Can covariant derivative tensorialness be derived?

[Pencilvester]

TL;DR Summary

Is the fact that the covariant derivative is tensorial something that can be derived, or is it insisted upon?

I'm building a mental framework for the Levi-Civita connection that is intuitive to me. I start by imagining an arbitrary manifold with arbitrary coordinates embedded in a higher dimensional Euclidean space, then if I take the derivative of an arbitrary coordinate basis vector with respect to an arbitrary coordinate, it gives a vector that sticks out of the tangent space. So I project the vector into the tangent space and I use the components of this projected vector to define the Christoffel symbols. I use the general process of differentiating, then projecting to define the covariant derivative. With this scaffolding I can derive the equation for $\Gamma^\lambda_{\mu \nu}$ as a function of the metric, as well as how to take the covariant derivative of any arbitrary tensor.

My question is, given this framework, is there a way to derive the fact that the covariant derivative is tensorial? i.e. can we derive the transformation law for the Christoffel symbols without insisting that it is a priori? Or do we simply declare that we want the covariant derivative to be tensorial, and thereby derive the CS transformation law? If it's the latter, I would guess that in any framework the tensorialness of the covariant derivative is simply insisted upon rather than derived— Is this correct?

 

[jbergman]

TL;DR Summary: Is the fact that the covariant derivative is tensorial something that can be derived, or is it insisted upon?

I'm building a mental framework for the Levi-Civita connection that is intuitive to me. I start by imagining an arbitrary manifold with arbitrary coordinates embedded in a higher dimensional Euclidean space, then if I take the derivative of an arbitrary coordinate basis vector with respect to an arbitrary coordinate, it gives a vector that sticks out of the tangent space. So I project the vector into the tangent space and I use the components of this projected vector to define the Christoffel symbols. I use the general process of differentiating, then projecting to define the covariant derivative. With this scaffolding I can derive the equation for $\Gamma^\lambda_{\mu \nu}$ as a function of the metric, as well as how to take the covariant derivative of any arbitrary tensor.

My question is, given this framework, is there a way to derive the fact that the covariant derivative is tensorial? i.e. can we derive the transformation law for the Christoffel symbols without insisting that it is a priori? Or do we simply declare that we want the covariant derivative to be tensorial, and thereby derive the CS transformation law? If it's the latter, I would guess that in any framework the tensorialness of the covariant derivative is simply insisted upon rather than derived— Is this correct?

The covariant derivative is tensorial? That's not the way I think of it. It's tensorial in the direction you want to take derivative in but not it it's other argument.

 

[renormalize]

The covariant derivative is tensorial? That's not the way I think of it. It's tensorial in the direction you want to take derivative in but not it it's other argument.

I'm confused by this statement. Are you referring here to the covariant derivative $\partial+\Gamma$ (schematically), which I view as tensorial by construction, or just the connection $\Gamma$ alone, which is a non-tensor?

 

[jbergman]

I'm confused by this statement. Are you referring here to the covariant derivative $\partial+\Gamma$ (schematically), which I view as tensorial by construction, or just the connection $\Gamma$ alone, which is a non-tensor?

I learned differential geometry from Math texts like Lee's Introduction to Riemannian Manifolds. And as he defines it a connection, $\nabla$, is $C^{\infty}(M)$-linear in the first argument and $\mathbb R$-linear in the second argument.

Tensorial objects are $C^{\infty}(M)$-linear in all their arguments. There are multiple ways to prove this.

I was looking into Wald to see what he says on this but didn't find anything with a quick glance.

We could be talking about different things, though.

 

[jbergman]

I learned differential geometry from Math texts like Lee's Introduction to Riemannian Manifolds. And as he defines it a connection, $\nabla$, is $C^{\infty}(M)$-linear in the first argument and $\mathbb R$-linear in the second argument.

Tensorial objects are $C^{\infty}(M)$-linear in all their arguments. There are multiple ways to prove this.

