# Can we get a Spin-like connection from a group reduction ${\rm GL}^+(4,\mathbb{R})/{\rm Spin}^c(3,1)$ or $\exp {\rm CL}(3,1)/{\rm Spin}^c(3,1)$?

I understand from 1 that reducing the structure group $${\rm GL}^+(4,\mathbb{R})$$ of the frame bundle $$FX$$ of a world manifold $$X^4$$ to the $${\rm SO}(3,1)$$ group entails the Levi-Civita connection as the (unique torsion-free) connection that preserves the reduction.

I am interested in a spin, or spin-like, connection.

I am working with the exponential map of the multivectors $$\mathbf{u}$$ of $${\rm CL}(3,1)$$. According to 2, it is isomorphic to this group:

$$\exp {\rm CL}(3,1) \to {\rm GL}^+(4,\mathbb{R})$$

It is also the case that $${\rm Spin}^c(3,1)$$ is in $$\exp {\rm CL}(3,1)$$ because $${\rm Spin}^c(3,1)$$ is merely the exponentials of bivectors + pseudoscalars (see this question 3).

Can we obtain the Spin connection as a structure group reduction $$\exp {\rm CL}(3,1)/{\rm Spin}^c(3,1)$$ or from $${\rm GL}^+(4,\mathbb{R})/{\rm Spin}^c(3,1)$$?

Previously, I asked for $${\rm GL}^+(4,\mathbb{R})/{\rm Spin}(4)$$ in this question and was told no on the grounds that $${\rm Spin}(4)$$ is not in $${\rm GL}^+(4,\mathbb{R})$$. Here I am asking for $${\rm Spin}^c(3,1)$$, and I stress that $${\rm Spin}^c(3,1)$$ is in $$\exp {\rm CL}(3,1)$$ (if not also in $${\rm GL}^+(4,\mathbb{R})$$ by virtue of the isomorphism mentioned in 2).

So it should be possible, no?

## 2 Answers

(In a previous answer, I focused on the existence and uniqueness of a reduction. But after some discussions with Anon21, I realized that he does not care about this aspect but is interested in the scope of arXiv:2002.01410. So I write this new answer.)

Can we obtain the Spin connection as a structure group reduction $$\exp \mathrm{CL}(3,1)/\mathrm{Spin}^c(3,1)$$ or from $$\mathrm{GL}^+(4,\mathbb{R})/\mathrm{Spin}^c(3,1)$$?

When we have a $$\exp Cl(3,1)$$ connection at hand, provided the existence of a reduction (see my previous answer), we indeed obtain a $$Spin^{c}(3,1)$$ connection in this way.

However, $$\exp Cl(3,1)$$ is not the structure group of the canonical Clifford bundle on a $$(3,1)$$-Riemannian manifold. The correct structure group is (at most) $$O(3,1)$$. So the $$Spin^{c}(3,1)$$ connection we obtained above is not what people usually talk about, but something else.

For what people usually talk about, we should first do the reduction $$GL(4,\mathbb{R})/SO(3,1)$$ so we obtain a $$SO(3,1)$$ connection, and then lift this $$SO(3,1)$$ connection to a $$Spin^c(3,1)$$ connection. This second process goes beyond the scope of arXiv:2002.01410 since $$Spin^c(3,1)$$ is not subgroup of $$SO(3,1)$$.

• The quotient would also induce a metric, right (since SO is in Spin, and $\exp CL$ isn't)? Commented Mar 28, 2023 at 12:52
• @Anon21 Yes. (But SO is not in Spin. SO is a quotient of Spin.)
– Leo
Commented Mar 28, 2023 at 13:08
• Ah Ok. Finally, if I say $\exp CL(3,1)$ is isomorphic to $\exp M(4,R)$, and therefore it generates GL+^(4,R) up to isomorphism. Then, if I further say taking $\exp CL(3,1)/Spin^c(3,1)$ is equivalent to taking $GL^+(4,R)/Spin^c(3,1)$ - am I taking it too far, or is it still reasonable? Commented Mar 28, 2023 at 13:12
• @Anon21 Yes, you can do it. You indeed obtained a $Spin^c(3,1)$ connection from a $GL^+(4,\mathbb{R})$ connection in this way (provided the reduction exists). My point is just that, this connection is not what people usually call a SpinC connection, although its structure group is indeed $Spin^c(3,1)$. A SpinC connection that people usually talk about is required to be related to the frame bundle in the precise way I described in the above answer.
– Leo
Commented Mar 28, 2023 at 13:36
• @Anon21 The trip matters. Different trips go to different destinations: the $Spin^c$ connection you get is different from a 'standard' one (there are many). So "you go from London to New York while the 'standard' goes from Paris to Los Angeles; just both destinations are in the US."
– Leo
Commented Mar 28, 2023 at 14:03

As long as you can find a structure group reduction at the bundle level, you can always obtain a structure group reduction at the bundle-with-connection level. For example, when the tangent/frame $$O(n)$$-bundle can be reduced to a $$Spin(n)$$-bundle, there's no essential difference between a Levi-Civita connection and a spin connection. They are just often written in different representations so look different at the first glance. What matters is whether such a bundle lifting exists, i.e. whether a spin structure exists.

