# Minimal spin connection, and the standard content of GR

In Sec. 12.1 of 'Superstring Theory', by Green, Schwarz, and Witten, the minimal spin connection is motivated as follows:

If we are to avoid modifying the standard content of general relativity, the two notions of the covariant derivative of a vector $V$ must be equivalent. This will be so, in the sense that $D_{\mu}V^{a} = e^{a\nu}D_{\mu}V_{\nu}$, if we define the spin connection so that the covariant derivative of the veilbein is zero, $D_{\mu}e^{a}_{\nu}=0$.

Until recently, I have always thought that this argument was quite sensible. But now I have doubts, due to the following considerations: The covariant derivative of a vector field $A$, say, in local Lorentz coordinates is given by

$$D_{\mu }A_{a} \equiv\partial _{\mu }A_{a}-\frac{1}{2}\omega _{\mu cd}\left( V^{cd}\right) ^{b}{}_{a}A_{b} =\partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b},$$

where $V^{cd}$ are the generators of the vector representation of the Lorentz group, given by $(V^{cd})^{a}{}_{b} = \eta ^{ca}\delta _{b}^{d}-\eta ^{da}\delta _{b}^{c}$. But then, using as well the standard expression for $D_{\mu }e^{a}{}_{\nu }$,

\begin{align} D_{\mu }A_{\nu } & = D_{\mu }\left( e^{a}{}_{\nu }A_{a}\right) \\ &=\left( D_{\mu }e^{a}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }D_{\mu }A_{a} \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }+\omega _{\mu }{}^{a}{}_{b}e^{b}{}_{\nu }\right) A_{a}+e^{a}{}_{\nu }\left( \partial _{\mu }A_{a}-\omega _{\mu }{}^{b}{}_{a}A_{b}\right) \\ &=\left( \partial _{\mu }e^{a}{}_{\nu }\right) A_{a}-\Gamma ^{\rho }{}_{\mu \nu }e^{a}{}_{\rho }A_{a}+e^{a}{}_{\nu } \partial _{\mu }A_{a} \\ &=\partial _{\mu }A_{\nu }-\Gamma ^{\rho }{}_{\mu \nu }A_{\rho } \\ &\equiv \nabla _{\mu }A_{\nu }, \end{align}

where $\Gamma ^{\rho }{}_{\mu \nu}$ are the Christoffel symbols. The derivation depends only on the antisymmetry of the spin connection, not on it being minimal. But if $D_{\mu }A_{\nu }=\nabla _{\mu }A_{\nu }$, as thus seems to be the case, then the covariant derivative of any tensor field (as a tensor product of vector fields) is the standard GR one, and thus the standard content of GR would seem to be unaltered. Am I doing something fundamentally wrong?

I suppose the vielbein is a vector bundle isomorphism $e: TM \to E$, for some vector bundle $E$. Your source seems to be using $D_\mu$ where you want to use $\nabla_\mu$. Thus $D_\mu A_\nu = \nabla_\mu A_\nu$ is a vacuous statement in the sense that $\nabla_\mu$ seems meaningless in the authors' notation. What you want to show is that $$D_\mu A^a = e^{a}{}_\nu D_\mu A^\nu. \tag{1}$$ Notice that in this expression, the left hand side contains the covariant derivative of an element of $E,$ while $D_\mu A^\nu$ is a covariant derivative of an element of $TM$. What you want to show is that the image of the latter under $e$ equals the former.

Your mistake is that you consider $D_\mu A_\nu$ to be something different from $\nabla_\mu A_\nu$, when really they are two different notations for the same thing. Thus your result is just the trivial identity $D_\mu A_\nu = D_\mu A_\nu$. Expanding on this note that \begin{align} D_\mu A_\nu &= D_\mu (e^a{}_\nu A_a) \end{align} is a covariant derivative on $TM^*$, and in expanding it you assume a connection on $TM^*$, namely the Levi-Civita one, which you call $\nabla$. Continuing: \begin{align} D_\mu A_\nu &= D_\mu (e^a{}_\nu A_a) \\ &= \partial_\mu (e^a{}_\nu A_a) - \Gamma^\rho{}_{\mu\nu}(e^a{}_\rho A_a), \end{align} where $\Gamma$ are the connection coefficients of your assumed connection (Christoffel symbols). What you do next is expand the first term over the contraction over the index $a$ in terms of the covariant derivative on $E$. This is no different than when we say that e.g. $\partial_\mu (A^aB_a) = B_a\nabla_\mu A^a + A^a\nabla_\mu B_a,$ and the result is guaranteed to be consistent from the standard Leibniz rule (naturality with respect to contraction), since $\nu$ is now a function index (incidentally, this is why the covariant derivative of $e^a{}_\nu$ is now defined regardless of view, below, since it is the image of $\partial_\nu$: $e(\partial_\nu) \in E$). Thus finally: \begin{align} D_\mu A_\nu &= D_\mu (e^a{}_\nu A_a) \\ &= \partial_\mu (e^a{}_\nu A_a) - \Gamma^\rho{}_{\mu\nu}(e^a{}_\rho A_a) \\ &= A_a\left(\partial_\mu e^a{}_\nu - \Gamma^\rho{}_{\mu\nu}e^a{}_{\rho} + \omega_\mu{}^a{}_b e^b{}_\nu\right) + e^a{}_\nu\left(\partial_\mu A_a - \omega_\mu{}^b{}_a A_b\right), \end{align} which is your result. As you can see it follows from you assuming that $D_\mu$ acts like $\nabla_\mu$ on $TM^*$, i.e. that $D_\mu = \nabla_\mu$ there, and the standard way of inducing a connection on $E^*$ from one on $E$. This assumption of yours is not wrong, because we want the covariant derivative on $TM$ (and $TM^*$) to be the Levi-Civita connection, but it makes the statement $D_\mu A_\nu = \nabla_\mu A_\nu$ trivial (since it is assumed).

