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I'm working on a theoretical framework where I explore different quotient spaces formed with GL$^+$(4,R) and various groups. Specifically, I'm interested in the types of geometry that arise from the following quotient spaces:

  1. GL$^+$(4,R)/SO(3,1)
  2. GL$^+$(4,R)/Spin(3,1)
  3. GL$^+$(4,R)/Spin$^c$(3,1)

For GL$^+$(4,R)/SO(3,1), the situation is relatively clear as it leads to the usual symmetric, non-degenerate metric tensors commonly used in theories of gravity. However, I encounter difficulties when trying trying to understand the GL$^+$(4,R)/Spin(3,1) and GL$^+$(4,R)/Spin$^c$(3,1) cases. In these scenarios, what geometry is described by these?

To provide some context, I am using the Majorana representation of the Dirac matrices to construct a representation of 4x4 matrices. Furthermore in this representation Spin$^c$(3,1) can be represented by exp (f+b), where f is a bivector and b a pseudo-scalar which is in GL$+$(4,R). My question is:

For instances GL+(4,R)/SO(3,1)xR spawns Weyl conformal geometry. But what do GL$^+$(4,R)/Spin(3,1) and GL$^+$(4,R)/Spin$^c$(3,1) spawn?

Any insights, references, or suggestions on how to approach these constructions would be greatly appreciated.

Anon21
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    To begin with, what is FX? – Moishe Kohan Nov 21 '23 at 21:25
  • Then there are no metrics for the 2nd and 3rd quotients. For the 2nd you can talk about an invariant complex volume form. – Moishe Kohan Nov 22 '23 at 00:20
  • @MoisheKohan Well, I am interested in the geometry spawned by GL+(4,R)/Spin^c(3,1). For instance GL+(4,R)/SO(3,1)xR spawns Weyl geometry with the conformal group. – Anon21 Nov 22 '23 at 01:13
  • Then you should edit your question appropriately. – Moishe Kohan Nov 22 '23 at 01:24
  • @MoisheKohan edited – Anon21 Nov 22 '23 at 01:41
  • How do you realize $Spin(3,1)$ and $Spin^c(3,1)$ as subgroups of $GL^+(4,\mathbb R)$? If this is via their standard action on $\mathbb R^4$, then there is no difference between cases 1. and 2, since the map $Spin(3,1)\to GL(4,\mathbb R)$ factorises through $SO(3,1)$ and hence these define the same subgroup of $GL^+(4,\mathbb R)$. (Probably the same applies to $Spin^c$.) – Andreas Cap Nov 22 '23 at 09:07
  • @AndreasCap Well I was thinking exp(a+x+f+v+b) for GL+(4,R) and exp(f+b) for Spin^c(3,1), where a is a scalar, x a vector, f a bivector, v a trivector and b a pseudoscalar of 3+1 geometric algebra. – Anon21 Nov 22 '23 at 12:06
  • @AndreasCap: $Spin(3,1)\cong SL(2,C)$ and the latter has standard embedding in $GL(4,R)$. Thus, an invariant complex volume form. – Moishe Kohan Nov 22 '23 at 14:09
  • @MoisheKohan Can you explain it akin to how GL+(4,R)/SO(3,1) produces inner products, GL+(4,R)/RxSO(3,1) produces conformal geometry (inner products with scale invariances)... what does GL+(4,R)/Spin(3,1) produces? Modified inner products... how modified? – Anon21 Nov 22 '23 at 15:29
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    Not an inner product of any kind but a complex volume form. I will add an answer in a day or so. – Moishe Kohan Nov 22 '23 at 15:43
  • @MoisheKohan Thank you so much! Looking forward! – Anon21 Nov 22 '23 at 16:08
  • @MoisheKohan : Then it should be a complex structure plus a complex volume form. – Andreas Cap Nov 22 '23 at 19:39
  • @AndreasCap: Of course. – Moishe Kohan Nov 22 '23 at 20:21

1 Answers1

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The group $Spin(3,1)$ is nothing but $SL(2, {\mathbb C})$. The latter has a natural embedding in $GL(4, {\mathbb R})$. The subgroup $SL(2, {\mathbb C})< GL(4, {\mathbb R})$ preserves two things:

  1. The standard complex structure on ${\mathbb R}^4\cong {\mathbb C}^2$.

  2. The complex volume form on ${\mathbb C}^2$, i.e. the alternating complex 2-form $dz_1 \wedge dz_2$.

Accordingly, suppose that $M$ is a smooth 4-dimensional manifold whose frame bundle admits a reduction to a vector bundle with the structure group $Spin(3,1)$. Then this reduction yields an almost complex structure $J$ on $M$, together with a complex volume form $\omega$ (where "complex" is understood with respect to the above almost complex structure $J$), i.e. $\omega$ is a nowhere vanishing complex 2-form. I can spell out the technical meaning of the above terms if you are interested. Things are a bit more intuitive if the almost complex structure $J$ is integrable, i.e. corresponds to a complex structure on the manifold $M$. Then, in local holomorphic coordinates, a form $\omega$ as above can be written as $f(z)dz_1 \wedge dz_2$, where $f(z)=f(z_1,z_2)$ is a smooth complex-valued function (which need not be holomorphic).