I was looking into Wald to see what he says on this but didn't find anything with a quick glance.

We could be talking about different things, though.

Reading around a bit, though, some people talk about $\nabla_{\mu}T_{\nu}$ as being tensorial which is true, because one has already supplied the non-tensorial argument to the covariant derivative.

 

[renormalize]

Reading around a bit, though, some people talk about $\nabla_{\mu}T_{\nu}$ as being tensorial which is true, because one has already supplied the non-tensorial argument to the covariant derivative.

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor. Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector. That said, your statement about Lee's definition seems to imply that the operator $\nabla_{\mu}$ by itself includes a non-tensorial part which somehow always drops out when it operates on a tensor. Can you point me to where in Lee (or another reference) is displayed the manifest form of that non-tensorial part and explains the mechanism of its disappearance when acting on arbitrary tensor fields? Thanks.

 

[jbergman]

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor. Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector. That said, your statement about Lee's definition seems to imply that the operator $\nabla_{\mu}$ by itself includes a non-tensorial part which somehow always drops out when it operates on a tensor. Can you point me to where in Lee (or another reference) is displayed the manifest form of that non-tensorial part and explains the mechanism of its disappearance when acting on arbitrary tensor fields? Thanks.

I will have to do some more digging to find a clear statement of this fact. Some references dance around it. This math.stackexchange.com post discusses it indirectly, https://math.stackexchange.com/ques...r-between-a-non-metric-torsion-free-connectio . And Wald talks about the difference of two connections being a tensor.

 

[Pencilvester]

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor.

I believe this directly answers my question, so thank you!

Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector.

Yes, though I did not express this fact in my original post, you correctly inferred my thinking, so thanks again!

 

[jbergman]

Also, if you accept the definition of a $(k,l)$ tensor field as $C^{\infty}(M)$ multilinear functions with $k$ vector fields and $l$ covector fields as arguments as described here, then it's clear that an affine connection is not a tensor as it is R-linear in one of it's arguments as described in https://en.m.wikipedia.org/wiki/Affine_connection.

 

[jbergman]

Also, if you accept the definition of a $(k,l)$ tensor field as $C^{\infty}(M)$ multilinear functions with $k$ vector fields and $l$ covector fields as arguments as described here, then it's clear that an affine connection is not a tensor as it is R-linear in one of it's arguments as described in https://en.m.wikipedia.org/wiki/Affine_connection.

And it is tensorial once you jave supplied an argument in the slot that is R-linear because the other slots are $C^{\infty}(M)$ linear.

 

[jbergman]

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor. Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector. That said, your statement about Lee's definition seems to imply that the operator $\nabla_{\mu}$ by itself includes a non-tensorial part which somehow always drops out when it operates on a tensor. Can you point me to where in Lee (or another reference) is displayed the manifest form of that non-tensorial part and explains the mechanism of its disappearance when acting on arbitrary tensor fields? Thanks.

To close the loop, see page 91 in Lee's "Introduction to Riemannian Manifolds". Lee writes,

Although the definition of a connection resembles the characterization of (1,2) tensor fields, ..., a connection on TM is not a tensor field because it is not linear over $C^{\infty}(M)$ in its second argument, but instead satisfies the product rule.

 

[jbergman]

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor. Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector. That said, your statement about Lee's definition seems to imply that the operator $\nabla_{\mu}$ by itself includes a non-tensorial part which somehow always drops out when it operates on a tensor. Can you point me to where in Lee (or another reference) is displayed the manifest form of that non-tensorial part and explains the mechanism of its disappearance when acting on arbitrary tensor fields? Thanks.

I see that you asked a more specific question than I answered. It is probably easiest to work this out oneself. I will try and write something up in index notation tonight.

