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Question: If there exists a covariant derivative, then why doesn't there also exist a "contravariant derivative"? Why are all or most forms of differentiation "covariant", or rather why do all or most forms of differentiation transform covariantly? What aspect of differentiation makes it intrinsically "covariant" and intrinsically "not contravariant"? Why isn't the notion of differentiation agnostic to "co/contra-variance"?

Motivation:
To me it is unclear (on an intuitive, i.e. stupid/lazy, level) how notions of differentiation could be restrained to being either "covariant" or "contravariant", since any notion of differentiation should be linear*, and the dual of any vector space is exactly as linear as the original vector space, i.e. vector space operations in the dual vector space still commute with linear functions and operators, they same way they commute with such linear objects in the original vector space.

So to the extent that the notion of linearity is "agnostic" to whether we are working with objects from a vector space or from its dual vector space, so I would have expected any notion of differentiation to be similarly "agnostic". Perhaps a better word would be "symmetric" -- naively, I would have expected that if a notion of "covariant differentiation" exists, then a notion of "contravariant differentiation" should also exist, because naively I would have expected one to exist if and only if the other exists.

However, it appears that no such thing as "contravariant derivative" exists (see here on Math.SE, also these two posts [a][b] on PhysicsForums), whereas obviously a notion of "covariant derivative" is used very frequently and profitably in differential geometry. Even differential operators besides the so-called "covariant derivative" seemingly transform covariantly, see this post for a discussion revolving around this property for the gradient. I don't understand why this is the case.

(* I think)

Chill2Macht
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    The word *covariant* in the phrase *covariant derivative* means only that the operation is invariant with respect to changes of coordinates, that is, that it "covaries" with changes of coordinates. – Mariano Suárez-Álvarez Aug 30 '16 at 01:02
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    Your question is somewhat unanswerable: in essence you are saying «well, there do exist fish that swim and I get that, but I don't understand why there are no fish that fly.» – Mariano Suárez-Álvarez Aug 30 '16 at 01:04
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    The reason that derivatives are naturally covariant is that the derivative is inherently (at a point) a transformation from one dual space to another. Transformations of the base space are contravariant, while transformations of the dual space are covariant. – Ben Grossmann Aug 30 '16 at 01:16
  • @Omnomnomnom OK this makes, it would also explain why differential forms live in the cotangent space and not the tangent space like I would have expected them to. When you say "base space", do you mean the manifold $X$, or its tangent bundle/the collection of all of its tangent spaces (I think those two descriptions are synonymous but I am not sure)? Would it be possible to define a linear operator which satisfies the product rule and which is a transformation of the dual space? Or is that just what derivations are already, so that derivations are the "contravariant derivative"? – Chill2Macht Aug 30 '16 at 02:14
  • @MarianoSuárez-Álvarez Haha good point. I guess maybe what I was trying to ask is that "I thought fish were things which have scales, but it turns out they can swim too. Why does having scales imply the ability to swim?" But that's also impossible to answer because having scales doesn't imply the ability to swim; lizards and snakes don't swim but have scales. (Here "scales" I think would be (1) linearity and (2) satisfying the product rule). Maybe a more answerable question would be "Are there things with scales which are not fish?" I am not sure if I should edit my question or post a new one. – Chill2Macht Aug 30 '16 at 02:21
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    IMO, the key idea here is that "covariant" and "contravariant" aren't really properties of the objects we're studying -- they're properties of how we represent them with coordinates. –  Aug 30 '16 at 02:39
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    Being covariant is so derivative nowadays... – Asaf Karagila Aug 30 '16 at 05:28
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    @Hurkyl: I don't understand your comment. On a manifold $M$, contravariant tensors are elements of $TM^{\otimes k}$, while covariant tensors are elements of $T^*M^{\otimes k}$. These are intrinsic notions, making no reference to coordinates. – Jesse Madnick Sep 01 '16 at 18:03
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    @JesseMadnick: The origin of the terms are about the action of changing bases on coordinate vectors, and that's how I interpreted the question: why do we put one particular action on the vectors produced by differentiation and not the other. I had the impression the question wouldn't have been asked if the OP had the concept of $TM$ and $T^*M$ as being bundles with their own intrinsic meaning. But I could very well have missed the OP's point. –  Sep 01 '16 at 18:20

5 Answers5

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The "covariant" in "covariant derivative" should really be "invariant". It is a misnomer, but we are stuck with it. It is not the same "covariant" as that of a "covariant vector", and therefore, there is no "contravariant derivative". Armed with this, Wikipedia should fill in the rest for you :)

