# Understand the need of affine connection

I can't understand why textbook says that directional derivative cannot be defined for general manifold, and a separate affine connection is needed.

Assume there are two vector fields $$X$$ and $$Y$$. In particular, you have a tangent vector at point $$p$$ called $$X_p$$. To find the derivative of $$Y$$ at $$p$$ in the direction of $$X_p$$, why can't you just operate $$X_p = a^i\partial_i$$ on $$Y_p = b^j \partial_j$$ component-wise?

By component-wise, I meant differentiate the component smooth function $$b^j$$ with respect to $$X_p$$. It seems that a smooth real valued function can still be differentiated by a vector field without connection?

• "component-wise"? Jul 3, 2019 at 17:42
• @LordSharktheUnknown I updated the question a bit to explain myself better. But to be honest, since I don't understand differential geometry well enough, I probably messed up the concepts in my mind. It would be appreciated if you can explain why affine connection is needed when we already got $X_p$ as a differentiation operator. Jul 3, 2019 at 17:50
• For your definition to make sense, you need to have "constant" vector fields $\partial_j$ ... which, of course, we have with the standard coordinates in Euclidean space but not otherwise. Jul 3, 2019 at 19:59
• @TedShifrin I will try asking in another way. I saw in the textbook that you can differentiate scalar field with respect to a vector field fine using the vector field as a differential operator, but you have to invent mechanism of connection in order to do it for vector field. Why is that the case? Jul 4, 2019 at 9:27
• Directional derivative of a function is just that, just the differentiable structure needed. You tell me how to differentiate a vector field on a surface in $\Bbb R^3$. You might read my diff geo text, linked in my profile, to learn some concrete stuff. Jul 4, 2019 at 16:41

The problem is that the vector field you get from componentwise directional derivative depends on choice of the coordinate system.

• Thanks for your answer, but I'm still confused. Are you talking about the dependence of coordinate system of the vector field being differentiated or the vector field used as differentiation direction? Maybe that's related with one of my comment? Why is scaler field can be differentiated without problem using just the directional vector field as an differential operator? Can you elaborate a bit more? Jul 4, 2019 at 9:50
• The problem concerns components of the vector field which is being differentiated. If these are $Y^i$ in one coordinate system, then in another one they take the form $\sum_k A^i_k Y^k$, where $A$ is the Jacobi matrix, which is not a constant. By Leibniz rule, the pointwise derivative of $Y$ in the two coordinate systems is not related by the $A$ matrix (as for any honest vector field), but contains also a term with derivatives of $A$. Jul 4, 2019 at 13:23
• Thank you for your answer. Can I ask if scalar field derivatives always agree independent on the coordinate system? If so, can I ask for a reference for such a proof? Jul 4, 2019 at 22:36

What you're describing is the contraction of exterior derivative of $$Y$$ by $$X$$, $$\textrm{d}Y(X)$$. This can be viewed as the trivial affine connection, as the connection matrix is $$0$$. In $$\mathbb{R}^n$$ this coincides with Levi-Civita connection of the Euclidean metric.

In fact, one can always pick coordinates at a point such that a connection matrix vanishes at that point (Riemann Normal coordinates in the case of the Levi-Civita connection).

Alternatively, if one were to naively differentiate 'Y with respect to X' like any function, one has $$X (Y)= X^i{\partial_i}(Y^j{\partial_j}) = X^i ({\partial_i}(Y^j)\partial_j +X^iY^j{\partial_i}({\partial_j}).$$ The issue here is that this is no longer a first order operation, as we are dealing with the second derivative of coordinates.

This leads to the definition another type of directional derivative, the Lie derivative $$\mathcal{L}_X$$. The intuition here is that $$\mathcal{L}_X$$ shows how the vector field $$Y$$ evolves over the integral curves of $$X$$, but one can show that $$\mathcal{L}_X(Y)=[X,Y]$$.

• How do you define $dY$ on a general smooth manifold? $Y$ is not a differential form Nov 8, 2022 at 12:00