 

[jbergman]

Given the general tensor field $T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$, the covariant derivative $\nabla_{\mu}T_{\nu\xi\pi\cdots}^{\alpha\beta\gamma\cdots}$ is defined so as to also be a tensor. Since in practice the covariant-derivative operator $\nabla_{\mu}$ always acts on something, never standing alone, it seems perfectly reasonable to say that, for practical purposes, $\nabla_{\mu}$ transforms like a covariant vector. That said, your statement about Lee's definition seems to imply that the operator $\nabla_{\mu}$ by itself includes a non-tensorial part which somehow always drops out when it operates on a tensor. Can you point me to where in Lee (or another reference) is displayed the manifest form of that non-tensorial part and explains the mechanism of its disappearance when acting on arbitrary tensor fields? Thanks.

Here is my own attempt to answer this question. It's a bit subtle at least for me. For simplicity I will focus on the case where $T^i$ is a vector field. In that case,

$$\nabla_j T^i = \partial_j T^i + T^k \Gamma^i_{kj}$$
and we know under a change of coordinates this transforms as a tensor like so
$$\frac{\partial\bar{T}^i}{\partial\bar{x}^j} + \bar{T}^k \bar{\Gamma}^i_{kj} =
\left(\frac{\partial T^r}{\partial x^s} + T^t \Gamma^r_{ts}\right)
\frac{\partial \bar{x}^i}{\partial x^r} \frac{\partial x^s}{\partial \bar{x}^j}$$

We also have additivity, i.e., which also transforms as we would expect.
$$\nabla_j (\bar{T}^i + \bar{U}^i) = \nabla_s (T^r + U^r) \frac{\partial \bar{x}^i}{\partial x^r} \frac{\partial x^s}{\partial \bar{x}^j} $$

However, for a smooth function $f$, $\nabla_j (f \cdot T^i)$ we run into a problem. Unlike a $(1,2)$ tensor, $$\nabla_j (f \cdot T^i) \ne f \cdot (\nabla_j T^i) $$ And if we look at the transformation of this, we will see we won't be able to pull out $f$.

This is because a covariant derivate depends on the value of $T^i$ in a small neighborhood and results in the covariant derivative obeying the leibniz rule, i.e.

$$\nabla_j (f \cdot T^i) = \partial_j (f \cdot T^i) + f \cdot T^k \Gamma^i_{kj} = T^i \partial_j f + f \cdot \partial_j T^i + f \cdot T^k \Gamma^i_{kj}$$

Now, to full understand why Tensors must have this rule is a bit hard to see. It's Proposition 12.22 in Lee's Smooth Manifold book, but I don't have a good geometric interpretation of why this must be so.

 

[renormalize]

$$\nabla_j f \cdot T^i \ne f \cdot \nabla_j T^i $$

Sorry, can you insert parentheses into this expression to make it clear exactly what terms are being acted on by each $\nabla_j$ operator?

 

[jbergman]

Sorry, can you insert parentheses into this expression to make it clear exactly what terms are being acted on by each $\nabla_j$ operator?

Updated my answer. The tldr is that for let's say a $A^i_j$ tensor we expect that,
$$\bar{f} \cdot \bar{A}^i_j = f \cdot \left(A^r_s \frac{\partial \bar{x}^i}{\partial x^r} \frac{\partial x^s}{\partial \bar{x}^j}\right)$$

And this doesn't hold for $\nabla$. Note, $\cdot$ is for multiplication not composition.

 

[renormalize]

It's Proposition 12.22 in Lee's Smooth Manifold book, but I don't have a good geometric interpretation of why this must be so.

Here's that proposition from John M. Lee, Introduction to Smooth Manifolds, 2nd ed., pg. 318:

This book is a graduate math text, and I'm admittedly a physicist and not a mathematician, but to my reading, 12.22 simply states that the product of a scalar field and a tensor field is itself a tensor, as is the product of two tensor fields. That is, it's the proposition that products of tensors are necessarily tensorial. It says nothing about the covariant derivative. (Indeed, per its index, the book never discusses or even introduces covariant differentiation anywhere in its 708 pages!) So I don't see how this proposition supports the notion that the covariant derivative $\nabla$ is nontensorial in its transformation properties. Instead, $\nabla$ is to be regarded, not as a tensor field, but as a tensor differential operator. It is tensorial because its operation on a rank $\left(\begin{array}{c} k\\ l \end{array}\right)$ tensor produces a rank $\left(\begin{array}{c} k+1\\ l \end{array}\right)$ tensor (Lee, Riemannian Manifolds: An Introduction to Curvature, Lemma 4.7, pg. 54).