SSD
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    But covariant derivatives obviously has the covariant nature: for all $(p, q)$ tensor, $\nabla X$ is a $(p, q+1)$ instead of a $(p+1, q)$ tensor. –  Aug 30 '16 at 01:31
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    Exactly. The name may be a misnomer due to history, sure, but why does $\nabla$ raise the covariant index by one, rather than the contravariant index? More generally, why should connections on a vector bundle $E$ be maps $\Gamma(E) \to \Gamma(E) \otimes \Gamma(T^*M)$ rather than to $\Gamma(E) \otimes \Gamma(TM)$? I haven't thought this through fully, but I think it comes down to how $df$ is a $(0,1)$-tensor that intrinsically defined, without reference to a metric. (By contrast, the $(1,0)$-tensor $\text{grad}(f)$ requires a metric.) – Jesse Madnick Aug 30 '16 at 05:52
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Let $S = C^\infty(X, \mathbb{R})$ be the $\mathbb{R}$-algebra of all smooth functions on the manifold X. There is a purely algebraic notion of Kähler differentials: there is an $S$-module $\Omega_{S/\mathbb{R}}$ presented by generators and relations as

  • For every $f \in S$, add a generator $\mathrm{d}f$
  • For every $r \in \mathbb{R}$, add the relation $\mathrm{d}r = 0$
  • For every $f,g \in S$, add the relation $\mathrm{d}(f+g) = \mathrm{d}f + \mathrm{d}g$
  • For every $f,g \in S$, add the relation $\mathrm{d}(fg) = f \mathrm{d}g + g \mathrm{d}f$

There are a few other ways to define it, but it's clear that $\Omega_{S / \mathbb{R}}$ is the "universal" way to differentiate things in $S$, if we presume that differentiation must be $\mathbb{R}$-linear and satisfy the Leibniz rule.

There is an obvious map $\Omega_{S / \mathbb{R}}$ to the global sections of $T^*X$, sending $\mathrm{d}f$ to $\mathrm{d}f$.

However, the Kähler differentials seem too big. However, for each point $P$, I will define an "evaluation map". Let $\Omega_{S/\mathbb{R}, P}$ be the $\mathbb{R}$-vector space you get by adding the further relations $f \mathrm{d}g = f(P) \mathrm{d}g$ for every pair of functions $f,g \in S$, and write $\mathrm{d}f|_P$ for the image of $\mathrm{d}f$ in this space.

Then, following the idea of this Mathoverflow answer, we can quickly show that $\Omega_{S / \mathbb{R}, x_0}$ is precisely the cotangent space at $x_0$. For any smooth $f$, we take the differential of a Taylor polynomial and get $$ \mathrm{d}f(x)|_{x_0} = f'(x_0) \mathrm{d} x|_{x_0}$$ (where $f'(x_0)$ is the linear functional $v \mapsto \nabla_v f(x_0)$) at which point it's clear that $\Omega_{S / \mathbb{R}, x_0} \cong T^*_{x_0} X$.

It's not hard to pass from $\mathbb{R}^n$ to manifolds. Consequently, we conclude that the exterior derivative is the "universal" way to differentiate smooth scalar fields in such a way that the derivative is completely determined by its "values" at points.


Reading more of the answers, I think they claim further that $\Omega_{S / \mathbb{R}}^{**}$ is isomorphic to the global sections of the cotangent bundle as $S$-modules (and that this implies the global sections have what is maybe a nicer but less general universal property) but I haven't followed the argument.

