I am tasked with showing the following (Exercise in Cartan For Beginners by Ivey and Landsberg):

For $X,Y,Z\in\Gamma(TM)$, define $$ \overline{R}(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. $$ Show that $\overline{R}$ is a tensor, i.e., it is $C^\infty(M)$-linear in all three factors, and that $\overline{R}=-\Theta$.


After looking into it, I've noticed our text has defined some things differently than the standard definitions, so I will go over these now. (Bear with me, I probably have some gaps in understanding which will become evident here.)

1. Let $M$ be a Riemannian manifold. We define the covariant derivative of a vector field $X=x^ie_i$ to be \begin{equation} \nabla X:=e_i\otimes(dx^i+x^j\overline{\eta}^i_j)\in\Omega^1(M,TM)=\Gamma(TM\otimes T^*M), \end{equation} where (I believe) $\overline\eta^i$, $\overline\eta^i_j$ are the pullbacks of the tautological forms and their associated connection forms on $M$ through a section of the orthonormal frame bundle. Further, for $Y\in\Gamma(TM)$ we define $$ \nabla_YX:=Y\lrcorner\nabla X=(dx^i+x^j\overline{\eta}^i_j)(Y)e_i. $$

2. If $\eta^i$, $\eta^j$ are the tautological forms on the orthonormal frame bundle $\mathcal{F}_{\text{on}}(M)$ to $M$, we define $\tilde{\Theta}^i_j:=d\eta^i_j+\eta^i_k\wedge\eta^k_j\in\Omega^2(\mathcal{F}_{\text{on}}(M))$, $$ \tilde{\Theta}:=\tilde{\Theta}^i_j\otimes\eta^j\otimes e_i\in\Omega^2(\mathcal{F}_{\text{on}}(M),\pi^*(\text{End}(TM))), $$ where $\pi:\mathcal{F}_{\text{on}}(M)\to M$ is the associated projection of the frame bundle $\mathcal{F}_{\text{on}}(M)$, and lastly define $\Theta\in\Omega^2(M,\text{End}(TM))$ to be the form such that $\tilde{\Theta}=\pi^*(\Theta)$.

If there is ambiguity here that needs explaining, please let me know and I will happily do my best to clear it up.


I think it should be simple enough to show that $\overline{R}$ is a tensor (let me know if it is not), so I'm putting most of my energy into showing $\overline{R}=-\Theta$.

1. There is a hint given with this exercise which states:

Differentiating $\nabla X=e_i\otimes(dx^i+x^j\overline{\eta}^i_j)$, with $Z$ in place of $X$, gives a relationship between $\Theta$ and the second derivatives of $Z$.

I don't quite know what is meant by "differentiate." It's a one form, so does it mean to take the exterior derivative? Does it mean to take the covariant derivative again? If the latter, How would I go about doing this?

2. I've attempted to dig into definitions to hopefully find something there, but I haven't found anything, and frankly I'm not very confident that I'm doing everything correctly. This is my attempt:

Let $X=x^ie_i$, $Y=y^ie_i$, and $Z=z^ie_i$. Note \begin{align*} \nabla_X\nabla_YZ&=\nabla_X\left((dz^i+z^j\eta^i_j)(Y)e_i\right)=\left(dz^j\wedge\eta^i_j+z^jd\eta^i_j+(dz^i+z^j\eta^i_j)\eta^i_j\right)(Y)(X)e_i,\\ \nabla_Y\nabla_XZ&=\nabla_Y\left((dz^i+z^j\eta^i_j)(X)e_i\right)=\left(dz^j\wedge\eta^i_j+z^jd\eta^i_j+(dz^i+z^j\eta^i_j)\eta^i_j\right)(X)(Y)e_i, \end{align*} so $$ \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ=-\left(dz^j\wedge\eta^i_j+z^jd\eta^i_j+(dz^i+z^j\eta^i_j)\eta^i_j\right)\big([X,Y]\big)e_i. $$ Now \begin{align*} \overline{R}(X,Y)Z&=-\left(dz^j\wedge\eta^i_j+z^jd\eta^i_j+(dz^i+z^j\eta^i_j)\eta^i_j\right)\big([X,Y]\big)e_i-(dz^i+z^j\eta^i_j)\big([X,Y]\big)e_i. \end{align*}

