Let M be an n-dim Riemannian manifold. Denote $$ F(M)=\{(x,e):x\in M, e=(e_1,e_2,...,e_m) \text{ is a frame at } x\} $$ its orthogonal frame bundle . For $(x,e)\in F(M)$ and for every geodesic $\gamma(t)$ with $\gamma(0)=x$, let $e(t)$ be the parallel translation of $e$ along $\gamma(t)$. So $\tilde{\gamma}(t)=(\gamma(t),e(t))$ is a curve in $F(M)$. $$\tilde{\gamma}'(0)=(\gamma'(0),0) \in T_{x,e}F(M).$$ We call the set of tangent vectors $\tilde{\gamma}'(0)\in T_{x,e}F(M)$ the Horizontal subspace at $(x,e)$.

Equipped $O(n)$ with a bi-invariant metric, there is a unique metric on $F(M)$ such that when restricting to a horizontal subspace, the projection $p:F(M)\to M$ is an isometry. Intuitively, how to construct this metric?

For constants $C_1,C_2$ such that the sectional curvature $|sec_M|\leqslant C_1$, the covariant derivative of the curvature tensor on M are bounded by $C_2$. Is there a $C$ depending on $C_1,C_2$ such that $|sec_{F(M)}|\leqslant C$?

I found in some books, they define the connection, curvature form of the frame bundle. But after reading them, I still don't know how to answer the above question.

  • $\begingroup$ If you know the curvature forms, then you know the sectional curvatures. You could also look up Rienannian submersions. $\endgroup$
    – Deane Yang
    Apr 29, 2018 at 14:33
  • $\begingroup$ @DeaneYang: Let $X,Y$be orthonormal vector fields on $M$, $\tilde{X},\tilde{Y}$ their lifts. By O'Neil's formula, $$sec_M(X,Y)=sec_{FM}(\tilde{X},\tilde{Y})+\frac34|[\tilde{X},\tilde{Y}]^V|$$. However, for vectors $Z_1,Z_2$ tangent to the fiber, I don't know how to compute $sec_M(X,Z_i)$ and $sec_M(Z_1,Z_2)$. $\endgroup$ Apr 30, 2018 at 14:00
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    $\begingroup$ I'm pretty sure that the fibers are totally geodesic, so it's just the sectional curvature of the orthogonal group. The mixed ones are the most confusing. One way is to use the structure equations: $$d\omega^i + \omega^i_j\wedge\omega^j = 0,\ \ d\omega^i_j + \omega^i_k\wedge\omega^k_j = 0$$ The $1$-forms $\omega^i, \omega^i_j$ form an orthonormal frame of $1$-forms on the bundle, and the equations above can be used to infer what the corresponding connection $1$-forms are. From there, you can figure out the curvature $2$-forms and therefore the sectional curvatures. $\endgroup$
    – Deane Yang
    Apr 30, 2018 at 14:36

1 Answer 1


Following Deane Yang's suggestion, it is not difficult to compute the Levi-Civita connection forms for the canonical orthonormal coframing of the orthogonal frame bundle. One then finds that such a $C$ depending on $C_1$ and $C_2$ with the desired properties does indeed exist and that there exist constants $a_0$, $a_1$, $a_2$, and $b$ depending only on the dimension $n$ such that $C$ can be taken to be $$ C = a_0 + a_1\,C_1 + a_2\,{C_1}^2 + b\,C_2\,.\tag1 $$ When $n\le 2$, one can take $a_0=0$, but, when $n>2$, one must take all four of these constants to be positive in order to get an estimate that holds for all metrics in dimension $n$.

The point is that, as Deane says, one has a canonical coframing on $F(M)$ given by the tautological $1$-forms $\omega_i$ and the corresponding Levi-Civita $1$-forms $\omega_{ij} = -\omega_{ji}$ satisfying the first structure equation of Cartan (with the Einstein summation convention assumed here and below) $$ \mathrm{d}\omega_i = -\omega_{ij}\wedge\omega_j\tag2 $$ and the second structure equation of Cartan $$ \mathrm{d}\omega_{ij} = -\omega_{ik}\wedge\omega_{kj} + \tfrac12R_{ijkl}\,\omega_k\wedge\omega_l\,.\tag3 $$ The functions $R_{ijkl}=-R_{jikl}=-R_{ijlk}=-R_{iklj}-R_{iljk}$ on $F(M)$ represent the components of the Riemann curvature tensor of the original metric pulled back to $F(M)$ and satisfy $$ \mathrm{d}R_{ijkl} =R_{pjkl}\,\omega_{pi}+R_{ipkl}\,\omega_{pj}+R_{ijpl}\,\omega_{pk} +R_{ijkp}\,\omega_{pl}+R_{ijklm}\,\omega_m\tag4 $$ for unique functions $R_{ijklm}$, which represent the components of the covariant derivative of the Riemann curvature tensor pulled back to $F(M)$.

