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I) Bundles

The tangent bundle $T\mathcal{M}$ forms a introductory pathway for the theory of bundles. Probably the reason to introduce the notion of bundles via tangent bundles relies on its simplicity given the notion: "the tangent bundle is a collection of all tangent spaces $\displaystyle T\mathcal{M} = \cup T_{p}\mathcal{M}$.

But, there is more on bundle theory: the principal bundles $P$. Now, a particular structure that the one can construct given a principal bundle $P$ is the so called associated bundle $A_{P}$.

Given a particular principal bundle called frame bundle $Fr(\mathcal{M})$, its associated bundle structure is the very tangente bundle $A_{Fr(\mathcal{M})} = T\mathcal{M}$

II) Connections

Historically the Levi-Civita Connection occurs when the one study two structures: the base manifold $\mathcal{M}$ and the tangent bundle $T\mathcal{M}$, and its local form (a.k.a. covariant derivative) is (with abuse of notation):

$$\nabla_{\mu} := \partial_{\mu} \pm \Gamma_{\mu \gamma}^{\nu} \tag{1}$$

the connection coeficients are the famous Christoffel symbols.

But, when the one starts to work with the Frame bundle structure, the notion of a connection map on $Fr(\mathcal{M})$ "changes" to the famous spin connection. Eventually, the tangent bundle is viewed as a associated bundle and then the covariant derivative changes to:

$$D_{\mu} := \partial_{\mu} + \omega_{\mu \gamma}^{\nu} \tag{2}$$

The $(2)$ occurs every time when the one is dealing with gauge theory and, in particular, with the Dirac operator $\gamma^{\mu}D_{\mu}$.

III) My Question

So, it seems that $T\mathcal{M}$ can be viewed as a "standalone" structure and as a associated bundle. Both of the cases, its a vector bundle, but the notion of the covariant derivative seems to "change".

My question is: how can the same structure $T\mathcal{M}$ render different notions of covariant derivatives ( $(1)$ and $(2)$ )?

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1 Answer 1

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The frame bundle of a Riemannian manifold is a principal $\text{O}(n)$-bundle. To get a spin connection, one must lift this to a principal $\text{Spin}(n)$-bundle. The standard representation of the complex Clifford algebra $\mathbb{C}l(n)$ induces a representation $\rho:\text{Spin}(n)\to\text{GL}(S_n)$, where $S_n$ is $\mathbb{C}^{2^{n/2}}$ if $n$ is even, and $\mathbb{C}^{2^{(n-1)/2}}$ if $n$ is odd. Assuming that this lift is possible (which is not always true), we get a principal $\text{Spin}(n)$-bundle $P$ and an associated vector bundle $E=P\times_\rho S_n$. Sections of $E$ are spinors. The Dirac operator is defined using Clifford multiplication $\gamma$ and the metric $g$, as the composition $$D:\Gamma(M,E)\xrightarrow{\nabla^S}\Gamma(M,T^*M\otimes E)\xrightarrow{g}\Gamma(M,TM\otimes E)\xrightarrow{\gamma}\Gamma(M,E)$$ Here, $\nabla^S$ is the spin connection. It is obtained from a principal connection $\omega^S$ on the bundle $P$, as the pullback of $\omega^\text{LC}$ on $\text{Fr}(E)$ which is the principal connection obtained from the Levi-Civita connection.

If we take the natural representation $\sigma:\text{O}(n)\to\text{GL}(n,\mathbb{R})$, then sure enough, $\text{Fr}(M)\times_{\sigma}\mathbb{R}^n\cong TM$. The connection induced by $\omega^\text{LC}$ is again just the Levi-Civita connection. Indeed, one can show that there is a bijective correspondence between principal connections on $\text{Fr}(E)$, and linear connections on $E$. In other words, your statement that the notion of a connection on the frame bundle "changes to a spin connection" because of the principal bundle structure is incorrect. You really have to do quite a lot of work, before you get to the spin connection and the Dirac operator. This does not simply happen from considering the frame bundle, instead of the vector bundle. Indeed, there are plenty of Riemannian manifolds which do not even admit a spin structure (i.e. a lift of the structure group from $\text{O}(n)$ to $\text{Spin}(n)$), and so they do not even have a spin connection.

Of course, the fact that $\omega^S$ is the pullback of $\omega^\text{LC}$ means that their components look similar enough, when comparing them in local coordinates. I would submit that it's very deceptive to rely on local coordinates for one's intuition, because it is clear that the spin connection $\nabla^S$ and the Levi-Civita connection $\nabla^\text{LC}$ act on very different objects. This does not become clear from only considering the local expressions of the operators.

Finally, the complex spinor representation $\rho$ is the conventional choice. Of course, one could choose a different representation, even a real representation. But either way, the resulting associated bundle will typically not coincide with the tangent bundle. In most sources that I have come across, one actually defines "the" spinor bundle of a Riemannian manifold $(M,g)$ to be the complex vector bundle associated to the representation $\rho$ I gave above (provided that it exists).

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  • $\begingroup$ Sorry to comment on an old post but I'm trying to understand how $\omega^\text{LC}$ is obtained. We start with the Levi-Civita connection $\nabla$ on $TM$ and because of the isomorphism $\text{Fr}(M)\times_{\sigma}\mathbb{R}^n\cong TM$, which I denote with $\psi$, we can pull this connection back to this associated vector bundle to get $\psi^*\nabla$. Then using the result that connections on a principal bundle and its associated vector bundle are in a bijective correspondence we can associate to $\psi^*\nabla$ a connection 1-form on the frame bundle which I denote by $\omega$. $\endgroup$
    – CBBAM
    Commented Jan 9 at 21:34
  • $\begingroup$ Have I understood this correctly? If so, what ensures that the $\psi^*\nabla$, the connection on the associated vector bundle $\text{Fr}(M)\times_{\sigma}\mathbb{R}^n$ is valued in the Lie algbera $\mathfrak{o}$(n)? $\endgroup$
    – CBBAM
    Commented Jan 9 at 21:34
  • $\begingroup$ That's ensured by the fact that the Levi-Civita connection is compatible with the metric, which means its holonomy is contained in $\text{O}(n)$. The structure group of $\text{Fr}(M)$ can be reduced to $\text{O}(n)$, and one gets the bundle of orthonormal frames $\text{OFr}(M)$, and you can take the connection to be on that bundle instead. $\endgroup$ Commented Jan 10 at 21:56
  • $\begingroup$ Thank you for your response. Is the compatibility with the metric preserved through the pullback? How did we transfer the metric on $TM$ to the associated vector bundle $\text{Fr}(M)\times_{\sigma}\mathbb{R}^n$? $\endgroup$
    – CBBAM
    Commented Jan 10 at 22:09

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