There is no "(semi)metric" of any kind on $M$ corresponding to such a reduction, since $Spin(3,1)< GL(4, {\mathbb R})$ does not preserve any nonzero real symmetric bilinear form on ${\mathbb R}^4$ (nor a complex symmetric bilinear form on on ${\mathbb C}^2$).

How useful it is, I am not sure.

I do not have a good geometric interpretation of a $Spin^c(3,1)$-structure. Algebraically speaking, $Spin^c(3,1)= S^1\cdot SL(2, {\mathbb C})$ (it is not a direct product since $S^1$, identified with the group of unitary scalar matrices of the form $Diag(e^{i\theta}, e^{i\theta})$, has nontrivial intersection with $SL(2, {\mathbb C})$). Accordingly, there exists a $Spin^c(3,1)$-invariant density, $|dz_1\wedge dz_2|$.

A reduction of the frame bundle to the structure group $Spin^c(3,1)$ does yield an almost complex structure on the 4-manifold $M$ and a real-valued density.

Moishe Kohan
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  • In the context of the quotient $GL^+(4,\mathbb{R})/\text{Spin}(3,1)$, it is clear that the Spin group encompasses Lorentz transformations since $\text{Spin}(3,1)$ is a double cover of the Lorentz group $SO(3,1)$. However, the resulting geometric structure from this quotient is described as an almost complex structure, and it's not immediately apparent how this structure relates to Lorentz transformations. In contrast, the quotient $GL^+(4,\mathbb{R})/SO(3,1)$ naturally leads to a Lorentzian metric that aligns directly with Lorentz transformations. – Anon21 Nov 25 '23 at 13:31
  • How does an almost complex structure, arising from the quotient $GL^+(4,R)/\text{Spin}(3,1)$, constrain to $\text{Spin}(3,1)$ transformations, particularly given that these transformations include Lorentz transformations? Furthermore, if we consider reducing $GL^+(4,R)/\text{Spin}(3,1)$ to $GL^+(4,R)/SO(3,1)$, should the almost complex structure transition to a Lorentzian metric? In what sense, then, can an almost complex structure be considered 'slightly less restrictive' compared to a metric, especially in the context of these group actions and their associated geometric structures?" – Anon21 Nov 25 '23 at 13:32
  • Could the isomorphism $Spin(3,1)$ ~ $SL(2,C)$ obscure the geometric intuition of the quotient space. I'm employing $CL(3,1)$, which corresponds to the algebra of 4x4 real matrices. Through this, the exponential map goes from $M(4,R)$ to $GL^+(4,R)$, allowing for a natural embedding of $Spin^c(3,1)$ in $GL^+(4,R)$; for example, $\exp(a+x+f+v+b)$ is in $GL^+(4,R)$ and $\exp(f+b)$ is in $Spin^c(3,1)$. Could it be that this representation holds a more intuitive and visualizable geometry than what is suggested by the $SL(2,C)$ representation, with a clearer association to SO(3,1) transformations? – Anon21 Nov 25 '23 at 13:52
  • The whole point is that $Spin(3,1)$ is not the Lorentz group and the embedding $Spin(3,1)\to SL(4,R)$ has nothing to do with the Lorentz group, thus you should not be trying to see Lorentz transformations here, you will find none.
    • – Moishe Kohan Nov 26 '23 at 04:36
  • Again, $Spin(3,1)$ does not include Lorentz transformations, you are making a mistake here. I do not really understand what you mean by "How does an almost complex structure, arising from the quotient +(4,)/Spin(3,1), constrain to Spin(3,1) transformations": The embedding of $Spin(3,1)$ in $GL(4, R)$ preserves a complex structure on $R^4$, that's all. As for the Lorentz metric, forget it: it is a wrong thing to look for in the context of $Spin(3,1)$. I would not call a complex structure here "slightly less restrictive", it is just a different restriction.
    • – Moishe Kohan Nov 26 '23 at 04:41
  • To me, the complex matrix representation is much clearer and cleaner than the one using the Clifford algebra (even though in the end, it will amount to the same real representation of $Spin(3,1)$), but maybe it is just a matter of taste. In particular, it yields an invariant geometric structure immediately, which is something I do not see when I look at the matter using the Clifford formalism. But maybe you will get something geometric for $Spin^c(3,1)$ using the Clifford algebra formalism, I just do not see what it would be.
    • – Moishe Kohan Nov 26 '23 at 04:49