# Are differential geometric and physics conventions for covariant derivatives consistent?

In a differential geometric setting, the covariant derivative can be defined as a map $$\nabla_X:\Gamma(TM)\to\Gamma(TM)$$, for any vector field $$X\in\Gamma(TM)$$, satisfying certain conditions. In other words, for any vector field, it maps vector fields into other vector fields. This definition is then readily extended to maps between arbitrary tensor fields. Given a local basis $$\{\partial_i\}_i$$ around some $$p\in M$$, it can be characterised via the Christoffel symbols as $$\nabla_i\partial_j\equiv \nabla_{\partial_i}\partial_j =\Gamma^k_{~~ij}\partial_k.\tag1$$ Similarly, we get local expressions such as $$\nabla_i Y = \nabla_i (Y^j \partial_j) = (\partial_i Y^j + Y^\ell Y^k \Gamma^j_{~\ell k})\partial_j.\tag2$$

So far, so good. My confusion arises when trying to match this with the notation used in more physical contexts. For example, consider these lecture notes ([29:46] on youtube). Here, they denote the covariant basis as $$\vec S_\alpha$$, and write the Christoffel symbol as $$\Gamma^\gamma_{~\alpha\beta}=\vec S^\gamma\cdot\partial_\beta \vec S_\alpha.\tag3$$ When writing this, they're assuming to be dealing with embedded surfaces, so taking the standard derivative still makes sense, and I can match this expression with (1) assuming the covariant derivative to be the projection of the standard derivative on the tangent surface.

However, from (3) they derive $$\vec S^\gamma\cdot(\nabla_\alpha\vec S_\beta) = \vec S^\gamma\cdot ( \partial_\alpha\vec S_\beta - \Gamma^\omega_{~\alpha\beta}\vec S_\omega ) = 0.\tag4$$ This appears now to be in direct contrast with (1), as $$\vec S_\alpha$$ in (4) should correspond to the local basis for the tangent space, $$\partial_i$$, in (1). In fact, following this notation, $$\nabla_\alpha \vec S_\beta$$ is normal to the surface, which seems in direct contrast with the definition of covariant derivative in (1), an an object mapping tangent vectors into tangent vectors. So what gives? Why are these two notations seemingly in contrast? Is there a more formal way to understand precisely what kind of object the covariant derivative is in the latter convention?

• By "embedded surfaces", do you mean a manifold embedded in Euclidean space? Also, by "standard derivative", do you mean the canonical connection in Euclidean space arising from the natural isomorphism? Commented Sep 26, 2021 at 9:49
• @VincentThacker I'm not completely sure what's the formalisation of the ideas underlying (3) and (4) here; that is, de facto, part of what I'm asking here I think. My understanding is that "embedded" is understood as you say, yes; "standard derivative" should mean the directional derivative taken in the embedding space (which is I think also the same as what you say). So what I write with $\partial_\alpha\vec S_\beta$ here
– glS
Commented Sep 26, 2021 at 10:24
• Alright, your equation (4) is evidently false. There is no reason, in general, for the covariant derivative of a vector to have zero dot product with all basis vectors. That will only be true if it were the zero vector (or normal to the surface). Either the notation is incorrect or you confused quantities defined on the manifold with those defined on the ambient (Euclidean) space. Commented Sep 26, 2021 at 10:46
• @VincentThacker that is my understanding as well. In fact, in the linked lectures, they say that $\nabla_a\vec S_b$ is normal to the surface, which seems in direct contrast with the covariant derivative as an object which by definition gives tangent vectors as output. This makes me think that there is a different way to understand "covariant derivative" in this context, which is pretty much what I'm asking. Still, I've seen these rules of how $\nabla_\alpha$ acts on objects with raised/lowered indices quite often in physics literature, and formulas like (4) seem a direct consequence of those
– glS
Commented Sep 26, 2021 at 10:51

I have read the same book as the one you linked, and I can say that it does have some technical inaccuracies that slowly build up to the confusion that you are having right now. I will try to list them in order. I will denote the standard covariant basis by $$\mathbf{e}_i$$, and I will use Greek letters for surface indices and English alphabet for ambient (Euclidean space) indices.

1. There is no such thing as "contravariant basis vectors". What the book refers to as "contravariant basis vectors" are actually basis covectors. In other words, they are one-forms that are linear maps from vectors to scalars. Therefore, expressions like $$\mathbf{e}^i \cdot\mathbf{e}_j = \delta^i_j$$ are incorrect because you can't take the dot product of an object that is not even a vector. Instead, the correct expression should be $$\mathbf{e}^i \left(\mathbf{e}_j\right) = \delta^i_j$$ where the one-form $$\mathbf{e}^i$$ is acting on the vector $$\mathbf{e}_j$$. The other expressions should all be modified accordingly.

