL_n(\mathbf{C})$, but only the first can be obtained as the fixed point of a skew-linear involution. | |
| May 22, 2015 at 11:03 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 9:47 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 9:40 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 9:13 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 7:52 | comment | added | Sergei Akbarov | @YCor: Pardon, I corrected the text in MSE (and here also). | |
| May 22, 2015 at 7:46 | comment | added | YCor | I don't understand "i.e." as the first 2 questions are not equivalent: in the first question you ask if the map from a certain set of involutions on $G$ to a certain set (which one?) of subsets of $G$ is a bijection (or an injection, if by definition the target set is by definition the image). In the purportedly equivalent second question, you map to a set of isomorphism classes, and anyway the second question is a surjectivity question. | |
| May 22, 2015 at 7:42 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 7:02 | comment | added | Sergei Akbarov | Does this give an answer in the case, when $K$ is a compact real Lie group and $G$ is its complexification? | |
| May 22, 2015 at 6:58 | comment | added | Peter Michor | On the Lie algebra level, yes. On the Lie group level, if $G$ is simply connected. See Ch III, Section 6 (p 178 ff) of Helgason: Differential Geometry, Lie groups, and symmetric spaces. | |
| May 22, 2015 at 6:53 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 6:25 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| May 22, 2015 at 6:16 | history | asked | Sergei Akbarov | CC BY-SA 3.0 |