# Riemannian Manifolds of Bounded Curvature

I am a complete newbie Riemannian Geometry with a particular application in mind so please excuse a lack of rigor in the question.

Suppose I have a manifold with sectional curvature everywhere negative and also lower bounded by $\kappa < 0$ (I think these are called Hadamard Manifolds). Between two point $x,y$ I consider two curves $\gamma_1(t), \gamma_2(t)$. I parallel transport a vector $v \in T_xM$ to $y$ along the two curves $\gamma_1, \gamma_2$ giving me the vectors $v_1, v_2 \in T_yM$. I wish to say something about difference $v_1 - v_2$, in particular to bound $\|v_1 - v_2\|$. My bound can depend upon the lengths of the curves $\gamma_1, \gamma_2$.

Please provide a reference that could help me prove the above quantitative bound. From the little I have read it seems the quantity I care about is fundamentally related to the curvature tensor. In that case can I find an upper bound with respect to a bound on the curvature tensor? If yes can I relate the bound to the sectional curvature?

A general reference to understand these notions of curvature would be appreciated too.

Thanks

• Parallel transport is an isometry, so $||v_1|| = ||v_2|| = ||v||$. This means that $||v_1 - v_2|| \leq 2||v||$. – Alec Payne May 1 '17 at 21:50
• I should have specified the kinds of bound I wish to achieve. So this bound does not depend on the curvature at all and if i move an infinitesimal amount the bound is not any better. I wish to have a bound that bounds the difference possibly with the curvature and the length of the curves. – user1189053 May 2 '17 at 1:57
• When the closed curve $\gamma_1^{-1}\gamma_2$ is null-homotopic, i.e., it bounds some disk $D$; the difference $v_1-v_2$ is estimated by the absolute value of the sectional curvature (-k) times the area of $D$, see Ambrose-Singer theorem. I ques, when the curvature is bounded above by some negative $K<0$ this area my be estimated further by the length of $\gamma_1^{-1}\gamma_2$. – valeri May 2 '17 at 9:13
• Thanks Valeri. Could you provide a reference to the version of Ambrose Singer Theorem you are referring to? Your answer seems very relevant to my query. In fact the precise setting I have is indeed that I am parallel transporting the vector across a geodesic triangle, i.e. $\gamma_1$ is a geodesic and $gamma_2$ is two geodesics, i.e. a geodesic to an intermediate point $p$ from $x$ and then a geodesic to $y$ from $p$. – user1189053 May 2 '17 at 14:15
• I think what you want is Theorem 13.6.4 of Andrew Pressley's "Elementary Differential Geometry." See this: math.stackexchange.com/questions/4479/… – Alec Payne May 2 '17 at 16:42

## 1 Answer

Sorry, (since I got this theorem myself and found out after that it is Ambrose-Singer) I mistaken exercise 4 in doCarmo p.105 for it. The proof looks like this: let $D$ be the image of a unit square with euclidean coordinates $0\leq s,t\leq 1$, and define $V(s,t)$ to be parallel along, say $s$-coordinate lines in $D$ and such that $D_t V(0,t)\equiv 0$. Then compute how the derivative of $V$ in $t$-direction changes along $s$-coordinate lines: it is $D_s D_t V = R(X,Y)V$ since $D_t D_s V \equiv 0$, and $s, t$ coordinate vectors commute - then $D_t V (1,t) = \int R(X,Y)V ds - D_t V(0,t)$, and further integral on $t$ of $D_t V(t,1)$ on $t$ from $0\leq t\leq 1$, gives the difference between $V(1,0)$ and $V(1,1)$ which is the difference between $V(0,0)$ and its parallel transport along the boundary of $D$ - check boundary conditions. To get scaling right note that $|X\wedge Y| ds dt$ equals the element of the area $dD$, where of course $X= \partial /\partial s$ and $Y= \partial /\partial t$. In other words, we may integrate area element times $R(A,B)V$ over unit square where $A,B$ - pair of unit orthogonal vectors tangent to $D$.

• Maybe you should add the isoperimetric inequality to get the kind of estimate the OP wants - see your comment above. – Sebastian Goette May 3 '17 at 19:25
• Sebastian Goette For the disk $D$ consisting of geodesics as in my comment above the isoperimetric estimate is directly following from Rauch type estimates for - in this case - Jacobi fields $\partial/\partial t (s,t)$ - easy. Otherwise - in general case - I am only guessing. – valeri May 3 '17 at 19:53