2. From the very beginning, the book uses an incorrect definition of $$\nabla_i\mathbf{e}_j$$ in Euclidean space as $$\nabla_i\mathbf{e}_j = \frac{\partial \mathbf{e}_j}{\partial x^i} -\Gamma^k_{ij}\mathbf{e}_k = \mathbf{0}$$ which is a correct evaluation but an incorrect definition. The book treated $$\mathbf{e}_j$$ as a $$(0,1)$$ tensor component with a lower index, which is most certainly wrong as $$\mathbf{e}_j$$ is a vector which is $$(1,0)$$.
The correct way to define the covariant derivative for a vector $$\mathbf{v}=v^i\mathbf{e}_i$$ is $$\nabla_i v^m = (\nabla\mathbf{v})\left(\mathbf{e}_i,\mathbf{e}^m\right) = (\nabla\mathbf{v})^{\;m}_i = \frac{\partial v^m}{\partial x^i}+\Gamma^m_{ik}v^k \\ \nabla_i \mathbf{v} = \left(\frac{\partial v^m}{\partial x^i}+\Gamma^m_{ik}v^k\right)\mathbf{e}_m$$ where it must be noted that the $$v^m$$ and $$v^k$$ are the components of $$\mathbf{v}$$ which are numbers. Therefore, the correct way to compute $$\nabla_i\mathbf{e}_j$$ is to let $$\mathbf{v}=\mathbf{e}_j$$ in the above definition. The components are therefore all zero except the $$j$$-th one which is equal to one. All of them are constant so the first term $$\partial v^m/\partial x^i$$ is zero. Applying the same logic to the second term, we see that the only non-vanishing term in the sum is the $$j$$-th one, which gives $$\Gamma^m_{ij} (1) = \Gamma^m_{ij}$$. Adding in the basis from the second line above, we have $$\nabla_i \left(\mathbf{e}_j\right) = \Gamma^m_{ij}\mathbf{e}_m = \partial_i\mathbf{e}_j$$ which is precisely that of the partial derivative. This is correct because the (Levi-Civita connection) covariant derivative in Euclidean space is simply the partial derivative because its affine nature supplies a canonical parallel transport. It is definitely not zero unlike what the book claims.

3. All of the above leads to the incorrect statement that $$\nabla_\alpha \mathbf{e}_\beta$$ is normal to the surface. It most certainly is not. The surface covariant derivative $$\nabla_\alpha$$ is simply the Euclidean partial derivative along the surface coordinate $$s^\alpha$$, with the normal component removed. The reason for the error is once again because the book used the incorrect definition $$\nabla_\alpha\mathbf{e}_\beta = \frac{\partial \mathbf{e}_\beta}{\partial s^\alpha} -\Gamma^\gamma_{\alpha\beta}\mathbf{e}_\gamma$$ which leads to the incorrect conclusion that $$\nabla_\alpha\mathbf{e}_\beta =\mathbf{0}$$ on the surface. In general, $$\partial_\alpha \mathbf{e}_\beta$$ lives in the ambient Euclidean space and therefore has components both normal and tangent to the surface. The normal component is given by the second fundamental form $$\mathbb{I}_{\alpha\beta}$$ multiplied by the unit normal vector $$\hat{\mathbf{n}}$$, while the tangential component is the surface covariant derivative $$\nabla_\alpha\mathbf{e}_\beta = \Gamma^\gamma_{\alpha\beta}\mathbf{e}_\gamma$$, which is exactly analogous to the correct definition I gave above for Euclidean space. This formula also precisely agrees with that on a pseudo-Riemannian manifold (where the ambient Euclidean space is not assumed to exist). In other words, we have $$\partial_\alpha\mathbf{e}_\beta = \Gamma^\gamma_{\alpha\beta}\mathbf{e}_\gamma + \mathbb{I}_{\alpha\beta}\hat{\mathbf{n}} = \nabla_\alpha \mathbf{e}_\beta + \mathbb{I}_{\alpha\beta}\hat{\mathbf{n}}$$

For more information on the first point, see this post. For more information on why the book's definition is wrong, see this post. Lastly, for the correct derivations, see this video which explains it very clearly.

• I don't quite get the first point. What's the problem in defining a dual basis to the covariant basis vectors (assuming we have a Riemannian metric)? I can understand that one can use differential 1-forms as well, but why can't we also define the "contravariant basis vector fields" via $\mathbf e^\alpha= g^{\alpha\beta}\mathbf e_b$ with $g^{\alpha\beta}$ components of the inverse of the metric?
– glS
Commented Sep 26, 2021 at 14:37
• your second point resonates though. If I understand, you are saying that the "covariant basis vectors" are still ultimately vector fields, and thus the covariant derivative should act on them via the standard rules (i.e. with the plus sign)
– glS
Commented Sep 26, 2021 at 14:38
• @glS See the post I linked: physics.stackexchange.com/a/334230 and the lines highlighted in NO!. My first point is not strictly related to your question, but I just wanted to correct some confusion that the book caused for myself. In any case, on manifolds, as far as I know the one-form definition of the contravariant basis is used as the standard. Commented Sep 26, 2021 at 14:39
• @glS Yes, that's correct. Someone else also ran into the same wrong expression, as shown on the third equation on this post (physics.stackexchange.com/q/281590). The point is that in Euclidean space, there is only one type of derivative (the ordinary partial derivative) because the affine connection is trivially supplied. It's just that on a surface embedded in this space, we take the derivative along the surface coordinate line $s^\alpha$ as we normally would, and split that into tangential and normal components. Commented Sep 26, 2021 at 14:42
• @bolbteppa I am fully aware of what you have just said. What I am saying is that the $i$ in $\mathbf{e}_i$ is not the same thing as a covariant index of a tensor component. $\mathbf{e}_i$ is most certainly not a $(0,1)$ tensor. It is a vector, with components and basis just like other vectors. On manifolds, $\mathbf{e}_i$ is by definition $\partial/\partial x^i$. Commented Sep 26, 2021 at 15:32