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May 20, 2021 at 9:46 comment added Ben McKay As I remarked above, I don't know of the Maurer-Cartan formalism being treated in the language of group schemes.
May 20, 2021 at 9:14 comment added user122276 Has the Maurer-Cartan formalism been treated in the "language of group schemes" - do you know a reference? The book referenced to above is written in "differential geometric language".
Apr 21, 2021 at 9:47 history edited Ben McKay CC BY-SA 4.0
added references on Maurer-Cartan form, as requested by OP
Apr 21, 2021 at 7:18 comment added user122276 it seems the Maurer-Cartan form is something that exists for a general $2n$-dimensional real Lie group - does the construction above generalize to other such Lie groups? Do you have references?
Apr 20, 2021 at 12:50 history edited Ben McKay CC BY-SA 4.0
added definition of the Maurer-Cartan form
Apr 20, 2021 at 12:37 comment added Ben McKay I cannot find a description in the literature of the Maurer-Cartan form in the context of group schemes, for example, so I am using a concept from differential geometry here: these $\omega_{\mu\bar\nu}$ are the matrix entries of the left invariant Maurer-Cartan form, for which I do not know of a description in the sort of algebraic language you are using.
Apr 20, 2021 at 12:36 comment added user122276 Yes - here you write down some 1-forms - which forms are the forms $\omega_{ij}$ - can you give an explicit formula? They should be explicit elements in $\Omega^1_{T_{SU(3)}}$ and then you must choose a presentation of $\Omega^1_{T_{SU(3)}}$.
Apr 20, 2021 at 12:09 comment added Ben McKay A complex-valued 1-form yields a complex valued function when you plug in a vector field: $\omega_{ij}\in\operatorname{Hom}_R(\operatorname{Der}_k(R),R\otimes_{\mathbb{R}}\mathbb{C})$.
Apr 20, 2021 at 11:55 comment added user122276 Since $T_{SU(3)}$ is trivial of rank 8 there should be an isomorphism of $R$-modules $Der_k(R) \cong R^8$. Hence $\omega_{ij}\in Hom_R(R^8,R)\cong R^8$. And $R^8$ is never isomorphic to $\mathbb{C}^4$.
Apr 20, 2021 at 11:52 comment added user122276 it seems to me the 1-forms you introduce take values in the complex numbers $\omega_{ij} \in \mathbb{C}$. Since $SU(3)$ is an affine variety we may write $SU(3):=Spec(R)$ where $R$ is the coordinate ring of $SU(3)$. A 1-form $\omega_{ij}$ gives canonically an element $w_{ij}\in Hom_R(Der_k(R),R)$ where $Der_k(R)$ is the module of derivations of $R$, hence $\omega_{ij}$ is never a complex number. I ask for a more detailed explanation of your notation....
Apr 17, 2021 at 16:30 history edited Ben McKay CC BY-SA 4.0
explained why it is not a complex Lie group
Apr 17, 2021 at 11:18 comment added user122276 The complex manifold $H_{\mathbb{C}}$ is a real Lie group - are the multiplication and inversion map holomorphic maps, ie is $H_{\mathbb{C}}$ a complex Lie group?
Apr 16, 2021 at 15:41 vote accept CommunityBot
Apr 16, 2021 at 15:36 history edited Ben McKay CC BY-SA 4.0
added references to establish non-Kaehlerity
Apr 16, 2021 at 14:59 comment added user122276 Let $H:=SU(3)$. $H$ is a real algebraic manifold and it seems to me you have constructed a holmorphic structure $H_{\mathbb{C}}$ on $H$ above. Are you claiming that $H_{\mathbb{C}}$ is a non-algebraic complex manifold?
Apr 16, 2021 at 11:53 comment added Ben McKay A 1-form is a linear map of tangent vectors to numbers. The map $\Omega$ is a vector-valued 1-form, so a map $\Omega \colon T_H \to \mathbb{C}^4$, which you can call $\phi$.
Apr 16, 2021 at 10:22 comment added user122276 let $H:=SU(3)$. It seems to me the map $\Omega$ should be a map $\Omega: T_H \rightarrow H \times \mathbb{C}^4$ with $\Omega(x)=(x,\phi(x))$ for some function $\phi$. Can you specify the map $\phi$? It seems you have chosen a trivialization $T_H \cong H\times \mathbb{C}^4$ and consider a map $\Omega \in End( H \times \mathbb{C}^4)$ which is invertible.
Apr 16, 2021 at 9:50 comment added Ben McKay Right: it is 8-dimensional. Fixed
Apr 16, 2021 at 9:49 history edited Ben McKay CC BY-SA 4.0
It is 4-complex dimensional, not 3.
Apr 16, 2021 at 9:47 comment added user122276 The real Lie group $SU(3)$ is 8-dimensional and its real tangent bundle is trivial of rank 8: $T_{SU(3)}\cong SU(3) \times \mathbb{R}^8$. It seems to me that $\Omega_{\epsilon}$ can not be inverted as you claim - it is a map from a real Lie group to a 6-dimensional real vector space. Is there a mistake?
Apr 16, 2021 at 9:38 history edited Ben McKay CC BY-SA 4.0
It is 4-complex dimensional, not 3.
Apr 15, 2021 at 17:27 comment added user122276 as a differential geometer you are used to using local coordinates, with background in algebra/geometry I use sheaves of functions and sheaves of sections of vector bundles. Hence an almost complex structure in my language is an endomorphism $J$ of the tangent bundle (which is a projective $R$-module of rank $2$) with square $=-Id$. If in a local trivialization $J$ is defined using "rational functions", $J$ is algebraic. If $J$ is locally defined using smooth functions it is smooth.
Apr 15, 2021 at 15:40 comment added Ben McKay From your comments, my impression is that my description of an example is not to your taste. I have never seen an almost complex structure described using the Serre-Swan theorem or in terms of vectors fields as forming a module, so I will need some idea of how you prefer to write almost complex structures if I am to provide a clearer description. Perhaps you can give an example of almost complex structures from your preferred perspective.
Apr 15, 2021 at 15:39 history edited Ben McKay CC BY-SA 4.0
more details
Apr 15, 2021 at 13:10 comment added user122276 I ask you to write this answer in terms of the tangent bundle $T_M$, its vector bundle of endomorphisms $End(T_M)$, possibly using the Serre-Swan theorem and the module of smooth sections of $T_M$.
Apr 14, 2021 at 16:30 history edited Ben McKay CC BY-SA 4.0
added 17 characters in body
Apr 14, 2021 at 16:23 history answered Ben McKay CC BY-SA 4.0