  • So the part where the obvious map from $\Omega_{S/\mathbb{R}}$ to the global sections of $T^*X$ is the part where we are forced to operate with the cotangent, rather than the tangent, space? The MathOverflow answer linked to seems to be talking about the operator being forced to be the derivative, rather then it being forced to exist on the cotangent space (I could be wrong, to be honest I don't think I understand it 100%). I want to accept this answer because it looks like it answers my question, but right now I don't really understand the answer, so I'm not sure what to do. – Chill2Macht Aug 30 '16 at 05:01
  • Assuming that $T^*X$ denotes the cotangent space -- that's what it looks like it denotes; I think I may have accidentally used the covariant derivative before, but all of the objects I was working with were in tangent spaces $TX$ so I'm not sure. I apologize for being so obtuse -- I should have mentioned in the question that I haven't studied graduate level Riemannian geometry yet -- I just completed an undergraduate course in differential geometry which strongly suggested elements of Riemannian geometry but didn't use the terms (e.g. 2nd fundamental form instead of Riemannian metric). – Chill2Macht Aug 30 '16 at 05:04
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    I am not sure if this answers the question. You showed that $d$ is the universal way to differentiate functions, but a covariant differentiation acts on vector fields and not scalar fields. –  Aug 30 '16 at 05:15
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    @William: The derivative of a multivariable function is covector-valued, so it's fine that it's the derivative! Think of vectors as being columns, but $\nabla f$ being a row vector. –  Aug 30 '16 at 05:20
  • If there is a linear isomorphism between $TX$ and $T^*X$, then why can't we define also define a "contravariant derivative" to be $(\nabla f)^T$ being a column vector? I don't understand why all of the same linearity properties and the Leibniz rule wouldn't work for such objects. I agree that $\Omega_{S\mathbb{R}}$ seems universal. From the point of view of just satisfying linearity and the Leibniz rule, what would be the drawback of switching to the dual module $(\Omega_{S\mathbb{R}})^*$? Is $\Omega_{S\mathbb{R}}$ not finite-dimensional, so we don't have a linear isomorphism? – Chill2Macht Aug 30 '16 at 06:10
  • You mention for instance $(\Omega_{S/\mathbb{R}})^{**}$ as being a different object from $\Omega_{S/\mathbb{R}}$ rather than treating it as being naturally linearly isomorphic, which would only _have_ to be the case if $\Omega_{S/\mathbb{R}}$ were finite-dimensional, so it seems like it might not be, i.e. it might be infinite-dimensional, which would explain why $\Omega_{S/\mathbb{R}} \not\simeq (\Omega_{S/\mathbb{R}})^*$. Admittedly I am not certain about the extent to which theorems about vector space duals extend to module duals. I have never looked into before. – Chill2Macht Aug 30 '16 at 06:13
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    @William: Modules can do weird things; if I follow, it has the opposite problem from what you think. For vector spaces, $V \to V^{**}$ is always injective, but it might not be surjective. Here, I believe $\Omega_{S / \mathbb{R}} \to \Omega_{S /\mathbb{R}}^{**}$ is surjective, but it's not injective! –  Aug 30 '16 at 06:16
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    @William: If you are applying an isomorphism to transpose between $TX$ and $T^*X$ (e.g. the canonical such isomorphism you have on a Riemann manifold), then there isn't really a difference between "covariant" and "contravariant"! –  Aug 30 '16 at 06:18
  • OK, and related to why $\nabla f$ can't be defined to be a column vector, why is the natural map from $\Omega_{S/\mathbb{R}}$ into $T^* X$ rather than $TX$? But really looking at this page it seems clear-ish https://ncatlab.org/nlab/show/K%C3%A4hler+differential#SmoothOrPlain that my problem just comes down to "Why does it make sense to define differential forms as the exterior algebra over the cotangent bundle instead of over the tangent bundle?" but this question (1) doesn't change the correctness of your answer, and (2) is one I've had before and should post as a new question sometime. – Chill2Macht Aug 30 '16 at 06:29
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    @William: The following cannot all be true, by considering the effects of scaling space by a factor of 2 (which doubles true vectors and leaves true scalars unchanged). **(1)** $v$ is a true vector. **(2)** $\nabla_v f(0)$ is a true scalar. **(3)** $\nabla_v f(0) = f'(0) \odot v$ for some bilinear product $\odot$. **(4)** $f'(0)$ is a true vector. –  Aug 30 '16 at 06:44
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    (although you can make them all hold if you replace $f'(0) \odot v$ with a trilinear product $f'(0) B v$, for the appropriate kind of gadget $B$ that transforms correctly) –  Aug 30 '16 at 06:49
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    @IlmariKaronen Ah, thanks! I has repeatedly having trouble with linking to mathoverflow losing most of the URL, and I guess the problem made it to the final edit. :( –  Aug 30 '16 at 12:41
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The reason why the covariant derivative makes a $(p,q+1)$-type tensor field out of a $(p,q)$ type tensor field is because for a tensor field $T$, $\nabla T$ is defined as $$ \nabla T(X,\text{filled arguments})=\nabla_XT(\text{filled arguments}), $$ and $\nabla_XT$ is $C^\infty(M)$-linear in $X$, so this relation defines a covariant tensor field - one that acts on vector fields. But why does it need be so?