The canonical metric on the frame bundle is then given by $$ g = \sum_{i=1}^n {\omega_i}^2 + \sum_{1\le i < j\le n}{\omega_{ij}}^2.\tag5 $$ If one uses the 'lexicographical' index ordering $$ 1 < 2<\cdots<n < 12 < 13\cdots <1n< 23 <\cdots < (n{-}1)n, $$ for the $g$-orthonormal coframing $$ \Omega = (\omega_1,\ldots,\omega_n,\omega_{12},\omega_{13},\ldots,\omega_{(n-1)1n}) = (\omega_\alpha) $$ and writes $\mathrm{d}\omega_\alpha = -\theta_{\alpha\beta}\wedge\omega_\beta$, where $\theta_{\beta\alpha}=-\theta_{\alpha\beta}$, then one finds that for $n <\alpha<\beta$, the $1$-form $\theta_{\alpha\beta}$ is a constant linear combination of the $\omega_\gamma$ where $n<\gamma$; for $\alpha \le n < \beta$, the $1$-form $\theta_{\alpha\beta}$ is a constant linear combination of terms of the form $R_{\gamma\delta}\omega_\epsilon$, where $\epsilon \le n < \gamma,\delta$ (which, in particular, implies Deane Yang's assertion that the $\mathrm{O}(n)$-fibers of $F(M)\to M$ are totally geodesic in the metric $g$); while for $\alpha<\beta\le n$, the $1$-form $\theta_{\alpha\beta}$ is a linear combination of the $\omega_{\gamma}$ where $\gamma>n$ with coefficients that are affine linear combinations of the $R_{\gamma\delta}$ (where $\gamma,\delta>n$).

Substituting this information into the curvature formulae for the Levi-Civita connection $\theta$ for $g$, i.e., $$ \Theta_{\alpha\beta} = \mathrm{d}\theta_{\alpha\beta} + \theta_{\alpha\gamma}\wedge\theta_{\gamma\alpha}\,,\tag6 $$ and using the given formula for the exterior derivatives of the $R_{ijkl}$, one finds that the coefficients of the $\Theta_{\alpha\beta}$ in the $\Omega$-coframing are linear combinations of terms that are either constants, constant multiples of $R_{ijkl}$, quadratic expressions in the $R_{ijkl}$ with constant coefficients, or constant multiples of the $R_{ijklm}$.

The existence of a constant $C$ bounding the sectional curvatures of $g$ that takes the form $(1)$ follows immediately from this.

Example: Take the case $n=2$. Then $F(M)\to M$ is an $\mathrm{O}(2)$-bundle and the structure equations on the $3$-manifold $F(M)$ become $$ \mathrm{d}\omega_1 = -\omega_{12}\wedge\omega_2\qquad \mathrm{d}\omega_2 = \omega_{12}\wedge\omega_1\tag{2'} $$ and $$ \mathrm{d}\omega_{12} = K\,\omega_1\wedge\omega_2\,,\tag{3'} $$ where I have written $K$ for $R_{1212}$, as is traditional. ($K$ is simply the Gauss curvature.) The equation for the covariant derivative of the Riemann curvature tensor simply becomes, in this case. $$ \mathrm{d}K = K_1\,\omega_1 + K_2\,\omega_2\tag{4'} $$ Now, for simplicity and to avoid confusion, I am going to write $\omega_3$ for $\omega_{12}$, etc., so that $$ \omega = \begin{pmatrix}\omega_1\\\omega_2\\\omega_3\end{pmatrix} = (\omega_\alpha) $$ becomes an orthonormal coframing for the metric $g = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2$ on $F(M)$ whose curvature we want to compute. We do this by first finding the unique skew-symmetric $3$-by-$3$ matrix $\theta = (\theta_{\alpha\beta})$ that satisfies $\mathrm{d}\omega = -\theta\wedge\omega$. Given the equations $(2')$ and $(3')$, we find that $$ \theta = (\theta_{\alpha\beta}) = \begin{pmatrix} 0 & (1{-}\tfrac12K)\omega_3 & -\tfrac12K\omega_2\\ -(1{-}\tfrac12K)\omega_3& 0 & \phantom{-}\tfrac12K\omega_1 \\ \tfrac12K\omega_2 & -\tfrac12K\omega_1& 0 \end{pmatrix}, $$ and the reader can verify that this has the general properties that I stated above for general $n$.