If correct so far (which I doubt) I don't quite know where to go next. I feel like I would be futile to spend all day working with arithmetic I'm not even confident in, so here I am. This is a homework assignment, so posting full answers would be unethical, but any comments/help/hints are greatly appreciated. Thanks in advance for even reading this post in its entirety.


Following Jesse Madnick's suggestion, I computed $$ \overline{R}(e_A)(e_B)e_C=-2d\eta^C_j(e_A)(e_B)e_C, $$ which seems to be on the right track. However, I've having trouble computing $\tilde{\Theta}=\tilde{\Theta}^i_j\otimes\eta^j\otimes e_i$; mostly, I don't understand how the tensor product $\tilde{\Theta}^i_j\otimes\eta^j$ acts on two vectors $e_A,e_B$, since $\tilde{\Theta}^i_j$ is two form and $\eta^j$ is a one form. Is it wrong to say $$ \theta^C_j\otimes\eta^j(e_A)(e_B)=\theta^C_j(e_A)(e_B)\eta^j(e_B)? $$ (It probably is.) Again, any clarification is greatly appreciated.

  • $\begingroup$ Let $E \to X$ be a vector bundle. We write $\Omega^k(X ;E)$ to mean $\Omega^k(X) \otimes \Gamma(E)$. In other words, if $\varphi = \alpha \otimes \sigma \in \Omega^k(X; E)$, then $\varphi(v_1, \ldots, v_k) = \alpha(v_1, \ldots, v_k) \cdot \sigma \in \Gamma(E)$ for vector fields $v_1, \ldots, v_k \in \Gamma(TX)$. In your case, $X = \mathcal{F}_{\text{on}}(M)$, etc. $\endgroup$ – Jesse Madnick Sep 22 '15 at 4:25
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    $\begingroup$ I love you. I will mark the question as answered now. $\endgroup$ – Blake Sep 22 '15 at 5:16
  • $\begingroup$ you're very welcome, Blake :-) $\endgroup$ – Jesse Madnick Sep 22 '15 at 5:20

Here are a few tips.

  • Once you show that $\overline{R}$ is $C^\infty(M)$-linear, then you know that $$\overline{R}(X,Y)Z = x^i y^j z^k \overline{R}(e_i, e_j)e_k.$$ Of course, $-\Theta$ is also $C^\infty(M)$-linear. So, it suffices to prove that $\overline{R}(e_i, e_j)e_k = -\Theta(e_i, e_j)e_k$.

In other words, you should try to compute both $\overline{R}(e_i, e_j)e_k$ and $-\Theta(e_i, e_j)e_k$.

  • In computing $-\Theta$, it might be helpful to remember that sections $s$ satisfy $\pi \circ s = \text{Id}$, so that $\Theta = s^*(\tilde{\Theta})$. Also useful are pullback rules like, say, $s^*(\alpha \wedge \beta) = s^*\alpha \wedge s^*\beta$ and $s^*(d\alpha) = d(s^*\alpha)$.

  • A formula that may or may not be useful is the following: for any $1$-form $\alpha$ and vector fields $X,Y$, we have: $$d\alpha(X,Y) = X\alpha(Y) - Y\alpha(X) - \alpha([X,Y]).$$

  • $\begingroup$ Thank you! I believe I've almost got it -- however I ran into a snag -- I'm not quite sure how the tensor product of a two form and one form acts on two vectors; you can see my edit for more information. If you have the time I would sure appreciate some clarification on this. Thanks again. $\endgroup$ – Blake Sep 21 '15 at 20:24

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