The geometric answer is that a covariant derivative is essentially a representation for a Koszul or principal connection, a device that allows for parallel transport of bundle data along curves. The reason it takes in vectors is because vectors are intrinsically tied to curves on your manifold. If your covariant derivative took in 1-forms as the directional argument instead of vectors, it would not represent a connection, because there is no way to canonically tie together curves and 1-forms without a tool like a metric tensor or a symplectic form.

Bence Racskó
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    This makes sense -- it also explains implicitly why differential forms are covariant instead of contravariant -- because they are designed to be functions _of_ or act _on_ the "natural" objects in the tangent space, i.e. being _functions_ of elements of the tangent space/bundle, they need to be in a different space, i.e. the cotangent bundle. I.e. one can argue by analogy with how bounded linear functionals form a separate space in functional analysis from the original Banach space on which they act -- this answer is both very elucidating and clarifying. Thank you! – Chill2Macht Aug 31 '16 at 18:42
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The "contravariant derivative" is usually called the gradient; for functions $f$, $$\langle \nabla f, v\rangle = df(v) = \nabla_v f$$ where $\nabla$ on the left is the gradient and on the right is the covariant derivative; here the gradient is still coordinate-free but transforms contravariantly (lives in tangent space, and not cotangent space).

You can define an analogous $(2,0)$ tensor that is the contravariant derivative of a vector field.

user7530
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    For the record, this convention is not universal -- I usually see $\nabla f$ reserved for the function $v \mapsto \nabla_v f$, *not* as the vector representing that function through the inner product. IMO, this is the better convention, both because of the similarity of the notations $\nabla$ and $\nabla_v$ and because it is independent of the inner product. –  Aug 30 '16 at 01:19
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    @hurkyl Fair enough; nabla as the gradient is extremely common in physics and optimization, at least. – user7530 Aug 30 '16 at 01:20
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    I'm somewhat confused by this answer, because this physics post says that the gradient is covariant: http://physics.stackexchange.com/questions/126740/gradient-is-covariant-or-contravariant, but you say that it is contravariant -- do you mean that the result of applying the gradient (i.e. the values of the gradient) are contravariant tensors? Although this answer might be saying that it is contravariant -- I am not sure http://physics.stackexchange.com/a/131956/115093, is that what you are arguing? – Chill2Macht Aug 30 '16 at 05:07
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    $\nabla f$ is covariant, no problem (at least in differential geometry context). But to cook up a contravariant one, the answerer uses the metric $\langle \nabla f, \cdot\rangle$ (so it becomes elements in $T^*M$). Note that this is the same as $df$ (see Hurkyl's answer). @William –  Aug 30 '16 at 05:23
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    @William References to physics won't really clarify things, since in physics we *always* have a metric and so the distinction between covariant and contravariant loses most of its meaning. A physicist will use either $\nabla_\mu$ or $\nabla^\mu \equiv g^{\mu\nu} \nabla_\nu$ as needed. –  Aug 30 '16 at 07:13
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    @Willliam As the answers to that question point out, the OP in the question you've linked is wrong; the gradient is contravariant, not covariant. – user7530 Aug 30 '16 at 14:18
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This is really hand-wavy, and maybe not even correct, but this is what I just thought of based on a new fact that I recently learned. Maybe the idea helps someone else.

On any differentiable manifold (it doesn't even have to be Riemannian), we have a canonical symplectic form on the cotangent bundle, see here or here. (This isn't something I knew previously, but is something I learned due to a comment by Moishe Cohen -- all credit goes to them.)

Anyway, this symplectic form on the cotangent bundle gives us a volume form on the cotangent bundle (I think, see here, I might be misinterpreting). Thus, canonically from the definition, any differentiable manifold has a volume form on its cotangent bundle, thus a notion of signed distance, signed area, signed volume, ..., i.e. volume+orientation, on the cotangent bundle.

These are exactly the ingredients we need for a meaningful notion of integration, and it occurs on the cotangent bundle, and not the tangent bundle, because the differentiable structure induces a symplectic form on the cotangent bundle, but not on the tangent bundle.

Thus the natural place for integration to occur on a differentiable manifold is in the cotangent bundle. And corresponding to any notion of integration should be a notion of differentiation, and since the former develops naturally only on the cotangent bundle, the latter should only develop naturally on the cotangent bundle. Thus why differential forms act on covectors, and why all derivatives are covariant, not contravariant -- because all integrals are covariant.

Chill2Macht
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