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 
 A: 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 $D$. 
