# Covariant derivative geometric interpretation

I'm having some trouble understanding what the covariant derivative means geometrically. I know the definition which states that for a tensor T with any number of indices: $\nabla_j T = \frac{\partial T}{\partial Z^k} +$ the Christoffel symbols contracted with each index of the tensor appropriately.

In the case of an invariant, I can visualize this as the normal partial derivative, so the geometric interpretation is clear. Also, for a first-order tensor $V^i$ such that vector $\textbf{R} = V^i \textbf{Z}_i$, I know that I can say $$\frac{\partial \textbf{R}}{\partial Z^k} = \nabla_k (V^i) \textbf{Z}_i$$ However, how can I understand the covariant derivative of higher order tensors and of vectors? What is the geometric meaning?

• The expression $\nabla_X T$ says (approximately) how tensor $T$ varies along direction $X$. – janmarqz Jan 11 '15 at 2:01
• What do you mean by approximately? – glaba Jan 11 '15 at 3:36
• this is because in case of standard calculus the directional derivative needs $||X||=1$ in order to compare with another directions. – janmarqz Jan 11 '15 at 3:51

## 3 Answers

I'll say a few words about how I think about covariant derivatives, which is really just expanding on janmarqz's comment (hopefully others will contribute their own viewpoints as well):

For me, the most important geometric idea behind a covariant derivative $\nabla$ is that given a curve $\gamma$ in a manifold $M$, $\nabla$ gives you an isomorphism between the tangent spaces $T_{\gamma(t_1)}M$ and $T_{\gamma(t_2)}M$ for any two points on the curve. Mathematically, this isomorphism $$P : T_{\gamma(t_1)}M \to T_{\gamma(t_2)}M$$ is the unique isomorphism with the property that for any $v \in T_{\gamma(t_1)}M$, there exists a vector field (which I'll call $v(t)$) along $\gamma$ such that $v(t_1) = v, v(t_2) = P(v)$, and $\nabla_{\gamma'(t)} v(t) = 0$ for all $t \in [t_1, t_2]$.

This isomorphism is called "parallel transport"; I like to picture a surface embedded in $\mathbb{R}^3$, such as the 2-sphere, and think of parallel transport along a curve $\gamma$ as "dragging" vectors along that curve. (Important remark: the isomorphism obtained depends on the choice of curve $\gamma$ in general.)

Of course, once you have an isomorphism of vector spaces, you get an isomorphism of any of the associated tensor spaces as well. So if $T$ is a $(k,l)$-tensor on $T_{\gamma(t_1)}M$, then we get a $(k,l)$-tensor $PT$ on $T_{\gamma(t_2)}M$.

Now the point is that once you have this "parallel transport" isomorphism, the covariant derivative $\nabla_X \mathcal{T}$ is a literal derivative in the following precise sense: Given a vector $X \in T_pM$, let $\gamma$ be any curve with $\gamma'(0) = X$, and let $P_t$ be the "parallel transport along $\gamma$" isomorphism $$P_t : T_{\gamma(t)}M \to T_{\gamma(0)}M \quad (= T_pM).$$ Then for any tensor field $\mathcal{T}$ on $M$, $$\nabla_X \mathcal{T} = \frac{d}{dt}\Big|_{t=0} \Big( P_t \big( \mathcal{T}(\gamma(t)) \big) \Big).$$ This is a very precise interpretation of the idea that $\nabla_X \mathcal{T}$ gives you the derivative of $\mathcal{T}$ in the direction of $X$.

• In a flat manifold, parallel transport of a tensor across any curve will result in the same vector, right? That would suggest the covariant derivative be 0, but it usually isn't. – glaba Jan 12 '15 at 20:14
• $\mathcal{T}$ is a tensor field, not a simple tensor; that is, the tensor $\mathcal{T}(p)$ might depend on the point $p \in M$. In particular we're parallel translating the tensor $\mathcal{T}(\gamma(t))$ back to the point $\gamma(0)$. -- As an aside, the only reason "result in the same vector" makes sense is because flatness gives you a canonical isomorphism between any two tangent spaces $T_pM$ and $T_qM$, which is parallel translation along any curve. Parallel translation can't literally result in the same vector since you wind up with vectors in different vector spaces. – mollyerin Jan 12 '15 at 20:59

@mollyerin gives a nice answer. Furthermore, depending on the rigor that you are looking for I think that Roger Penrose's book Road to Reality gives a nice overview of the geometric interpretation of covariant derivatives. It should be noted that the book attempts to present topics such as the covariant derivative and other mathematical subjects using layman's terms.

• This book gave very good geometric insight into these concepts. – glaba Jan 11 '15 at 23:48

Let me add that in order to find curves of minimal length between to positions in a surface for example, one sets an equation like $$\nabla_{C'}C'=0,$$ which leads you to the classical expression $\ddot{u}^k+\Gamma^k{}_{st}\dot{u}^s\dot{u}^t=0$, where $C'=\dot{u}^s\partial_s$ is the tangent field for $C$.

This is pretty geometrical.