To clarify: we seek to define a connection on $E$, but this does not define a new connection on $TM$. The connection on $TM$ remains the Levi-Civita connection, whether we use the symbol $D_\mu$ or $\nabla_\mu$.

Here follows one way look at why $(1)$ must be true: Thus the statement $(1)$ seeks an induced connection on $E$ (given one on $TM$, the Levi-Civita) that is natural with respect to $e$, in the sense that $e$ then commutes with the covariant derivatives. What this means is that it does not matter when we map the covariant derivative of a vector (and in extension any tensor) to $E$ (corresponding tensor bundle), before or after the derivative is applied, which is obviously desirable if our purpose is that $E$ and $TM$ describe the same physical content. In this context, one would use $(1)$ to define $D_\mu V^a$: \begin{align}\begin{split} D_\mu A^a &= e^{a}{}_\nu D_\mu A^\nu \\ \partial_\mu A^a + \omega_\mu{}^a{}_bA^b &= e^a{}_\nu\left(\partial_\mu A^\nu + \Gamma^\nu{}_{\mu\rho}A^\rho \right) \\ \end{split} &&\iff \omega_\mu{}^a{}_b &= e_b{}^\rho \left(-\partial_\mu e^a{}_\rho + e^a{}_\sigma\Gamma^\sigma{}_{\mu\rho}\right), \end{align} where we used the fact that $e^a{}_\nu e_a{}^\mu = \delta^\mu_\nu$. Using this fact again, and contracting with $e^b{}_\nu$ on both sides yields $$\partial_\mu e^a{}_\nu - e^a{}_{\rho}\Gamma^\rho{}_{\mu\nu} + \omega_\mu{}^a{}_be^b{}_\nu = 0,$$ but I want to emphasize that, a priori, we cannot equate the left hand side with $D_\mu e^a{}_\nu$, since $e$ is neither a tangent vector nor an element of $E$. However, this way of thinking makes obvious that $(1)$ is certainly required to ensure that working on $E$ and transforming back to $TM$ yields the same result as working on $TM$.

Here is how it follows from your standard expression: Your "standard expression" for $D_\mu e^a{}_\nu$ seems to imply we should rather consider $e^a{}_\nu$ as an element of $E \otimes TM^*$, with $D_\mu$ acting on $(S,X) \in E\otimes TM^*$ according to a Leibniz rule of sort: $D_\mu(S,X) = (D_\mu S,X) + (S,D_\mu X)$. Any vector bundle isomorphism $TM\to E$ may, of course, rather be considered this way, and vice versa. With $e^a{}_\nu \in E \otimes TM^*$ considered this way, we find \begin{align} \tag{2} D_\mu A^a = A^\nu D_\mu e^a{}_\nu + e^a{}_\nu D_\mu A^{\nu}, \end{align} and $(1)$ is true if and only if the first term vanishes. Since this must hold for arbitrary vector fields we have $D_\mu e^a{}_\nu = 0$. Notice the similarity with metric compatibility, and the naturality with respect to contraction. Only this time, I feel it is less natural to consider this second approach, since we have no reason a priori to let $D_\mu$ act on $e^a{}_\nu$. Add to that that it appears to lead to a lot of confusion (not only in this question).

In conclusion: Either way of looking at it leads to the same result, once you accept that $(1)$ must be true, but I feel it is more natural to do so in the context of vector bundle isomorphisms. In the second approach, compare $(1)$ and $(2)$, and the result follows. But try to understand where $(1)$ comes from.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Dec 1 '17 at 15:18
• @ACuriousMind: Please take a look at the page to which you link. The MathJax formatting has been lost. Do you think that that is conducive to further discussion? I don't. I expect your act to have effectively killed this thread. I actually feel intellectually malested. – John Fredsted Dec 1 '17 at 17:37
• @JohnFredsted See this meta post for ways to activate MathJax in chat. The only thing not conducive to further discussion here is your immediate assumption that I don't know what I'm doing - please assume good faith. – ACuriousMind Dec 1 '17 at 17:41
• I am tapping out. – John Fredsted Dec 1 '17 at 17:46