I'm interested in finding a global (coordinate-free) description of the Levi-Civita connection on a (possibly infinite-dimensional) Riemannian manifold X.

I'm not looking for a description of this object as a differential operator.

Instead, I'm looking for a splitting of the natural map $\alpha = (\pi_{T(TX)}, D\pi_{TX}): T(TX) \to TX \oplus TX$, where $\pi_{T(TX)}$ is the structure map of the double tangent bundle and $D\pi_{TX}$ is the map on tangent bundles induced by the structure map $\pi_{TX} : TX \to X$.

Noting that $TX \oplus TX= (\pi_{TX})^* (TX)$, a splitting of $\alpha$ is the analogue, for the vector bundle $TX$, of the standard notion of a connection on a principle bundle: it's a way of lifting tangent vectors on the manifold $X$ up to tangent vectors on the bundle $TX$.

Lang (in GTM 160, Differential and Riemannian Manifolds) explains how to obtain this splitting using the metric spray, which is a map $F: TX \to T(TX)$ that splits both of the above maps $T(TX) \to TX$ (and satisfies another "quadratic" condition). Lang gives a global description of $F$ as the vector field on $TX$ corresponding, under the metric, to the 1-form -dK, where $K(v) = (1/2)\langle v,v\rangle$ is the kinetic energy functional on $TX$. However, he doesn't really give a coordinate-free extension of F to the desired splitting. From studying the discussion in Lang, it seems to me that there is a unique splitting $H: T(TX) \to TX \oplus TX$ satisfying $F(v+w) = (F(v) + H(w,v)) + (F(w) + H(v,w))$ and such that in any local chart U on X, H has the form $H(x, v, w) = (x, v, w, B(x, v, w))$ (as a map $U\times E \times E \to (U \times E) \times (E\times E)$) with $B(x, -, -)$ a symmetric bilinear mapping. Here E is the Hilbert space on which X is modeled.

The parentheses in the expression $(F(v) + H(w,v)) + (F(w) + H(v,w))$ are important: inside the parentheses, + means addition in the fibers of the map $D\pi_{TX}$, whereas outside the parentheses, + means addition in a fiber of $\pi_{T(TX)}$. Note that H itself is definitely not symmetric, so I don't think it's clear from the global formula that H exists.

Establishing existence of the map H seems to depend on the rather ugly change-of-coordinate formulas for the "quadratic part" of the spray F, given by Lang.

Lang mentions that the book Symmetric Spaces (Loos, 1969) gives some discussion of this material in terms of second-order jet bundles, and I suspect that may be what I'm looking for. However, this book is hard to come by. I can't find any previews on-line, and it's not in our library. Lang also mentions Pohl's paper "Differential geometry of higher order" (Topology 1 1962 169--211) but I couldn't see anything about the Levi-Civita connection in there.

Does anyone know if Loos has what I'm looking for? Are there other discussions of these ideas in the literature? Does anyone have other suggestions for how to think about the splitting H?

I'll point out, as motivation, that the splitting $H$ gives a decomposition of $T(TX)$ as a direct sum $\pi^* (TX) \oplus \pi^* (TX)$ of bundles over $TX$ (because the kernel of $\alpha$ is isomorphic to $\pi^* (TX)$, and so this is one way to think about the standard fact that $TX$ is an orientable manifold, with a Riemannian metric inherited from the one on $X$.

  • 2
    $\begingroup$ Dan, why not go via the differential operator description? In finite dimensions, if $nabla$ is a covariant derivative in $TX$, the horizontal lifts of a vector in $X$ tangent to a curve $\gamma$ are the sections of $\gamma^* TX$ in the kernel of $\gamma^*\nabla$. Moreover, there's a coordinate-free (Koszul) formula for the L-C covariant derivative. Does something go wrong in infinite dimensions? $\endgroup$
    – Tim Perutz
    Apr 24, 2010 at 2:56
  • 2
    $\begingroup$ @Tim, perhaps the problem is that you can't necessarily find a horizontal section of $\gamma^*TX$ with arbitrary initial condition. In finite dimensions this always works because you're solving an ODE, but in infinite dimensions this becomes a PDE. I remember reading somewhere that the Koszul formula didn't necessarily imply existence of the Levi-Civita connection in infinite dims, but this was a throw-away remark and I never thought about it again until now... $\endgroup$
    – Joel Fine
    Apr 24, 2010 at 7:59
  • 1
    $\begingroup$ Joel, point taken. I think one can solve initial-value problems in Hilbert space (by finite-dimensional approximation), so maybe we're OK in Hilbert manifolds? But now I see why one might look to jet bundles and the like. $\endgroup$
    – Tim Perutz
    Apr 24, 2010 at 13:56
  • $\begingroup$ (I mean IVPs for 1st order linear ODE in Hilbert space.) $\endgroup$
    – Tim Perutz
    Apr 24, 2010 at 14:23
  • $\begingroup$ Tim and Joel: Lang does explain how to obtain the L-C covariant derivative associated to a Riemannian metric on a Hilbert manifold, and he gives the global Kozul formula. So Tim is right that this works on Hilbert manifolds. (It does sound, from Lang's book, that there's an issue on general Banach manifolds: there metrics don't make sense, but the correspondence between sprays and covariant derivatives breaks down.) $\endgroup$
    – Dan Ramras
    Apr 24, 2010 at 18:41

1 Answer 1


See my response (number 4) to the MO question:

Exponential map and covariant derivative

There is a Math Review article by Kuranishi there of the paper "Sprays" by Ambrose, Singer, and Palais (in which sprays were first defined). I think the approach taken there (and described in Kuranishi's review) is pretty much what Dan Ramras is asking for.

  • $\begingroup$ This does sound promising! I'll need to look at your paper more carefully. I've just learned this material from Lang's book. He says something about a discussion in terms of jet bundles that appears in Loos's book Symmetric Spaces (Vol. I) but I looked there and couldn't figure out what part Lang was referring to. Maybe Lang was talking about the something along the lines of the discussion in terms of kth order tangent vectors in Kuranishi's review? $\endgroup$
    – Dan Ramras
    Jul 28, 2010 at 21:47
  • $\begingroup$ Incidentally, paper where this question arose is now available at arxiv.org/abs/1006.0063. I was hoping to find a cleaner way to phrase the discussion on p. 5. $\endgroup$
    – Dan Ramras
    Jul 28, 2010 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.