Now, we compute the curvature by computing the matrix $$ \Theta = \mathrm{d}\theta + \theta \wedge\theta = (\Theta_{\alpha\beta}), $$ and, using $(2')$, $(3')$, and $(4')$, we find that $$ \begin{pmatrix} \Theta_{23} \\ \Theta_{31} \\ \Theta_{12}\end{pmatrix} = \begin{pmatrix} \tfrac14K^2& 0 & -\tfrac12K_2 \\ 0 & \tfrac14K^2 & \phantom{-}\tfrac12K_1\\ -\tfrac12K_2 & \phantom{-}\tfrac12K_1 & (K{-}\tfrac34K^2)\end{pmatrix} \begin{pmatrix} \omega_2{\wedge}\omega_3 \\ \omega_3{\wedge}\omega_1 \\ \omega_1{\wedge}\omega_2\end{pmatrix} $$ The Riemann curvature tensor of $g$ has now been shown to be $$ \mathrm{Riem}(g) = \sum_{\alpha<\beta}\Theta_{\alpha\beta}\otimes \omega_\alpha{\wedge}\omega_\beta = \Theta_{23}\otimes \omega_2{\wedge}\omega_3 +\Theta_{31}\otimes \omega_3{\wedge}\omega_1 +\Theta_{12}\otimes \omega_1{\wedge}\omega_2\,. $$ It follows that the components of the Riemann curvature tensor of $g$ in this $g$-orthonormal coframing are linear combinations of $K$, $K^2$, $K_1$, and $K_2$. Thus, there is a bound on the sectional curvature of $g$ of the form claimed above in terms of $C_1$, an upper bound for $|K|$, and $C_2$, an upper bound for $\sqrt{{K_1}^2+{K_2}^2}$. (Note that, we do not need the constant $a_0$ in the case $n=2$. However, when $n>2$, constant terms do of course, show up, because the fibers of $F(M)\to M$, which are totally geodesic, are copies of $\mathrm{O}(n)$, which is not flat when $n>2$.)

  • $\begingroup$ Thank you for your answer! I still have a quetion. For $\alpha \leqslant n$, let $\alpha=i$, we have $$ -\sum_{\beta>\alpha} w_{\alpha \beta}\wedge w_{\beta}=dw_{\alpha}=dw_i=-\sum_{j=1}^nw_{ij}\wedge w_j. $$ If $\beta>n$, then $w_{\alpha\beta} \wedge w_{\beta}$ are not of the form $w_{ij}\wedge w_j$. So from this, for $\beta \leqslant n$, let $\beta=j$, we should have $w_{\alpha \beta}=w_{ij}.$ Why do you say "with coeffients that are affine linear combinations of $R_{\gamma \delta}$"? $\endgroup$ May 5, 2018 at 13:40
  • $\begingroup$ @mathmetricgeometry: I do not understand your question. I think that you must be misunderstanding the difference between the $\theta_{\alpha\beta}$ and the $\omega_{\alpha\beta}$, which are not the same thing at all. I'll add an example at the end of my answer to illustrate exactly what I mean. Maybe that will clarify everything. $\endgroup$ May 5, 2018 at 17:11
  • $\begingroup$ @mathmetricgeometry: You seem to have deleted your comment that said that you understood my answer. Have you developed doubts or have further questions? $\endgroup$ May 11, 2018 at 11:37
  • $\begingroup$ I don't know why my comment has been deleted, I understand your answer. Thank you very much! $\endgroup$ May 20, 2018 at 7:45

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