So you should really care about bundles instead of connections. Let me first narrate the general story and then go to your concrete question.

Given a base space $$X$$ and a Lie group homomorphism $$G\stackrel{f}{\to}H$$, we can always go from a $$G$$-bundle to an $$H$$-bundle in an obvious way. But the reduction of an $$H$$-bundle to a $$G$$-bundle does not always exist; even when it exists, the reduction may not be unique. Recall that $$G$$-bundle isomorphism classes are classified by homotopy classes of maps $$X\to BG$$, where $$BG$$ is $$G$$'s classifying space. Therefore, our question is a standard obstruction problem in algebraic topology asking whether a map $$X\to BH$$ can be lifted to a map $$X\to BG$$ via $$BG\stackrel{f}{\to} BH$$.

From this point of view, we can immediately reach a nice theorem stating a sufficient condition for the existence and uniqueness of reductions:

Theorem: For any base space, if $$G\stackrel{f}{\to}H$$ is also a homotopy equivalence, every $$H$$-bundle has a unique reduction to a $$G$$-bundle.

In other words, in this case we have a one-to-one correspondence between $$G$$-bundle isomorphism classes and $$H$$-bundle isomorphism classes. A common situation of this sort is that $$f$$ is injective and $$\mathrm{im}f$$ contains a maximal compact subgroup of $$H$$. This is because any Lie group is homeomorphic to the product of its maximal compact subgroup with some $$\mathbb{R}^n$$. This includes the standard example of $$O(n,\mathbb{R})\to GL(n,\mathbb{R})$$, which means you can always have a positive-definite metric on a smooth manifold (you can prove this fact more easily via partition of unity, though).

For the general cases, we should dive into the obstruction theory in algebraic topology. Some general remarks are as follows. The obstruction for a reduction is given by a certain characteristic class of the $$H$$-bundle. When this characteristic class vanishes, different reductions are classified by a certain cohomology group of $$X$$ (non-canonically). A textbook example is $$Spin(n)\stackrel{1}{\to} SO(n)\stackrel{2}{\to} O(n)$$. For 2, the obstruction is the first Stiefel-Whitney class $$w_1$$; when $$w_1=0$$, different orientations are classified by $$H^0(X,\mathbb{Z}_2)$$. For 1, the obstruction is the second Stiefel-Whitney class $$w_2$$; when $$w_2=0$$, different spin structures are classified by $$H^1(X,\mathbb{Z}_2)$$. In finding obstructions, the above theorem is still useful allowing us to focus on maximal compact subgroups.

Now come to your monomorphism $$Spin^{\mathbb{C}}(3,1)\stackrel{g}{\to} GL^+(4,\mathbb{R})$$. A maximal compact subgroup of $$GL^+(4,\mathbb{R})$$ is isomorphic to $$SO(4)$$ while a maximal compact subgroup of $$Spin^{\mathbb{C}}(3,1)$$ is isomorphic to $$U(2)$$. We can use the above theorem to reduce $$g$$ to $$U(2)\stackrel{\tilde{g}}{\to}SO(4)$$. Here $$\tilde{g}$$ can be described as multiplying $$SU(2)\stackrel{\mathrm{id}}{\to}SU(2)$$ with $$U(1)\to SU(2)$$ and then factoring out a diagonal $$\mathbb{Z}_2$$ center.

It is now clear that only the $$U(1)\to SU(2)$$ part is questionable for the existence of a reduction. For a general base space, there are apparently a huge number of obstructions so a reduction is often impossible. However, in your question, your base space seems to be a 4-manifold. In this case, because $$BSU(2)$$ is 3-connected, there is only one obstruction which is the second Chern class $$c_2$$. Anyways, there's an obstruction, so a reduction is not always possible. But if you consider still lower dimensional manifolds, then a reduction is always possible. Moreover, in this case, the different reductions are classified by $$H^2(X,\mathbb{Z})$$ which eventually gives the first Chern class of the resulting $$U(1)$$-bundle.

A final remark. This is not related to the question itself but I would like to point out that the Clifford bundle with fiber $$Cl_{\mathbb{R}}(p,q)$$ has the (universal) structure group $$O(p,q)$$. So be careful. If what you would like to consider is a Clifford bundle with fiber $$Cl_{\mathbb{R}}(3,1)$$, the structure group is $$O(3,1)$$ instead of $$GL(4,\mathbb{R})$$. In particular, it is not the structure group of the tangent/frame bundle.

If in your mind, $$GL^+(4,\mathbb{R})$$ is indeed the structure group of the oriented tangent/frame bundle, then the homomorphism $$Spin^{\mathbb{C}}(3,1)\stackrel{g'}{\to} GL^+(4,\mathbb{R})$$ you should consider is no longer the embedding you described. This $$g'$$ factors through $$Spin^{\mathbb{C}}(3,1)\to SO(3,1)\to GL^+(4,\mathbb{R})$$.