 

[Pencilvester]

However, for a smooth function $f$, $\nabla_j (f \cdot T^i)$ we run into a problem. Unlike a $(1,2)$ tensor, $$\nabla_j (f \cdot T^i) \ne f \cdot (\nabla_j T^i) $$ And if we look at the transformation of this, we will see we won't be able to pull out $f$.

I suppose this is a semantics issue that depends on your definition of “tensorial”— if you want this property to be a part of your definition, which, given tensors’ linearity, doesn’t seem unreasonable, then I see your point. But if by “tensorial” you simply mean that it transforms like a tensor and (for a rank $(k,l)$ tensor) has the property that it maps an (ordered) set of $l$ vectors and $k$ dual vectors to the real numbers, which also seems reasonable to me, then you’d consider the covariant derivative of a tensor and/or a scalar to be tensorial.

 

[jbergman]

Here's that proposition from John M. Lee, Introduction to Smooth Manifolds, 2nd ed., pg. 318:
View attachment 348853
This book is a graduate math text, and I'm admittedly a physicist and not a mathematician, but to my reading, 12.22 simply states that the product of a scalar field and a tensor field is itself a tensor, as is the product of two tensor fields. That is, it's the proposition that products of tensors are necessarily tensorial. It says nothing about the covariant derivative. (Indeed, per its index, the book never discusses or even introduces covariant differentiation anywhere in its 708 pages!) So I don't see how this proposition supports the notion that the covariant derivative $\nabla$ is nontensorial in its transformation properties. Instead, $\nabla$ is to be regarded, not as a tensor field, but as a tensor differential operator. It is tensorial because its operation on a rank $\left(\begin{array}{c} k\\ l \end{array}\right)$ tensor produces a rank $\left(\begin{array}{c} k+1\\ l \end{array}\right)$ tensor (Lee, Riemannian Manifolds: An Introduction to Curvature, Lemma 4.7, pg. 54).

I think in some sense we are arguing semantics. But I just want to point out again, that on p. 91 in Lee's newer book, "Introduction to Riemannian Manifolds", he states that $\nabla$ is not a $(1,2)$ tensor field. That's all I was claiming. I agree that it maps tensors to tensors.

However, in contrast the Riemannian Curvature tensor is a tensor field.

The point I was trying to show, and it probably needed more elaboration, is that given a connection, $\nabla$, $f(p)(\nabla_j T^i)(p)\ne \nabla_j (f T^i)(p)$ which we would expect if it was a tensor.

 

[jbergman]

I suppose this is a semantics issue that depends on your definition of “tensorial”— if you want this property to be a part of your definition, which, given tensors’ linearity, doesn’t seem unreasonable, then I see your point. But if by “tensorial” you simply mean that it transforms like a tensor and (for a rank $(k,l)$ tensor) has the property that it maps an (ordered) set of $l$ vectors and $k$ dual vectors to the real numbers, which also seems reasonable to me, then you’d consider the covariant derivative of a tensor and/or a scalar to be tensorial.

Agree on the semantics issue, however, I do want to point out that $C^{\infty}(M)$ linearity is important for tensor fields, because at each $p \in M$, $T^i$ should be R-linear at a point and this includes covectors which should have the property that $\omega(a v) = a\omega(v)$. Carefully working this through requires not just that $T^i$ transforms as a tensor but also that $fT^i$ transforms also in a way that $f$ can be pulled out. This doesn't happen with the covariant derivative.