Let $(M,g)$ be a Riemannian manifold and let $R(X,Y)Z$ be its $(3,1)$ Riemann curvature tensor given by


Let the input vectors $X,Y,Z$ come from a compact subset of the tangent bundle. Then is the metric norm of the vector $R(X,Y)Z$ bounded?

If not, what extra conditions should we impose on $X,Y,Z$ to make $R$ bounded?

EDIT: The input vector fields $X,Y,Z$ won't have to be smooth or continuous, assume that the vector fields just come from a compact subset of the tangent bundle.

EDIT II: Since the Riemann curvature tensor does NOT depend on the values of vector fields $X,Y,Z$,but just depends on the pointwise values, I can re-ask my question by saying will $R(X,Y)Z$ be bounded if I just work with uniformly bounded values of $X,Y,Z$?


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    $\begingroup$ Hmmmm . . . your added condition that $X, Y, Z$ etc. needn't be confinuous may force we to edit my answer, which assumes all quantities are continuous. Cheers! $\endgroup$ Jun 25, 2015 at 21:58
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    $\begingroup$ Thanks for your answer, Robert! I think your proof goes through with minor modification, where you bound $X^\mu, Y^\nu, Z^\rho$ first, and then use the continuity to bound the coefficients of $R$. $\endgroup$ Jun 25, 2015 at 22:30
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    $\begingroup$ Since the Riemann curvature tensor depends only on the pointwise values of its arguments, in edit II, I assumed that we only have pointwise values of $X,Y,Z$ and all these pointwise values are uniformly bounded. $\endgroup$ Jun 25, 2015 at 22:40
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    $\begingroup$ Yes, I think you are correct. I may write up a second answe. covering the more general case. Thanks! $\endgroup$ Jun 25, 2015 at 22:41

2 Answers 2


This somehow 'depends'. If, as usual, in Riemannian Geometry, you are looking at a smooth ($C^\infty$) manifold with smooth metric, then the answer is, for sure, yes. Only if you are reducing the differentiability assumptions on the manifold and the metric you may run into trouble at a certain point.

What you are looking at (e.g. in local coordinates) is an expression with components from the metric and derivatives of the metric up to second order, combined with bounded coefficients from the vector fields. This is, locally, clearly bounded if the single factors entering the expressions are (e.g. if the metric is $C^2$). Then it is a matter of mathematical craftmanship to go from local to compact (if the vector field arise from a compact set, then this happens over a compact subset of the manifold in question, of course).

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    $\begingroup$ Your answer thoughtfully covers the case $X, Y, Z$ not necessarily continuous, which mine did not. Well done; endorsed! $\endgroup$ Jun 25, 2015 at 22:33
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    $\begingroup$ @RobertLewis Thanks for the endorsement. I have to admit I wasn't even aware of the fact that (non) continuity of the vector fields was an issue for the OP. But if it is, then the thing to note is that the curvature tensor is, as the name suggests, a tensor field. As such, it depends only on pointwise evaluation of it's arguments, and this in a mulitlinear fashion, which immediately implies that $$|R(X, Y)Z |\le ||R|| |X|_\infty |Y|_\infty |Z|_\infty$$ -- whatever $||R|| $ might be. But it should be clear that $||R||$ only depends on $M$ and the metric, not on the vector fields. $\endgroup$
    – Thomas
    Jun 26, 2015 at 5:43
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    $\begingroup$ @Thomas: thanks for your answer, upvoted from me too! Yes I seemed to have overlooked/forgotten the fact that $R$, being a tensor field, depends only on the pointwise values of $X,Y,Z$ itself, rather than the values of vector fields themselves. And so $R_p$ is finite for every $p$ because $T_p{M}$ is finite-dimensional, and $R_p$ being a continous function of $p$, attains a maximum on a compact set. I think I just re-wrote your argument, but is it correct? $\endgroup$ Jun 26, 2015 at 12:51
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    $\begingroup$ @Mathmath If $R_p$ is continuous, then yes. Continuity of $R$ is a consequence of the metric tensor being of class $C^2$. $\endgroup$
    – Thomas
    Jun 26, 2015 at 13:10

NB: in the following, it is assumed that $X, Y, Z$ and $R$ are continuous.

If $C \subset TM$ is the compact subset from which $X$, $Y$ and $Z$ are taken, then $C \times C \times C$ is compact (Tychonoff's theorem), and the map

$R(X, Y)Z: C \times C \times C \to TM \tag{1}$

is continuous; this follows by virtue of the fact that in any local coordinate system $w_\alpha$ on $M$, with corresponding tangent basis vector fields $e_\alpha$ where

$e_\alpha = \dfrac{\partial}{\partial w_\alpha}, \tag{2}$

we may write

$R(X, Y)Z = e_\beta R^\beta_{\mu \nu \rho} X^\mu Y^\nu Z^\rho, \tag{3}$


$X = X^\mu e_\mu \tag{4}$

and so forth. (The Einstein summation convention is understood as being operative in (3) and (4), as throughout this answer.) Since all of the coefficients $R^\beta_{\mu \nu \rho}$, $X^\mu$ and so forth are continuous, as well as the vector fields $e_\alpha$, the continuity of $R(X, Y)Z$ follows. Note that we are in essence drawing on the fact that a composition of continuous maps is itself continuous.

Denoting the metric norm by $\Vert \cdot \Vert_g$, we see in a manner similar to the above that

$\Vert \cdot \Vert_g: TM \to \Bbb R, \tag{5}$

which takes tangent vectors $X \in TM$ to their metric norms $\Vert X \Vert_g = (g((X, X))^{1/2}$, is also continuous. It then further follows that the composed map

$\Vert R(X, Y)Z \Vert_g: C \times C \times C \to TM \to \Bbb R \tag{6}$

is continuous, again since it is a composition of continuous functions.

It now follows from the compactness of $C \times C \times C$ that $\Vert R(X, Y)Z \Vert_g$ is bounded on this set, since a continuous function on a compact set is always bounded.

Note Added Friday 26 June 2015 1:47 PM PST: In the light of our colleague Thomas well-framed answer and comments, as well as those of our OP Mathmath, I don't suppose it's necessary to say much more, so I will forgo adding another answer to address the case in which $X$, $Y$, $Z$ need not be continuous; but I would like to take this opportunity add a few (hopefully brief) words on the subject. I was indeed aware that a bound should exist whether or not the continuity of $X$, $Y$, $Z$ binds, that that it would most likely take the form

$\Vert R(X, Y)Z \Vert_g \le \Vert R \Vert \Vert X \Vert_g \Vert Y \Vert_g \Vert Z \Vert_g, \tag{7}$

where $\Vert R \Vert$ may be taken to be some constant depending only on the tensor field $R$. In fact, (7) holds pointwise, without any reference to continuity at all. Whilst thinking about how to best write-up an explanation of (7), it occurred to me that in the case of $X$, $Y$, $Z$ continuous one can altogether dispense with the details of multilinearity and argue, more or less directly, from the fact that $R(X, Y)Z$ is a polynomial in the components of $X$, $Y$, $Z$ with continuous coefficients the $R^\alpha_{\mu \nu \rho}$; thus, $\Vert R(X, Y)Z \Vert_g$ is a continuous mapping on the compact set $C \times C \times C$. And having spied that ball, I picked it up and ran with it, forgetting in my haste that continuity of $X$, $Y$, $Z$ is not essential here. I must confess that I am not used to considering vector fields sans continuity, so I was pre-disposed to read it into the hypotheses of this question; but still, I think it worth noting that the general continuity argument can, in certain situations, replace the more linearity-specific methods which depend on estimates such as (7). End of Note.


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